TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Students get through the TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Answer the following questions.

Question 1.
Define the term chemotherapy.
Answer:
The specific treatment of a disease by using medicine is known as chemotherapy.

Question 2.
Define therapeutic index? What is its use?
Answer:
It is defined as the ratio between the maximum tolerated dose of a drug (above which it becomes toxic) and the minimum curative dose (below which it becomes ineffective). Higher the values of therapeutic index safer is the drug.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 3.
Explain the term, target molecules or drug targets as used in medicinal chemistry.
Answer:
Drugs interact with macromolecules like proteins, carbohydrates, lipids, nucleic acids. Hence they are called target drugs. Proteins perform several roles in the body. Enzymes are crucial in communication systems and are called receptors.

Question 4.
Give examples of drugs that are grouped based on the biological effect that they produce on the recipient.
Answer:

  1. Antibiotic drugs: Amoxycillin, ampicillin, cefixime, cefpodoxime, erythromycin, tetracycline, etc.
  2. Antihypertensive drugs: propranolol, atenolol, metoprolol succinate, amlodipine, etc.

Question 5.
Streptomycin and erythromycin are classified in the same group. Justify the statement.
Answer:
Both inhibit the protein synthesis (target process) in bacteria and are classified in the same group. However, the mode of action is different. Streptomycin inhibits the initiation of protein synthesis while erythromycin prevents the incorporation of new amino acids to the protein. These drugs are grouped based on the biological system/process that targets the recipient.

Question 6.
Write a short note on enzymes as drug targets.
Answer:
In all living systems, the biochemical reactions are catalysed by enzymes. Hence, these enzyme actions are highly essential for the normal functioning of the system. If their normal enzyme activity is inhibited, then the system will be affected. This principle is usually applied to kill many pathogens.
In enzyme catalysed reactions, the substrate molecule binds to the active site of the enzyme by means of the weak interaction such as hydrogen bonding, Van der waals force etc., between the amino acids present in the active site and the substrate.
When a drug molecule that has a similar geometry (shape)as the substrate is administered, it can also bind to the enzyme and inhibit its activity. In other words, the drug acts as an inhibitor to the enzyme catalyst. These types of inhibitors are often called competitive inhibitors.
In certain enzymes, the inhibitor molecule binds to a different binding site, which is commonly referred to as an allosteric site and causes a change in its active site geometry (shape). As a result, the substrate cannot bind to the enzyme. These types of inhibitors are called allosteric inhibitors.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 7.
How do drugs interact with targets? (or) Give a brief account of drug-target interaction.
Answer:
Drugs interact with macromolecules like carbohydrates, proteins, nucleic acids lipids present in the cell. The macromolecules perform various functions in the body, eg: Proteins perform several roles in the body.

  1. Proteins that act as biological catalysts are called enzymes.
  2. Proteins that are crucial to a communication system in the body are called receptors.
  3. Proteins that carry polar molecules across the membranes are called carrier proteins.

In their catalytic activity enzymes perform two major functions.

  1. The first function of an enzyme is to hold the substrate molecule for a chemical reaction.
  2. The second function of the enzyme is to provide functional groups which will attack the substrate to carry out the chemical reaction.

Question 8.
How do drugs interact with enzymes? (or) Give a brief account of drug-enzyme interaction.
Answer:
Drugs that inhibit any two of the activities mentioned in Q7 are called enzyme inhibitors. Enzyme inhibitors can block the binding site thereby preventing the binding of the substrate to the active site and hence inhibit the catalytic activity of enzymes.
Drugs inhibit the attachment of natural substrate on active site in two ways:

  1. Drugs that compete with the natural substrate for their attachment on the active sites of enzymes are called competitive inhibitors.
  2. Same drugs, however, do not bind on the active site but bind to different sites of the enzyme which is called the allosteric site.

This binding of the drug (inhibitor) at the allosteric site changes the shape of the active site of the enzyme in such a way that the natural substrate cannot recognize it. Such enzymes are called noncompetitive inhibitors.
If the bond between an enzyme and the drug (inhibitor) is a strong covalent bond that cannot be broken easily then the enzyme is blocked permanently. The body then degrades the enzyme-drug (inhibitor) complex and synthesis the new enzyme.

Question 9.
Explain the terms (i) Competitive inhibitors, (ii) Allosteric inhibitors.
Answer:
(i) Competitive inhibitors:
In enzyme-catalyzed reactions, the substrate molecule binds to the active site of the enzyme by means of the weak interaction such as hydrogen bonding, Van der Waals force etc., between the amino acids present in the active site and the substrate. When a drug molecule that has a similar geometry (shape)as the substrate is administered, it can also bind to the enzyme and inhibit its activity. In other words, the drug acts as an inhibitor to the enzyme catalyst. These types of inhibitors are often called competitive inhibitors.
(ii) Allosteric inhibitors:
In certain enzymes, the inhibitor molecule binds to a different binding site, which is commonly referred to as an allosteric site and causes a change in its active site geometry (shape). As a result, the substrate cannot bind to the enzyme. These types of inhibitors are called allosteric inhibitors.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 10.
Explain the term (i) Chemical messengers, (ii) Receptor, (iii) Antagonists, (iv) Agonists.
Answer:
(i) Chemical messengers: In the body, messages between two neurons (nerve cells) or that between neurons and muscles are communicated through a certain chemicals in substances called chemical messengers.
(ii) Receptors: Receptors are proteins that are crucial to the communication system in the body.
(iii) Antagonists: Drugs that bind to the receptor site and inhibit its natural function are called antagonists. These are useful when blocking of the message is required.
(iv) Agonists: Drugs that mimic (imitate) the natural chemical messengers by switching on the receptor are called agonists. These are useful when there is a lack of natural chemical messengers.

Question 11.
Mention the various types of chemical messengers and explain how they act?
Answer:
There are two types of chemical messengers.
(i) Hormones (ii) Neurotransmitters.

  1. Hormones are a group of biomolecules
    which are produced in the ductless (endocrine) glands. These enter into the bloodstream and travel to different parts of the body activating all the receptors. which recognize them for message
    transfer. They are not deactivated quickly. Adrenaline is an example of a hormone.
  2. Neurotransmitters: Nerves transfer messages through neurotransmitters. These bind to the receptor (target) for a very short time to transfer the message to it and depart quickly unchanged after transferring the message. The receptor then forwards the message inside the cell. After leaving the active site, neurotransmitters undergo degradation and lose their capacity to transfer messages.

Question 12.
With reference to which classification has the statement “ranitidine, is an antacid” been given?
Answer:
This statement refers to the classification of drugs according to pharmacological effect because drugs that will be used to counteract the excess acid in the stomach will be called antacids.

Question 13.
List two major classes of antibiotics with an example of each class.
Answer:

(a) Bactricidal (b) Bacteriostatic
(i) Penicillin (i) Erythromycin
(ii) Aminoglycosides (ii) Tetracycline
(iii) Oxoflavin (iii) Chloroampinicol

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 14.
What are tranquilizers? How do they act? Give two examples.
Answer:
Tranquilizers are neurologically active drugs. They are used in the treatment of stress anxiety, sleep disorders, and mental diseases. They act on the central nervous system by blocking the neurotransmitter dopamine in the main Dizephan, alprazolam is the examples.

Question 15.
Give two examples for each (i) Anti-inflammatory drugs (ii) Antipyretics (iii) nonsteroidal Anti-inflammatory drugs.
Answer:
(i) AntiinflammatorydrugsAcetaminophen or paracetamol, ibuprofen, aspirin.
(ii) Antipyretics: Acetylsalicylic acid (aspirin), Acetaminophen or Paracetamol.
(iii) Nonsteroidal anti-inflammatory drugs: Ibuprofen.

Question 16.
Mention an important difference between non-narcotic analgesics and narcotic analgesics.
Answer:
Non-narcotic analgesics are non-addictive drugs while narcotic analgesics are addictive drugs.

Question 17.
Give two examples of narcotic analgesics.
Answer:
Morphine and codeine are examples of narcotic analgesics.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 18.
Give two examples for local anesthetics.
Answer:
Procaine and Lidocaine.

Question 19.
Name the anesthetics used for major surgical procedures.
Answer:
Propofol and Isoflurane.

Question 20.
Give examples of antacids. How do antacids function in case of acidity?
Answer:
Milk of magnesia, sodium bicarbonate, aluminum hydroxide, ranitidine, cimetidine, omeprazole, rabeprazole are examples of antacid. They neutralize the acid in the stomach that causes acidity.

Question 21.
What are antihistamines? Give two examples.
Answer:
Antihistamines are drugs that provide relief to allergic effects, eg: cetirizine, levocetirizine.

Question 22.
What are antimicrobials? Give two examples.
Answer:
Drugs that are used to cure diseases caused by microbes or microorganisms such as bacteria, viruses, fungi, etc are called antimicrobials. These include antibacterials, antifungal, and viral agents.
eg: penicillin, erythromycin.

Question 123
How does (i) β lactams and (ii) macrolides function as antimicrobials?
Answer:
(i) β – lactam inhibits cell wall biosynthesis.
(ii) Macrolides target bacterial ribosomes and prevent protein production.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 24.
Give examples for β – lactam and macrolides antimicrobials.
Answer:
β – lactams: Penicillin, ampicillin, cephalosporins, carbapenems, and monobactams.
Macrolides: Erythromycin, azithromycin.

Question 25.
What are the uses of β-lactam and macrolide antimicrobials?
Answer:
Beta lactams: They are used to treat skin infections, dental infections ear infections, respiratory tract infections.
Macrolides: They are used to treat respiratory tract infectious, genital, gastrointestinal tract and skin infectious.

Question 26.
Give examples for ‘Fluoroquinolones’ and mention their uses.
Answer:
Clinafloxacin, ciprofloxacin, levofloxacin are examples of fluoroquinolones. They are used to treat urinary tract infections, skin infections, and respiratory infections.

Question 27.
How do tetracyclines class of antibiotics function? Mention their uses.
Answer:
Doxycycline, minocycline, oxytetracycline are examples of the tetracyclines group of antibiotics. They inhibit the bacterial protein synthesis via interaction with the 30S subunit of the bacterial ribosome.
They are used in the treatment of peptic ulcers, infection of the respiratory tract, etc.

Question 28.
What are aminoglycosides? Give examples.
Answer:
Aminoglycosides are a type of antibiotics that bind to the 30S subunit of the bacterial ribosome, thus stopping bacteria from making proteins.
eg: kanamycin, gentamycin neomycin.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 29.
What are food additives? Give examples.
Answer:
The substances which are not naturally a part of the food and are added to improve the quality of food are called food additives. The substances enhance the nutritive, sensory, and practical value of the food. They also increase the shelf life of food.
Important categories of food additives:

  1. Aroma compounds
  2. Food colors
  3. Preservatives
  4. Stabilizers
  5. Artificial Sweeteners
  6. Antioxidants
  7. Buffering substances
  8. Vitamins and minerals

Question 30.
What are food preservatives?
Answer:
Chemical substances which are used to protect food against bacteria, yeasts and molds are called preservatives, eg: sodium metabisulphite (sodium meta sulfite), sodium benzoate etc.

Question 31.
Name the preservative used in the preparation of pickles and vegetables.
Answer:

  1. Acetic acid for pickles
  2. Sodium meta-sulphate for fresh vegetables and fruits.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 32.
Name the chemicals which are used as emulsifiers.
Answer:
Sucrose esters with palmitic and steric esters are used as emulsifiers.

Question 33.
Name the physical methods used in the preservative of food.
Answer:

  1. Heat treatment (pasteurization and sterilizations)
  2. Cold treatment (chilling and freezing)
  3. Drying (dehydration)
  4. Irradiation is used to preserve food.

Question 34.
What are antioxidants? Give examples.
Answer:
Antioxidants are substances that retard the oxidative deteriorations of food. Food containing fats and oils is easily oxidized and turns rancid. To prevent the oxidation of the fats and oils, chemical BHT(butylhydroxytoluene), BHA(Butylated hydroxyanisole) are added as food additives. They are generally called antioxidants.

Question 35.
How do antioxidants prevent the oxidation of food?
Answer:
They readily undergo oxidation by reacting with free radicals generated by the oxidation of oils, thereby stop the chain reaction of oxidation of food. Sulphur dioxide and sulphites are also used as food additives. They act as anti-microbial agents, antioxidants, and enzyme inhibitors.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 36.
Define “Total Fatty Matter (TFM). What is its use?
Answer:
It is defined as the total amount of fatty matter that can be separated from a sample after splitting with mineral acids. The higher the TFM quantity in the soap better is its quality.

Question 37.
What is an anionic detergent?
Answer:
Anionic detergent: They are so named because a large part of their molecules are anions and it is the anionic part of the molecule is involved in the cleansing actions. These are sodium salts of sulphonated long-chain alcohols or hydrocarbon. For example, sodium lauryl sulphate, sodium dodecyl benzene sulphonate, etc. Anionic detergents are used is household work and in toothpastes.

Question 38.
What is cationic detergents?
Answer:
Cationic detergent: They are so-called because a large part of their molecule are cations and it is the cationic part of the molecule is involved in the cleansing action cationic detergents are quaternary ammonium salts of amines, with acetates, chlorides, or bromides as anions. Cetyl trimethyl ammonium bromide is a cationic detergent and is used in hair conditioners. Cationic detergents have germicidal properties and are expensive, therefore they are of limited use.

Question 39.
What is Non-ionic detergent?
Answer:
Non- ionic detergents: These detergents do not contain any ion. These are esters of high molecular weight alcohols. One such detergent is formed when stearic acid reacts with polyethylene glycol. Liquid dishwashing detergent are nonionic type.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 40.
Explain the terms monomer and polymer and polymerization.
Answer:
Monomer: Simple and reactive molecules from which polymers are prepared either by addition or condensation are called monomers, eg: Vinyl chloride, ethene, formaldehyde, acrylonitrile, phenol, etc.
Polymers: These are compounds of high molecular mass formed by the combination of a large number of simple molecules called monomers. The process by which monomers are converted to polymers is called polymerization.

Question 41.
What are synthetic and natural polymers? Give two examples for each type.
Answer:
Synthetic polymers are man-made polymers prepared in the laboratory, eg: polyethylene, Teflon, nylon, etc.
Natural polymers are naturally occurring polymers, mostly in plants, animals, etc. eg: Protein natural rubber, etc.

Question 42.
In which classes, the polymers are classified on the basis of molecular forces.
Answer:
On the basis of intermolecular forces of attraction operating between different polymers chains, polymers are classified as

  1. Elastomers
  2. Fibres
  3. Thermoplastic polymers
  4. Thermo setting polymers

Question 43.
Write names of monomers of the following polymers and classify them as addition or condensation polymers.
(a) Teflon, (b) Bakelite, (c) Natural rubber
Answer:

Polymer Type Monomer
Teflon Addition Tetrafluoro ethene
Bakelite Condensation phenol and formaldehyde
Natural rubber Addition cis – isoprene

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 44.
What is the role of benzoyl peroxide in the polymerization of ethene?
Answer:
Benzoyl peroxide is an initiator. It forms a free radical.

Question 45.
What are LDPE and HDPE? How are they prepared?
Answer:
LDPE is low-density polyethylene. It is obtained by the polymerization of ethene under high pressure of 1000 – 2000 atm at 350 – 570 K in the presence of an initiator. HDPE is called high-density polyethylene. It is obtained by the polymerization of ethene in the presence of Ziegler – Natta catalyst at 333 – 343 K under 6-7 atm.

Question 46.
Write the structure of the monomers of the following polymers, (i) PVC, (ii) Polypropene, (iii) PAN, (iv) Nylon – 6.
Answer:
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 1

Question 47.
Give examples each of (i) addition polymers (ii) condensation polymer, (iii) copolymer.
Answer:
(i) Polythene, PVC
(ii) Buna – S, Buna – N
(iii) Nylon 6, Nylon 6, 6

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 48.
Write the names of the structure of monomers of the following polymers, (i) Buna – S, (ii) Neoprene.
Answer:
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 2

Question 49.
What is the repeating unit in the condensation polymer by combining HOOC—CH2—CFE —COOH (succinic acid) and H2NCH2CH2NH2 (ethylenediamine).
Answer:
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 3

Question 50.
Differentiate between molecular structure and behavior of thermoplastic and thermosetting plastic. Give one example of each type.
Answer:

Thermoplastic polymers Thermosetting polymers
These polymers soften and melt on heating. These polymers do not soften on heating but rather becomes hard. On prolonged heating, these start burning.
These polymers can be remoulded, recast and reshaped. These polymers cannot be remoulded or reshaped
They are less brittle and soluble in same organic solvents. They are move brittle and insoluble in organic solvents.
These polymers, usually have linear structures. These polymers have three dimensional cross linked structures.
eg: Polyethylene, PVC, Teflon, Nylon etc. eg: Bakelite, urea formaldehyde, resin, terelene etc.

Question 51.
What are the monomeric repeating units of Nylon 6, Nylon 6, 6?
Answer:
The monomeric repeating unit of nylon 6 is TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 4 which is derived from caprolactam.
The monomer repeating unit of nylon 6, 6 is derived from two monomers hexamethylenetetramine and adipic acid.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 5

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 52.
Name a synthetic polymer that is an amide.
Answer:
Nylon 6, 6.

Question 53.
Mention which of the following are addition polymers, (i) Terelene, (ii) Nylon 6, 6 (iii) Neoprene, (iv) Teflon.
Answer:
Neoprene and teflon are addition polymers.

Question 54.
What are biodegradable polymers?
Answer:
Polymers that disintegrate by themselves in biological systems during & certain period of time by oxidation are called biodegradable polymers, eg: PHBV. i.e., poly-β- hydroxybutyrate – co-β hydroxy valerate.

Question 55.
How are polymer classification based on forces operating between their molecule?
Answer:
Classification based as sources:

  1. Natural polymer: Polymers are found in nature, mostly in plants, and animals are called natural polymers, eg: proteins and natural rubber, cellulose, silk.
  2. Synthetic polymers: These are man-made polymers prepared in the laboratory, eg: polythene, PVC, teflon, nylon etc.
  3. Semi-synthetic polymers: Polymers which are obtained by making some modifications in natural polymer by artificial means, eg: Nitrocellulose, cellulose diacetate, viscose rayon etc.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 56.
Give the preparation of bakelite and its uses.
Answer:
Preparation of bakelite:
Monomer: Phenol and formaldehyde
Type of polymerization: Condensation polymerization.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 16
Uses: Navolac is used in paints. Soft bakelites are used for making glue for binding laminated wooden planks and in vanishes, Hard bakelites are used to prepare combs, pens etc…

Question 57.
Give the preparation and use of melamine.
Answer:
Preparation of melamine:
Monomer: Melamine and formaldehyde
Type of polymerisation: Condensation polymerisation
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 17
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 18
Uses: It is used for making unbreakable crockery.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 58.
Give examples for biodegradable polymer.
Answer:

  1. Polyhydroxy butyrate (PHB)
  2. Polyhydroxy butyrate-co-A-hydroxyl valerate (PHBV)
  3. Polyglycolic acid (PGA), Polylactic acid (PLA)
  4. Poly (∈ caprolactone) (PCL)

Question 59.
How is decron obtained from ethylene glycol and terephthalic acid?
Answer:
Decron is obtained by condensation polymerisation of ethylene glycol and terephthalic acid. The reaction is carried out at 420 – 460 K in the presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 6

Question 60.
How is urea-formaldehyde prepared?
Answer:
It is formed by the condensation polymerization of the monomers urea and formaldehyde.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 7

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 61.
How is PHBV prepared? Give equation mention its uses.
Answer:
It is the co-polymer of the monomers 3-hydroxybutyric acid and 3-hydroxypentanoic acid. In PHBV, the monomer units are joined by ester linkages.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 8

Question 62.
Identify the type of polymer where A and B are monomers.
—A—B—B—A—A—A— B—A.
Answer:
Co-polymer.

Question 63.
Why is bakelite a thermosetting polymer?
Answer:
Due to high degree of cross linking, bakelite cannot be reshaped on heating and hence, bakelite is a thermosetting polymer.

Question 64.
Give a brief account of vulcanization of rubber.
Answer:
Vulcanization is heating natural rubber with sulphur and an appropriate additive to improve its physical properties. On vulcanization sulfur forms cross-links at the reactive sites of the double bond and thus rubber gets stiffened.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 9
In this process, cross-linking of cis 1,4 polyisoprene chains through disulfide (—S—S) bond occurs.

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 65.
Describe the preparation of neoprene and mention its uses.
Answer:
Neoprene is formed by the free radical polymerization of the monomer, 2-chloro buta-1,3 diene(chloroprene).
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 10
It is used is the manufacture of chemical containers and conveyor belts.

Question 66.
What is the name of the polymer formed from the monomers acrylonitrile and butal, 3-diene? How it is prepared?
Answer:
Buna – N is a copolymer formed by the polymerization of acrylonitrile and butal, 3-diene.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 11

Question 67.
What type of polymer Buna-S is? Give its method of preparation.
Answer:
Preparation of Buna-S:
It is a copolymer. It is obtained by the polymerization of buta-1,3-diene and styrene in the ratio 3:1 in the presence of sodium.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 12

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

Question 68.
Name the polymer formed by the copolymerization of glycine and aminocaproic acid. How it is prepared?
Answer:
Nylon 2 – Nylon 6.
Preparation: By condensation polymerization of the monomers glycine and aminocaproic acid.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 13

Choose the correct answer:

1. Which one of the following antacids is an antihistamine?
(a) Ranitidine
(b) Lansoprazole
(c) Terfen adine
(d) Luminal
Answer:
(a)

2. Which of the following is / are nurologically active drug?
(a) Aspirin
(b) Phenelzine
(c) Heroin
(d) all the above
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

3. Antiseptic chloroxylenol is:
(a) 4 – chloro, 3, 5 dimethyl phenol
(b) 3 – chloro, 4, 5 dimethyl phenol
(c) 4 – chloro, 2, 5 dimethyl phenol
(d) 5 – chloro, 3, 4 dimethyl phenol
Answer:
(a)

4. Structurally a biodegradable detergent should contain a:
(a) normal alkyl chain
(b) branched alkyl chain
(c) phenyl side chain
(d) cyclohexyl side chain
Answer:
(a)

5. Which of the following statements is not correct?
(a) Some antiseptics can be added to soap.
(b) Dilute solutions of disinfectants can be used as antiseptic.
(c) Disinfectants are antimicrobial drugs.
(d) Antiseptic medicine can be infected.
Answer:
(d)

6. The most useful classification of drugs for medicinal chemists is:
(a) on the basis of chemical structure
(b) on the basis of drug action
(c) on the basis of molecular targets
(d) on the basis of active drug.
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

7. A compound that causes general anti¬depressant action on the central nervous system belongs to the class of:
(a) analgesics
(b) tranquilizers
(c) narcotic analgesics
(d) antihistamines
Answer:
(b)

8. Compound which is added to soap to inpart antiseptic properties is:
(a) sodium lauryl sulphate
(b) sodium do decyl benzene sulphonate
(c) resin
(d) bithional
Answer:
(d)

9. Glycerol is added to soap. Its function is:
(a) as a filler
(b) to increase lathering
(c) to prevent rapid drying
(d) to make soap granules
Answer:
(c)

10. Polyethylene glycols are used in the preparation of which type of detergents?
(a) Cationic detergents
(b) Anionic detergents
(c) Non-ionic detergents
(d) Soaps
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

11. Which of the following is employed as anti-histamine?
(a) Omeprazole
(b) Chloroampinicol
(c) Diphenylhydramine
(d) Norethindrone
Answer:
(c)

12. Tincture of iodine is:
(a) aqueous solution of I2
(b) solution of I2 in KI
(c) alcoholic solution of I2
(d) aqueous solution of KI
Answer:
(c)

13. Which among the following is not an antibiotic?
(a) Penicillin
(b) Oxytocin
(c) Erythromycin
(d) Tetracyclin
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

14. Match entries in column I with appropriate entries in column II.
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 14
TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life 15
Answer:
(c)

15. Which of the following is not an antimicrobial?
(a) Salvarsan
(b) Sulphanilamide
(c) Prontsil
(d) Paracetamol
Answer:
(d)

16. Which of the following is not a semi synthetic polymer?
(a) Cis poly isoprene
(b) Cellulose nitrate
(c) Cellulose acetate
(d) Vulcanised rubber
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

17. Which of the following polymers is prepared by condensation polymerisation?
(a) Styrene
(b) Nylon 6,6
(c) Teflon
(d) Rubber
Answer:
(b)

18. Which of the following is a chain growth polymer?
(a) Starch
(b) Nucleic acid
(c) Polystyrene
(d) Proteins
Answer:
(c)

19. Terelene is a condensation polymer of ethylene glycol and :
(a) benzoic acid
(b) pthalic acid
(c) salicylic acid
(d) terepthalic acid
Answer:
(d)

20. Which one of the following is a copolymer formed by condensation polymerisation?
(a) Terelene
(b) Buna – S
(c) Buna – N
(d) Neoprene
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

21. Bakelite is obtained from phenol by the reaction with:
(a) HCHO
(b) (CH2OH)2
(c) CH3CHO
(d) CH3COCH3
Answer:
(a)

22. Which of the following statements is not true?
(a) B una – S is a copolymer of butadiene and styrene.
(b) Natural rubber is a 1,4-polymer of isoprene.
(c) In vulcanisation the formation of sulphur bridges between different chains makes rubber harder and stronger.
(d) Natural rubber has trans configuration at every double bond.
Answer:
(d)

23. Teflon, styron and neoprene are all:
(a) copolymers
(b) condensation polymers
(c) homo polymers
(d) monomers
Answer:
(c)

24. Which of the following sets contain only thermoplastics?
(a) Polythene, bakelite, nylon-6
(b) Glyptal, melane, PAN
(c) PVC, PMMA, Polystyrene
(d) Polypropylene urea formaldehyde, teflon
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 15 Chemistry in Everyday Life

25. Which of the following sets contain only co-polymers?
(a) SBR, Glyptal, Nylon 6,6
(b) Nylon 6, Butyl rubber, Neoprene
(c) Poly ethylene, polyester, PVC
(d) Melmac, Bakelite, Teflon.
Answer:
(a)

26. Which of the following are not thermosetting polymers?
(1) Bakelite
(2) Polystyrene
(3) PVC
(4) Melmac
(a) 1, 2
(b) 2,3
(c) 2, 4
(d) 3, 4
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Students get through the TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be hydrolysed to simple molecules. Their general molecular formula is Cn(H2O)n where n = 3 to 7. There are two types. Those which contain an aldehyde group are called aldoses and those which contain a ketonic group are called ketoses. They are further classified as trioses, tetroses, pentose, hexoses etc according to as they contain 3, 4, 5, 6 carbon atom respectively.

Question 2.
Classify the following as monosaccharides and disaccharides:
Ribose: 2 – deoxy ribose; maltose, galactose, fructose, and lactose.
Answer:
Monosaccharides: Ribose, 2-Deoxyribose, galactose and fructose.
Disaccharides: Maltose and lactose.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 3.
What are the products of hydrolysis of (i) sucrose (ii) lactose.
Answer:
Both sucrose and lactoses are disaccharides. Sucrose on hydrolysis gives one molecule of each of glucose and fructose.
Question 1. What are monosaccharides? Answer: Monosaccharides are carbohydrates which cannot be hydrolysed to simple molecules. Their general molecular formula is C<sub>n</sub>(H<sub>2</sub>O)<sub>n</sub> where n = 3 to 7. There are two types. Those which contain an aldehyde group are called aldoses and those which contain a ketonic group are called ketoses. They are further classified as trioses, tetroses, pentose, hexoses etc according to as they contain 3, 4, 5, 6 carbon atom respectively. Question 2. Classify the following as monosaccharides and disaccharides: Ribose: 2 - deoxy ribose; maltose, galactose, fructose, and lactose. Answer: Monosaccharides: Ribose, 2-Deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose. Question 3. What are the products of hydrolysis of (i) sucrose (ii) lactose. Answer: Both sucrose and lactoses are disaccharides. Sucrose on hydrolysis gives one molecule of each of glucose and fructose. Lactose on hydrolysis gives one molecule of glucose and galactose. Question 4. How do you explain the presence of all six carbon atoms in glucose in a straight chain? Answer: Since, n - hexane has six carbon atoms connected in a straight chain, therefore glucose also has six carbon atoms connected in a straight chain. Question 5. The letter ‘D’ or ‘L’ before the name of a stereo isomer indicates the correlation of configuration of that particular stereo isomers. This refers to their relationship with one of the isomers of glyceraldehyde. Predict whether the compound is has D or L configuration. Answer: Since the configuration of the ‘OH’ group at the penultimate chiral carbon (last but one or C<sub>5</sub>) is towards left, therefore the given compound has ‘L’ configuration. Question 6. How do you explain the presence of five - OH groups in a glucose molecule? Answer: On acetylation with acetic anhydride in presence of pyridine, or a few drops of cone. H<sub>2</sub>SO<sub>4</sub>, it is converted into glucose penta acetate. This confirms the presence of five OH groups. Question 7. Why does compound (A) given below does not form an oxime? Answer: Glucose penta acetate does not have a free OH group at carbon atom 1(C<sub>1</sub>) and hence cannot be converted into the open chain form having a free CHO group therefore, glucose penta acetate does not form an oxime. Question 8. How do you explain the presence of an aldehyde group in glucose molecule? Answer: Glucose reacts with hydroxyl amine, (NH<sub>2</sub>OH) to form an oxime, and adds one molecule of hydrogen cyanide (HCN) to give a cyano hydrin. Therefore, glucose contains a carbonyl group which can be either an aldehyde or ketone. On mild oxidation with bromine water, it gives a carboxylic acid i.e., gluconic acid containing same six carbon atoms present in glucose. This indicates that the carbonyl group present in glucose is an aldehyde group. Question 9. How do you distinguish 1° and 2° alcoholic groups present in glucose? Explains with reactions. Answer: Consider the structure of glucose The ‘OH’ group present on the terminal carbon (C<sub>6</sub>) is called the primary alcoholic group while all other four ‘OH’ groups present in C<sub>2</sub>, C<sub>3</sub>, C<sub>4</sub> and C<sub>5</sub> are called secondary alcoholic groups. 1° alcoholic groups are readily oxidised to carboxylic acids but 2° alcoholic groups undergo oxidation under drastic conditions. For example glucose on oxidation cone. HNO<sub>3</sub> gives a dicarboxylic acid, saccharic acid, having the same number of carbon atoms as in glucose. This indicates that glucose contains one 1° alcoholic group while the remaining four are 2° alcoholic groups. Question 10. Write the reactions of D glucose which can’t be explained by its open chain structure. Reactions of D- glucose which can’t be explained by its open chain structure. Answer: (i) Glucose does not form NaHSO<sub>3</sub> additions product, aldehyde-ammonia, 2-4 DNP derivative and does not respond to schiff reagent test. These are the characteristic reactions of aldehyde. (ii) Glucose reacts with NH<sub>2</sub>OH to form an oxime but glucose penta acetate does not. This shows that the aldehyde group is absent in glucose penta acetate. (iii) D(+) glucose exists in two isomeric forms i.e., α glucose and β glucose. Both these undergo mutarotation. (iv) D(+) glucose forms two isomeric methyl glycosides. (Aldehydes normally react with two moles of methanol per mole of aldehyde where as one mole of glucose reacts with one mole of methanol to form a mixture of two methyl D - glucosides) The two methyl glucosides behave as acetates. Question 11. Write the cyclic structure of glucose. Answer: The cyclic structures of a -D glucose and β -D glucose are as follows: Question 12. What is the structure feature characterising reducing sugars? Answer: The main structural feature of reducing sugars is the presence of an aldehyde group (— CHO) such as in glucose, mannose, galactose or a ketol group (—CO—CH<sub>2</sub>OH) as present in fructose. Question 13. Fructose contains a keto group but it still reduces Tollen’s reagent. Explain. Answer: Under basic conditions of Tollen’s reagent fructose undergoes a rearrangement as shown below: The same equilibrium is obtained even of one strands with D(+) mannose or D(-) fructose, i.e., Fructose gives an equilibrium mixture of fructose, glucose and mannose. Since both glucose and mannose contain —CHO group, they reduce Tollen’s reagent. Question 14. Explain the term glycosidic linkage with an example. Answer: The ethereal or oxygen linkage through two monosaccharides are joined together by the loss of a molecule of water to form a molecule of a disaccharide is called a glycosidic linkage. The glycosidic linkage of two α - D glucose in maltose is given below: Question 15. What happens when D - glucose is treated with the following reagents (i) HI (ii) Bromine water (iii) HNO<sub>3</sub>. Answer: Question 16. Glucose and fructose give the same osazone. Give reason. Answer: During osazone formation the reaction occurs only at C<sub>1</sub> and C<sub>2</sub>. As glucose and fructose differ from each other only in the arrangement of atoms C<sub>1</sub> and C<sub>2</sub> they give the same osazones. Question 17. Name two components of starch. How do they differ from each other structurally? Answer: Starch is a polymer of a - glucose and consists of two components amylose and amylopectin. Amylose is a long unbranched chain with 200 - 1000 a D(+) glucose units held by C<sub>1</sub> - C<sub>4</sub> glycosidic linkage. Amylopectin is a branched chain polymer of a D glucose units in which chain is formed by C<sub>1</sub> - C<sub>4</sub> glycosidic linkage branching occurs by C<sub>1</sub> - C<sub>6</sub> glycosidic linkage. Question 18. Name the reaction which proves the presence of carbonyl group in fructose. Answer: Fructose reacts with hydroxylamine (NH<sub>2</sub>OH) and hydrogen cyanide (HCN). These reactions . prove the presence of carbonyl group in fructose. Question 19. Explain the reaction which indicates the presence of a carbonyl group in fructose. Answer: Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C<sub>2</sub>. This confirms the presence of keto group. Question 20. Oxidation of fructose with nitric acid gives glycolic acid and tartaric acid. What information that this reaction gives to establish the structure of fructose. Answer: On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose. This shows that a keto group is present in C<sub>2</sub>. It also shows the presence of 1° alcoholic groups at C<sub>1</sub> and C<sub>6</sub>. Question 21. Explain why sucrose is called ‘invert’ sugar. Answer: Sucrose (+ 66.6°) and glucose (+ 52.5°) are dextrorotatory compounds while fructose is levo rotatory (- 92.4°). During hydrolysis of sucrose the optical rotation of the reaction,mixture changes from dextro to levo. Hence, sucrose is also called as invert sugar. Question 22. Write the open chain structure of D(+) fructose and indicate the asymmetric carbon atoms present in it. Answer: Question 23. Briefly discuss the cyclic structure of fructose. Answer: (a) Mechanism of cyclic structure; (b) Cyclic structure of fructose Like glucose, fructose also forms cyclic form. Unlike glucose it forms a five membered ring similar to furan. Hence it is called furanose form. When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form. Question 24. Explain why sucrose is called a non reducing sugar. Answer: In sucrose, C<sub>1</sub> of α-D-glucose is joined to C<sub>2</sub> of -D-fructose. The glycosidic bond thus formed is called α-1,2 glycosidic bond. Since, the both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non-reducing sugar. Question 25. Based on its cyclic structure, explain why lactose is a reducing sugar. Answer: On hydrolysis, it yields galactose and glucose. Here, the β-D-galactose and β-D-glucose are linked by β -1,4 glycosidic bond as shown in the figure. The aldehyde carbon is not involved in the glycosidic bondbence; it retains its reducing property and is called a reducing sugar. Question 26. Maltose is a reducing sugar. Explain. Answer: Maltose consists two molecules of α-D-glucose units linked by an α -1,4 glycosidic bond between anomeric carbon of one unit and C-4 of the other unit. Since one of the glucose has the carbonyl group intact it is also acts as a reducing sugar. Question 27. Mention the uses of cellulose. Answer: Cellulose is used extensively in manufacturing paper, cellulose fibres, rayon, explosive, (Gun cotton - Nitrated ester of cellulose) and so on. Humans cannot use cellulose as food because our digestive systems do not contain the necessary enzymes (glycosidases or cellulases) that can hydrolyse the cellulose. Question 28. Briefly explain the structure of cellulose. Answer: Cellulose is a straight chain polysaccharide. The glucose molecules are linked by β(1, 4) glycosidic bond. Question 29. Following two amino acids lysine and glutamine dipeptide linkage. What are the two possible dipeptides? Answer: Question 30. Give the. important uses of carbohydrates. Answer: (i) Carbohydrates, widely distributed in plants and animals, acts mainly as energy sources and structural polymers. (ii) Carbohydrate is stored in the body as glycogen and in plant as starch. (iii) Carbohydrates such as cellulose which is the primary components of plant cell wall, is used to make paper, furniture (wood) and cloths (cotton). (iv) Simple sugar glucose serves as an instant source of energy. (v) Ribose sugars are one of the components of nucleic acids. (vi) Modified carbohydrates such as hyaluronate (glycosaminoglycans) act as shock absorber and lubricant. Question 31. What is glycogen? How is it different from starch? Answer: Glycogen is a carbohydrates stored in animal body. Starch is a mixture of two components a water soluble compounds called amylose (15 - 20%) and water insoluble amylopectin (80 - 85%) chemically, amylose is a long unbranched chain with 200 - 1000 a - D (+) glucose units held by C<sub>1</sub> - C<sub>4</sub> glycosidic linkage. But both glycogen and amylopectin are branched polymers of α - D glucose rather glucogen is more highly branched than amylopectin. Whereas amylopectin chain consist of 20 - 25 glucose units glucose chain consists of 10 - 14 glucose units. Question 32. What is the basic structural difference between starch and cellulose? Answer: Starch contains amylose and amylopectin. Amylose is a linear polymer of α D glucose whereas cellulose is a linear polymer of β - D - glucose. In amylose, C-1 of one glucose unit is connected to C-4 of the other through glycosidic linkage. Cellulose is a straight chain polysaccharides composed of only β -D glucose units which are joined by glycosidic linkages between C-1 of one glucose unit and C-4 of the next glucose unit. Question 33. What are essential and non essential amino acids? Give two examples of each type. Answer: α - amino acids which are required for health and growth of human beings but are not synthesised by the human body are known as essential amino acids, eg: valine, leucine, phenylalanine etc., On the other hand, α amino acids which are needed for health and growth of human being and are synthesised by the body are called non essential amino acids, eg: glycine, alanine, aspartic acid etc. Question 34. Give examples for fibrous and globular proteins. Answer: Fibrous protein: Keratin in skin, hair, nails, and wood, collagen in tendous. Fibroin in silk, myosin in muscles. Globular protein: All enzymes, hormones like insulin, from pancreas, thyroglobulin from thyroid gland antibodies, haemoglobin. Question 35. What is isoelectic point? Explain with a suitable examples. Answer: The pH at which there is no net migration of the amino acid under the influence of an applied electrified is called isoelectric point. Question 36. What is peptide bond? Answer: The covalent bond —NH—CO— formed between —NH<sub>2</sub> group of one amino acid and —COOH group of the other with the elimination of a molecule of water is called a peptide bond. Question 37. How are epimers differ from anomers? Answer: Carbohydrates which differ in configuration at the glycosidic carbon (i.e., C<sub>1</sub> in aldoses and C<sub>2</sub> in ketoses) are called anomers while those differ in configuration at any carbon other than glycosidic carbon are called epimers. For example α - D glucose and β-D glucose are anomers since they differ in configuration at C<sub>1</sub> while glucose and mannose are called epimers since they differ in configuration at C<sub>2</sub> (other than glycosidic carbon C<sub>1</sub>). In other words glucose and mannose are C<sub>2</sub> epimers. Similarly glucose and galactose are C<sub>4</sub> epimers. Question 38. Differentiate between globular and fibrous proteins. Answer: Question 39. Describe (i) primary structure (ii) secondary structure (iii) tertiary structure (iv) quaternary structure of proteins. Answer: (i) Primary structure: Proteins may contain one or more polypeptide chains. Each polypeptide chain has a large number of α - amino acids which are linked to one another in a specific sequence. The specific sequence in which the various α - amino acids present in a protein are linked to one another is called its primary structure. Any change in the sequence of α - amino acids creates a different protein. (ii) Secondary structure: It refers to shape in which a long polypeptide chain exists. A protein may assume a - helix structure or β - pleated structure. The α - helix structure results due to regular coiling of polypeptide chain which is stabilized by intramolecular hydrogen bonding. In β pleated sheet structure, all peptide chains are stretched to a nearly maximum extention and then arranged side by side and held together by intramolecular hydrogen bonding. (iii) Tertiary structure: The tertiary structure of proteins represent overall folding of the polypeptide chains, i.e., further folding of the secondary structure. The main force which stabilises 2° and 3° of proteins are hydrogen bonds, disulphide linkages, Vanderwaals forces of attraction and electrostatic force of attraction. (iv) Quaternary structure: Same of the proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these sub units with respect to each other is called quaternary structure. Question 40. What are the common types of secondary structure of protein? Answer: The conformations which the polypepetide chains assume as a result of hydrogen bonding is called the secondary structure of protein. The two types are (i) α helix and (ii) β pleated structure Question 41. What types of bonding helps in stabilising α - helix structure of proteins? Answer: The α helix structure of proteins is stabilised by intramolecular hydrogen bonding between C = O of one amino acid residue and —N—H of the fourth amino acid residue in the chain. Question 42. What is the effect of denaturation on the structure of proteins? Answer: During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation the globular protein (soluble in H<sub>2</sub>O) are converted to fibrous proteins (insoluble in H<sub>2</sub>O) and their biological activity is lost. Question 43. Mention the causes for the denaturation of proteins. Answer: The denaturation of proteins occurs when the protein is exposed to higher temperature the presence of certain such as urea, alteration of pH, ionic strength etc. Question 44. Mention the importance of protein. Answer: (i) All biochemical reactions occur in the living systems are catalysed by the catalytic proteins called enzymes. (ii) Proteins such as keratin, collagen acts as structural back bones. (iii) Proteins are used for transporting molecules (Haemoglobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters). (iv) Antibodies help the body to fight various diseases. (v) Proteins are used as messengers to coordinate many functions. Insulin and glucagon controls the glucose level in the blood. (vi) Proteins act as receptors that detect presence of certain signal molecules and activate the proper response. (vii) Proteins are also used to store metals such as iron (Ferritin) etc. Question 45. What are enzymes? What is the most important reason for their specific action? Answer: Enzymes are biomolecule& which catalyst biological reactions. Chemically they are globular proteins (water soluble) which have very high molecular near ranging from 15,000 to 1,000,000 mol<sup>-1</sup>. Enzymes are highly specific in their actions. The specificity is due to the presence of active sites of definite size and shape an their surfaces so that only specific substrates can fit into them. This specific binding leads to the formation of enzyme substrates complex which accounts for high specificity. Question 46. Name the enzyme that converts (i) carbonic acid to CO<sub>2</sub> and H<sub>2</sub>O (ii) hydrolysis of surcose to fructose and glucose. (iii) hydrolysis of lactose to glucose and galactose. Answer: (i) Carbanic anhydrase enzyme catalyses the interconversion of carbonic acid to water and carbon-dioxide. (ii) Sucrose enzyme catalyses the hydrolysis of sucrose to fructose and glucose. (iii) Lactase enzyme hydrolysis the lactose to glucose and galactose. Question 47. Explain the mechanism of enzyme action. Answer: Mechanism of Enzymatic action: The general scheme is represented as Step 1: Binding of enzyme (E) to the substrate (S) to form complex (ES) is Enzyme - substrate complex. Step 2: Product formation in the complex Step 3: Release of enzyme product complex and leaving the enzyme as unchanged. Question 48. What are lipids? How are they classified? Answer: Lipids are the principal components of cell membranes including cell walls. They act as energy source for living systems. Classification of lipids: Based on their structures Lipids can be classified as simple lipids, compounds lipids and derived lipids. Simple lipids can be further classified into fats, which are esters of long chain fatty acids with glycerol (triglycerides) and waxes which are the esters of fatty acids with long chain monohydric alcohols (Bees wax). Compounds lipids are the esters of simple fatty acid with glycerol which contain additional groups. Based on the groups attached, they are further classified into phospholipids, glycolipids and lipoproteins. Phospholipids contain a phospho-ester linkage while the glycolipids contain a sugar molecule attached. The lipoproteins are complexes of lipid with proteins. Question 49. How are vitamins classified? Answer: Vitamins are classified depending on their solubility in water. (i) Water soluble Vitamins: These include Vitamin B - complex (B<sub>1</sub>, B<sub>2</sub>, B<sub>5</sub>, i.e., nicotinic acid, B<sub>5</sub>, B<sub>12</sub> and vitamin C) (ii) Water insoluble Vitamins (or fat soluble Vitamins): These include Vitamins D, A, E and K. They are stored in liver and in fat storing tissues. Question 50. What are coenzymes and prothetic group? Answer: coenzymes are loosely held to the protein and can be easily separated by dialysis. Prothetic group: These are tightly held to the protein molecule by covalent bonds. They can be separated only by hydrolysis. Question 51. What are nucleic acids? Mention their two important functions? Answer: Nucleic acids are polynucleotides i.e., they are formed by the condensation of thousand molecules of nucleotides with the elimination of water molecules. Nucleic acids are two types: deoxyribonucleic acid (DNA) and ribonucleiqacid (RNA). The two main functions of nucleic acids are: (i) DNA is responsible for transmission of heredity effects from one generation to another. This is due to the unique property of replication during cell division as a result of which two identical DNA strands are transmitted to daughter cells. (ii) DNA and RNA are responsible for protein synthesis needed for growth and maintenance of our body. Actually proteins are synthesised by various RNA molecules (rRNA, mRNA and tRNA) in the cell but the message for the synthesis of a particular protein is given by DNA molecules. Question 52. Write the structure of nucleic acids. Answer: Primary structure: The primary structure of nucleic acid refers to the sequence in which the four nitrogen bases (A, G, C and T in DNA or A, G, C and U in RNA) are attached to the sugar - phosphate back bone of a nucleotide chain. Secondary structure: DNA consists of two strands of polynucleotides coiled around each other in form of a double helix. The back bone of each strand consists of sugar - phosphate units and the base unit of each strand are pointed into the interior of the helix and are linked together through hydrogen bonds. Whereas G and C are held by three H - bonds, A and T are held by H - bonds. Question 53. Write down the structure of sugar present in DNA. Answer: DNA contains β D - 2 deoxyribose as sugar. In structure in the furanose form is Question 54. What purine and pyramidine bases are present in DNA and RNA? Answer: Purines: adenine and guanine Pyramidine: Cytosine and thymine are present in DNA. Cytosine and uracil is present in RNA. Question 55. What is a nucleoside? Answer: A nucleoside consists of two components, i.e., a nitrogenous base (purine or pyramidine)' and a five carbon sugar (ribose or deoxy ribose). It is obtained when the nitrogeneous base is attached, to C<sub>1</sub> of sugar by a β - linkage. Thus, in general, a nucleoside may be represented as base - super. Examples of . nucleosides are adenosine, cytidine etc., Question 56. What is a nucleotide? Answer: A nucleotide consists of all the three basic components of nucleic acids, i.e., a nitrogenous purine or a pyramidine base, a five carbon ' ribose or deoxyribose and phosphoric acid. A nucleotide is obtained when a nitrogenous base is attached to C<sub>1</sub> of the sugar by a β linkage and a nucleotide thus obtained when C<sub>5</sub>—OH of the sugar is esterified with phosphoric acid. Thus,in general, nucleotides may be represented as base - sugar - phosphate. Question 57. What type of linkage holds together the monomers of DNA? Answer: The monomers of DNA are polydeoxy ribo nucleotides. These are held together by H bonds. There are three H bonds between guanine and cytosine (G ≡ C) and two between adenine and thymine (A = T). Question 58. The two strands in DNA are not identical but complimentary explain. Answer: The two strands in DNA molecule are hold together through H — bonds between purine base of one strand and pyramidine base of the other and vice versa. Because of different sizes and geometries of the base, the only possible pairing in DNA are G(guanine) and C(cytosine) through two hydrogen bonds i.e., bases (C ≡ G) and between A(adenine) and thymine (T) through two hydrogen bonds (i.e., A = T) due to this base - pairing principle the sequences of bases in one strands automatically fixes the sequence of bases in the other strand. Thus the two strands are not identical but complimentary. Question 59. Mention the functional difference DNA and RNA. Answer: Question 60. What is DNA finger printing? Explain. Answer: The DNA finger print is unique for every person. By using this method, individual specific variation in human DNA can be detected. Question 61. When RNA is hydrolysed, there is not relationship among the quantities of different bases obtained. What does it suggest about the structure of RNA? Answer: A DNA molecule has two strands in which four complimentary bases pair each other i.e., cytosine (C) always pairs with guanine (G) while Thymine (T) always pair with adenine(A). Therefore when DNA molecule is hydrolysed, the molar amount of cytosine is equal to that of guanine and that of adenine is always equal to that of thymine. Since RNA has no such relationship between the quantities of the four bases (C, G, A and U) obtained, therefore, based on the base pairing principle, i.e., A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.
Lactose on hydrolysis gives one molecule of glucose and galactose.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 2

Question 4.
How do you explain the presence of all six carbon atoms in glucose in a straight chain?
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 3
Since, n – hexane has six carbon atoms connected in a straight chain, therefore glucose also has six carbon atoms connected in a straight chain.

Question 5.
The letter ‘D’ or ‘L’ before the name of a stereo isomer indicates the correlation of configuration of that particular stereo isomers. This refers to their relationship with one of the isomers of glyceraldehyde. Predict whether the compound is has D or L configuration.
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 4
Since the configuration of the ‘OH’ group at the penultimate chiral carbon (last but one or C5) is towards left, therefore the given compound has ‘L’ configuration.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 6.
How do you explain the presence of five – OH groups in a glucose molecule?
Answer:
On acetylation with acetic anhydride in presence of pyridine, or a few drops of cone. H2SO4, it is converted into glucose penta acetate.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 5
This confirms the presence of five OH groups.

Question 7.
Why does compound (A) given below does not form an oxime?
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 6
Answer:
Glucose penta acetate does not have a free OH group at carbon atom 1(C1) and hence cannot be converted into the open chain form having a free CHO group therefore, glucose penta acetate does not form an oxime.

Question 8.
How do you explain the presence of an aldehyde group in glucose molecule?
Answer:
Glucose reacts with hydroxyl amine, (NH2OH) to form an oxime, and adds one molecule of hydrogen cyanide (HCN) to give a cyano hydrin. Therefore, glucose contains a carbonyl group which can be either an aldehyde or ketone. On mild oxidation with bromine water, it gives a carboxylic acid i.e., gluconic acid containing same six carbon atoms present in glucose. This indicates that the carbonyl group present in glucose is an aldehyde group.

Question 9.
How do you distinguish 1° and 2° alcoholic groups present in glucose? Explains with reactions.
Answer:
Consider the structure of glucose
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 7
The ‘OH’ group present on the terminal carbon (C6) is called the primary alcoholic group while all other four ‘OH’ groups present in C2, C3, C4 and C5 are called secondary alcoholic groups. 1° alcoholic groups are readily oxidised to carboxylic acids but 2° alcoholic groups undergo oxidation under drastic conditions. For example glucose on oxidation cone. HNO3 gives a dicarboxylic acid, saccharic acid, having the same number of carbon atoms as in glucose. This indicates that glucose contains one 1° alcoholic group while the remaining four are 2° alcoholic groups.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 8

Question 10.
Write the reactions of D glucose which can’t be explained by its open chain structure. Reactions of D- glucose which can’t be explained by its open chain structure.
Answer:
(i) Glucose does not form NaHSO3 additions product, aldehyde-ammonia, 2-4 DNP derivative and does not respond to schiff reagent test. These are the characteristic reactions of aldehyde.
(ii) Glucose reacts with NH2OH to form an oxime but glucose penta acetate does not. This shows that the aldehyde group is absent in glucose penta acetate.
(iii) D(+) glucose exists in two isomeric forms i.e., α glucose and β glucose. Both these undergo mutarotation.
(iv) D(+) glucose forms two isomeric methyl glycosides. (Aldehydes normally react with two moles of methanol per mole of aldehyde where as one mole of glucose reacts with one mole of methanol to form a mixture of two methyl D – glucosides)
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 9
The two methyl glucosides behave as acetates.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 11.
Write the cyclic structure of glucose.
Answer:
The cyclic structures of a -D glucose and β -D glucose are as follows:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 10

Question 12.
What is the structure feature characterising reducing sugars?
Answer:
The main structural feature of reducing sugars is the presence of an aldehyde group (— CHO) such as in glucose, mannose, galactose or a ketol group (—CO—CH2OH) as present in fructose.

Question 13.
Fructose contains a keto group but it still reduces Tollen’s reagent. Explain.
Answer:
Under basic conditions of Tollen’s reagent fructose undergoes a rearrangement as shown below:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 11
The same equilibrium is obtained even of one strands with D(+) mannose or D(-) fructose, i.e., Fructose gives an equilibrium mixture of fructose, glucose and mannose. Since both glucose and mannose contain —CHO group, they reduce Tollen’s reagent.

Question 14.
Explain the term glycosidic linkage with an example.
Answer:
The ethereal or oxygen linkage through two monosaccharides are joined together by the loss of a molecule of water to form a molecule of a disaccharide is called a glycosidic linkage. The glycosidic linkage of two α – D glucose in maltose is given below:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 12

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 15.
What happens when D – glucose is treated with the following reagents (i) HI (ii) Bromine water (iii) HNO3.
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 13

Question 16.
Glucose and fructose give the same osazone. Give reason.
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 14
During osazone formation the reaction occurs only at C1 and C2. As glucose and fructose differ from each other only in the arrangement of atoms C1 and C2 they give the same osazones.

Question 17.
Name two components of starch. How do they differ from each other structurally?
Answer:
Starch is a polymer of a – glucose and consists of two components amylose and amylopectin. Amylose is a long unbranched chain with 200 – 1000 a D(+) glucose units held by C1 – C4 glycosidic linkage.
Amylopectin is a branched chain polymer of a D glucose units in which chain is formed by C1 – C4 glycosidic linkage branching occurs by C1 – C6 glycosidic linkage.

Question 18.
Name the reaction which proves the presence of carbonyl group in fructose.
Answer:
Fructose reacts with hydroxylamine (NH2OH) and hydrogen cyanide (HCN). These reactions . prove the presence of carbonyl group in fructose.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 19.
Explain the reaction which indicates the presence of a carbonyl group in fructose.
Answer:
Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C2. This confirms the presence of keto group.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 15

Question 20.
Oxidation of fructose with nitric acid gives glycolic acid and tartaric acid. What information that this reaction gives to establish the structure of fructose.
Answer:
On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 16
This shows that a keto group is present in C2. It also shows the presence of 1° alcoholic groups at C1 and C6.

Question 21.
Explain why sucrose is called ‘invert’ sugar.
Answer:
Sucrose (+ 66.6°) and glucose (+ 52.5°) are dextrorotatory compounds while fructose is levo rotatory (- 92.4°). During hydrolysis of sucrose the optical rotation of the reaction,mixture changes from dextro to levo. Hence, sucrose is also called as invert sugar.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 22.
Write the open chain structure of D(+) fructose and indicate the asymmetric carbon atoms present in it.
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 17

Question 23.
Briefly discuss the cyclic structure of fructose.
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 18
Like glucose, fructose also forms cyclic form. Unlike glucose it forms a five membered ring similar to furan. Hence it is called furanose form. When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 19TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 19

Question 24.
Explain why sucrose is called a non reducing sugar.
Answer:
In sucrose, C1 of α-D-glucose is joined to C2 of -D-fructose. The glycosidic bond thus formed is called α-1,2 glycosidic bond. Since, the both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non-reducing sugar.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 20

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 25.
Based on its cyclic structure, explain why lactose is a reducing sugar.
Answer:
On hydrolysis, it yields galactose and glucose. Here, the β-D-galactose and β-D-glucose are linked by β -1,4 glycosidic bond as shown in the figure. The aldehyde carbon is not involved in the glycosidic bondbence; it retains its reducing property and is called a reducing sugar.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 21

Question 26.
Maltose is a reducing sugar. Explain.
Answer:
Maltose consists two molecules of α-D-glucose units linked by an α -1,4 glycosidic bond between anomeric carbon of one unit and C-4 of the other unit. Since one of the glucose has the carbonyl group intact it is also acts as a reducing sugar.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 22

Question 27.
Mention the uses of cellulose.
Answer:
Cellulose is used extensively in manufacturing paper, cellulose fibres, rayon, explosive, (Gun cotton – Nitrated ester of cellulose) and so on. Humans cannot use cellulose as food because our digestive systems do not contain the necessary enzymes (glycosidases or cellulases) that can hydrolyse the cellulose.

Question 28.
Briefly explain the structure of cellulose.
Answer:
Cellulose is a straight chain polysaccharide.
The glucose molecules are linked by β(1, 4) glycosidic bond.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 23

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 29.
Following two amino acids lysine and glutamine dipeptide linkage. What are the two possible dipeptides?
Answer:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 24

Question 30.
Give the. important uses of carbohydrates.
Answer:

  1. Carbohydrates, widely distributed in plants and animals, acts mainly as energy sources and structural polymers.
  2. Carbohydrate is stored in the body as glycogen and in plant as starch.
  3. Carbohydrates such as cellulose which is the primary component of the plant cell wall is used to make paper, furniture (wood) and cloths (cotton).
  4. Simple sugar glucose serves as an instant source of energy.
  5. Ribose sugars are one of the components of nucleic acids.
  6. Modified carbohydrates such as hyaluronate (glycosaminoglycans) act as shock absorber and lubricant.

Question 31.
What is glycogen? How is it different from starch?
Answer:
Glycogen is a carbohydrates stored in animal body. Starch is a mixture of two components a water soluble compounds called amylose (15 – 20%) and water insoluble amylopectin
(80 – 85%) chemically, amylose is a long unbranched chain with 200 – 1000 a – D (+) glucose units held by C1 – C4 glycosidic linkage. But both glycogen and amylopectin are branched polymers of α – D glucose rather glucogen is more highly branched than amylopectin. Whereas amylopectin chain consist of 20 – 25 glucose units glucose chain consists of 10 – 14 glucose units.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 32.
What is the basic structural difference between starch and cellulose?
Answer:
Starch contains amylose and amylopectin. Amylose is a linear polymer of α D glucose whereas cellulose is a linear polymer of β – D – glucose. In amylose, C-1 of one glucose unit is connected to C-4 of the other through glycosidic linkage.
Cellulose is a straight chain polysaccharides composed of only β -D glucose units which are joined by glycosidic linkages between C-1 of one glucose unit and C-4 of the next glucose unit.

Question 33.
What are essential and non essential amino acids? Give two examples of each type.
Answer:
α – amino acids which are required for health and growth of human beings but are not synthesised by the human body are known as essential amino acids, eg: valine, leucine, phenylalanine etc.,
On the other hand, α amino acids which are needed for health and growth of human being and are synthesised by the body are called non essential amino acids, eg: glycine, alanine, aspartic acid etc.

Question 34.
Give examples for fibrous and globular proteins.
Answer:
Fibrous protein: Keratin in skin, hair, nails, and wood, collagen in tendous. Fibroin in silk, myosin in muscles.
Globular protein: All enzymes, hormones like insulin, from pancreas, thyroglobulin from thyroid gland antibodies, haemoglobin.

Question 35.
What is isoelectic point? Explain with a suitable examples.
Answer:
The pH at which there is no net migration of the amino acid under the influence of an applied electrified is called isoelectric point.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 25

Question 36.
What is peptide bond?
Answer:
The covalent bond —NH—CO— formed between —NH2 group of one amino acid and —COOH group of the other with the elimination of a molecule of water is called a peptide bond.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 26

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 37.
How are epimers differ from anomers?
Answer:
Carbohydrates which differ in configuration at the glycosidic carbon (i.e., C1 in aldoses and C2 in ketoses) are called anomers while those differ in configuration at any carbon other than glycosidic carbon are called epimers.
For example α – D glucose and β-D glucose are anomers since they differ in configuration at C1 while glucose and mannose are called epimers since they differ in configuration at C2 (other than glycosidic carbon C1). In other words glucose and mannose are C2 epimers. Similarly glucose and galactose are C4 epimers.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 27

Question 38.
Differentiate between globular and fibrous proteins.
Answer:

Globular protein Fibrous protein
Have almost spherical shape due to folding of polypeptide chains. Polypeptide chains consists of thread like molecule which tend to lie side by side to form fibres.
Soluble in water. Insoluble in water.
Sensitive to small changes of temper­ature and pH. Therefore they undergo denatura- tion on heating or an treatment with acid or base. Stable to moderate change of temperature and pH.
The interaction are
(i) disulphide bridging
(ii)  intermolecular hydrogen
(iii)  vanderwaals attraction
(iv)  dipolar attraction
Held together by disulphide bridges and weak intermolecular hydrogen bonds.

Question 39.
Describe (i) primary structure (ii) secondary structure (iii) tertiary structure (iv) quaternary structure of proteins.
Answer:
(i) Primary structure: Proteins may contain one or more polypeptide chains. Each polypeptide chain has a large number of α – amino acids which are linked to one another in a specific sequence. The specific sequence in which the various α – amino acids present in a protein are linked to one another is called its primary structure. Any change in the sequence of α – amino acids creates a different protein.
(ii) Secondary structure: It refers to shape in which a long polypeptide chain exists. A protein may assume a – helix structure or β – pleated structure. The α – helix structure results due to regular coiling of polypeptide chain which is stabilized by intramolecular hydrogen bonding. In β pleated sheet structure, all peptide chains are stretched to a nearly maximum extention and then arranged side by side and held together by intramolecular hydrogen bonding.
(iii) Tertiary structure: The tertiary structure of proteins represent overall folding of the polypeptide chains, i.e., further folding of the secondary structure. The main force which stabilises 2° and 3° of proteins are hydrogen bonds, disulphide linkages, Vanderwaals forces of attraction and electrostatic force of attraction.
(iv) Quaternary structure: Same of the proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these sub units with respect to each other is called quaternary structure.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 40.
What are the common types of secondary structure of protein?
Answer:
The conformations which the polypepetide chains assume as a result of hydrogen bonding is called the secondary structure of protein. The two types are

  1. α helix and
  2. β pleated structure

Question 41.
What types of bonding helps in stabilising α – helix structure of proteins?
Answer:
The α helix structure of proteins is stabilised by intramolecular hydrogen bonding between
C = O of one amino acid residue and —N—H of the fourth amino acid residue in the chain.

Question 42.
What is the effect of denaturation on the structure of proteins?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, the globular protein (soluble in H2O) are converted to fibrous proteins (insoluble in H2O) and their biological activity is lost.

Question 43.
Mention the causes for the denaturation of proteins.
Answer:
The denaturation of proteins occurs when the protein is exposed to the higher temperature the presence of certain such as urea, alteration of pH, ionic strength etc.

Question 44.
Mention the importance of protein.
Answer:

  1. All biochemical reactions that occur in the living systems are catalysed by the catalytic proteins called enzymes.
  2. Proteins such as keratin, collagen acts as structural backbones.
  3. Proteins are used for transporting molecules (Haemoglobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters).
  4. Antibodies help the body to fight various diseases.
  5. Proteins are used as messengers to coordinate many functions. Insulin and glucagon control the glucose level in the blood.
  6. Proteins act as receptors that detect the presence of certain signal molecules and activate the proper response.
  7. Proteins are also used to store metals such as iron (Ferritin) etc.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 45.
What are enzymes? What is the most important reason for their specific action?
Answer:
Enzymes are biomolecules & which catalyst biological reactions. Chemically they are globular proteins (water-soluble) that have very high molecular near ranging from 15,000 to 1,000,000 mol-1.
Enzymes are highly specific in their actions. The specificity is due to the presence of active sites of definite size and shape on their surfaces so that only specific substrates can fit into them. This specific binding leads to the formation of enzyme substrates complex which accounts for high specificity.

Question 46.
Name the enzyme that converts (i) carbonic acid to CO2 and H2O (ii) hydrolysis of sucrose to fructose and glucose. (iii) hydrolysis of lactose to glucose and galactose.
Answer:
(i) Carbonic anhydrase enzyme catalyses the interconversion of carbonic acid to water and carbon dioxide.
(ii) Sucrose enzyme catalyses the hydrolysis of sucrose to fructose and glucose.
(iii) Lactase enzyme hydrolysis the lactose to glucose and galactose.

Question 47.
Explain the mechanism of enzyme action.
Answer:
Mechanism of Enzymatic action: The general scheme is represented as
Step 1: Binding of the enzyme (E) to the substrate (S) to form a complex (ES) is Enzyme – substrate complex.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 28
Step 2: Product formation in the complex
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 29
Step 3: Release of enzyme product complex and leaving the enzyme as unchanged.
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 30

Question 48.
What are lipids? How are they classified?
Answer:
Lipids are the principal components of cell membranes including cell walls. They act as an energy source for living systems. Classification of lipids: Based on their structures Lipids can be classified as simple lipids, compounds lipids and derived lipids. Simple lipids can be further classified into fats, which are esters of long-chain fatty acids with glycerol (triglycerides) and waxes which are the esters of fatty acids with long-chain monohydric alcohols (Beeswax). Compounds lipids are the esters of simple fatty acid with glycerol which contain additional groups. Based on the groups attached, they are further classified into phospholipids, glycolipids and lipoproteins. Phospholipids contain a phospho-ester linkage while the glycolipids contain a sugar molecule attached. The lipoproteins are complexes of lipid with proteins.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 49.
How are vitamins classified?
Answer:
Vitamins are classified depending on their solubility in water.

  1. Water-soluble Vitamins: These include Vitamin B – complex (B1, B2, B5, i.e., nicotinic acid, B5, B12 and vitamin C)
  2. Water-insoluble Vitamins (or fat-soluble Vitamins): These include Vitamins D, A, E and K. They are stored in the liver and in fat-storing tissues.

Question 50.
What are coenzymes and prosthetic groups?
Answer:
coenzymes are loosely held to the protein and can be easily separated by dialysis.
Prothetic group: These are tightly held to the protein molecule by covalent bonds. They can be separated only by hydrolysis.

Question 51.
What are nucleic acids? Mention their two important functions?
Answer:
Nucleic acids are polynucleotides i.e., they are formed by the condensation of thousand molecules of nucleotides with the elimination of water molecules.
Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). The two main functions of nucleic acids are:

  1. DNA is responsible for the transmission of heredity effects from one generation to another. This is due to the unique property of replication during cell division as a result of which two identical DNA strands are transmitted to daughter cells.
  2. DNA and RNA are responsible for protein synthesis needed for the growth and maintenance of our body. Actually, proteins are synthesised by various RNA molecules (rRNA, mRNA and tRNA) in the cell but the message for the synthesis of a particular protein is given by DNA molecules.

Question 52.
Write the structure of nucleic acids.
Answer:
Primary structure: The primary structure of nucleic acid refers to the sequence in which the four nitrogen bases (A, G, C and T in DNA or A, G, C and U in RNA) are attached to the sugar-phosphate backbone of a nucleotide chain.
Secondary structure: DNA consists of two strands of polynucleotides coiled around each other in form of a double helix. The backbone of each strand consists of sugar-phosphate units and the base unit of each strand are pointed into the interior of the helix and are linked together through hydrogen bonds. Whereas G and C are held by three H – bonds, A and T are held by H – bonds.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 53.
Write down the structure of sugar present in DNA.
Answer:
DNA contains β D – 2 deoxyribose as sugar. In structure in the furanose form is
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 31

Question 54.
What purine and pyrimidine bases are present in DNA and RNA?
Answer:
Purines: adenine and guanine Pyrimidine: Cytosine and thymine are present in DNA.
Cytosine and uracil is present in RNA.

Question 55.
What is a nucleoside?
Answer:
A nucleoside consists of two components, i.e., a nitrogenous base (purine or pyramidine)’ and a five-carbon sugar (ribose or deoxyribose). It is obtained when the nitrogenous base is attached, to C1 of sugar by a β – linkage. Thus, in general, a nucleoside may be represented as a base – super. Examples of nucleosides are adenosine, cytidine etc.,

Question 56.
What is a nucleotide?
Answer:
A nucleotide consists of all the three basic components of nucleic acids, i.e., a nitrogenous purine or a pyrimidine base, a five-carbon ribose or deoxyribose and phosphoric acid. A nucleotide is obtained when a nitrogenous base is attached to C1 of the sugar by a β linkage and a nucleotide is thus obtained when C5—OH of the sugar is esterified with phosphoric acid. Thus, in general, nucleotides may be represented as a base – sugar-phosphate.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 57.
What type of linkage holds together the monomers of DNA?
Answer:
The monomers of DNA are polydeoxy ribo nucleotides. These are held together by H bonds. There are three H bonds between guanine and cytosine (G ≡ C) and two between adenine and thymine (A = T).

Question 58.
The two strands in DNA are not identical but complimentary explain.
Answer:
The two strands in the DNA molecule are held together through H — bonds between the purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the base, the only
possible pairing in DNA is G(guanine) and C(cytosine) through two hydrogen bonds i.e., bases (C ≡ G) and between A(adenine) and thymine (T) through two hydrogen bonds (i.e., A = T) due to this base – pairing principle the sequences of bases in one strand automatically fixes the sequence of bases in the other strand. Thus the two strands are not identical but complementary.

Question 59.
Mention the functional difference DNA and RNA.
Answer:

DNA RNA
DNA has unique property of replication. RNA does not replicate.
DNA controls transition of hereditary effects. RNA controls the synthesis of proteins.

Question 60.
What is DNA fingerprinting? Explain.
Answer:
The DNA fingerprint is unique for every person. By using this method, individual-specific variation in human DNA can be detected.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 61.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does it suggest about the structure of RNA?
Answer:
A DNA molecule has two strands in which four complementary bases pair each other i.e., cytosine (C) always pairs with guanine (G) while Thymine (T) always pair with adenine(A). Therefore when a DNA molecule is hydrolysed, the molar amount of cytosine is equal to that of guanine and that of adenine is always equal to that of thymine. Since RNA has no such relationship between the quantities of the four bases (C, G, A and U) obtained, therefore, based on the base-pairing principle, i.e., A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.

Choose the correct answer:

1. Glucose does not react with:
(a) Br2 / H2O
(b) NH2 OH
(c) (CH3 CO)2O
(d) NaHSO3
Answer:
(d)

2. The letter ‘D’ in D – glucose signifies:
(a) Configuration at all chiral carbons
(b) Dextro rotatory
(c) That it is a monosaccharide
(d) Configuration at the penultimate chiral carbon.
Answer:
(d)

3. The term anomer of glucose refers to:
(a) isomers of glucose that differ in configuration at carbon one and four (C-1 and C-4).
(b) a mixture of D – glucose and L – glucose
(c) enantiomers of glucose
(d) isomers of glucose that differ in configuration at C-1.
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

4. The helical structure of proteins is stabilised by:
(a) dipeptide bonds
(b) hydrogen bonds
(c) ether bonds
(d) peptide bonds
Answer:
(b)

5. The vitamins absorbed from intestine along with fats are:
(a) A, D
(b) A, B
(c) A, C
(d) D, B
Answer:
(a)

6. The human body does not produce:
(a) enzymes
(b) DNA
(c) vitamins
(d) hormones
Answer:
(c)

7. Adenosine is an example of:
(a) nucleotide
(b) nucleoside
(c) purine base
(d) pyramidine base
Answer:
(b)

8. Which of the following statements is not true about glucose?
(a) It is aldohexose.
(b) On heating with HI it forms n-hexane
(c) It is present in furanose form
(d) It does not give 2-4 DNP test.
Answer:
(c)

9. Which of the following reactions of glucose can be explained only by its cyclic structure?
(a) Glucose forms Penta acetate
(b) Glucose reacts with hydroxylamine to form an oxime.
(c) Pentaacetate of glucose does not react with hydroxylamine.
(d) Glucose is oxidised by nitric acid to gluconic acid.
Answer:
(c)
Hint: Due to the absence of free ‘OH’ group at C1 cyclic structure of glucose Penta acetate cannot revert to open-chain aldehyde form and hence cannot form Penta acetate.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

10. Three cyclic structures of monosaccharides are given below, which are anomers?
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 32
(a) I and II
(b) II and III
(c) I and III
(d) III is anomer of I and II
Answer:
(a)
Hint: Structure (I and II) differ only in the position of OH at C1 and hence anomers.

11. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be :
(a) primary structure of proteins
(b) secondary structure of proteins
(c) tertiary structure of proteins
(d) quaternary structure of proteins
Answer:
(a)

12. RNA and DNA contains four bases each which of the following bases is not present in RNA.
(a) Adenine
(b) Uracil
(c) Thymine
(d) Cytosine
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

13. In fibrous proteins, polypeptide chains are held together by:
(I) Vander waals forces
(II) disulphide linkages
(III) electrostatic forces of attraction
(IV) hydrogen bonds
(a) I and II
(b) II and IV
(c) III and IV
(d) IV only
Answer:
(b)

14. Which of the following does not exhibit the phenomenon of muta rotation?
(a) (-) Fructose
(b) (+) sucrose
(c) (+) Lactose
(d) (+) Maltose
Answer:
(b)
Hint: Only monosaccharides i.e., (-) fructose and reducing disaccharides i.e., (+) lactose and (+) maltose show mutarotation. Since (+) sucrose is a non reducing sugar, it does not show mutarotation.

15. Which of the following statements is not true regarding (+) lactose?
(a) (+) lactose, C12H22O11 contains 8 (—OH) groups.
(b) On hydrolysis, (+) lactose gives equal amounts of D(+) glucose and D(+) galactose.
(c) (+) Lactose is a β – glycoside formed by the union of a molecule of D(+) glucose and a molecule of D(+) galactose.
(d) (+) Lactose is a reducing sugar and does not exhibit mutarotation.
Answer:
(d)

16. The vitamin which is neither soluble in water nor in fat is:
(a) biotin
(b) phylloquinone
(c) thiamine
(d) ergocalciferol
Answer:
(a)

17. Which of the following is not a fat soluble vitamins?
(a) Vitamin B complex
(b) Vitamin D
(c) Vitamin E
(d) Vitamin A
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

18. Which of the following structure represent the peptide chain?
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 33
Answer:
(c)

19. The tripeptide is written as glycine – alanine – glycine. The correct structure of the tripeptide is:
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 34
Answer:
(c)
TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 35

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

20. The hormone which controls the processes of the burning of fats, proteins and carbohydrates and liberates energy in the body is:
(a) thyroxine
(b) adrenaline
(c) insulin
(d) cortisone
Answer:
(c)

21. The secondary structure of a protein refers to:
(a) fixed configuration of polypeptide backbone.
(b) a- helical backbone
(c) hydrophobic interactions
(d) sequence of amino acids
Answer:
(b)

22. Which of the following statements about “denaturation” given below are correct?
(i) denaturation of proteins causes loss of secondary and tertiary structure of proteins.
(ii) denaturation leads to the conversion of double strand DNA to a single strand.
(iii) denaturation affects the primary structure which gets distorted.
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (i), (ii) and (iii)
Answer:
(c)

23. The pyramidine bases present in DNA are :
(a) Cytosine and adenine
(b) Cytosine and guanine
(c) Cytosine and thymine
(d) Cytosine and uracil
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

24. Nitrogen base that is found in RNA but not in DNA is:
(a) uracil
(b) thymine
(cl cytosine
(d) adenine
Answer:
(a)

25. RNA and DNA are chiral molecules, then chirality is due to:
(a) D – sugar component
(b) L – sugar component
(c) Chiral bases
(d) Chiral phosphate ester unit
Answer:
(a)

26. In DNA, the complimentary bases are:
(a) Adenine and guanine; thymine and cytosine
(b) Uracil and adenine; cytosine and guanine
(c) Adenine and thymine; guanine and cytosine
(d) adenine and thymine; guanine and uracil
Answer:
(c)

27. Assertion: Fructose does not contain an aldehyde group but still reduces Tollen’s reagent.
Reason: In presence of base, fructose undergoes rearrangement to form glucose and mannose.
(a) Both assertion and reason are correct and the reason is a correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not a correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

28. Assertion: Glycosides are hydrolysed in acidic conditions.
Reason: Glycosides are acetals.
(a) Both assertion and reason are correct and the reason is a correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not a correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a)

29. Match the entries in column I with appropriate entries in column II and choose the correct option.

Column I Column II
(A) Glycosidic linkage (1) Monomeric units of nucleic acids
(B) peptide bond (2) Sugar heterocyclic base combination
(C) Nucleoside (3) linkage between two amino acid units
(D) Nucleotide (4) Linkage between two monosaccharide units.

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules 36
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 14 Biomolecules

30. The two forms of D – glucopyranose are called:
(a) Enantiomers
(b) Anomers
(c) Epimers
(d) Diastereomers
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Students get through the TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Answer the following questions.

Question 1.
Classify the following aliphatic or aromatic nitro compounds.
(i) 1, 2, dimethyl-1-nitropropane
(ii) 2-nitro-1 -methyl benzene
(iii) 1, 3, 5 trinitro benzene
(iv) 2-phenyl-1-nitro ethane
(v) 2-methyl-2-nitro propane
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 1
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 2

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 2.
What are nitro compounds? How are they classified? Give one example for each type.
Answer:
(i) Nitro compounds are considered as the derivaties of hydrocarbons. If one of the hydrogen atom of hydrocarbon is replaced by the —NO2 group, the resultant organic compound is called a nitrocompound.
(ii) Examples for primary nitro compound.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 3
(iii) Example for secondary nitro compound.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 4
(iv) Example for a tertiary nitro compound.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 5

Question 3.
Write the structural formula of the isomers of the following compound and indicate the type of isomerism involved, (i) 1-nitrobutane,
(ii) nitroethane.
Answer:
(i) (a) 1-nitrobutane and 2-methyl-1-nitropropane are- chain isomers. These two isomers differ in the length of carbon chain and hence exhibit chain isomerism.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 6
(b) 1-nitrobutane, 2-nitrobutane and 2-methyl 2-nitropropane are position isomers. They differ in the position of the functional group and hence exhibit positfon isomerism.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 7
(c) 1-nitrobutane and butyl nitrite are functional isomers. They differ in the nature of the functional group and hence functional isomerism.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 8
(ii) Nitromethane and methyl nitrite exhibit tautomerism.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 9

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 4.
Methyl nitrite and nitro methane exhibits tautomerism. How will you distinguish between these form?
Answer:

Nitro form Aci-form
Less acidic More acidic and also called pseudoacids (or) nitronic acids.
Dissolves in NaOH slowly. Dissolves in NaOH instantly.
Decolourises FeCl3 solution. With FeCl3 gives reddish-brown colour
Electrical conductivity is low. Electrical conductivity is high.

Question 5.
Between 2-nitrobenzene, and 2-methyl – 2- nitro propane which does not exhibit tautomerism? Why?
Answer:
Tautomerism is exhibited by nitro alkanes which contain one or more ‘α’ hydrogen atoms. Then 1° and 2° nitro alkanes exhibit tautomerism while 3° nitro alkanes which do not have a ‘α’ hydrogen atom does not exhibit tautomerism.
2-nitrobutane has a hydrogen atom and hence exhibit tautomerism. While 2-methyl -2- nitropropane does not contain an a hydrogen atom and hence does not exhibit tautomerism.

Question 6.
Explain the acidic nature of nitroalkanes.
Answer:
Primary and secondary nitro alkanes show acidic nature because of the presence of a hydrogen atom.
Because of the presence of electron withdrawing nitro group, the ‘α’ hydrogen atom is abstracted by a base.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 10
Hence, they are acidic. When the number of alkyl groups attached to α- carbon acidity decreases because of +1 effect of alkyl groups.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 11

Question 7.
How will you prepare nitro benzene from (i) ethyl bromide (ii) methane, (iii) α – chloroacetic acid (iv) tert-butyl amine, (v) acetaldoxine.
Answer:
(i) Alkyl bromides (or) iodides on heating with ethanolic solution of potassium nitrite gives nitroethane.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 12
(ii) Gaseous mixture of methane and nitric acid passed through a red hot metal tube to give nitromethane.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 13
(iii) α – chloroacetic acid when boiled with aqueous solution of sodium nitrite gives nitromethane.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 14TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 15
(iv) tert-butyl amine is oxidised with aqueous KMnO4 to give tert – nitro alkanes.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 16
(v) Oxidation of acetaldoxime and acetoneoxime with trifluoroperoxy acetic acid gives nitroethane (1°) and 2- nitropropane (2°) respectively.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 17

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 8.
Give the equation for the reduction of nitromethane in (a) acid medium and (b) neutral medium.
Answer:
(a) Nitromethane on reduction in acid medium gives methylamine.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 18
(b) Reduction under neutral condition using Zn / NH4OH or Zn /CaCl2, hydroxylamine is formed.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 19

Question 9.
Complete the following equation. Identify A, B, and C.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 20
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 21
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 22
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 23

Question 10.
How does ethyl nitrite react with (i) Sn / HCl and (ii) HCl / H2O.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 24

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 11.
What is Nef carbonyl synthesis? Give equation.
Answer:
The formation of an aldehyde by the alkaline hydrolysis of a nitro alkane is known as Nef carbonyl synthesis.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 25

Question 12.
How are the following conversion made?
(i) Benzene to mcta dinitro benzene, (ii) Nitro benzene to nitroso benzene, (iii) Nitro benzene to azo benzene, (iv) Nitro benzene to hydrazo benzene.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 26
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 27

Question 13.
Identify the reagents used in the following conversions. Write complete equations.
(i) nitrobenzene to aniline
(ii) metadinitrobenzene to metanitroaniline
(iii) nitrobenzene to 3-nitrobenzene sulphonic acid.
Answer:
(i) Any reducing agent: Ni (or) Pt or LiAlH4
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 28
(ii) Ammonium poly sulphide
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 29
(iii) Conc. H2SO4
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 30

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 14.
Explain why nitro group in nitro benzene is meta directing.
Answer:
The structure of nitro benzene is a resonance hybrid of the following cannonical structures.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 31
As a result of electron withdrawing nature of the NO2 group, the electron density at ortho and para position deefeases compafed to meta position i.e., the meta position is relatively electron rich compound to ortho and para positions. Hence the electrophile attacks the meta position.

Question 15.
How will you effect the following conversions?
(i) Benzene to m-chloro nitro benzene,
(ii) Benzene to 1, 2, 3 tri nitro benzene
(iii) m-chloro nitro benzenetto m-chloro aniline,
(iv) 1, 3, di nitrobenzene to 3- nitro aniline.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 32

Question 16.
Give the IUPAC names of the following:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 33
Answer:
(i) N -phenyl benzenamine
(ii) N, N -diphenyl benzenamine
(iii) N -methyl-N-phenylethanamine
(iv) prop -2- en-1-amine
(v) hexane-1, 6-diamine
(vi) N -methyl-propan-1-amine
(vii) N, N -diethyl-butan-1-amine
(viii) N -ethyl- N-methyl propan-2-amine
(ix) N, N -dimethyl-benzenamine
(x) phenyl-methanamine

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 17.
Write the structures of the chain isomers of Butan-1-amine.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 34

Question 18.
Explain ‘metamerism’ with a suitable example of amines.
Answer:
Aliphatic amines having the same molecular formula but different alkyl groups on either side of the nitrogen atoms show metamerism,
eg: CH3CH2NHCH2CH3 (diethyl amine) and CH3NHCH2CH2CH3 (methyl-n-propyl amine)
CH3NHCH(CH32)2 (Isopropyl methylamine) are metamers.

Question 19.
What are all the possible isomers of an amine having the molecular formula C3H9N and C4H11N.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 35
These amines exhibit functional isomerism.
Amine (C4H11N): CH3CH2CH2CH2NH2 (Butan-1-amine) exhibit chain as well as position isomerism.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 36

Question 20.
Write the structure of the following:
(i) 4N-dimethyl pentan-2-amine
(ii) 2 (N, N-dimethyl) butanamine
(iii) 2-aminoethanol
(iv) 4-aminobutanoicacid
(v) N-methyl-2-nitro pentanamine
(vi) prop-2-en-l -amine.
(vii) N-ethyl-N-methylbenzenamine
(viii) 3-methylbenzenamine
(ix) 2-methoxybenzenamine
(x) N-phenylbenzenamine
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 37
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 38

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 21.
How is ethanamine prepared from (i) nitroethane, (ii) ethanenitrile, (iii) acetamide, (iv) ethyl bromide?
Answer:
(i) Reduction of nitroethane using H2 in the presence of Ni or Sn/HCl or Pd/H2.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 39
(ii) Reduction of ethanenitrile using sodium amalgam in ethanol.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 40
(iii) Reduction of acetamide using LiAlH4.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 41
(iv) By Gabriel’s phthalimide synthesis.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 42

Question 22.
Identify the products:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 43
Answer:
(i) C6H5NH2 (aniline)
(ii) CH3NHCH3 (N-methyl methanamine)
(iii) C6H5NH2 (aniline)
(iv) C6H5CH2NH2 (benzyl amine)

Question 23.
Using sodium azide (NaN3) convert methyl bromide to methylamine.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 44

Question 24.
How will you prepare aniline from (i) chlorobenzene and (ii) phenol?
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 45

Question 25.
What happens when
(i) Vapours of ethanol and ammonia are passed over alumina.
(ii) Ethanamide is treated with LiAlH4. Give equation.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 46

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 26.
Explain Why?
(i) Amines have higher boiling points than hydrocarbons of comparable molecular mass.
(ii) Among isomeric amines 3° amines have the lowest melting point.
(iii) The boiling point of amines are lower than those of alcohols and acids of comparable molecular mass.
(iv) Aliphatic amines with maximum six carbon atoms are soluble in water to some extent, while aromatic amines are insoluble in water.
Answer:
(i) This is due to the reason that amines being polar form inter molecular hydrogen bonding (except tertiary amines which do not have a H atom linked to N atom) and exist as associated molecules.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 47
(ii) The degree of association depends on the extent of hydrogen bonding. Since 1° amines have two, 2° amines have one while 3° amines have no hydrogen linked to nitrogen. Therefore, among isomeric amine 1° amines have highest while 3° amine have the lowest boiling points.
(iii) The electronegativity of nitrogen is lower than oxygen, amines form weaker H—bonding compared to alcohol and carboxylic acids.i.e., The extent of association is less compared to that of alcohols or carboxylic acids. Hence amines will have lower boiling point compared to alcohol or carboxylic acid of comparable molecular mass.
(iv) All the three classes of amines (1°, 2° and 3°) form H bonds with water as a result lower aliphatic amines are soluble in water. As the size of the alkyl group increases (with increase in molecular mass), the solubility decreases due to a corresponding increase in hydrocarbon part of the molecule. The borderline solubility is reached with amines of about six carbon atoms in the molecule. Aromatic amines, on the other hand are insoluble in water. This is due to larger hydrocarbon part which tend to retard the formation of H-bonds.

Question 27.
Account for the fact that among ethyl amines, the decreasing order of basicity in aqueous solution is (CH3CH2)2NH > (CH3CH2)3N > CH3CH2NH2.
Answer:
The basicity of an amine in aqueous solution primarily depends upon the stability of the ammonium cation or the conjugate acid formed by accepting a proton from water. The stability of the ammonium cation in turn depends on the combination of the following factors.
(i) +1 effect of alkyl groups.
(ii) Extent of hydrogen bonding with water molecules.
(iii) Steric effects of the alkyl groups.
In the case of alkyl groups bigger than ‘CH3’ group i.e., ethyl, propyl, etc there will be steric hindrance to hydrogen bonding. As a result, stability due to +1 effect predominates over the stability due to H—bonding and hence 3° amines because more basic than 1° amines. In all overall decreasing basic strength of ethyl amine follows the sequence (CH3CH2)2NH > (CH3CH2)3N > CH3CH2NH2 > NH3.

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 28.
How does nitrous acid (a mixture of sodium nitrite and dil. HCl) react with (i) CH3NH2 (ii) (CH3)2NH and (iii) (CH3)3N? Give equations.
Answer:
Aliphatic primary amines react with nitrous acid give an alcohol and nitrogen gas is produced. Methyl amine is an exception.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 48
Secondary amines will form nitroso alkyl amine.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 49
Tertiary amines react with nitrous acid and form nitrites.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 50

Question 29.
How does nitrous acid react with the following? Give equations.
(i) ethylamine, (ii) di-ethylamine, (iii) triethyl amine, (iv) aniline, (v) N-methyl aniline, (vi) N, N- dimethylaniline.
Answer:
(i) Ethylamine reacts with nitrous acid to give ethyl diazonium chloride, which is unstable and it is converted to ethanol by liberating N2.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 51
(ii) A yellow dye of N-nitrosodiethylamine is formed.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 52
(iii) Triethylammoniumnitrate which is soluble in water formed.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 53
(iv) Aniline reacts with nitrous acid at low temperature (273 – 278 K) to give benzene diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 54
This reaction is known as diazotisation.
(v)
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 55
This reaction is known as Libermann’s nitroso test.
(vi) Aromatic tertiary amine reacts with nitrous acid at 273K to give p-nitroso compound.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 56

Question 30.
Explain why amino group is ortho para directing in electrophilic substitution reactions.
Answer:
The NH2 group is a strong activating group. In aniline the NH2 is directly attached to the benzene ring, the lone pair of electrons on the nitrogen is in conjugation with benzene ring which increases the electron density at ortho and para position, thereby facilitating the electrophilic attack at ortho and para.

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 31.
Aniline gives 2, 4, 6 tribromo aniline when treated with bromine water, but not a mono bromo aniline. Explain why?
Answer:
Aniline is a more powerful activating group. The lone pair of electrons enters into conjugation with the benzene ring, as a result both ortho and para position become rich in electron density. At the same time Br+ is also a strong electrophile. Hence, all the three positions are attacked by the electrophile.

Question 32.
Why is it necessary to acetylate aniline to get a mono bromo aniline?
Answer:
Acetylation of aniline reduces the activity of the amino group. The activity of the acetylated amine is reduced because the lone pair of electron on nitrogen is delocalised by the neighbouring carbonyl group by resonance. Hence, it is not easily available for conjugation.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 57
The acetyl amino group is less activating than amino group.

Question 33.
Explain why direct nitration of aniline gives a mixture of ortho, meta and para isomers.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 58
Nitration of aniline involves acidic medium. So protonation of aniline takes place forming mainly m-nitro aniline. Hence, direct nitration of aniline forms the following products.

Question 34.
How will you convert aniline?
(i) to p-bromo aniline
(ii) to 2,4,6 tribromo aniline
(iii) to para nitro aniline
(iv) to benzene diazonium chloride
(v) to sulphanilic acid
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 59
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 60
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 61

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 35.
What is zwitter ion? Explain with an example.
Answer:
Zwitter ion is a compound in which both acidic and basic groups are present in the same molecule, eg: sulphanilic acid. It has an acidic (SO2OH) and a basic group (NH2). Thus a proton from the acid leaves and protonates the basic group. As a result it exists as a dipolar ion.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 62

Question 36.
How will you distinguish between primary, secondary and tertiary amine?
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 63
Heisenberg reagent: Benzene sulphoanyl chloride in the presence of excess aqueous KOH solution.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 63

Question 37.
Complete the following reactions:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 65
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 66

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 38.
Give examples for (i) Sandmeyer’s reaction (ii) Gattermann reaction.
Answer:
(i) Sandmeyer’s reaction: Benzene diazonium chloride solution on reaction with cuprous chloride in hydrochloric acid (CuCl / HCl) or cuprous bromide in hydrochloric acid (CuBr / HCl) gives corresponding chloro benzene or bromo benzene. This reaction is known as Sandmeyer’s reaction. Aryl cyanides can also be prepared by this method.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 67
(ii) Gattermann reaction: This reaction involves the treatment of benzene diazonium chloride with Cu / HCl or Cu / HBr respectively instead of cuprous halide and hydrogen halide.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 68

Question 39.
How does the following reagents react with benzene diazonium chloride? Give equations.
(i) HBF4 and the product formed is heated.
(ii) HBF4 and die product formed is heated with an aqueous solution of sodium nitrite in the presence of copper.
(iii) HBF4 and the product is heated with CH3COOH.
(iv) Cuprous cyanide in the presence of KCN.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 69

Question 40.
Accomplish die following conversions.
(i) Nitrobenzene to benzene,
(ii) 4-Nitro aniline to 1, 2, 3 -tribromo benzene,
(iii) p-toludine to 2-bromo-4-methyl aniline,
(iv) m-nitro aniline to m-chloroaniline,
(v) p-nitroaniline to p-iodomtro benzene,
(vi) benzyl chloride to 2-phenyl ethanamine.
Answer:
(i) Nitrobenzene to benzene:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 70
(ii) 4 -nitro aniline to 1, 2, 3 – tribromo benzene.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 71
(iii) p-toludine to 2-bromo 4-methyl aniline.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 72
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 73
(vi) Benzyl chloride to 2- phenyl ethylamine.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 74

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 41.
Give equations for the following reactions.
(i) Ethyl bromide is treated with KCN
(ii) Heating acetamide with P2O5
(iii) Heating acetaldoxime with P2O5
(iv) Treating methyl magnesium bromide with cyanogen chloride.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 75
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 76

Question 42.
Write the IUPAC names of the following:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 77
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 78
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 79

Question 43.
What is the reducing agent used in the following reduction reactions? Give equations.
(i) Ethanenitrile to ethanamine
(ii) Benzonitrile to benzylamine
(iii) Ethane nitrile to Acetaldimine hydrochloride which on hydrolysis gives acetaldehyde.
Answer:
(a) Complete reduction or strong reduction when reduced with H2 in the presence of Pt or Ni as catalyst or by using LiAlH4, or sodium and alcohol alkyl or aryl cyanides yield primary amines.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 93
This specific type of reduction of alkyl or aryl cyanides using LiAlH4 or Na/ alcohol is named as mendices reduction.
(b) Partial reduction or mild reduction: When SnCl2 / HCl is used as a reducing agent at room temperature alkyl or aryl cyanides are reduced to amine hydrochloride which on subsequent hydrolysis gives aldehyde. This type of reduction is named as Stephen’s reduction.
SnCl2 + 2HCl → SnCl4 + 2 [H]
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 80

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 44.
Give examples for (i) Thrope nitrile condensation and (ii) Levine and Hauser acetylation.
Answer:
(i) Thrope nitrile condensation: Self condensation of two molecules of alkyl nitrile (containing α—H atom) in the presence of sodium to form iminonitrile.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 81
(ii) Levine and Hauser acetylation: The nitriles containing α-hydrogen also undergo condensation with esters in the presence of sodamide in ether to form ketonitriles. This reaction is known as “Levine and Hauser” acetylation. This reaction involves replacement of ethoxy (OC2H5) group by methylnitrile (-CH2CN) group and is called as cyanomethylation reaction.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 82

Question 45.
Write the electronic structure of alkyl cyanides and isocyanides.
Answer:
The electronic structure of alkyl cyanides is
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 83
The carbon atom of CN group is sp hybridised. CN group contain one σ and two pi bonds.
The electronic structure of isocyanides is
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 84
Structurally isocyanides are represented as a resonance hybrid of the following two forms.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 85
The dipolar form (I) makes higher contribution.

Question 46.
Name the reagents used in the following reactions.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 86
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 87

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 47.
What happens when methyl isocyanide is:
(a) treated with dilute HCl.
(b) treated with sodium and alcohol
(c) heated at 250°C
(d) treated with sulphur in the presence of ozone.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 88

Question 48.
Mention the uses of (a) Nitro alkanes,
(b) Nitrobenzene,
(c) Cyanides and isocyanides
Answer:
(a) Nitroalkanes:

  1. Nitromethane is used as a fuel for cars.
  2. Chloropicrin (CCl3NO2) is used as an insecticide.
  3. Nitroethane is used as a fuel additive and precursor to explosive and they are good solvents for polymers, cellulose ester, synthetic rubber and dyes etc.,
  4. 4% solution of ethylnitrite in alcohol is known as sweet spirit of nitre and in used as diuretic.

(b) Nitrobenzene:

  1. Nitrobenzene is used to produce lubricating oils in motors and machinery.
  2. It is used in the manufacture of dyes, drugs, pesticides, synthetic rubber, aniline and explosives like TNT, TNB.

(c) Cyanides and isocyanides;

  1. Alkyl cyanides are important intermediates in the organic synthesis of larger number of compounds like acids, amides, esters, amines etc.
  2. Nitriles are used in textile industry in the manufacture of nitrile rubber and also as a solvent particularly in perfume industry.

Question 49.
An organic compound (A) having the molecular formula C2H7N is treated with nitrous acid to give (B) of molecular formula C2H6O which answers the iodoform test. Identify A and B and explain the reactions.
Answer:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 89
Since, ‘B’ is an alcohol formed from A by the action of nitrous acid, A should be a primary amine which contains ‘NH2’ group.
C2H5N—NH2=C2H5 i.e. The alkyl group is C2H5 and hence A is C2H5NH2 ethyl amine. B is ethyl alcohol and it is confirmed by the fact that it answer iodo form test.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 90

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

Question 50.
An organic compound (A) with molecular formula C6H7N gives (B) with HNO2 / HCl at 273K. The aqueous solution of (B) on heating gives compound (C) which gives violet colour with neutral FeCl3. Identify the compounds A, B and C. Write the equations.
Answer:
A = C6H5NH2 (aniline);
B = C6H2N2Cl (benzene diazonium chloride)
C = C6H5OH (phenol)
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 91
(i) Since ‘C’ gives a violet colouration with neutral FeCl3 it should be phenol.
(ii) Since phenol is obtained by boiling aqueous solution of (B), it should be benzene diazonium chloride.
(iii) Hence, A should be aniline.
Equation:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 92

Choose the correct answer:

1.1, 2 dimethyl- 1-nitropropane. Choose the
incorrect statement about this compound.
(a) It is a primary nitro compound.
(b) It is also called nitroneopentane.
(c) The ‘NO2’ group is attached to a secondary carbon atom.
(d) It is an aliphatic compound.
Answer:
(c)

2. 1-nitrobutane and 2-methyl-1-nitro propane:
(a) chain isomers
(b) position isomers
(c) functional isomers
(d) metamers
Answer:
(a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 94

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

3. Which of the following nitro compounds does not exhibit tautomerism?
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 95
Answer:
(d)
Hint: Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atoms.

4. The incorrect statement between nitroform and acidform of nitromethane is:
(a) nitroform is less acidic whereas aciform is more acidic.
(b) nitroform dissolves in NaOH slowly while aciform dissolves in NaOH instantly.
(c) Both decolorise FeCl3 solution
(d) They are tautomers
Answer:
(c)
Hint: Nitroform decolorises FeCl3 solution while aciform gives a reddish brown colour with FeCl3 solution.

5.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 96
Identify A and B.
(a) A = CH3CH2NO2 ; B = CH3CH2NH2
(b) A- CH3NO2 ; B = CH3NH2
(c) A = CH3CHO ; B = CH3CH2OH
(d) A = CH3CH2NH2 ; B = CH3CH2OH
Answer:
(a)

6.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 97
Identify A and B.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 98
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 99
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

7. The products obtained by the reduction of nitromethane in acid medium and neutral medium are:
(a) methylamine and N-methylhydroxylamine
(b) methylamine and ethanol
(c) N-methylhydroxylamine and methylamine
(d) N-methylhydroxylamine in both cases.
Answer:
(a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 100

8. An amine is boiled with HCl and H2O. Which of the following will give acetone:
(a) CH3NH2
(b) (CH3)2CHNO2
(c) (CH3)3CNO2
(d) both (b) and (c)
Answer:
(b)
Hint: CH3NH2 gives CH3COOH (CH3)2
CHNO2 gives (CH3)2CO
(CH3)3 CNO2 no reaction

9. The correct IUPAC name for CH2=CH—CH2NH—CH3 is:
(a) Allylmethyl amine
(b) 2-amino-4-pentane
(c) 4 amino pent-l-ene
(d) N-methyl prop-2 en-1 amine
Answer:
(d)

10. Amongst the following the strongest base in aqueous medium is:
(a) CH3NH2
(b) NC—CH2NH2
(c) (CH3)2NH
(d) C6H5NH—CH3
Answer:
(c)
Hint: 2° amines are more basic than 1° amines, i. e., (CH3)2NH is more basic than CH3NH2. Due to -I effect of CN group NC—CH2NH2 is less basic than even CH3NH2. C6H5NHCH3 is less basic than both C6H5NH2 and (CH3)2NH due to delocalisation of lone pair of electrons on the N atom into the benzene ring.

11. In order to prepare a 1 ° amine from an alkyl halide with simultaneous addition of one CH2 group in the carbon chain, the reagent used a source of nitrogen is:
(a) Sodium amide, NaNH2
(b) Sodium azide, NaN3
(c) Potassium cyanide, KCN
(d) Potassium phthalimide C6H4(CO2) NK
Answer:
(c)
Hint: Cyanides on reduction gives 1: amines with a CH2 group
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 101

12. The correct increasing order of basic strength for the following compounds is:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 102
(a) II < III < I
(b) III < I < II
(c) III < II < I
(d) II < I < III
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

13. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is:
(a) Hoffmann bromamide reaction
(b) Gabriel’s phthalimide synthesis
(c) Sandmeyer’s reaction
(d) Reaction with NH3
Answer:
(b)

14. Which of the following compounds is the weakest Bronsted base?
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 103
Answer:
(c)
Hint: Amines (a) and (b) have a stronger tendency to accept a proton and hence stronger Bronsted bases than phenol (c) and alcohol (d) since phenol is more acidic than alcohol, phenol (c) has the least tendency to accept a proton and hence the weakest Bronsted base.

15. Which of the following compounds cannot be prepared by Sandmeyer’s reaction?
(I) Chlorobenzene
(II) Bromobenzene
(III) Iodo benzene
(IV) Fluoro benzene
(a) (IV)
(b) (III)
(c) (I) and (II)
(d) (III) and (IV)
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

16. The products of the following reaction are:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 104
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 105
Answer:
(c)
Hint: —NHCOCH3 is an o/p directing group.

17. An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’. ‘C’ when treated with Br2 in the presence of KOH produces ethylamine. The compound ‘A’ is:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 106
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 107

18. In a set of reactions, meta mono benzoic acid gave a product ‘D’. Identify D.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 108
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 109
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 110
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 111

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

19.
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 112
Answer:
(b)
Hint: Diazonium salts of benzylamine is not stable. It decomposes insitu, to form benzyl alcohol.

20. Predict the product:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 113
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 114
Answer:
(d)

21. Aniline in a set of following reactions yielded a coloured product Y. The structure of Y would be:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 115
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds 116

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

22. Butylamine (I), diethyl amine (II) and N, N diethyl amine, (III) have the same molar mass. The increasing order of their boiling points is:
(a) III < II < I
(b) I < II < III
(c) II < III < I
(d) III < I < II
Answer:
(a)
Hint: Extent of H-bonding increases as the number of H-atoms increases and hence boiling point increases in the order 3°<2°< 1°.

23. Assertion: Ammonolysis of alkyl halides is not a suitable method for the preparation of primary amines.
Reason: Ammonolysis of alkyl halides mainly produces 2° amines.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c)

24. Assertion: Gabriel phthalimide reaction can be used to prepare aryl and arylalkyl amines.
Reason: Aryl halides are as reactive as alkyl halides towards nucleophilic substitution reactions.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d)
Hint: Correct assertion: Gabriel phthalimide reaction can be used to prepare alkyl and aryl alkyl primary amines.
Correct reason: Alkyl and aralkyl halides undergo nucleophilic substitution reaction but aryl halides do not.

TN Board 12th Chemistry Important Questions Chapter 13 Organic Nitrogen Compounds

25. Assertion: Aniline does not undergo Friedel Craft’s reaction.
Reason: Friedel crafts reaction is an electrophilic substitution reaction.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true and reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(b)
Hint: Correct explanation: AlCl3 forms a salt with aniline (C6H5NH2+ AlCl3) which deactivates the benzene ring thereby preventing Friedel craft’s reaction.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Students get through the TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Answer the following questions.

Question 1.
Write the IUPAC names of the following:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 1
Answer:
(i) 3-phenyl prop-2-en-l-al
(ii) cyclohexanecarbaldehyde
(iii) 3-oxopentan-l-al
(iv) But-2-en-l-al

Question 2.
Give the structure of the following compounds.
(i) 4 – Nitropropiophenone
(ii) 2 – Hydroxycyclopentane carbaldehyde
(iii) phenyl acetaldehyde
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 2

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 3.
Write the IUPAC names of
(i) Diacetone alcohol
(ii) Crotonaldehyde
Answer:
(i) 4-Hydroxy-4-methylpentan-2-ol
(ii) But-2-en-l-al

Question 4.
Write the structure of
(i) 3 – oxopentanal
(ii) 1 – phenylpentan – 1 – one
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 3

Question 5.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2CO(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO(P)
Answer:
(i) 4-methyl pentanal
(ii) 6-chloro-4-ethylhexan-3 -one
(iii) But-2-en-l-al
(iv) Pentane 2,4, dione
(v) 3, 3, 5 Trimethylhexan – 2 – one
(vi) 3, 3, Dimethyl butanoic acid
(vii) Benzene 1,4, dicarbaldehyde

Question 6.
Draw the structures of the following compounds.
(i) 3-methyl butanal
(ii) 4-nitro propiophenone
(iii) p-methyl benzaldehyde
(iv) 4-methylpent-3-en-2-one
(v) 4-chloropentan-2-one
(vi) 3-Bromo-4-phenyl pentanoic acid
(vii) p, p’ dihydroxy benzo phenone
(viii) Hex-2-en-4 yonic acid
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 4
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 5

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 7.
Write IUPAC names of the following aldehydes and ketones. Wherever possible give also common names.
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CH(Br)CH2CH(CH3)CHO
(iii) CH3(CH2)4CHO
(iv) Ph—CH=CH.CHO
(v)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 6
(vi) PhCOPh
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 193

Question 8.
Give the product of ozonolysis of the following alkenes.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 7
Answer:
To identify the product of ozonolysis, draw a line in between C=C bond and ‘O’ on either side of the double bond
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 8
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 9

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 9.
Identify the products of hydration of the following:
(i) ethyne
(ii) prop-1-yne
(iii) Hex-1-yne
(iv) Diphenyl acetylene
Answer:
Hydration means addition of water. The reagent used is mercuric sulphate and dilute sulphuric acid at 333K. Ketones are formed on hydration of alkynes. MarkovnikofFs mle is followed in water addition.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 10
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 11

Question 10.
How are the following compounds formed by distillation of calcium salts of their carboxylic acids?
(i) Formaldehyde,
(ii) Acetaldehyde,
(iii) Acetone,
(iv) Butan-2-one,
(v) Cyclopentanone,
(vi) Benzaldehyde
Answer:
(i) Calcium formate alone is heated.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 12
(ii) By dry distillation of calcium acetate and calcium formate.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 13
(iii) Dry distillation of calcium acetate alone.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 14
(iv) Calcium acetate and calcium propionate are distilled together.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 15
(v) Cyclic ketones are formed when calcium salts of dibasic acids are heated.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 16
(vi) By dry distillation of a mixture of calcium benzoate and calcium formate.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 17

Question 11.
Complete the following equations:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 18
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 19
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 20

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 12.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 21
(i) Identify the product.
(ii) Name the reaction.
(iii) What is the intermediate formed in the reaction.
Answer:
(i) CH3CHO (acetaldehyde)
(ii) Stephen’s reaction
(iii) CH3—CH=NH (an imine)
Complete reaction:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 22

Question 13.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 23
(i) Name the reagent i.e., X used for the above reduction.
(ii) Why other reducing agents like H2, in the presence of a catalyst are not used.
(iii) Write the IUPAC name of the product formed.
Answer:
(i) Diisobutyl aluminium hydride (DIBAL – H) is used as a reducing agent.
X = H2O
(ii) It selectively reduces alkyl cyanides to imines, which on hydrolysis gives aldehydes. The double bond in the nitrile is not reduced. The other reducing agents reduce the double bond also.
(iii) hex-4-enal is the IUPAC name of the product.

Question 14.
How is benzaldehyde prepared from (i) methyl benzene, (ii) benzene, (iii) benzyl alcohol. Give equations.
Answer:
(i) Oxidation of methyl benzene, using chromyl chloride as oxidising agent gives benzaldehyde. Acetic anhydride and CrO3 can also be used for reaction.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 24
(ii) Benzene is treated with CO and HCl Benzaldehyde is formed.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 25
This reaction is known as Gattermann- koch reaction.
(iii) Oxidation of benzyl alcohol using PCC (pyridinium chromo chromate) gives benzaldehyde.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 26

Question 15.
Identify the product of the following reactions. Write the complete equation.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 27
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 28
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 29

Question 16.
Give examples for Friedel-crafts acetylation reaction.
Answer:
Refer Question 15 (ii) and (iii). These reaction are called Friedel-crafts reaction. When acetyl chloride is used it is known as acetylation reaction. When benzoyl chloride is used, it is called benzoylation reaction.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 17.
The boiling points of aldehydes and ketones are high compared to hydrocarbons and ethers of comparable molecular mass. Explain why?
Answer:
It is due to the weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 30

Question 18.
Explain nucleophilic addition reaction with an example.
Answer:
The carbonyl carbon carries a small degree of positive charge. Nucleophile such as CN can attack the carbonyl carbon and uses its lone pair to form a new carbon – nucleophile ‘σ‘ bond, at the same time two electrons from the carbon – oxygen double bond move to the most electronegative oxygen atom. This results in the formation of an alkoxide ion. In this process, the hybridisation of carbon changes from sp2 to sp3.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 31
The tetrahedral intermediate can be protonated by water or an acid to form an alcohol.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 32
eg: Addition of HCN
Attack of CN on carbonyl carbon followed by protonation gives cyanohydrins.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 33

Question 19.
Explain why the carbonyl group in aldehydes and ketones is polar.
Answer:
The carbonyl group of aldehydes and ketones contains a double bond between carbon and oxygen. Oxygen is more electronegative than carbon and it attracts the shared pair of electron which makes the carbonyl group as polar and hence aldehydes and ketones have high dipole moments.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 34

Question 20.
What is meant by the following terms? Give an example in each case.
(i) Cyanohydrin,
(ii) Acetal,
(iii) Semicarbazone,
(iv) Aldol
(v) Hemiacetal,
(vi) Oxime
(vii) Ketal,
(viii) Imine,
(ix) 2,4, DNP derivative
(x) Schiff’s base
Answer:
(i) gem – Hydroxy nitriles i.e., compounds containing hydroxyl and cyano groups on the same carbon atom are called cyano hydrins. These are produced by the addition of HCN to aldehydes and ketones in weakly basic medium.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 35
(ii) gem – dialkoxy alkanes in which two alkoxy groups attached to the terminal carbon atom are called acetals. They are produced by the action of an aldehyde with two equivalent of monohydric alcohol in the presence of dry HCl gas.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 36
(iii) Semi carbazones are derivatives of aldehydes and ketones and are produced by the action of semicarbazides on them in a weakly acidic medium.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 37
(iv) Aldols are β – hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones (containing an alpha hydrogen atom) in the presence of dilute, aqueous NaOH.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 38
(v) α – alkoxy alcohols are called hemi acetals. They are produced by the addition of one molecule of a mono hydric alcohol to an aldehyde in the presence of dry HCl gas.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 39
(vi) Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 40
(vii) gem-dialkoxy alkanes in which two alkoxy groups present on the same carbon atom are called ketals.
They are produced when a ketone is heated with ethylene glycol in the presence of dry HCl gas.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 41
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 42
(viii) Compounds containing >C=N group are called imines. These are produced when aldehydes and ketones react with ammonia and its derivatives.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 43
When ‘Z’ = H, alkyl, aryl,—NH, —OH, C6H5 etc..,
(ix) 2, 4, dinitro phenyl hydrazones (i.e., 2, 4, DNP derivative) are produced when aldehydes or ketones react with 2,4 dinitro phenyl hydrazine in weakly acidic medium.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 44
(x) Aldehydes and ketones react with primary aliphatic or aromatic amines in the presence of acid form azomethenes or schifTs bases. Schiffs bases may also be regarded as imines.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 45

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 21.
How does acetaldehyde react with (i) hydroxyl amine, (ii) hydrazine, (iii) phenyl hydrazine, (iv) semicarbazide? Give equations.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 46
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 47

Question 22.
Give equations for the reaction between
(i) acetaldehyde and sodium bisulphite
(ii) acetone and sodium bisulphite
Answer:

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 48

Question 23.
Write the structure of the prodcuts formed when acetone reacts with (i) hydrazine, (ii) phenyl hydrazine, (iii) semicarbazide.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 49
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 50

Question 24.
What is urotropine? How it is formed? Write its structure.
Answer:
Hexamethelenetetramine is called urotropine. It is formed when formaldehyde reacts with ammonia.
6HCHO + 4NH3 → (CH2)6N4 + 6H2O
Structure:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 51

Question 25.
Mention the uses of urotropine.
Answer:

  1. Urotropine is used as a medicine to treat urinary infection.
  2. Nitration of Urotropine under controlled condition gives an explosive RDX (Research and development explosive). It is also called cyclonite or cyclotri methylenetrinitramine.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 26.
What is popoffs rule? How it is used to predict the oxidation products of unsymmetrical ketones.
Answer:
It states that during the oxidation of an unsymmetrical ketone, a (C—CO) bond is cleaved in such a way that the keto group stays with the smaller alkyl group.
Eg:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 52

Question 27.
How will you prepare
(i) diacetone amine from acetone.
(ii) aldimine from acetaldehyde.
(iii) hydrobenzamide from benzaldehyde.
Answer:
(i)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 53
Two molecules of acetone, combine with ammonia to form diacetoneamine.
(ii)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 54
(iii) With ammonia, benzaldehyde forms a complex condensation product called hydrobenzamide.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 55

Question 28.
What is a haloform reaction? Explain with suitable example.
Answer:
Aldehydes and ketones containing CH3CO group (methyl carbonyl group), on treatment with excess halogen in the presence of an alkali produce a haloform (chloro form, bromo form, iodoform). This reaction is known as haloform reaction.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 56

Question 29.
What is iodoform test? Explain.
Answer:
Compounds containing CH3CO (methyl carbonyl) group or compounds on oxidation give compounds containing CH3CO group, when treated with iodine and NaOH gives a yellow precipitate is called iodoform. This is known as iodoform test.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 30.
How will you distinguish between by means of a chemical test.
(i) Propanal and propanone
(ii) acetaldehyde and benzaldehyde
(iii) Ethanal and propanal
Answer:
All these pairs of compounds can be distinguished by iodoform test.
(i) Propanone (CH3 CO CH3) will answer iodoform test. Propanal will not undergo iodo form test.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 57
(ii) Acetaldehyde (CH3CHO) will answer iodoform test, whereas benzaldehyde does not answer iodoform test
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 58
(iii) Ethanal (acetaldehyde) will answer iodoform test, but not propanal acetaldehyde
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 59
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 60

Question 31.
Write a short note an aldol condensation.
Answer:
The carbon attached to carbonyl carbon is called α – carbon and the hydrogen atom attached to α – carbon is called α – hydrogen. In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α – hydrogen add together to give β – hydroxyl aldehyde (aldol) or β – hydroxyl ketone (ketol). The reaction is called aldol condensation reaction. The aldol or ketol readily loses water to give α, β – unsaturated compounds which are aldol condensation products.
Acetaldehyde when warmed with dil.NaOH gives β – hydroxyl butraldehyde (acetaldol)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 61

Question 32.
Give the mechanism of aldol condensation.
Answer:
Mechanism: The mechanism of aldol condensation of acetaldehyde takes place in three steps.
Step 1: The carbanion is formed as the α – hydrogen atom is removed as a proton by the base.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 62
Step 2: The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 63
Step 3: The alkoxide ion formed is protonated by water to form aldol.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 64
The aldol rapidly undergoes dehydration on heating with acid to form α – β unsaturated aldehyde.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 65

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 33.
Give the oxidation products obtained when pentan-2-one is oxidised by (conc.HNO3).
Answer:
During these oxidations, rupture of the carbon-carbon bonds occur on either side of the keto group giving a mixture of carboxylic acid, each containing less a number of carbon atom. Than the original ketone.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 66

Question 34.
What is crossed aldol condensation? Give an example.
Answer:
Aldol condensation can also takes place between two different aldehydes or ketones or between one aldehyde and one ketone such an aldol condensation is called crossed or mixed aldol condensation. This reaction is not very useful as the product is usually a mixture of all possible condensation products and cannot be separated easily.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 67

Question 35.
What happens when benzaldehyde reacts with (i) acetaldehyde in the presence of dilute NaOH (ii) acetone in the presence of dilute NaOH.
[OR] Explain with examples Claisen – Schmidt condensation.
Answer:
Benzaldehyde condenses with aliphatic aldehyde or methyl ketone in the presence of dil. alkali at room temperature to form unsaturated aldehyde or ketone. This type of reaction is called Claisen – Schmidt condensation.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 68

Question 36.
Write a short note on Cannizaro reaction.
Answer:
In the presence of concentrated aqueous or alcoholic alkali, aldehydes which do not have α – hydrogen atom under go self oxidation and reduction (disproportionation) to give a mixture of alcohol and a salt of carboxylic acid. This reaction is called Cannizaro reaction.
Benzaldehyde on treatment with concentrated NaOH (50%) gives benzyl alcohol and sodium benzoate.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 69
This reaction is an example of disproportion reaction.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 37.
Write the mechanism of Cannizaro reaction.
Answer:
Cannizaro reaction involves three steps.
Step 1: Attack of OH on the carbonyl carbons.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 70
Step 2: Hydride ion transfer
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 71
Step 3: Acid – base reaction

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 72
Cannizaro reaction is a characteristic of aldehyde having no α – hydrogen.

Question 38.
Give an example for crossed Cannizaro reaction.
Answer:
When Cannizaro reaction takes place between two different aldehyde (neither containing an α hydrogen atom), the reaction is called as cross cannizaro reaction.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 73
In crossed cannizaro reaction more reactive aldehyde is oxidized and less reactive aldehyde is reduced.

Question 39.
Give an example for benzoin condensation.
Answer:
Benzaldehyde reacts with alcoholic KCN to form benzoin
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 74

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 40.
Complete the following equation.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 74
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 76
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 77
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 78
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 79
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 80

Question 41.
How will you convert ethanal into following compounds?
(i) Butane 1, 3 diol
(ii) But -2-enal
(iii) But-2-enoic acid.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 81

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 42.
Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Acetophenone and benzophenone
(ii) Phenol and benzoic acid
(iii) Pentan -2- one and pentan – 3-one
(iv) Benzaldehyde and acetophenone
Answer:
(i) Acetophenone and benzophenone: Acetophenone being a methyl ketone, when treated with I2 / NaOH (NaOI) gives a yellow precipitate of iodoform but benzophenone does not give a yellow precipitate.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 82
(ii) Phenol and benzoic acid: NaHCO3 test: Benzoic acid being a stronger acid than carbonic acid (H2CO3) decomposes NaHCO3 to evolve CO2, but phenol being weaker than carbonic acid does not.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 83
FeCl3 Test: Phenol gives a violet colouration with neutral ferric chloride solution, while benzoic acid gives a buff coloured precipitate of ferric benzoate.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 84
(iii) Pentan-2-one and pentan-3-one:
Sodium bisulphite test: 2-Pentanone being a methyl ketone when treated with a saturated solution of sodium bisulphite gives a white precipitate of 2 – pentanone sodium bisulphite addition compound where as 3 – pentanone does not.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 85
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 86
(Note: Sterically hindered ketones, do not undergo this reaction, eg: acetophenone)
Iodoform Test: 2 – pentanone, being a methyl ketone, when treated NaOI (I2 / NaOH) gives a yellow precipitate of iodoform but 3-pentanone does not.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 87
(iv) Benzaldehyde and acetophenone:
AgNO3 test: Benzaldehye reduces Tollen’s reagent to give a silver mirror (Ag) but acetophenone does not reduce Tollen’s reagent.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 88
Iodoform test: Acetophenone gives a yellow precipitate when treated with I2 / NaOH but benzaldehyde does not.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 89

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 43.
What happens when benzaldehyde is treated with
(i) Br2 in the presence of FeBr3
(ii) a mixture of conc.H2SO4 and conc.HNO3
(iii) conc.H2SO4
(iv) chlorine
(v) chlorine in the presence of FeCl3
(vi) methyl bromide in the presence of anhydrous AlCl3.
Answer:
(i) m-bromobenzaldehyde is formed.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 90
(ii) m-nitrobenzaldehyde is formed.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 91
(iii) m- benzaldehyde sulphonic acid is formed:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 92
(iv) Benzoyl chloride is formed:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 93
(v) m-chloro benzaldehyde is formed:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 94

Question 44.
Mention the uses of (i) formaldehyde (ii) urotropine (iii) acetaldehyde (iv) acetone (v) benzaldehyde (vi) acetophenone and benzophenone.
Answer:
Formaldehyde:

  1. 40% aqueous solution of formaldehyde is called formalin. It is used for preserving biological specimens.
  2. Formalin has hardening effect, hence it is used for tanning.
  3. Formalin is used in the production of thermosetting plastic known as bakelite, which is obtained by heating phenol with formalin.

Urotropine:

  1. Urotropine is used as a medicine to treat urinary infection.
  2. Nitration of Urotropine under controlled condition gives an explosive RDX (Research and development explosive). It is also called cyclonite or cyclotri methylene trinitramine.

Acetaldehye:

  1. Acetaldehyde is used for silvering of mirrors.
  2. Paraldehyde is used in medicine as a hypnotic.
  3. Acetaldehyde is used in the commercial preparation of number of organic compounds like acetic acid, ethyl acetate etc.,

Acetone:

  1. Acetone is used as a solvent, in the manufacture of smokeless powder (cordite).
  2. It is used as a nail polish remover.
  3. It is used in the preparation of sulphonal, a hypnotic.
  4. It is used in the manufacture of thermosoftening plastic Perspex.

Benzaldehyde is used:

  1. as a flavoring agent
  2. in perfumes
  3. in dye intermediates
  4. as starting material for the synthesis of several other organic compounds like cinnamaldehyde, cinnamic acid, benzyl chloride etc.

Aromatic Ketones:

  1. Acetophenone has been used in perfumery and as a hypnotic under the name hyphone.
  2. Benzophenone is used in perfumery and in the preparation of benzhydrol drop.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 45.
Give equation for the following reactions.
(i) Nitration of acetophenone
(ii) Bromination of benzophenone
(iii) Friedel – crafts alkylation of benzo phenone
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 95

Question 46.
Explain the structure of carboxylic acid group.
Answer:
A carboxyl group is made up of a carboxyl group joined to a hydroxyl group.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 96
In the carboxyl group, the carbon atom is attached to two carbon atoms: one by a double bond and the other by a single bond which in turn linked to a hydrogen atom by a single bond. The remaining free valency of the carbon atom of the carboxyl is satisfied by a H atom or an alkyl group. Thus, the structure of carboxylic acids.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 97
where ‘R’ is ‘H’ or alkyl group.
The carbon atom and the two oxygen atoms are sp2 hybridised. The two sp2 hybridesed orbitals of the carboxyl carbon overlap with one sp2 hybridised orbital of each oxygen atom while the third sp2 hybridised orbital of carbon overlaps with either a ‘s’ orbital of H – atom or a sp3 hybridised orbital of ‘C’ atom of the alkyl group to form three a bonds. Each of the two oxygen atoms are left with one unhybridised orbital which is perpendicular to the a bonding skeleton.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 98
All the three ‘p’ orbitals being parallel, overlap to form n bond which is partly delocalised between carbon and oxygen atoms on one side and carbon and oxygen of the ‘OH’group on the other side.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 99
In other words, RCOOH may be represented as a resonance hybrid of the following two canonical structures.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 100
As a result of resonance (i) the C—O single bond length in carboxylic acids is shorter than the normal C—O single bond in alcohols and ethers and (ii) C=O bond length in carboxylic acids is slightly larger than the normal C=O bond length in aldehydes and ketones.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 47.
How is acetic acid prepared from (i) ethanol (ii) methyl cyanide (iii) ethyl acetate (iv) methyl magnesium bromide (v) acetyl chloride (vi) acetic anhydride? Give equation.
Answer:
(i) Oxidation of ethanol by acidified K2Cr2O7.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 101
(ii) Acid hydrolysis of methyl cyanide.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 102
(iii) Acid hydrolysis of ethylacetate
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 103
(iv) Methyl magnesium bromide reacts with carbondioxide and the product formed on hydrolysis gives acetic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 104
(v) Hydrolysis of acetyl chloride
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 105
(vi) Hydrolysis of acetic anhydride
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 106

Question 48.
Identify X and Y
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 107
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 108
Answer:
(i) X = C6H5COOMg Br; Y = C6H5COOH
(ii) X = C6H5COOH; Y = C6H5COCl
(iii) X = C6H5COOH; Y = C6H5COONa

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 49.
Explain why?
(i) Carboxylic acids have higher boiling point than aldehydes, ketones, or even alcohols of comparable molecular mass.
(ii) Lower aliphatic / carboxylic acid are miscible with water while higher carboxylic acids are immiscible with water.
Answer:
(i) Carboxylic acids have higher boiling point ’ than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more association of carboxylic acid molecules through intermolecular hydrogen bonding.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 109
(ii) Lower aliphatic carboxylic acids (up to four carbon) are miscible with water due to the formation of hydrogen bonds with water. Higher carboxylic acid are insoluble in water due to increased hydrophobic interaction of hydrocarbon part. The simplest aromatic carboxylic acid, benzoic acid is insoluble in water.

Question 50.
How will you convert
(i) Ethyl benzene to benzoic acid
(ii) Isopropyl benzene to benzoic acid
(iii) p-nitro toluene to p-nitrpbenzoic acid
(iv) o-xylene to phthalic acid
(v) But-2-ene to ethanoic acid
(vi) Benzonitrile to benzoic acid
(vii) Ethyl magnesium iodide to propanoic acid
(viii) Ethyl benzoate to benzoic acid
(ix) Benzamide to benzoic acid
(x) Propan – 2 – one to butanoic acid.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 110
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 111
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 112
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 113

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 51.
Which of the following compounds will undergo aldol condensation? Which the Cannizaro reaction, and which neither? Write the structure of the expected products of aldol condensation and Cannizaro reaction.
(i) Methanal
(ii) 2 – methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vii) Phenyl acetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-dimethyl butanal
Answer:
I. 2 – methyl pentanal
Cyclohexanone
1 – phenyl propanone and phenyl acetaldehyde
contain one or more a hydrogen and hence undergo aldol condensation.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 114
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 115
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 116
II. Methanal, benzaldehyde, and 2,2 dimethyl butanal do not contain a hydrogen atoms and hence undergo cannizaro reaction. The reaction and structures are:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 117
III. Benzophenone is a ketone with no a hydrogens while butan-1-ol is an alcohol. Both of these neither undergo aldol condensation nor cannizaro reaction.

Question 52.
Give equations for reactions of acetic acid with the following reagents.
(i) Na (ii) NaOH (iii) Na2CO3 (iv) PCl5 (v) SOCl2 (vi) C2H5OH (vii) LiAlH4 (viii) Red P and HI (ix) Sodalime (x) NH3 followed by heating (xi) Cl2 / Red P.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 118
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 119
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 120
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 121

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 53.
Show how each of the following could be converted to benzoic acid:
(i) Ethyl benzene,
(ii) acetophenone,
(iii) benzophenone,
(iv) phenyl ethene.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 122
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 123
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 124

Question 54.
What is esterification? Give an example.
Answer:
The formation of an ester by the action of carboxylic acid and alcohol in the presence of an acid is called esterification.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 125
In these reaction the C—OH bond in acids is cleaved.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 55.
Give a brief accounts of decarboxylation reaction.
Answer:
Removal of CO2 from carboxyl group in called as decarboxylation. Carboxylic acids lose carbon di oxide to form hydrocarbon when their sodium salts are heated with soda lime (NaOH and CaO in the ratio 3:1)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 126

Question 56.
Suggest a suitable reagent to bring about the following conversions. Give equations.
(i) Acetic acid to acetyl chloride
(ii) Benzoic acid to ethyl benzoate
(iii) m – nitro benzoic acid to m – nitro methyl benzoate
(iv) Ethanoic acid to ethanol
(v) Acetic acid to ethane
Answer:
(i) PCl5 or SOCl2
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 127
(ii) C2H5OH in the presence of dil H2SO4
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 128
(iii) Methyl alcohol in the presence of dil. H2SO4
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 129
(iv) Lithium aluminium hydride
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 130
(v) HI in the presence of Red phosphorus
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 131

Question 57.
Give the products of the following:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 132
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 133
Answer:
All these reactions are decarboxylation reactions. In these reactions the COONa group is replaced by ‘H’.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 134

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 58.
What is HVZ reaction? Give an example.
Answer:
Carboxylic acids having an α – hydrogen are halogenated at the α – position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to form α halo carboxylic acids. This reaction is known as Hell-Volhard-Zelinsky reaction (HVZ reaction) The α-Halogenated acids are convenient starting materials for preparing α – substituted acids, eg:
Substitution reaction in the hydrocarbon part α – Halogenetion:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 135

Question 59.
Explain why the ‘COOH’ group in benzoic acid in meta directing.
Answer:
The structure of benzoic acid is a resonance hybrid of the following canonical structure.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 136
Because of -M effect of carboxylic acid, the ortho and para positions becomes less in electron density relative to the meta position, i.e., meta-position becomes rich in electron density compared to ortho and para positions. Hence, the electrophile attacks the meta position, i.e., the COOH group deactivates the benzene ring and is the meta directing group.

Question 60.
How are the following compounds prepared from benzoic acid?
(i) m- Bromo benzoic acid
(ii) m- nitro benzoic acid
(iii) m-sulpho benzoic acid.
Answer:
(i) Bromination of benzene gives m – bromo benzoic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 137
(ii) Nitration of benzene using nitrating mixture gives m – nitro benzoic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 138
(iii) Sulphonation of benzene, using fuming sulphuric acid gives m – sulpho benzoic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 139

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 61.
Formic acid is a reducing agent. Substantiate this statement with examples.
Answer:
Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 140
(i) Formic acid reduces Tollens reagent (ammonical silver nitrate solution) to metallic silver.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 141
(ii) Formic acid reduces Fehlings solution. It reduces blue coloured cupric ions to red coloured cuprous ions.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 142

Question 62.
Give the tests for carboxylic acid group in an organic compound.
Answer:

  1. In an aqueous solution, carboxylic acid turns blue litmus red.
  2. Carboxylic acids give brisk effervescence with sodium bicarbonate due to the evolution of carbon dioxide.
  3. When a carboxylic acid is warmed with alcohol and conc.H2SO4 it forms an ester, which is detected by its fruity odour.

Question 63.
Give a brief account of the acidity of carboxylic acids.
Answer:
Carboxylic acids undergo ionization to produce H+ and carboxylate ions in an aqueous solution. The carboxylate anion is stabilized by resonance which makes the carboxylic acid to donate the proton easily.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 143
The resonance structure of carboxylate ion are given below.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 144
The strength of carboxylic acid can be expressed in terms of the dissociation constant(Ka):
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 145
The dissociation constant is generally called acidity constant because it measures the relative strength of an acid. The stronger the acid, the large will be its Ka value.
The dissociation constant of an acid can also be expressed in terms of pKa value.
pKa = log Ka
A stronger acid will have higher Ka value but smaller pKa value, the reverse is true for weaker acids.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 64.
Briefly discuss the effect of substituents on the acidity of carboxylic acids.
Answer:
Effect of substituents on the acidity of carboxylic acid.
(i) Electron releasing alkyl group decreases the acidity: The electron-releasing groups (+1 groups) increase the negative charge on the carboxylate ion and destabilize it and hence the loss of proton becomes difficult. For example, formic acid is more stronger than acetic acid,
eg:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 146
(ii) Electron withdrawing substituents increases the acidity: The electron-withdrawing substituents decrease the negative charge on the carboxylate ion and stabilize it. In such cases, the loss of proton becomes relatively easy.
Acidity increases with increasing electronegativity of the substituents. For example, the acidity of various halo acetic acids follows the order F—CH2—COOH > Cl—CH2—COOH > Br—CH2—COOH > I —CH3—COOH
Acidity increases with increasing number of electron – withdrawing substituents on the a – carbon.
For example
Cl3C—COOH > Cl2CH—COOH > ClCH2COOH > CH3COOH
The effect of various, electron withdrawing groups on the acidity of a carboxylic acid follows the order,
—NO2 > —CN >—F >—Cl >—Br >—I >Ph
The relative acidities of various organic compounds are
RCOOH > ArOH > H2O > ROH >RC ≡ CH

Question 65.
Fluorine is more electronegative than chlorine even their p- fluoro benzoic acid in weaker than p- chloro benzoic aicd. Explain.
Answer:
Since halogens are far more electronegative than carbon and also possess lone pairs of electrons, they exert both -1 and + M effects. Now, the fluorine, the lone pairs of electrons are present in ‘2p’ orbitals and in chlorine, they are present in ‘3p’ orbitals. Since ‘2p’ orbitals of fluorine and chlorine are of almost equal size, the + M effect is more pronounced in p fluoro benzoic acid than in p-chloro benzoic effect.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 147

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 66.
Explain why p – nitrobenzoic acid has a higher Ka value than benzoic acid.
Answer:
Higher the Ka value, the stronger is the acid. Thus, p-nitro benzoic acid is a stronger acid than benzoic acid. This is due to

  1. Because of — I and -M effect of NO2 group the electron density in O—H bond decreases. As a result, the —O—H bond becomes weak and hence, p – nitrobenzoic acid easily loses a proton than benzoic acid.
    TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 148
  2. Due to -1 and -M effect of nitrogroup dispersal of negative charge occurs and hencep -nitro benzoate ion becomes more stable than benzoate ion.
    TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 149

Question 67.
Give the IUPAC names of the following compounds,
(i) PhCH2CH2COOH
(ii) (CH3)2C=CHCOOH CH
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 150
Answer:
(i) 3 -phenyl propanoic acid
(ii) 3 -methyl but – 2-enoic acid
(iii) 2 -methylcyclopentane carboxylic acid
(iv) 2, 4, 6 -trinitro benzoic acid or 2, 4, 6-trinitro benzene carboxylic acid.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 68.
Explain the mechanism for the reaction.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 151
Answer:
Esterification of carboxylic acids with alcohols is a nucleophilic acyl substitution.
In such reactions the nucleophile first adds to a carbonyl group to give a tetrahedral intermediate which then readily loses the leaving group to give the substitution products. The mechanism involves the following steps.
Step 1: Protonation of the carbonyl group:
In the presence of mineral acids (H2SO4 or HCl), the carbonyl group of the carboxylic acid accepts a proton to form protonated carboxylic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 152
Step 2: Nucleophilic attack of the alcohol molecule. As a result of protonation, the carbonyl carbon becomes electrophilic (i.e. more electropositive) and hence readily undergoes nucleophilic attack by lone pairs of electrons on the oxygen atom of the alcohol molecule to form a tetrahedral intermediate (II).

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 153
Step 3: Loss of a molecule of water and a proton. The tetrahedral intermediate (II) undergoes a proton transfer to give another tetrahedral intermediate (III). During this proton transfer, the OH group gets converted to – OH2 group which being a good leaving group is lost as a neutral water molecule. The protonation ester (IV), thus formed, finally loses a proton to give the ester (V).
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 154

Question 69.
Give the uses of (i) formic acid (ii) acetic acid (iii) benzoic acid (iv) acetyl chloride (v) acetic anhydride (vi) ethyl acetate.
Answer:
Formic acid: It is used

  1. for the dehydration of hides
  2. as a coagulating agent for rubber latex
  3. in medicine for the treatment of gout
  4. as an antiseptic in the preservation of fruit juice.

Acetic acid: It is used

  1. as table vinegar
  2. for coagulating rubber latex
  3. for manufacture of cellulose acetate and poly vinylacetate.

Benzoic acid: It is used

  1. as food preservation either in the pure form or in the form of sodium benzoate
  2. in medicine as an urinary antiseptic
  3. for manufacture of dyes.

Acetyl chloride: It is used

  1. as acetylating agent in organic synthesis
  2. in detection and estimation of —OH, — NH2 groups in organic compounds.

Acetic anhydride: It is used

  1. acetylating agent
  2. in the preparation of medicine like asprin and phenacetin
  3. for the manufacture plastics like cellulose acetate and poly vinyl acetate.

Ethyl acetate is used

  1. in the preparation of artificial fruit essences
  2. as a solvent for lacquers
  3. in the preparation of organic synthetic reagent like ethyl acetoacetate.

Uses
Acetamide is used in the preparation of primary amines.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 70.
A compound with molecular formula, C4H10O3 on acetylation with acetic anhydride gives a compound with molecular weight 190. Find out the number of hydroxyl groups present in the compound.
Answer:
During acetylation one ‘H’ atom (at mass = 1 amu) of the OH group is replaced by an acetyl group (CH3CO – molar mass = 43 amu)
—OH + (CH3CO)2O → —O—COCH3 + CH3COOH
In other words acetylation of each OH group increases the molecular mass by 43 – 1 = 42 amu. Now that the molecular mass of the compound C4H10O3 = 106 amu, while that of the acetylated product is 190 amu. Therefore the number of ‘OH’ groups present in the compound is \(\frac{190-106}{42}\) =2.

Question 71.
Explain (i) Perkins’ reaction (ii) Knoevenagal reaction.
Answer:
(i) Perkins’ reaction:
When an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the acid corresponding to the anhydride, condensation takes place and an α, β unsaturated acid is obtained. This reaction is known as Perkin’s reaction.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 155
(ii) Knoevenagal reaction:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 156
Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid, Pyridine act as the basic catalyst.

Question 72.
Give names of reagents which bring about the following conversions.
(i) Hexan-1-ol to hexanal
(ii) Cyclohexanol to cyclohexanone
(iii) p-fluorotoluene to p-fluorobenzaldehyde,
(iv) Ethanenitrile to ethanal
(v) allyl alcohol to propenal
(vi) But-2-ene to ethanal
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 157
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 158

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 73.
Bring out Che following conversions.
(i) Benzyl alcohol to phenyl ethanoic acids.
(ii) 3 – nitrobromo benzene to – 3 – nitrobromo benzoic acid
(iii) Methyl acetophenone to benzene 1, 4, dicarboxylic acid
(iv) cyclohexene to hexane 1, 6 dicarboxylic acid.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 159

Question 74.
Identify A – E in the following reactions:
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 160

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 75.
Write a short note on the electrophilic substitution reaction of benzaldehyde.
Answer:
The ‘CHO’ group in benzaldehyde deactivates the benzene ring due to its —M effect. As a result the electron densities in ortho and para position are decreased compared to the meta position, i.e, The meta position is electron rich compared to ortho and para position. Hence electrophilic substitution reaction occur at meta position.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 161

Question 76.
Which acid of each pair shown here would you expect to be stronger?
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 162
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 163
Answer:
(i)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 164
(ii) Due to much stronger -1 effect of F over Cl FCH2 COO ion is much more stable than ClCH2COO ion and hence FCH2COOH is a stronger acid than ClCFI2COOH.
(iii)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 165
Inductive effect decreases with increasing distance, therefore, -I effect of is somewhat stronger in 3-fluorobutanoic acid than in 4-fluorobutanoic acid. In other words, CH3CHCH2COCT ion is in more stable than
FCH2CH2CH2COO ion and hence TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 166 is a stronger acid than FCH2CH2CH2COOH.
Therefore due to greater stability of CF3—C6H4—COO (p) ion over CH3—C6H4COOH (p) is a stronger acid than H3CC6H4—COOH (p).
(iv)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 167

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 77.
Identify A to E in the following reactions:
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 168
(Note: NaBH4 reduces COCl to CH2OH, but does not reduce NO2 to NH2 group.)

Question 78.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene
(ii) Benzoic acid to benzaldehyde
(iii) Ethanol to 3 – hydroxy butanal
(iv) Benzene to m – nitrobenzene
(v) Benzaldehyde to benzophenone
(vi) bromo benzene to 1-phenyl ethanol
(vii) Benzaldehyde to 3 phenylpropan-1-ol
(viii) Benzaldehyde to a – hydroxy phenyl acetic acid
(ix) Benzoic acid to p-nitrobenzyl alcohol.
Answer:
(i) Propanone to propene:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 169
(ii) Benzoic acid to benzaldehyde
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 170
(iii) Ethanol to 3 – hydroxy butanal:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 171
(iv) Benzene to m – dinitrobenzene:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 172
(v) Benzaldehyde to benzophenone:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 173
(vi) Bromobenzene to 1 – phenyl ethanol:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 174
(vii) Benzaldehyde to 3 – phenylpropan – 1 – ol:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 175
(viii) Benzaldehyde to a – hydroxyphenyl acetic acid:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 176
(ix) Benzoic acid to m-nitrobenzyl alcohol:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 177

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 79.
What is transesterification? Give an example.
Answer:
Esters of an alcohol can react with another alcohol in the presence of a mineral acid to give the ester of second alcohol. The interchange of alcohol portions of the esters is termed transesterification.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 178

Question 80.
Give an example for Claisen condensation.
Answer:
Esters containing at least one α – hydrogen atom undergo self-condensation in the presence of a strong base such as sodium ethoxide to form β – keto ester. This reaction is known as Claisen condensation.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 179

Question 81.
How are the following compounds prepared?
(i) acetyl chloride and ethyl chloride from ethyl acetate
(ii) acetamide from acetyl chloride
(iii) Acetamide from acetic anhydride.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 180

Question 82.
Complete the following equation:
(i) CH3COCl + CH3NH2
(ii) CH3COCl + (CH3)2 NH →
(iii) (CH3CO)2O + PCl5
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 181

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 83.
How are the following compounds prepared from acetamide?
(i) acetic acid (ii) sodium acetate (iii) methyl cyanide (iv) methyl amine (v) ethyl amine
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 182

Question 84.
Write the structure and IUPAC names of the following acid derivatives.
(i) acetyl chloride (ii) propionyl chloride (iii) Benzoyl chloride (iv) acetic anhydride (v) propionic anhydride (vi) benzoic anhydride (vii) methyl acetate (viii) Ethyl acetate (ix) phenyl acetate (x) acetamide (xi) propionamide (xii) Benzamide.
Answer:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 183
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 184

Question 85.
Briefly explain amphoteric nature of acid amide.
Answer:
Amides behave both as weak acid as well as weak base and thus show amphoteric character. This can be proved by the following reactions.
Acetamide (as base) reacts with hydrochloric acid to form salt.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 185
Acetamide (as acid) reacts with sodium to form sodium salt and hydrogen gas is liberated.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 186

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

Question 86.
An unknown aldehyde (A) on reaction with alkali gives a β – hydroxy aldehyde which loses water to form an unsaturated aldehyde, 2-butanal. Another aldehyde (B) undergoes a disproportionation reaction in the presence of conc.alkali to form products (C) and (D). (C) is an aryl alcohol with formula C7H8O. Identify (A) and (B).
(ii) Write the sequence of chemical reaction involved.
(iii) Name the product, when (B) reacts with zinc amalgam and hydrochloric acid.
Answer:
Since the aldehyde (A) oft reaction With alkali gives a β – hydroxy aldehyde (i.e, an aldol), which loses water to form 2 – butanal (an unsaturated aldehyde) ‘A’ must be acetaldehyde.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 187
Since aldehyde (B) on treatment with conc.alkali undergo disproportionation (i.e., Cannizaro reaction) to give two products (C) and (D). One of these products must be an alcohol and the other must be the corresponding acid. Further, since, the product (C) is an aryl alcohol with molecular C7H5O, (C) must be benzyl alcohol, (i.e, C6H5 aryl group) OH is alcoholic group C6H5O. C7H8O—C6H6O = CH2. i.e, C6H5CH2OH. (D) must be sodium benzoate and aldehyde (B) must be benzaldehyde.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 188
Thus aldehyde (A) is acetaldehyde, (B) is benzaldehyde.
(ii) The sequence of reaction is already explained above.
(iii) When (B) reacts with Zn—Hg / HCl, toluene obtained.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 189

Question 87.
A compound ‘X’ (C2H4O) on oxidation gives ‘Y’ (C2H4O2). X undergoes a haloform reaction. On treatment with HCN, (X) forms (Z) which on hydrolysis gives 2 – hydroxy propanoic acid.
(i) Write down the structures of ‘X’ and ‘Z’.
(ii) Name the product when ‘X’ is treated with dilute NaOH.
(iii) Write down the equation for the reaction involved.
Answer:
(a) Since the MF C2H4O of X corresponds to the general formula CnH2nO, characteristic of aldehydes or ketones. (X) may either be an aldehyde or ketone. Since ketone must have at least ‘3’ carbon atoms but X (C2H4O) has only ‘2’ two. Therefore (X) must be an aldehyde. The only possible aldehyde is CH3CHO (MF C2H4O) (acetaldehyde).
If (X) is acetaldehyde, than the compound (Z) is acetic acid (CH3COOH).
(b) (X) is acetaldehyde is confirmed by the following reaction.
(i) It undergoes haloform reaction
(i.e.,)
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 190
(ii) On treatment with KCN, acetaldehyde forms a cyanohydrin (Z) which on hydrolysis gives 2 – hydroxy propanoic acid.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 191
(iii) In the presence of dilute NaOH, acetaldehyde undergoes aldol condensation to form 3- hydroxy butanal commonly called aldol.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 192

Choose the correct answer:

1. The IUPAC name of the compound is
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 194
(a) Mesityl oxide
(b) 4- methyl pent – 3 – en – 2 – one
(c) 4 – methyl pent – 2 – en – 4 – one
(d) 2 – methyl pent – 4 – one
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

2. Ozonolysis of 2-methyl but-2-ene followed by treatment with Zn/H2O gives:
(a) ethanal
(b) propanone
(c) propanal and prop – 2 – one
(d) ethanal and propan – 2 – one
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 195

3. Identify the product ‘Y’ in the following reaction sequence.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 196
(a) pentane
(b) cyclobutane
(c) cyclopentane
(d) cyclopentanone
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 197

4. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having molecular mass 44 u. The alkene is:
(a) 2 – butene
(b) ethene
(c) propene
(d) 1 – butene
Answer:
(a)
Hint: The aldehyde with molecular mass 44 u is CH3CHO (acetaldehyde). Therefore the symmetrical alkene is 2 – butene.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 198

5. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:
(a) a vinyl group
(b) an isopropyl group
(c) an acetylene triple bond
(d) two ethylenic double bonds
Answer:
(a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 199

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

6. Identify the compound ‘X’ in the following reaction:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 200
Answer: (a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 201

7. The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds.
I. CH3CHO,
II. (CH3)2CO and
III. PhCOPh is:
(a) III > II > I (b) II > I > III
(c) I > III > II (d) I > II > III
Answer:
(d)
Hint: Reaction between RMgX with carbonyl compound is nucleophilic addition. The greater the positive charge on the carbonyl carbon, the faster is the rate of nucleophilic addition.

8. A carbonyl compound reacts with hydrogen cyanide to form a cyanohydrin which on hydrolysis forms a racemic mixture of a – hydroxy acids. The carbonyl compound is:
(a) formaldehyde
(b) acetaldehyde
(c) acetone
(d) diethyl ketone
Answer:
(b)
Hint: Acetaldehyde cyanohydrin TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 202 on hydrolysis gives lactic acid TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 203
This contains a chiral carbon and exists as a racemic mixture. All other compounds are symmetrical and hence, their cyanohydrins and their corresponding α – hydroxy acids are not optically active because they do not contain chiral carbon.

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

9. In a set of reactions, acetic acid yielded the product D
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 204
The structure of ‘D’ would be:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 205
Answer:
(a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 206

10. (CH3)2C=CHCOCH3 can be oxidised to (CH3)2C=CHCOOH by:
(a) Chormic acid
(b) NaOI
(c) Cu at 300°C
(d) KMnO4
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 207

11. A compound ‘A’ (C5H10C312) on hydrolysis gives C5H10O which reacts with NH2OH, forms iodoform but does not give Fehling’s test A is:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 208
Answer:
(a)
Hint: Since the compound ‘A’ on hydrolysis gives C5H10O does i ot reduce Fehlings solution but reacts with NH2OH, forms iodoform, it must be methyl ketone.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 209

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

12. Acetophenone, when reacted with a base, C2H5ONa, yields a stable compound that has the structure:

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 210
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 211
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 212

13.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 213
The structure of ‘B’ is:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 214
Answer: (a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 215

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 216

14. Self-condensation of two moles of ethyl acetate in presence of sodium ethoxide yields:
(a) ethyl propionate
(b) ethyl butyrate
(c) acetoacetic ester
(d) methyl acetoacetate
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 217

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

15. m- chlorobenzaldehyde on reaction with conc. KOH at room temperature gives:
(a) potassium-m-chloro benzoic acid and m-hydroxy benzaldehyde.
(b) m-hydroxy benzaldehyde and m-chloro benzyl alcohol.
(c) m-chloro benzyl alcohol and m-hydroxy benzyl alcohol.
(d) potassium-m-chloro benzoate and m-chloro benzyl alcohol.
Answer: (d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 218

16. Consider the following compounds:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 219
Which will give the iodoform test?
(a) only I
(b) both I and II
(c) only II
(d) all
Answer:
(c)
Hint: Although TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 220 contains a CH3CO group linked to carbon, it does not undergo iodoform test. This is because iodination occurs at the more reactive CH2 group rather than terminal CH3 which is essential for iodoform test to occur
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 221

17. Fehling solution will oxidise:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 222
(a) All
(b) only I and IV
(c) only II and IV
(d) only III and IV
Answer:
(d)
Hint: Fehling solution oxidises only aliphatic aldehydes

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

18.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 223
The product ‘Z’ is:
(a) benzaldehyde
(b) benzoic acid
(c) benzene
(d) toluene
Answer:
(b)
Hint:
X = Benzene
Y = Toluene
Z = Benzoic acid

19. Match entries of column I with appropriate entries of column II.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 224
(a) (A) – (p)- (B) – (s)- (C) – (q); (D) – (r)
(b) (A) – (q); (B) – (s); (C) – (p); (D) – (r)
(c) (A) – (r); (B) – (s); (C) – (p); (D) – (q)
(d) (A) – (r); (B) – (p); (C) – (q); (D) – (s)
Answer:
(c)

20. Assertion: Fehling solution oxidizes acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.
Reason: The C—H bond in benzaldehyde is stronger than acetaldehyde.
(a) Both assertion and reason are correct and the reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but the reason is not the correct explanation of assertion.
(c) Assertion is true but reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

21. Assertion: Carboxylic acids contain a carbonyl group but do not give characteristic reactions of the carbonyl group.
Reason: Due to resonance, the electrophilic nature of the carboxyl cation is greatly reduced as compared to carbonyl carbon in aldehydes and ketones.
(a) Both assertion and reason are correct and the reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but the reason is not the correct explanation of assertion.
(c) Assertion is true but the reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(a)

22. Among the following acids which has the lowest pKa value?
(a) CH3COOH
(b) HCOOH
(c) (CH3)2CHCOOH
(d) CH3CH2COOH
Answer:
(b)
Hint: HCOOH is the strongest acid and hence the lowest pKa value.

23. Which of the following presents the correct order of the acidity in the given compounds?
(a) FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH
(b) CH3COOH > BrCH2COOH > ClCH2COOH > FCH2COOH
(c) FCH2COOH > CH3COOH > BrCH2COOH > ClCH2COOH
(d) BrCH2COOH > ClCH2COOH > FCH2COOH > CH3COOH
Answer:
(a)

24. The correct acidity order of the following is
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 225
(a) III > IV > II > I
(b) IV > III > I > II
(c) III > II > I > IV
(d) II > III > IV > I
Answer: (a)
Hint: Carboxylic acids are stronger than phenols. Further, electrons donating groups decrease the acidity of phenol/carboxylic acid while electron-withdrawing groups increase the acidity of phenol/carboxylic acid. Thus, the over all order.
III > IV > II > I

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

25. Which one of the following pairs gives effervescence with aq.NaHCO3
I. CH3COCI
II. CH3COCH3
III. CH3COOCH3
IV. CH3COOCOCH3
(a) I and II
(b) I and IV
(c) II and III
(d) I and III
Answer:
(b)
Hint: Both CH3COCl and CH3COOCOCH3 react with water to produce acids and hence give effervescence with NaHCO3.

26. Propionic acid with Br2/P yields a dibromo product. Its structure would be:
(a) CH(Br)2—CH2COOH
(b) CH2(Br)CH2COBr
(c) CH3C(Br)2COOH
(d) CH2(Br)CH(Br)COOH
Answer:
(c)
Hint:Bromination occurs at a position.

27. When benzoic acid is treated with LiAlH4, it forms:
(a) benzaldehyde
(b) benzyl alcohol
(c) benzene
(d) toluene
Answer:
(b)

28. Sodium ethoxide reacts with ethanoyl chloride. The compound that is produced in the above reaction is:
(a) 2-butanone
(b) ethyl chloride
(c) ethyl ethanoate
(d) diethyl ether
Answer:
(c)
Hint: CH3COCl + C2H5ONa → CH3COOC2H5 + NaCl

TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids

29. Consider the following:
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 226
The correct decreasing order of their reactivity towards hydrolysis is:
(a) II > IV > I > III
(b) II > IV > III > I
(c) I > II > III > IV
(d) IV > II > I > III
Answer:
(a)
Hint: Electron donating groups decrease while electron-withdrawing groups increase the reactivity of acid chlorides towards hydrolysis.
TN Board 12th Chemistry Important Questions Chapter 12 Carbonyl Compounds and Carboxylic Acids 227
The reactivity depends on the positive charge on the cabonyl carbon.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Students get through the TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Answer the following questions.

Question 1.
Classify the following as primary, secondary and tertiary alcohols.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 1
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 2
Answer:
Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv), (v)
Tertiary alcohol: (vi)

Question 2.
Identify allyl alcohols in the above (Q.No. 1) examples.
Answer:
(ii) and (vi).

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 3.
Name the following compounds according to IUPAC system:
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 3

Question 4.
Write IUPAC names of the following:
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 4
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 5

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 5.
Write the structures of the compounds whose IUPAC names are as follows:
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 6
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 7

Question 6.
Complete the following reactions:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 8
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 9
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 10

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 7.
Using suitable Grignard reagent and a carbonyl compound, how will you prepare
(i) phenyl methanol,
(ii) butan-2-ol,
(iii) 2-methylhexan-2-ol,
(iv) propan-2-ol.
Answer:
(i) phenyl methanol:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 11
(ii) butan-2-ol:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 12
(iii) 2-methylhexan-2-ol:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 13
(iv) propan-2-ol:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 14

Question 8.
Give equation for the preparation of 2-methyl-2-propanol using a suitbale Grignard reagent and a carbonyl compound.
Answer:
Tertiary containing at least two identical groups can be prepared by the addition of a Grignard reagent to an ester other the formic ester followed by acid hydrolysis.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 15

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 9.
Name the reducing agents that will convert a carbonyl compound to an alcohol. Give an example for each.
Answer:
Raney Ni, / Sodium analgen in H2O (Na—Hg/H2O), H2 in the presence of Pt.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 16
where ‘R’ is an alkyl group.

Question 10.
Why is LiAlH4 is used to reduce crotonaldehyde?
Answer:
Li crotonaldehyde is an unsaturated aldehyde (CH3CH=CHCHO). Lithium aluminium hydride selectively reduces ‘CHO’ group to ‘CH2OH’ group without affecting the C=C.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 17

Question 11.
Name the product formed when sodium borohydride is used as a reducing agent for the reduction of a compound containing both carbonyl and carboxyl group, (or) Identify the product:
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 18

Question 12.
Complete the following reactions:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 19
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 20
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 21

Question 13.
How is glycerol prepared from triglycerides or fats?
Answer:
The alkaline hydrolysis of these fats gives glycerol and the reaction is known as saponification.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 22

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 14.
Complete the following equations:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 23
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 24
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 25

Question 15.
Give mechanism of reaction between alcohols and halogen acids.
Answer:
Primary alcohols react by SN2 mechanism while secondary and tertiary alcohol react by SN1 mechanism.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 26

Question 16.
How will you distinguish between
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 27
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 28
(ii) Secondary alcohol will give a blue colour as shown above whereas a tertiary alcohol will not produce any colour.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 29

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 17.
Give the structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl2, (b) HBr, (c) SOCl2 (i) Butan-l-ol and (ii) 2-methylbutan-2-ol.
Answer:
(a) with HCl-ZnCl2 (Lucas reagent):
[2-methylbutan-2-ol being a tertiary alcohol reacts with Lucas reagent to produce turbidity immediately due to formation of insoluble tert-alkyl-halide, while butan-l-ol being a primary alcohol does not react with Lucas reagent at room temperature.]
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 30
(b) with HBr: Both alcohols produce the corresponding alkyl bromides.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 31
(c) with SOCl2: Both alcohols react to form their corresponding alkyl chlorides.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 32

Question 18.
Arrange the following compounds in increasing order of boiling points, pentan-1 -ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol. Give reason for your answer.
Methanol < ethanol < propan-1-ol < butan-2- ol < butan-1-ol < pentan-1-ol.
Answer:
Boiling point increases regularly as the molecular mass increases due to the corresponding increase in their vander waals forces of attraction. The boiling point increases in the order.
methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol.
Among isomeric alcohol, 2° alcohols have lower boiling point than 1° alcohols due to corresponding decrease in the extent of hydrogen bonding because of steric hindrance. Thus the boiling point of butan-2-ol is lower than butan-1-ol.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 19.
Explain why propanol has higher boiling point than that of hydrocarbon butane?
Answer:
The molecuels of butane are held together by weak vander waals forces of attraction while these of propanol are held together by stronger hydrogen bonding.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 33
Therefore boiling point of propanol is higher than that of butane.

Question 20.
Write the mechanism for the reaction
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 34
Answer:
The reaction follows SN1 mechanism.
Step 1: Formation of protonated alcohol.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 35
Step 2: Protonated alcohols forms carbocation.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 36
Step 3: Elimination of a proton to form alkane.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 37

Question 21.
Give mechanism for the formation of prop-1 – ene from propan-1-ol.
Answer:
The reaction is
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 38
Mechanism:
Step 1: Formation of protonated alcohol.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 39
Step 2: Formation of carbocation.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 40
Step 3: Elimination of a proton.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 41

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 22.
State Saytzeff rule.
Answer:
During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

Question 23.
Name the reagents in the following reactions:
(i) oxidation of a primary alcohol to a carboxylic acid.
(ii) oxidation of a primary alcohol to an aldehyde.
(iii) benzyl alcohol to benzoic acid.
(iv) butan-2-one to butan-2-ol.
Answer:
(i) Acidified potassium dichromate or neutral or acidic or alkaline potassium permanganate (followed by hydrolysis with dilute H2SO4).
(ii) Pyridinium chloro chromate(PCC) in CH2Cl2 or pyridinium dichromate (PDC) in CH2Cl2.
(iii) Acidified or alkaline KMnO4 (followed by hydrolysis with dil H2SO4).
(iv) Ni/H2 or NaBH4 or LiAlH4.

Question 24.
What happen when,
(i) vapours of ethanol is passed over heated copper at 573 K.
(ii) vapours of propan-2-ol is passed over heated copper at 573 K.
(iii) vapours of 2-methyl propan-2-ol is passed over heated copper at 573 K. Give equation.
Answer:
(i) Ethanal is formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 42
(ii) Propanone is formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 43
(iii) 2-methyl prop-1-ene is formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 44

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 25.
What is nitroglycerine? How it is formed? Give equation.
Answer:
Glycerotrinitrate (1,2,3, trinitroxy propane) is nitroglycerine. It is formed by nitration of glycerine in the presence of conc. H2SO4.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 45

Question 26.
How is acrolein formed from glycerol?
Answer:
When glycerol is heated with dehydrating agents like conc. H2SO4, KHSO4, or P2O5 acrolein is formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 45

Question 27.
Mention the cause for the acidic nature of alcohols.
Answer:
The acidic character of alcohols is due to the electronegative oxygen atom which withdraws electrons of the O—H bond towards itself. As a result, the O—H bond becomes weak and hence a proton can be easily abstracted by a base, i.e., ionization of alcohol to yield alkoxide ion and H+ ions are facilitated.
ROH ⇌ RO + H+

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 28.
Mention the uses of (i) methanol, (ii) ethanol, (iii) ethylene glycol, (iv) glycerol.
Answer:
(i) Uses of methanol:
(a) Methanol is used as a solvent for paints, varnishes, shellac, gums, cement, etc.
(b) In the manufacture of dyes, drugs, perfumes, and formaldehyde.

(ii) Uses of ethanol:
(a) Ethanol is used as an important beverage.
(b) It is also used in the preparation of

  • Paints and varnishes.
  • Organic compounds like ether, chloroform, iodoform, etc.,
  • Dyes, transparent soaps.

(c) As a substitute for petrol under the name power alcohol used as fuel for airplane.
(d) It is used as a preservative for biological specimens.

(iii) Uses of ethylene glycol:
(a) Ethylene glycol is used as an antifreeze in an automobile radiators.
(b) Its dinitrate is used as an explosive with DNG.

(iv) Uses of glycerol:
(a) Glycerol is used as a sweetening agent in confectionery and beverages.
(b) It is used in the manufacture of cosmetics and transparent soaps.
(c) It is used in making printing inks and stamp pad ink and lubricant for watches and clocks.
(d) It is used in the manufacture of explosives like dynamite and cordite by mixing it with china clay.

Question 29.
Alcohols are easily protonated in comparison to phenols. Explain why.
Answer:
In phenols, the lone pair of electrons on the oxygen atom is delocalized over the benzene ring due to resonance and hence, not easily available for protonation. In contrast, in alcohols, the lone pair of electrons on the oxygen atom is localized due to the absence of resonance and hence are easily available for protonation.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 30.
Identify the major product obtained in the following reaction:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 47
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 48

Question 31.
Identify the product of the reaction:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 49
Answer:
In the presence of a weak nucleophile such as ethanol, neopentyl bromide ionizes to form first a 1° carbocation which rearranges to form a more stable 3° carbocation. This is then attacked by the weak nucleophile ethanol followed by a less proton to yield ethyl neopentyl ether.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 50

Question 32.
Show how 2-methyl propan-1-ol is prepared by the reaction of a suitable Grignard reagent.
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 51
This part that has been enclosed in the box comes from methanal while the remaining part comes from the Grignard reagent. Therefore the Grignard reagent should be isopropyl magnesium bromide. Thus,
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 52

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 33.
Write the structure of the products of the following reactions.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 53
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 54
(ii) NaBH4 (sodium bromohydride) is a reducing agent. It reduces aldehydes and ketones but not esters.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 55
(ii) NaBH4 reduces, -CHO group to CH2OH group. Thus,
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 56

Question 34.
What is esterification? Give an example.
Answer:
Formation of an ester TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 57 from an alcohol and an organic acid in the presence of a catalyst is known as esterification. In the reaction, the ‘OH’ of the carboxylic acid and ‘H’ of the alcoholic group are removed as water.
Eg:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 58

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 35.
Why are alcohols are weaker acids than water?
Answer:
In water, ‘O’ atom is attached to two ‘H’ atom, but in alcohols, ‘O’ atom is attached to one ‘H’ atom and one ‘R’ group (alkyl group). Since ‘R’ group has an electron-donating inductive effect (+I), it increases the electron density in the O—H bond compared to that O—H bond in water. In other words, —O—H bond in alcohols is stronger than the —O—H bond in water. As a result, proton removal from alcohol is more difficult than in water. Hence, alcohols are weaker acids than water.

Question 36.
The acidic nature of alcohols is in the order 1° alcohol > 2° alcohol > 3° alcohol. Explain.
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 59
Alkyl (R) groups are electron releasing (i.e., +I effect) the electron density in the O—H bond in tertiary alcohols is maximum followed by secondary alcohol, while in primary alcohols, it is minimum. This means that O—H bond in tertiary alcohol is the strongest and hence most difficult to break followed by O—H bond in the secondary alcohols, while the O—H bond in primary alcohols is the weakest and hence most easy to break. Thus the acidic strength is in the order.
Primary > Secondary > Tertiary

Question 37.
Arrange the following compounds in increasing order against each.
(i) CH3CH2OH, CF3CH2OH. CCl3CH2OH (acid strength)
(ii) 2-methyl-2-propanol, 1-butanol, 2-butanol (reactivity towards sodium)
Answer:
(i) Due to -I effect of halogens, the electron density in the O—H bond decreases. As a result, the O—H bond strength becomes weak and proton removal is easier compared to CH3CH2OH. Further, since fluorine has a stronger -I effect, than chlorine, CF3CH2COOH is a stronger acid than CCl3CH2COOH. Hence, CF3CH2OH is the strongest acid while CH2CH2OH is the weakest acid. Hence increasing order of acid strength is
CH3CH2OH < CCl3CH2OH < CF3CH2OH.
(ii) It is an acid-base reaction since alcohols are acidic and sodium is a strong base. As such, the reactivity of these alcohols towards sodium increases as the acidic character of alcohols increases. Since the acidic character increases in the order 3° < 2° < 1°. Therefore the reactivity towards alcohol increases in the same order.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 60

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 38.
How will you convert phenol to (i) benzene, (ii) aniline, (iii) phenylacetate, (iv) phenyl benzoate, (v) anisole?
Answer:
(i) Benzene: Phenol is converted to benzene on heating with zinc dust. In this reaction, the hydroxyl group which is attached to the aromatic ring is eliminated.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 61
(ii) Aniline: Phenol on heating with ammonia in presence of anhydrous ZnCl2 gives aniline.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 62
(iii) Phenyl acetate:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 63
(iv) Phenyl benzoate:

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 64
(v) Anisole:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 65

Question 39.
Identify A and B in the following:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 66
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 67
(ii) A= HNO2 (273 -278) K
B = H2O
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 68
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 69

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 40.
What is picric acid? How it is prepared? 2,4,6 trinitrophenol is called picric acid.
Answer:
Nitration with cone. HNO3 and conc.H2SO4 gives picric acid.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 70

Question 41.
Give equation for the following:
(i) nitration of phenol,
(ii) sulphonation of phenol,
(iii) nitrosation of phenol.
Answer:
(i) Nitration of phenol: Phenol can be nitrated using 20% nitric acid even at room temperature, a mixture of ortho and para nitro phenols are formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 71
(ii) Sulphonation of phenol: Phenol reacts with cone. H2SO4 at 280 K to form o-phenol sulphonic acid as the major product. When the reaction is carried out at 373 K the major product is p-phenol sulphonic acid.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 72
(iii) Nitrosation of phenol: phenol can be readily nitrosated at low temperature with HNO2.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 73

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 42.
What happens when phenol is treated with
(i) acidified potassium dichromate,
(ii) hydrogen in the presence of nickel as a catalyst,
(iii) bromine water,
(iv) bromine in the presence of CCl4.
Answer:
Phenol treated with
(i) acidified potassium dichromate (oxidation):
Phenol undergoes oxidation with air or acidified K2CrO7 with conc. H2SO4 to form 1, 4-benzoquinone.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 74
(ii) hydrogen in the presence of nickel as catalyst (reduction): Phenol on catalytic hydrogenation gives cyclohexanol.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 75
(iii) bromine water (halogenation): phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromo phenol.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 76
(iv) bromine in the presence of CCl4: If the reaction is carried out in CS2 or CCl4 at 278 K, a mixture of ortho and para bromo phenols are formed.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 77

Question 43.
Explain why the phenolic group is ortho para directing.
Answer:
The lone pair of electrons on the oxygen atom of the phenolic group enter into conjugation with the benzene ring due +M effect of the phenolic group. As a result, the ortho and para position become electron-rich compounds to the meta position. Hence the electrophilic attack the ortho and para position.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 78

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 44.
Write a short note on Reimer-Tiemann reaction.
Answer:
On treating phenol with CHCl3 / NaOH, a -CHO group is introduced at the ortho position.
This reaction proceeds through the formation of substituted benzal chloride intermediate.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 79

Question 45.
How will you distinguish between alcohol and phenol?
Answer:

  1. Phenol reacts with benzene diazonium chloride to form a red-orange dye, but ethanol has no reaction with it.
  2. Phenol gives purple coloration with neutral ferric chloride solution, alcohols do not give such coloration with FeCl3.
  3. Phenol reacts with NaOH to give sodium phenoxide. Ethyl alcohol does not react with NaOH.

Question 46.
Mention the uses of phenol.
Answer:

  1. About half of the world production of phenol is used for making phenol-formaldehyde resin. (Bakelite).
  2. Phenol is a starting material for the preparation of
    (a) drugs such as phenacetin, salol, aspirin, etc.
    (b) phenolphthalein indicator.
    (c) explosive like picric acid.
  3. It is used as an antiseptic-carbolic lotion and carbolic soaps.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 47.
Give the IUPAC names of the following ethers.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 80
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 81
Answer:
(i) 1-ethoxy-2-methyl propane
(ii) 2-chloro-l-methoxy ethane
(iii) 4-nitroanisole
(iv) 1-methoxy propane
(v) 4-ethoxy-1, 1, dimethyl cyclohexane
(vi) ethoxy benzene

Question 48.
Ethers are soluble in water. Give reason.
Answer:
The oxygen of ether can also form hydrogen bonds with water and hence they are miscible with water. Ethers dissolve a wide range of polar and non-polar substances.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 82

Question 49.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis.
(i) 1 -propoxypropane
(ii) Ethoxy benzene
(iii) 2-methyl-2-methoxy propane
(iv) 1-methoxyethane
Answer:
(i)
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 83
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 84
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 85

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 50.
Give the mechanism for the conversion of (i) ethanol to ethoxyethane, (ii) dehydration of 1-propanol to 1-proproxy propane.
Answer:
(i)
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 86
[Acid catalyzed 1° alcohol to ethers occurs via SN2 reaction, involving a nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.]
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 87

Question 51.
Explain why the boiling point of ethers is slightly higher than that of alkanes and lower than that of alcohols of comparable molecular mass.
Answer:
Due to weak dipole-dipole interactions, the boiling point of ethers is only slightly higher than those of u-alkanes having comparable molecular masses.
The boiling points of ethers are less than that of alcohols is that ethers do not form hydrogen bonds.

Question 52.
Mention the uses of (i) Diethyl ether and (ii) Anisole.
Answer:
Uses of Diethyl ether:

  1. Diethyl ether is used as a surgical anesthetic agent in surgery.
  2. It is a good solvent for organic reactions and extraction.
  3. It is used as a volatile starting fluid for diesel and gasoline engines.
  4. It is used as a refrigerant.

Uses of anisole:

  1. Anisole is a precursor to the synthesis of perfumes and insecticide pheromones
  2. It is used as a pharmaceutical agent.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 53.
Discuss the mechanism of the reaction
CH3OCH2CH3 + HI → CH3I + CH3CH2OH.
Answer:
This reaction is an SN2 reaction. The cleavage of ethers by halogen acids occurs by the following mechanism.
Step 1: Ethers being Lewis acids, undergo protonation to form oxonium ions.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 88
This reaction takes place with HBr or HI because they are sufficiently acidic.
Step 2: Iodide ion is a good nucleophile. Due to steric hindrance, it attacks the smaller alkyl group of oxonium ion formed in step 1 and displaces the alcohol molecule by SN2 mechanism as shown below:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 89
Step 3: when excess HI is used, ethanol formed reacts with another molecule of HI to form ethyl iodide.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 90

Question 54.
Explain why ‘OCH3’ in the anisole group is ortho-para directing.
Answer:
The lone pair of electrons on the oxygen atom of the alkoxy group enters into the conjugation with the benzene ring due to its +M effect.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 91
As a result, the ortho and para position become rich in electron density compared to the meta position. Hence, the electrophile attacks ortho and para position.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

Question 55.
Briefly account of dipolar nature of C—O bonds in ethers.
Answer:
Because of the greater electronegativity of oxygen and carbon, the C—O bond is slightly polar thus have a dipole moment. Since the two C—O bonds in ethers are inclined to each other at an angle of 110°, the two dipoles do not cancel each other. As a result, ethers have a dipole moment of 1.15 to 1.3D
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 92

Question 56.
Give the equation for the reaction between diethyl ether with an excess of oxygen.
Answer:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 93

Question 57.
Give examples for (i) phthalein fusion and (ii) coupling reaction.
Answer:
(i) Phthalein fusion: On heating phenol with phthalic anhydride in presence of conc. H2SO4, phenolphthalein is obtained.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 94
(ii) Coupling reaction: Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy azobenzene (a red-orange dye).
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 95

Choose the correct answer:

1. The IUPAC name of the compound is:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 96
(a) 2-chloro-5- hydroxyhexane
(b) 2-hydroxy-5-chlorohexane
(c) 5-chlorohexan-2-ol
(d) 2-chlorohexan-5-ol
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

2. The IUPAC name of the compound is:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 97
(a) 1-methoxy-1-methyl ethane
(b) 2-methoxy-2-methyl ethane
(c) 2-methoxypropane
(d) isopropyl-methyl ether
Answer:
(c)

3. Arrange the following compounds in increasing order of boiling point:
I – propan-1-ol
II – butan-1-ol
III – butan-2-ol
IV – pentan-1-ol
(a) I, III, II, IV
(b) I, II, III, IV
(c) IV, III, II, I
(d) IV, II, III, I
Answer:
(a)
Hint: The boiling point increases as the molecular mass increases. Further among isomeric alcohols 1° alcohols have higher boiling points than 2° alcohols.

4. Acid catalysed hydration of alkenes except ethene, leads to the formation of:
(a) primary alcohol.
(b) secondary or tertiary alcohol.
(c) a mixture of primary and secondary alcohol.
(d) a mixture of secondary and tertiary alcohol.
Answer:
(b)
Hint: Hydration of ethene gives a 1° alcohol while all others give a either a 2° alcohol or 3° alcohol.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 98

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

5. Which of the following Grignard reagent is suitable for the preparation of 3-methyl-2- butanol?
(a) 2-butanone + methyl magnesium bromide
(b) Acetone + ethyl magnesium bromide
(c) Acetaldehyde + isopropyl magnesium bromide
(d) Ethyl propionate + methyl magnesium bromide
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 99

6. Which of the following esters shown, after reduction with LiAlH4 and aqueous workup, will yield two molecules of only a single alcohol?
(a) C6H5COOC6H5
(b) CH3CH2COOCH2CH3
(c) CH3COOCH3
(d) C6H5COOCH2C6H5
Answer:
(d)
Hint: Acyl and arylalkyl groups contain the same number of carbon atoms.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 100

7. Which of the following will exhibit highest boiling point?
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 101
Answer:
(b)
Hint: Among isomeric alcohols, the alcohol with no branching has the highest boiling point.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

8. Consider the reactions:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 102
The mechanisms of reactions (i) and (ii) are respectively:
(a) SN1 and SN2
(b) SN1 and SN1
(c) SN2 and SN2
(d) SN2 and SN1
Answer:
(c)

9.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 103
which of the following compounds will be formed?
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 104
Answer:
(a)
Hint: Nucleophilic attack occurs on the smaller alkyl group.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 105

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

10. The reaction of TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 106with RMgX leads to the formation of:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 107
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 108

11.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 109
X is:
(a) bromine in benzene
(b) bromine in water
(c) potassium bromide solution
(d) bromine in CCl4 at 0° C
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 110

12. Phenol can be distinguished from ethanol by the following reagent except.
(a) sodium
(b) I2 / NaOH
(c) neutral FeCl3
(d) Br2 / H2O
Answer:
(a)
Hint: Na reacts with both ethapol and phenol.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

13. An ether is more volatile than an alcohol having the same molecular formula. This is due to:
(a) intermolecular hydrogen bonding in alcohols
(b) dipolar character of ethers
(c) alcohols having resonance structures
(d) intermolecular hydrogen bonding in ethers
Answer:
(a)

14. Mark the correct increasing order of reactivity of the following compounds with HBr / HI.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 111
(a) I < II < III
(b) II < I < III
(c) II < III < I
(d) III < II < I
Answer:
(c)
Hint: All the three benzyl alcohols react with HBr or HI through the formation of intermediate carbocation. Obviously more stable the carbocation, more reactive is the alcohol. Now, electron withdrawing group i.e, NO2, Cl, etc., decrease the stability of carbocations. Since, NO2 is a stronger, electron withdrawing group, then Cl, stability of carbocation increases in the order.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 112
Therefore, the reactivity of the benzyl alcohols increases in the same orders, i.e., II < III < I.

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

15. Which of the following reagents can be used to oxidize primary alcohols to aldehydes?
1. CrO3 in acidic medium,
2. KMnO4 in acidic medium,
3. Pyridinium chlorochromate,
4. Heat in the presence of Cu at 573 K.
(a) 1, 3, 4
(b) 1, 3
(c) 1, 4
(d) none of the above
Answer:
(a)
Hint: KMnO4 will oxidise the aldehyde formed to carboxylic acid.

16. Assertion: Bond angles in ethers is slightly more than the tetrahedral angle.
Reason: There are repulsions between two bulky ‘R’ groups.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is wrong, but reason is correct.
(d) Both assertion and reason are wrong.
Answer:
(c)
Hint: Correct assertion: Bond angles in ethers are slightly more than the tetrahedral bond angle.

17. Assertion: Bromination of benzene does not require Lewis acid is catalyst.
Reason: Lewis acid polarises the bromine molecule in Friedel craft’s reaction.
(a) Assertion and reason both are correct, and reason is the correct explanation of assertion.
(b) Assertion and reason both are correct, but reason is not the correct explanation of assertion.
(c) Assertion is true, but reason is wrong.
(d) Both assertion and reason are wrong.
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

18. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 113
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 114
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 115
Answer:
(b)

19. Match the entries in column I with appropriate entries in column II and choose the correct option using the codes given:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 116
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 117
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

20. In the following sequence of reactions,
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 118
(a) CH3CH2CH2OH
(b) CH3CH(OH)CH3
(c) CH3CH2CH2CH2OH
(d) (CH3)3CCH2OH
Answer:
(b)
Hint: Since alkenes give Markovnikoff addition products, therefore 1° alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, ‘Z’ is 2-propanol.
i.e.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 119

21. Consider the following reaction:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 120
The compound ‘Z’ is:
(a) CH3CH2OCH2CH3
(b) CH3CH2OSO3H
(c) CH3CH2OH
(d) CH2=CH2
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 121

22. Formation of tert-butyl ether by reaction of sodium tert-butoxide and methyl bromide involves:
(a) elimination reaction
(b) electrophilic addition reaction
(c) nucleophilic addition reaction
(d) nucleophilic substitution reaction
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

23.
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 122
Answer:
(a)

24. The red coloured compound formed during Victor-Meyer’s test is:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 123
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers 124

TN Board 12th Chemistry Important Questions Chapter 11 Hydroxy Compounds and Ethers

25. Following compounds are given:
(i) CH3CH2OH
(ii) CH3COCH3
(iii) CH3CH(OH)CH3
(iv) CH3OH
Which of the above compounds on being warmed with iodine solution andNaOH, will give iodoform?
(a) (i), (iii) and (iv)
(b) only (i)
(c) (i), (ii) and (iii)
(d) (i) and (ii)
Answer:
(c)
Hint: Compounds containing CH3CO group or CH3CH(OH) group on warming with I2 and NaOH will form iodoform. Hence, (i), (ii) and (iii) will give iodoform.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Students get through the TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Answer the following questions.

Question 1.
Give three examples of adsorption.
Answer:

  1. Charcoal adsorbs ammonia.
  2. Silica gel adsorbs water.
  3. Charcoal adsorbs colorants from sugar.

Question 2.
Distinguish between adsorption and absorption.
Answer:
Adsorption is a surface phenomenon while absorption is a bulk phenomenon, i.e., the adsorbate molecules are distributed throughout the adsorbent.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 3.
What are adsorbent and adsorbates? Give examples.
Answer:
Adsorbent is the material on which adsorption take place. Adsorbed substance is called adsorbate.
Examples for adsorbates: The gaseous molecules like He, Ne, O2, N2, SO2 and NH3 and solutions of NaCl or KCl.
Examples for adsorbents: Silica gel, metals like Ni, Cu, Pt, Ag and Pd and certain colloids.

Question 4.
What is an interface?
Answer:
The surface of separation of two phases where the concentration of adsorbed molecules is high is known as interface.

Question 5.
Mention the characteristics of adsorption.
Answer:

  1. Adsorption can occur in all interfacial surfaces i.e., the adsorption can occur in between gas-solid, liquid solid, liquid- liquid, solid- solid and gas-liquid.
  2. Adsorption is always accompanied by decrease in free energy. When ΔG reaches zero, the equilibrium is attained.
  3. Adsorption is a spontaneous process.
  4. When molecules are adsorbed, there is always a decrease in randomness of the molecules.
    We know, ΔG = ΔH – T ΔS where ΔG is change in free energy.
    ΔH is change in enthalpy and ΔS is change in entropy.
    Hence, ΔH = ΔG + TΔS Adsorption is exothermic as there is an interaction between adsorbate and adsorbent.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 6.
What do you understand by the term ‘sorption’?
Answer:
The term represents simultaneous adsorption and absorption.

Question 7.
What is the term used for sorption of gases on metal surfaces?
Answer:
Occlusion.

Question 8.
Give three examples for chemical adsorption or chemisorption.
Answer:

  1. Adsorption of O2 on tungsten.
  2. Adsorption of H2 on nickel.
  3. Adsorption of ethyl alcohol vapours on nickel.

Question 9.
Give two examples for physical adsorption.
Answer:

  1. Adsorption of N2 on mica.
  2. Adsorption of gases on charcol.

Question 10.
Explain the nature of the force of attraction that exists between an adsorbent and an adsorbate in (i) physical adsorption and (ii) chemical adsorption.
Answer:
The forces of attraction that exists between adsorbate and adsorbent in
(i) Physical adsorption are (a) vander waals force of attraction, (b) dipole-dipole interaction and (c) dispersion forces.
(ii) In chemisorption gas molecules are held to the surface by formation of chemical bonds. Since strong bonds are formed, nearly 400 kJ mole is given out as heat of adsorption.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 11.
Why is adsorption exothermic?
Answer:
When adsorption occurs, entropy decreases, i.e., ΔS = negative. As ΔG = ΔH – TΔS, for a spontaneous process, AG must be negative. This can only be so, if ΔH is negative, i.e., the process is exothermic.

Question 12.
How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid?
Answer:

  1. The smaller the size of the particles of the adsorbent, greater is the surface area, greater is the adsorption.
  2. At constant temperature, adsorption first increases with increase in pressure and then attains equilibrium at high pressures.
  3. In physical adsorption, it decreases with increase in temperatures, but in chemisorption first, it increases and then decreases.

Question 13.
What are adsorption isotherms and isobars? Explain.
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsoiption isotherm.
The plot of the amount of adsorption verses temperature at constant pressure is called adsorption isobar. The adsorption isobars of physisorption and chemisorption as shown in figure.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 1
In physical adsorption, \(\frac{x}{m}\) decreases with increases in T, But in chemical adsorption, \(\frac{x}{m}\) increases with rise in temperature and then decreases. The increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy.
The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 14.
Derive Freundlich adsorption isotherm.
Answer:
A plot between the amount of adsorbate adsorbed and the pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.
In order to explain these isotherms, various equations were suggested as follows: Freundlich adsorption isotherm.
According to Freundlinch,
\(\frac{x}{m}\) = Kp\(\frac{1}{n}\)
where V is the amount of adsorbate or adsorbed on ‘m’ gm of adsorbent at a pressure of p. K and n are constants.
Value n is always less than unity.
This equation is applicable for adsoiption of gases on solid surfaces. The same equation becomes \(\frac{x}{m}\) = Kc\(\frac{1}{n}\) when used for adsorption in solutions with c as concentration.
This equation quantitively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at a constant temperature.
Taking log on both sides of the equation
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 2
Hence the intercept represents the value of log K and the slope \(\frac{b}{q}\) gives \(\frac{1}{n}\).
This equation explains the increase of \(\frac{x}{m}\) with an increase in pressure. But experimental values show the deviation at low pressure.

Question 15.
Explain the application of adsorption (i) in the use of gas masks, (ii) in the use of softening hard water, (iii) in decolourisation of sugar.
Answer:
(i) Gas masks are devices containing suitable adsorbents so that poisonous gases present in the atmosphere are preferentially adsorbed and the air for breathing is purified.
Activated charcoal is one of the best adsorbents.
(ii) Permutit is employed for this process which adsorbs Ca2+ and Mg2+ ions in its surface, there is an ion exchange as shown below it occurs on the surface.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 3
Exhausted permutit is regenerated by adding a solution of common salt.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 4
(iii) Sugar prepared from molasses is decolourised to remove coloured impurities by adding animal charcoal which acts as decolourising material.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 16.
What are adsorption indicators?
Answer:
In the precipitation titrations, the endpoint is indicated by an external indicator that changes its colour after getting adsorbed on precipitate. It is used to indicate the endpoint of the titration.

Question 17.
Distinguish between homogeneous catalysis and heterogeneous catalysis with suitable examples.
Answer:
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 5

Question 18.
Name the catalysts used in the following:
(i) Haber’s process in the manufacture of ammonia.
(ii) Oxidation of ammonia.
(iii) Hydrogenation of alkenes.
(iv) Decomposition of H2O2.
(v) Formation of acetophenone from benzene and ethanoyl chloride. Also, give equations for the reactions.
Answer:
(i) The catalyst used is Iron,
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 6
(ii) The catalyst is platinum gauze.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 7
(iii) The catalyst is finely divided nickel.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 8
(iv) The catalyst is platinum
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 9
(v) The catalyst is anhydrous aluminium chloride
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 10

Question 19.
Mention the characteristics of catalysts.
Answer:

  1. For a chemical reaction, the catalyst is needed in very small quantities. Generally, a pinch of catalyst is enough for a reaction in bulk.
  2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
  3. A catalyst itself can not initiate a reaction. It means it can not start a reaction that is not taking place. But, if the reaction is taking place at a slow rate it can increase its rate.
  4. A solid catalyst will be more effective if it is taken in a finely divided form.
  5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.
  6. In an equilibrium reaction, the presence of catalyst reduces the time for attainment of equilibrium and here it does not affect the position of equilibrium and the value of the equilibrium constant.
  7. A catalyst is highly effective at a particular temperature called as optimum temperature.
  8. The presence of a catalyst generally does not change the nature of products
    eg: 2SO2 + O2 → 2SO3
    This reaction is slow in the absence of a catalyst, but fast in the presence if Pt catalyst.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 20.
What are promoters? Explain with an example.
Answer:
In a catalysed reaction the presence of a certain substance increases the activity of a catalyst. Such a substance is called a promoter. For example in the Haber’s process of manufacture ammonia, the activity of the iron catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter. In the same way, Al2O3 can also be used as a promoter to increase the activity of the iron catalyst.

Question 21.
Explain the action of anhydrous aluminium chloride in Friedel-Crafts reaction.
Answer:
The mechanism of Friedel crafts reaction is given below
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 11
The action of catalyst is explained as follows:
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 12
It is an intermediate.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 13

Question 22.
Give the mechanism of the following reactions based on intermediate formation theory.
Answer:

  1. Thermal decomposition of KClO3 in the presence of MnO2.
    Steps in the reaction
    2KClO3 → 2KCl + 3O2
    can be given as
    2KClO3 + 6MnO2 → 6MnO3 + 2KCl
    It is an intermediate,
    6MnO3 → 6MnO2 + 3O2
  2. The reaction between H2 and O2 in the presence of copper.
    2Cu + \(\frac{1}{2}\) O2 → Cu2O It is an intermediate.
    Cu2O + H2 → H2O + 2Cu
  3. Oxidation of HCl by air in the presence of CuCl2.
    2CuCl2 → Cl2 + Cu2Cl2
    2Cu2Cl2 + O2 → 2Cu2OCl2
    It is an intermediate.
    2Cu2OCl2 + 4HCl → 2H2O + 4CuCl2
    This theory describes
    (a) the specificity of a catalyst and
    (b) the increase in the rate of the reaction with an increase in the concentration of a catalyst.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 23.
Mention the limitations of the intermediate compound formation theory of catalysis.
Answer:

  1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promotors).
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
Explain the salient features of the adsorption theory of catalysis.
Answer:
According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.
The various steps involved in a heterogeneous catalysed reaction are given as follows:

  1. Reactant molecules diffuse from bulk to the catalyst surface.
  2. The reactant molecules are adsorbed on the surface of the catalyst.
  3. The adsorbed reactant molecules are activated and form an activated complex which is decomposed to form the products.
  4. The product molecules are desorbed.
  5. The product diffuses away from the surface of the catalyst.

Question 25.
What are active centres? How does the presence of active centres influence the rate of catalysed reaction?
Answer:
The surface of a catalyst is not smooth. It bears steps, cracks and comers. Hence the atoms on such locations of the surface are coordinatively unsaturated. So, they have much residual force of attraction. Such sites are called active centres. So, the surface carries high surface free energy.
The presence of such active centres increases the rate of reaction by adsorbing and activating the reactants.
The adsorption theory explains the following

  1. Increase in the activity of a catalyst by increasing the surface area. An increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
  2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
  3. A promoter or activator increases the number of active centres on the surfaces.

Question 26.
What is the role of a promoter in heterogeneous catalysis?
Answer:
A promoter increases the number of active centres on the surfaces.

Question 27.
What are catalytic poisons? How do they affect a catalysed reaction?
Answer:
Substances that destroy the activity of catalysts are known as catalytic poisons. They affect the catalytic reaction by occupying the active centres of the catalyst as a result active centres will not available for adsorption by reactants.

Question 28.
Give some examples of enzyme catalyses. Some common examples for enzyme catalysis.
Answer:

  1. The peptide glycyl L-glutamyl L-lyrosin is hydrolysed by an enzyme called pepsin.
  2. The enzyme diastase hydrolyses starch into maltose.
    2(C6H10O5)n + nH2O → nC2H22O11
  3. The yeast contains the enzyme zymase which converts glucose into ethanol.
    C6H12O6 + H2O → 2C2H5OH + 2CO2
  4. The enzyme micoderma aceti oxidises alcohol into acetic acid.
    C2H5OH + O2 → CH3COOH + H2O
  5. The enzyme urease present in soya beens hydrolyses the urea.
    TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 14

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 29.
Briefly outline the characteristics of enzyme catalysed reaction.
Answer:

  1. An enzyme catalyst is specific in nature, i.e., one catalyst cannot catalyse more than one reaction.
  2. The enzyme catalysed reactions are very fast compared to other catalysed reactions.
  3. The rate of an enzyme-catalysed reaction depends on temperature. At an optimum temperature, the rate of the reaction becomes maximum. Beyond the optimum temperate, the rate decrease.
  4. The enzyme activity is maximum at a particular pH called optimum pH. The favourable pH range for enzymatic reactions is 5-7.
  5. The enzymatic activity is increased by the presence of containing substances, known as coenzymes.
  6. The activity of enzymes can also be inhibited or poisoned due to the presence of certain substances.

Question 30.
Explain phase transfer catalysis with an example.
Answer:
Substitution of Cl and CN in the following reaction.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 15
R-Cl = 1-chlorooctane
R-CN = 1-cyanooctane
By direct heating of two-phase mixture of organic 1 -chlorooctane with aqueous sodium cyanide for several days, 1-cyanooctane is not obtained. However, if a small amount of quaternary ammonium salt like tetraalkyl ammonium chloride is added, a rapid transition of 1 -cyanooctane occurs in about 100% yield after 1 or 2 hours. In this reaction, the tetraalkyl ammonium cation, which has hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reaction with 1-chloroocatne as shown below:
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 16
So phase transfer catalyst speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Name the dispersed phase and dispersed medium in the following: fog, dust, sharing cream, milk, paints.
Answer:

Example Dispersed phase Dispersion medium
Fog Liquid Gas
Dust Solid Gas
Sharing cream Gas Liquid
Milk Liquid Liquid
Paint Solid Liquid

Question 32.
Give the name of the colloid and one example for (i) a gas dispersed in solid, (ii) a liquid dispersed in a solid, (iii) a solid dispersed in a solid.
Answer:
(i) A gas dispersed in a solid is known as a solid foam. Pumice stone, rubber, bread are examples.
(ii) A liquid dispersed in a solid is known as a gel. Butter and cheese are examples.
(iii) A solid dispersed in another solid is known as a solid sol. Pearls, coloured glass, alloys are examples of solid sols.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 33.
Briefly outline the principle involved in the preparation of colloidal solution by dispersion methods.
Answer:
In the dispersion method, the particles whose size is higher than that of colloidal particles are disintegrated to that of the colloidal particles subjecting it to a mechanical disintegration in a colloid mill or by electro-dispersion.

Question 34.
Explain how colloidal particles of metals are prepared by electro dispersion method.
Answer:
An electrical arc is struck between electrodes dispersed in the water surrounded by ice. When a current of 1 amp /100V is passed an arc produced forms vapours of metal which immediately condense to form a colloidal solution. By this method colloidal solution of many metals like copper, silver, gold, platinum, etc. can be prepared Alkali hydroxide is added as a stabilising agent for the colloidal solution.

Question 35.
Explain peptisation with an example.
Answer:
The conversion of a freshly precipitated substance into a colloid by the addition of an electrolyte (peptising agent) is known as peptisation. For example, freshly precipitated silver chloride is converted into a colloidal solution by the addition of HCl.

Question 36.
Explain the principle involved in the preparation of colloids by the condensation method.
Answer:
Particles whose size are lesser than the colloidal particles are brought together to the size of colloidal particles.

Question 37.
What is the principle involved in the purification of a colloidal solution by dialysis?
Answer:
The separation of crystalloids (electrolytes) from colloids is based upon the principle that particles of crystalloid pass through the animal membrane (bladder) or parchment paper, or cellophane sheet whereas those of colloid do not.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 38.
Explain why? (i) colloidal particles are not visible to the naked eye, (ii) colloidal particles pass through ordinary filter paper, (iii) the sky appears blue, (iv) finest gold sol is red in colour.
Answer:
(i) No particle is visible to the naked eye if its diameter is less than half the wavelength of light used. The shortest wavelength of light is about 4000Å or 400 mμ. Hence no particle of diameter less than 200 mμ can be seen. The size of the colloidal particles is less than 200 mμ.
(ii) The size of the colloidal particles are smaller than the pores in the filter paper and hence, they pass through the filter paper.
(iii) This is due to the Tyndall effect. The colloidal particles absorb light energy and then scatter in all directions.
(iv) The colour of the colloidal solution depends as the wavelength of scattered radiation by dispersed particles. The wavelength of light further depends on the size and nature of particles. The finest gold sol is red in colour. As the size of particles increase, it becomes purple, then blue and finally golden yellow.

Question 39.
Explain the Tyndall effect.
Answer:
Colloids have optical properties. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark, in a perpendicular direction.
But when light passes through a colloidal solution, it is scattered in all directions. The colloidal particles absorb a portion of light and the remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear. This phenomenon is known as the Tyndall effect.

Question 40.
Give an account of the Brownian movement.
Answer:
The zig-zag random motion of colloidal particles, when viewed through ultramicroscope, is known as Brownian movement. The reason for Brownian movement is that the colloidal sol particles are continuously bombarded with the molecules of the dispersion medium and hence they follow a zigzag, random, continuous movement.
Brownian movement enables us,
I. to calculate Avogadro number.
II. to confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with an increase in temperature.
III. to understand the stability of colloids: As the particles are in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 41.
Explain the term electrophorosis.
Answer:
The movement of colloidal particles towards cathode or anode depending on the charge of the particles under the influence of electric current is known as electrophorosis.
This experiment is used to detect the charge on the colloidal particles. If the colloidal particles move towards the cathode, it is negatively charged. If they are positively charged, they migrate towards anode.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 42.
Give examples for positively charged and negatively charged colloids.
Answer:

Positively charge colloids Negatively charge colloids
Ferric hydroxide Ag, Au & Pt
Aluminium hydroxide Arsenic sulphide
Basic dyes Clay
Haemoglobin Starch

Question 43.
Account for the charge on colloidal particles.
Answer:
The colloidal particles, in whichever method is prepared, contain traces of electrolytes. It preferentially adsorbs either positive or negative ions from the solution. If the particles adsorb positive ions from the solution, it becomes a positively charged colloid. If it adsorbs a negative ion from the solution, it becomes a negatively charged colloid.

Question 44.
Account for the stability of a colloid.
Answer:
Since the colloidal particles are either positive or negatively charged, they do not come close together. They repel each other. Thus, the charge on the colloidal particles is responsible for its stability.

Question 45.
Explain the term coagulation or flocculation.
Answer:
The conversion of a colloid to a precipitate is known as coagulation or precipitation. Coagulation results in the removal of charges on the colloid by any one of the following:

  1. Addition of electrolytes
  2. electrophoresis
  3. mixing oppositely charged sols
  4. boiling.

These methods help remove charges from the particles. Thus, the particles come closer, form an aggregate of particles and finally settles down as a precipitate due to gravity.

Question 46.
The peptising agent is added to convert a precipitate into a colloidal solution. Give reason.
Answer:
Ions (either positive or negative) of the peptising agent (electrolyte) are adsorbed on the particles of the precipitate. They repel and hit each other breaking the particles of the precipitate into colloidal size.

Question 47.
Explain what is observed (i) when a beam of light is passed through a colloidal solution.
(ii) an electrolyte, NaCl is added to ferric hydroxide solution,
(iii) electric current is passed through a colloidal solution.
Answer:
(i) Scattering of light by colloidal particles take place and the path of light becomes visible.
(ii) The positively charged colloidal particles of Fe(OH)3 gets coagulated by the oppositely charged Cl ions provided by NaCl.
(iii) On passing electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge Cl get coagulated.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 48.
What happens when gelatin is mixed with gold sol?
Answer:
Gold sol is lyophobic. Gelatin forms lyophilic sol and acts as a protective colloid. On adding gelatin to gold sol, the latter becomes more stable.

Question 49.
A colloid is formed by adding FeCl3 in excess of hot water. What will happen if an excess ferric chloride is added to this colloid?
Answer:
On adding FeCl3 to water, a colloidal sol of hydrated ferric oxide is formed. This is a positively charged colloid as it adsorbs Fe+3 ions on its surface. On adding NaCl, Cl ions bring about coagulation.

Question 50.
How does the boiling of a colloid, help the coagulation of a colloidal solution?
Answer:
Boiling vessels in increased collisions between the colloidal particles. The sol particles combine and settle down as a precipitate.

Question 51.
Explain the protective action of lyophilic colloids.
Answer:
The addition of a lyophobic colloid into lyophilic colloid protects the latter from coagulation. Hence, lyophilic colloids act as protective colloids.

Question 52.
Define gold number. Mention its use.
Answer:
‘Gold number’ as a measure of protecting the power of a colloid. The gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl solution. Smaller the gold number greater the protective power.

Question 53.
What are emulsions? Mention various types of emulsions.
Answer:
Emulsions are a colloidal solutions in which a liquid is dispersed in another liquid.
Generally, there are two types of emulsions.

  1. Oil in Water (O/W)
  2. Water in Oil (W/O)

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

Question 54.
Give one example each for (i) oil in water and (ii) water in oil emulsions.
Answer:
(i) In Oil in Water (O/W) emulsions, the dispersed phase is oil and the dispersion medium is water.
eg: Milk is an emulsion of liquid fat dispersed in water and vanishing cream.
(ii) Emulsion of Water in Oil (W/O) in which water is the dispersed phase and oil is the dispersion medium.
eg: Cod liver oil, butter and cold cream.

Question 55.
What is emulsification?
Answer:
The process of preparation of emulsion by the dispersal of one liquid in another liquid is called Emulsification.

Question 56.
What is an emulsifier or emulsification agent?
Answer:
It is a substance added to have a stable emulsion. The type of emulsion formed depends upon the nature of the emulsifying agent. For example, the presence of soluble soaps as an emulsifying agent favours the formation of oil in water type of emulsions, whereas insoluble soaps (containing non alkali metal atoms) favours the formation of water in oil emulsions.

Question 57.
Mention the properties of the emulsion.
Answer:

  1. Emulsions exhibit all the properties like the Tyndall effect, Brownian movement, Electrophoresis, Coagulation on the addition of electrolytes.
  2. Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic precipitation by adding large amounts of the electrolytes to precipitate out the dispersed phase or by chemical destruction of emulsifying agent.
  3. Emulsions can be diluted by adding any amount of dispersion medium.

Choose the correct answer:

1. Which of the following is true with respect of adsorption?
(a) ΔG < 0, ΔS > 0, ΔH < 0
(b) ΔG < 0, ΔS < 0, ΔH < 0
(c) ΔG > 0, ΔS > 0, ΔH < 0
(d) ΔG < 0, ΔS < 0, ΔH > 0
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

2. Which of the following statements is incorrect regarding physisorption?
(a) It occurs because of vander Waals forces.
(b) More easily liquefiable gases are adsorbed readily.
(c) Under high pressure, it results in multimolecular layer on the adsorbent surface.
(d) Enthalpy of adsorption in low and positive.
Answer:
(d)
Hint: Enthalpy of adsorption is always negative.

3. Physical adsorption of a gaseous species may change to chemical adsorption with
(a) decrease in temperature
(b) increase in temperature
(c) increase in surface area of the adsorbent
(d) decrease in surface area of the adsorbent
Answer:
(b)
Hint: At high temperatures, gaseous species may dissociate into atoms that are chemisorbed, eg: N2 an iron surface at ≥ 773 K.

4. In physisorption, adsorbent does not show specificity for any particular gas because:
(a) involved van der Waals forces are universal
(b) gases involved behave like ideal gases
(c) enthalpy of adsorption is low
(d) it is a reversible gas
Answer:
(a)

5. Which of the following is an example of adsorption?
(a) water on silica gel
(b) water on calcium carbide
(c) hydrogen on finely divided nickel
(d) oxygen on the metal surface
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

6. On the basis of data given below predict which of the following gases show least adsorption on a definite amount of charcoal
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 17
(a) CO2
(b) SO2
(c) CH4
(d) H2
Answer:
(d)
Hint: Higher the critical temperature, the greater is the adsorption. H2 has the least critical temperature and hence least adsorbed.

7. Which of the following are the characteristics of chemisorption?
(i) High heat of adsorption
(ii) Irreversibility
(iii) Low activation energy
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Answer:
(a)

8. The plot of \(\frac{x}{m}\) (along Y-axis) versus log C (along X-axis) in the Freundlich adsorption isotherm is a horizontal line parallel to X-axis when,
(a) n = 0
(b) n = 1
(c) n = ∞
(d) such a plot is impossible
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 18
i.e., constant parallel to log C axis.
[Note: This is the equation for adsorption of solids in solution.]

9. Match the following:

A Fog 1 Gel
B Milk 2 Foam
C Cheese 3 Emulsion
D Soap lather 4 Aerosol

Which of the following is the correct option?
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 19
Answer:
(b)

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

10. Which of the following will show the Tyndall effect?
(a) Aqueous solution of soap below critical miscelle concentration.
(b) Aqueous solution of soap above critical miscelle concentration.
(c) Aqueous solution of sodium chloride
(d) Aqueous solution of sugar
Answer:
(b)

11. Method by which lyophobic sol can be protected:
(a) addition of oppositely charged sol
(b) addition of an electrolyte
(c) addition of lyophilic sol
(d) by boiling
Answer:
(c)

12. Freshly prepared precipitate sometimes gets converted to colloidal solution by:
(a) coagulation
(b) electrolysis
(c) diffusion
(d) peptisation
Answer:
(d)

13. Which of the following electrolytes will have maximum coagulating value for AgI / Ag+ Sol?
(a) Na2S
(b) Na3PO4
(c) Na2SO4
(d) NaCl
Answer:
(b)
Hint: AgI / Ag+ is a positively charged sol. Electrolytes whose negative ion (anion) has the least negative charge will require the maximum amount and will have the maximum coagulating value. Hence, Cl in NaCl.

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

14. A colloidal system having a solid substance as a dispersed phase and liquid as dispersion medium is classified as:
(a) solid sol
(b) gel
(c) emulsion
(d) sol
Answer:
(d)

15. Which of the following process is responsible for the formation of the delta at places where the river meets the sea?
(a) Emulsification
(b) Colloid formation
(c) Coagulation
(d) Peptisation
Answer:
(c)
Hint: Formation of the delta-shaped heap of sand clay etc where the river falls into the sea due to coagulation of sand/clay by NaCl present in seawater.

16. When an excess of a very dilute aqueous solution of KI is added to a very dilute aqueous solution of silver nitrate, the colloidal particles of silver iodide are associated with which of the following Helmholtz double layer?
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 20
Answer:
(d)
Hint:
As excess KI has been added, I ions are adsorbed on AgI forming a fixed layer (giving negative charge). It then attracts the counter ion (K+) from the medium forming a second layer, (diffused layer).

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

17. The ratio of a number of moles of AgNO3, Pb(NO3)2 and Fe(NO3)3 required for coagulation of a definite amount of colloidal sol of silver iodide prepared by mixing AgNO3 with an excess of KI will be:
(a) 1:2:3
(b) 3:2:1
(c) 6:3:2
(d) 2:3:6
Answer:
(c)
Hint:
With excess KI, colloidal particles will be [AgI]I.
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 21
Molar ratio required for coagulation of same amount of [AgI]I is 1 : \(\frac{1}{2}\) : \(\frac{1}{3}\) = 6 : 3 : 2.

18. Gelatin is mostly used in making ice creams in order to:
(a) prevent the formation of colloidal sol.
(b) enrich fragrance.
(c) prevent crystallisation and stabilise the mix.
(d) modify the taste.
Answer:
(c)
Hint: Gelatin is a protective colloid. It stabilises the mix and prevents crystallisation.

19. Which of the following has a minimum gold number?
(a) Starch
(b) Sodium oleate
(c) Gum arabic
(d) Gelatin
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry

20. Match the following:
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 22
Which of the following is the correct option?
TN Board 12th Chemistry Important Questions Chapter 10 Surface Chemistry 23
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Students get through the TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Answer the following questions.

Question 1.
Define the following terms:
(i) Resistivity,
(ii) conductivity,
(iii) cell constant,
(iv) specific conductance,
(iv) molar conductivity,
(vi) equivalent conductance.
Answer:
(i) Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross-sectional area and are separated by unit distance.
(ii) The reciprocal of specific resistance \(\left(\frac{1}{\rho}\right)\) is called specific conductance or conductivity.
(iii) The ratio between the distance between the two electrodes in a conductivity cell to their area of cross-section is called cell constant.
(iv) The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension.
(v) Molar conductance is defined by the conducting power of all the ions produced by 1 mole of the electrolyte.
(vi) Equivalent conductance is defined as the conductance of ‘V’ m3 of the electrolyte solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are 1 meter apart.

Question 2.
Give the relationship between resistance and specific resistance of an electrolytic conductor.
Answer:
R = ρ × \(\frac{l}{a}\)
where ‘R’ is the resistance, ρ is the specific resistance or resistivity, ‘l’ is the distance between the electrodes in a conductivity cell, and ‘a’ is the area of cross-section of electrodes.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 3.
Mention the units of (i) resistance, (ii) specific resistance, (iii) conductance, (iv) specific conductance, (v) molar conductance and (vi) equivalent conductance.
Answer:

Property Unit
Resistance ohm
Specific resistance ohm metre (Dm)
Conductance Siemens (s)
Specific conductance ohm-1m-1 or mhom-1 (Sm-1)
Molar conductance Sm2mol-1
Equivalent conductance Sm-1

Question 4.
Give the relationship between molar conductance and specific conductance of an electrolyte.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 1

Question 5.
Give the relationship between equivalent conductance and specific conductance of an electrolyte.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 2

Question 6.
Mention the factors which influence electrolytic conductance.
Answer:

  1. A solvent of a higher dielectric constant shows high conductance in solution.
  2. Conductance is inversely proportional to the viscosity of the medium, i.e., conductivity increases with the decrease in viscosity.
  3. If the temperature of the electrolytic solution increases, conductance also increases. An increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.
  4. Molar conductance of a solution increases with an increase in dilution. This is because, for a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, the degree of dissociation increases with dilution.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 7.
Explain how does the molar conductance of the electrolytes varies with the concentration of the electrolyte.
Answer:
The molar conductance of a strong electrolyte increases on dilution. The variation of molar conductance with dilution is given by an empirical formula
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 3
where Λm0 is called limiting molar conductivity and C is the concentration of the electrolyte.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 4
For strong electrolytes such as KCl, NaCl, etc., the plot, Λm vs √C, gives a straight line as shown in the graph. It is also observed that the plot is not a linear one for weak electrolytes.
For strong electrolyte, at a high concentration, the number of constituent ions of the electrolyte in a given volume is high and hence the attractive force between the oppositely charged ions is also high. Moreover, the ions also experienced a viscous drag due to greater solvation. These factors attribute to the low molar conductivity at high concentrations. When the dilution increases, the ions are far apart and the attractive forces decrease. At infinite dilution the ions are so far apart, the interaction between them becomes insignificant, and hence, the molar conductivity increases and reaches a maximum value at infinite dilution.
For a weak electrolyte, at high concentration, the plot is almost parallel to the concentration axis with a slight increase in conductivity as the dilution increases. When the concentration approaches zero, there is a sudden increase in the molar conductance and the curve is almost parallel to Λm0 axis. This is due to the fact that the dissociation of the weak electrolyte increases with the increase in dilution (Ostwald dilution law). Λm0 values for strong electrolytes can be obtained by extrapolating the straight line. But the same procedure is not applicable for weak electrolytes, as the plot is not a linear one, A°m values of the weak electrolytes can be determined using Kohlraush’s law.

Question 8.
Flow much charge is required for the following reactions?
(i) 1 mole of Al3+ to Al
(ii) 1 mole of Cu2+ to Cu
(iii) 1 mole of MnO4 to Mn2+
Answer:
(i) Al3+ +3e → Al
Reduction of 1 mole of Al3+ requires 3 moles of electrons.
∴ Charge on 3 moles of electrons
= 3 × 96500 C
= 289500 coulomb
(ii) Cu2+ + 2e → Cu
Reduction of 1 mole of Cu2+ requires 2 mole of electrons
= 2 × 96500 C
= 193000 coulomb
(iii) MnO4 + 8H+ + 5e → Mn2+ + 4H2O
Reduction of MnO4 ions require 5 mole of electrons
= 5 × 96500 C
= 482500 Coulomb

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 9.
A solution of CuSO4 is electrolyzed for 10 minutes with a current of 1.5 amperes. What mass of copper deposited at the cathode?
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 5
∴ Mass of copper deposited at the cathode = 0.296 g

Question 10.
How many hours does it take to reduce 3 mole of Fe3+ to Fe2+. With 2 ampere of current? (F = 96500 coulomb)
Answer:
The required equation is Fe3+ + e → Fe2+
3 mole of Fe3+ requires 3 mole of electrons.
∴ Required charge = 3 × 96500 coloumbs
= 289500 coloumbs
charge = current × time
289500 = 2 × t
t = \(\frac{289500}{2}\)
= 1.475 × 105 sec
= 40 hour.

Question 11.
In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold salt and the second cell contains copper sulfate solution.
If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also, calculate the magnitude of the current in amperes.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 6
Let Z be the electrochemical equivalent of copper
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 7

Question 12.
The specific conductance of a decinormal solution of KCl is 0.00112 ohm-1 cm-1. The resistance of a cell containing the solution was found to be 56 ohms. What is the value of the cell constant?
Answer:
Specific conductance = \(\frac{Cell constant}{Resistance}\)
Cell constant = Specific conductance × Resistance
= 0.0112 × 56
= 0.6272 cm-1

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 13.
1.0 N solution of salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq.cm in area was found to offer a resistance of 50 ohms. Calculate the equivalent conductivity of the solution.
Answer:
Given: l = 2.1 cm; a = 4.2 sq.cm; R = 50 ohm
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 8
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 9

Question 14.
The specific conductivity of 0.02 M KCl solution at 25° C is 2.768 × 10-3 ohm-1 cm-1. The resistance of this solution at 25° C, when measured with a particular cell, was 250.2 ohm. The resistance of 0.01 M CuSO4 solution at 25° C measured with the same cell was 8331 ohm. Calculate the molar conductivity of the copper sulfate solution.
Answer:
Cell constant = Specific conductivity of KCl × Resistance = 2.768 × 10-3 × 250.2
For 0.01 M CuSO4
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 10

Question 15.
At 291 K, the molar conductivities at infinite dilution of NH4Cl, NH4OH, and NaCl are 129.8, 217.8, and 108.9 S cm2 respectively. If the molar conductivity of a cent normal solution of NH4OH is 9.33 S cm2, what is the percentage dissociation of NH4OH at this dilution? Also, calculate the dissociation constant of NH4OH.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 11
∴ degree of dissociation (α)
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 12
Calculation of dissociation constant
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 13

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 16.
The conductivity of a saturated solution of AgCl at 288 K. is found to be 1.382 × 10-6 Scm-1. Find its solubility. Given ionic conductances of Ag+ and Cl ion at infinite dilution are 61.9 S cm2 mol-1 and 76.3 S cm2 mol-1 respectively.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 14

Question 17.
Define (i) electrode potential and (ii) emf of a cell.
Answer:
(i) Electrode potential: The potential difference set up between the metal and its ions in the solution is called the electrode potential, (or) electrode potential may be defined as the tendency of an electrode to lose or gain electrons when it is in contact with a solution of its own ions.
(ii) emf of a cell; The difference between the electrode potentials of the two half cells in a galvanic cell is known as the cell potential (or) It is the driving forces that permit the flow of electrons from anode to cathode.

Question 18.
Explain the terms (i) oxidation potential and (ii) reduction potential of an electrode.
Answer:
(i) Oxidation potential: when an electrode is negatively charged with respect to the solution, i.e., it acts as an anode, the potential difference is called oxidation potential.
M ⇌ Mn+ + ne
It refers to the tendency for oxidation to occur at the electrode.
(ii) Reduction potential: When an electrode is positively charged with respect to solution, i.e., it acts as the cathode, the potential difference developed is called reduction potential
Mn+ + ne ⇌ M
It refers to the tendency for reduction to occur at the electrode.

Question 19.
What is a reference electrode? Mention its use.
Answer:
A reference electrode is an electrode whose electrode potential value is known. It is used to determine, electrode potentials of various electrodes.

Question 20.
Mention the characteristics of electrode chemical series.
Answer:

  1. The negative sign of the standard reduction potential indicates that the electrode when joined with SHE, acts as an anode and oxidation occurs at this electrode. Similarly, the sign of standard reduction potential indicates that the electrode when joined with SHE acts as and reduction occurs on this electrode.
  2. Greater the value of standard reduction potential move easily the substance (element or ions) is reduced or in other words, stronger oxidizing agent it is. Similarly, the lower the value of standard reduction potential of the substance greater is the extent of oxidation half-reaction and the substance acts as a reducing agent.
  3. Reactivity of metals: A metal that has a high negative value (or small positive value) of standard reduction potential loses electrons is said to be chemically active. The chemical reactivity decreases from top to bottom of the series.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 21.
Given the standard reduction potentials
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 15
Arrange these metals in their increasing order of reducing character.
Answer:
Ag < Hg < Cr < Mg < K.
The lesser the reduction potentials greater is their reducing character.

Question 22.
Write down the electrode reactions and the net cell reaction for the following cells. Which electrode would be the positive terminal in each cell.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 16
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 17
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 18

Question 23.
The standard EMF of the cell:
Ni | N12+ || Cu2+ | Cu is 0.59 V. The standard reduction potential of the copper electrode is 0.34 V. Calculate the standard reduction potential of the nickel electrode.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 19

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 24.
A cell is set up between copper and silver electrodes as Cu | Cu2+ || Ag+ | Ag.
Given: TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 41. Calculate the emf of the cell.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 20

Question 25.
Two half cells are Al3+ | Al and Mg2+ | Mg. The reduction potentials of these half cells are -1.66 and -2.36 V. Construct the galvanic cell, write the cell reaction and calculate the emf of the cell.
Answer:
For the construction of a galvanic cell,
= (Electrode with lesser value of reduction potential – Electrode with higher value of reduction potential )
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 21
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 22

Question 26.
A cell is prepared by dipping a copper rod in 1M CuSO4 solution and a nickel rod in 1M NiSO4 solution. The standard reduction potentials of copper and nickel electrodes are 0.34 V and – 0.25 V respectively.
(a) How will you represent the cell?
(b) What will be the cell reaction?
(c) What will be the standard emf of the cell?
(d) Which electrode will be positive?
Answer:
The cell is represented as
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 23
(d) The cathode is a copper electrode. Because reduction occurs at this electrode.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 27.
Predict whether zinc and silver react with 1M H2SO4 to give hydrogen gas or not. Given the standard reduction potentials of zinc and silver electrodes are -0.76 and 0.80 V respectively.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 24
Answer:
The cell is
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 25
The emf of the cell is positive. Hence the cell reaction is spontaneous. Thus, zinc will liberate hydrogen gas from 1M H2SO4.
Similarly,
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 26
The cell is
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 27
The emf of the cell is negative. Hence the reaction will not occur, i.e., Ag will not displace hydrogen from 1M H2SO4.

Question 28.
Can a nickel sulphate be used to stir a solution of copper sulphate. Support your answer with a reason.
[OR]
Can a solution of 1M copper sulphate be stored in a vessel made of nickel.
Given:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 28
Answer:
The reaction,
Ni + CuSO4 → NiSO4 + Cu
should not occur, if CuSO4 solution to be stored in a nickel vessel.
The cell would be Ni | Ni2+ || Cu2+ | Cu
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 29
The emf is positive. This means copper sulfate reacts with nickel, Hence CuSO4 cannot be stored in a nickel vessel.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 29.
Give the relationship between the electrode potential and the concentration of the electrode.
Answer:
This relationship is given by Nernst equation. For an electrode reaction,
Mn+ + ne → M
the electrode potential is given by the relationship,
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 30

Question 30.
For general cell reactions
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 31
Write the Nernst equation.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 32
where is the number of electrons involved in the cell reaction?

Question 31.
Calculate the electrode potential of copper electrode, when a copper wire is dipped in 0.1 M CuSO4 solution ECu2+/Cu0 = 0.34.
Answer:
The electrode reaction is Cu2+ + 2e → Cu
Applying Nernst equation
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 33
[Note: Decreasing concentration of the electrolyte decreases the electrode potential]

Question 32.
The following reaction takes place in a galvanic cell.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 34
Calculate the emf of the cell.
Answer:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 35
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 36

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 33.
Give the relationship between the free energy charge of the cell reaction and the emf of the cell.
Answer:
ΔG° = -n FEcell

Question 34.
Give the relationship between the free energy change for the cell reaction and the equilibrium constant.
Answer:
ΔG° = – 2.303 RT log Kc

Question 35.
(a) Calculate the standard free energy change and maximum work obtainable by a galvanic cell, where the cell reaction
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 37
(b) Also calculate the equilibrium constant for the reaction.
Answer:
(a) The cell is
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 38
Thus the maximum work that can be obtainable from the cell = 212.3 kJ mol-1
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 39

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 36.
Describe the nature of anode, cathode, the cell reaction, and salient features of the Leclanche cell.
Answer:
Anode: Zinc container
Cathode: Graphite rod in contact with Mn02
Electrolyte: ammonium chloride and zinc chloride in water.
Emf of the cell is about 1.5 V.
Cell reaction:
Oxidation at anode:
Zn(s) → Zn2+(aq) + 2e … (1)
Reduction at cathode:
2NH4+(aq) + 2e → 2NH3(aq) + H2(g) … (2)
The hydrogen gas is oxidised to water by MnO2
H2(g) + 2MnO2(s) → Mn2O3(s) + H2O (l) …(3)
Equation (1) + (2) + (3) gives the overall redox reaction
Zn(s) + 2NH4+ (aq) + 2MnO2(s) → Zn2+(aq) + Mn2O3(s) + H2O (l) + 2NH3 ….. (4)
The ammonia produced at the cathode combines with Zn2+ to form a complex ion [Zn(NH3)4]2+(aq). As the reaction proceeds, the concentration of NH4+ will decrease and the aqueous NH3 will increase which lead to the decrease in the emf of cell.

Question 37.
Describe the nature of anode, cathode the cell reaction, and salient features of the mercury button cell.
Answer:
Anode: Zinc amalgamated with mercury.
Cathode: HgO mixed with graphite.
Electrolyte: Paste of KOH and ZnO.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 40
Cell emf: about 1.35V.
Uses: It has a higher capacity and longer life. Used in pacemakers, electronic watches, cameras etc…

Question 38.
Explain the functioning of lead storage batteries.
[OR]
Explain the chemical reaction involved in the process of charging and recharging in a lead storage battery.
Answer:
Anode: Spongy lead
Cathode: Lead plate bearing PbO2
Electrolyte: 38% by mass of H2SO4 with density 1.2g / mL.
Oxidation occurs at the anode:
Pb(s) → Pb2+(aq) + 2e … (1)
The Pb2+ ions combine with SO4-2 to from PbSO4 precipitate.
Pb2+(aq) + SO42 (aq) → PbSO4(s) …….. (2)
Reduction occurs at the cathode:
PbO2(s) + 4H+(aq) + 2e → Pb2+(aq) + 2H2O (l) ……. (3)
The Pb2+ ions also combine with SO42- ions from sulphuric acid to form PbS04 precipitate.
Pb2+(aq) + SO42-(aq) → PbSO …… (4)
Overall reaction:
Equation (1) + (2) + (3) + (4) ⇒
Pb(s) + PbO2 (s) + 4H+(aq) + 2SO42-(aq) → 2PbSO4(s) + 2H2O (l)
The emf of a single cell is about 2 V. Usually six such cells are combined in series to produced 12 volt.
The emf of the cell depends on the concentration of H2SO4. AS the cell reaction uses SO22 ions, the concentration H2SO4 decreases. When the cell potential falls to about 1.8V, the cell has to be recharged.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Question 39.
Discuss the nature of the anode, cathode, and the cell reaction of the Lithium-ion battery. How does it act as a secondary cell?
Answer:
Anode: Porus graphite.
Cathode: transition metal oxide such as CoO2. Electrolyte: Lithium salt in an organic solvent At the anode oxidation occurs:
Li(s) → Li+(aq) + e
At the cathode reduction occurs:
Li+ + CoO2 (s) + e → Li CoO2 (s)
Overall reactions:
Li(s) + CoO2 → Li CoO2 (s)
Both electrodes allow Li+ ions to move in and out of their structures.
During discharge, the Li+ ions produced at the anode move towards the cathode through the non-aqueous electrolyte. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li+ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
Uses: Used in cellular phones, laptop computer digital camera, etc…

Question 40.
Mention the various methods of protecting metals from corrosion.
Answer:
This can be achieved by the following methods.

  1. Coating metal surface by paint.
  2. Galvanizing by coating with another metal such as zinc, zinc is a stronger oxidizing agent than iron and hence it can be more easily corroded than iron, i.e., instead of iron, the zinc is oxidized.
  3. Catholic protection: In this technique, unlike galvanizing the entire surface of the metal to be protected need not be covered with a protecting metal instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded.

Passivation: The metal is treated with strong oxidizing agents such as concentrated HNO3. As a result, a protective oxide layer is formed on the surface of the metal.
Alloy formation: The oxidizing tendency of iron can be reduced by forming its alloy with other more anodic metals.
eg: stainless steel – an alloy of Fe and Cr.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

Choose the correct answer:

1. In the electrolytic cell, the flow of electrons is from:
(a) cathode to anode in the solution
(b) cathode to the anode through external supply
(c) cathode to the anode through the internal supply
(d) anode to the cathode through the internal supply
Answer:
(b)
Hint: In electrolytic cells, electrons enter the cathode and leave at the anode. Thus in the external supply, electrons flow from cathode to anode. In the solution, only ions flow and not the electrons.

2. Which of the statements about solutions of electrolytes is not correct?
(a) the conductivity of the solution depends upon the size of the ions.
(b) conductivity depends upon the viscosity of the solution.
(c) conductivity does not depend upon the solvation of ions present in the solution.
(d) the conductivity of the solution increases with temperature.
Answer:
(c)
Hint: Greater the solvation of ions, lesser is the conductivity. Hence, (c) is an incorrect statement.

3. A dilute solution of Na2SO4 is electrolyzed using platinum electrodes. The products at the cathode and anode are:
(a) O2 and H2
(b) SO2, Na
(c) O2, Na
(d) S2O82-, H2
Answer:
(a)
Hint: H20 is more readily reduced than Na+. It is more readily oxidised than S04 2. Hence, the electrode reactions are:
at cathode: 2H2O + 2e → H2 + 2OH
at anode: 2H2O → O2 + 4H+ + 4e.

4. The quantity of charge required to obtain one mole of aluminum from Al2O3 is:
(a) 1F
(b) 6F
(c) 3F
(d) 2F
Answer:
(c)
Hint: Al2O3 → 2Al.
i.e., 2Al2O3 + 6e → 2Al or
Al2O3 + 3e → Al.
Hence, to obtain one mol of Al, the charge required is 3F.

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

5. Electrolysis of dilute aqueous solution of NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H2 gas at the cathode is (1 Faraday = 96500 C mol-1):
(a) 9.65 × 104 sec
(b) 19.3 × 104 sec
(c) 28.95 × 104 sec
(d) 38.6 × 104 sec
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 42
Thus,
0.5 mol of H2 is liberated by IF = 96500 C 0.01 mol of H2 will be liberated by charge
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 43

6. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At mass of Al = 27 g mol-1):
(a) 8.1 × 104 g
(b) 2.4 × 105 g
(c) 1.3 × 104 g
(d) 9.0 × 103 g
Answer:
(a)
Hint: Al3+ + 3e → Al
1 mol of Al = 27 g of Al is deposited by 3F or 3 × 96500 C of electricity.
Quantity of electricity passed
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 44

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

7. What current is to be passed for 0.02 A for deposition of a certain weight of metal which is equal to electrochemical equivalent?
(a) 4 A
(b) 100 A
(c) 200 A
(d) 2 A
Answer:
(a)
Hint: Electrochemical equivalent is the weight deposited by 1 coulomb.
Q = I × t
1 = I × 0.25 or I = 4A

8. Acertain current liberates 0.504 g of hydrogen in 2 hours. How many grams of oxygen can be liberated by the same current in the same time?
(a) 2.0 g
(b) 0.4 g
(c) 4.0 g
(d) 8.0 g
Answer:
(c)
Hint: H2O → H2 + \(\frac{1}{2}\) O2
Thus if one mole of H2 is liberated, O2 liberated = \(\frac{1}{2}\) mole
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 45

9. Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.3 Sm-1. If the resistance of 0.4M solution of the same electrolyte is 260Ω the molar conductivity is:
(a) 6250 Sm2 mol-1
(b) 6.25 × 10-4 Sm2 mol-1
(c) 625 × 10-4 Sm2 mol-1
(d) 62.5 Sm2 mol-1
Answer:
(b)
Hint:
specific conductance = observed conductance × cell constant
K = \(\frac{1}{R}\) × cell constant
or cell constant = K × R
= 1.3 Sm-1 × 50Ω
= 65 m-1
For 0.4 M solution, specific conductance
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 46

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

10. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:
(a) increase in the number of ions
(b) increase in ionic mobility
(c) 100% dissociation of electrolyte at normal dilution
(d) increase in both i.e., number of ions and ionic mobility of ion.
Answer:
(b)
Hint: A strong electrolyte is completely ionized at all concentrations. Hence, the number of ions remains the same. However, on dilution, interionic forces decrease, and hence ionic mobility increases. Therefore, equivalent conductance increases.

11. In the electrolysis of which solution, OH ions are discharged in preference to Cl ions?
(a) dilute NaCl
(b) very dilute NaCl
(c) fused NaCl
(d) solid NaCl
Answer:
(b)
Hint: In very dilute NaCl solution, the following reactions take place on electrolysis.
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 47

12. The charge in coulombs on lg ion of N-3 is:
(a) 28.9 × 105 C
(b) 2.89 × 105 C
(c) 2890 C
(d) 28900 C
Answer:
(b)
Hint: Charge on 1 gm ion of N-3
= 3 × 1.6 × 10-19 C
One g ion = 6.023 × 1023 ion
Charge on 1 gm ion of N-3
= 3 × 1.6 × 10-19 × 6.023 × 1023
= 2.89 × 105 C

13. A solution of copper sulfate is electrolyzed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode is: (atomic mass of Cu = 63.5 g mol-1)
(a) 29.6 g
(b) 0.296 g
(c) 2.96 g
(d) 1.564 g
Answer:
(b)
Hint: Current = 1.5 amperes
Time = 10 minutes
= 10 × 60 = 600 s
Quantity of electricity passed = i × t
= 1.5 × 600 = 900 C
The reaction occurring at cathode is
Cu2+ + 2e → Cu
i.e., 2F or 2 × 96500 C deposit
= 1 mol = 63.5 g of Cu
900 C will deposit = \(\frac{63.5}{2 \times 96500}\) × 900 = 0.296 g

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

14. The limiting molar conductance of HCl, CH3COONa, and NaCl are respectively 425, 90, and 125 mho cm2 mol-1 at 25° C. The molar conductivity of 0.1 M CH3COOH solution is 7.8 mho cm2 mol-1 at the same temperature. The degree of dissociation of 0.1 M acetic acid at this temperature is:
(a) 0.1
(b) 0.02
(c) 0.15
(d) 0.03
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 48

15. Degree of dissociation of pure water is 1.9 × 10-9. Molar conductances of H+ and OH ions at infinite dilution are 200 S cm2 mol-1 and 350 Scm2 mol-1 respectively.
(a) 3.8 × 10-7 Scm2 mol-1
(b) 5.7 × 10-7 Scm2 mol-1
(c) 9.5 × 10-7 Scm2 mol-1
(d) 1.045 × 10-6 Scm2 mol-1
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 49

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

16. When measured against a standard calomel electrode, an electrode is found to have a standard reduction potential of 0.100 V. If the standard reduction potential of the calomel electrode is 0.244 V, the standard reduction potential of the same electrode against the standard hydrogen electrode will be:
(a) – 0.144 V
(b) + 0.100 V
(c) – 0.344 V
(d) – 0.100 V
Answer:
(b)
Hint: Standard electrode potential is a fixed quantity for a given electrode. Hence, its standard electrode potential against SHE will remain the same. Viz., (0.100 V). (emf of the cell formed will change)

17. On the basis of the following E° values, the strongest oxidising agent is:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 50
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 51
Answer:
(c)
Hint: The strongest oxidizing agent is a substance that is reduced move easily, i.e., it has the highest reduction potential. Reduction potentials of [Fe(CN)6]-3 and Fe3+ are 0.355 V and 0.77 V. Hence, Fe3+ is reduced more easily and hence is the strongest oxidizing agent.

18. Standard electrode potentials for Sn4+ / Sn2+ couple is +0.15 V and that for Cr3+/Cr couple is – 0.74 V. These two couples in their standard states are connected to make a cell. The cell potential will be:
(a) +1.83 V
(b) +1.19 V
(c) +0.89 V
(d) +0.18 V
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 52

19. Cr2O72- + I → I2 + Cr3+;
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 53
(a) 0.54 V
(b) – 0.54 V
(c) +0.18 V
(d) -0.18 V
Answer:
(a)
Hint: In the given reaction, I is oxidized to I2 and Cr2O72+ ions have been reduced to Cr3+. The cell is
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 54

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

20. Given the standard electrode potentials K+/K = -2.93 V; Ag+ / Ag = 0.80 V; Mg22+ / Hg = 0.79; Mg2+/Mg = -2.37 V; Cr2+/Cr = -0.74 V.
Arrange these metals in their increasing order of reducing power.
(a) Ag < Hg < Cr < Mg < K
(b) K < Mg < Cr < Hg < Ag
(c) Hg < Cr < Mg < K < Ag
(d) Mg < Cr < Hg < Ag < K
Answer:
(a)
Hint: Lower the reduction potential, more easily it is oxidized and hence greater is its reducing power.

21. The reduction potential of hydrogen half cell will be negative if:
(a) pH2 = 1 atm and [H+] = 1.0 M
(b) pH2 = 2 atm and [H+] = 1.0 M
(c) pH2 = 2 atm and [H+] = 2.0 M
(d) pH2 = 1 atm and [H+] = 2.0 M
Answer:
(b)
Hint: For hydrogen electrode,
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 55

22. 2Hg → Hg2+; E° = 0.855 V
Hg → Hg2+; E° = 0.799 V
Equilibrium constant for the reaction
Hg + Hg2+ ⇌ Hg22+ at 27° C is:
(a) 89
(b) 82.3
(c) 79
(d) none of these
Answer:
(c)
Hint:
E° for the reaction = 0.855 – 0.799 = 0.056 V
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 56

TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry

23. Standard free energies of formation of (in kJ mol-1) at 298 K are -237.3, -394.2, and -8.2 for H2O(l), CO2(g), and pentane (g) respectively. The value of E°ell for the pentane-oxygen fuel cell is:
(a) 1.968 V
(b) 2.0968 V
(c) 1.0968 V
(d) 0.968 V
Answer:
(c)
Hint: The balanced equation for pentane oxygen cell reaction will be:
TN Board 12th Chemistry Important Questions Chapter 9 Electro Chemistry 57

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Students get through the TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Answer the following questions.

Question 1.
State the formula and the name of the conjugate base of each of the following acids.
(a) H3O+,
(b) HSO4,
(c) HF,
(d) NH4+,
(e) CH3NH3+,
(f) CH3PO4,
(g) CH3COOH,
(h) H2PO4,
(i) HS,
(j) HCO3.
Answer:
(a) H2O – water
(b) SO4-2 – sulphate ion
(c) F – fluoride ion
(d) NH3 – ammonia
(e) CH3NH2 – methyl amine
(f) H2PO4 – dihydrogen phosphate ion
(g) CH3COO – acetate ion
(h) HPO4-2 – hydrogen phosphate ion
(i) S-2 – sulphide ion
(j) CO3-2 – carbonate ion

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 2.
State the formula and name of the conjugate acid of each of the following:
(a) OH,
(b) CH3COOH,
(c) HPO4-2,
(d) CO3-2,
(e) NH3,
(f) CH3NH2,
(g) HS,
(h) CH3COO
(i) H2PO4,
(j) Cl.
Answer:
(a) H3O – water
(b) CH3COOH2+ – protonated acetic acid
(c) H2PO4 – dihydrogen phosphate ion
(d) HCO3 – bicarbonate ion
(e) NH4+ – ammonium ion
(f) CH3NH3+ – protonated methyl amine
(g) H2S – hydrogen sulphide
(h) CH3COO – acetic acid
(i) H3PO4 – phosphoric acid
(j) HCl – hydrochloric acid

Question 3.
Which of the following behave both as Bronsted acids as well as Bronsted bases?
NH3, HSO4, H2SO4, HCO3; H3PO4, HS, H2O.
Answer:
NH3, HSO4 , HCO3 ; HS and H2O behave both as Bronsted acids and Bronsted bases.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 1

Question 4.
Classify Lewis acids and bases from the following:
H3O+, CH3NH2, BF3, OH, Cu2+, S2-, CH3COO.
Answer:
Lewis acids: H3O+, BF3, Cu2+.
Lewis bases: CH3NH2, OH, S2-, CH3COO.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 5.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base.
(a) OH, (b) F, (c) H+, (d) BCl3
Answer:
(a) OH is a Lewis base because it acts as an electron-pair donor.
(b) F is also a Lewis base.
(c) H+ is a Lewis acid because it acts as an electron pair acceptor.
(d) BCl3 is an electron-deficient compound electron pair acceptor, hence Lewis acid.

Question 6.
Explain how the [H+] concentration or [OH] concentration, decides a solution as acidic, neutral, or alkaline.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 2

Question 7.
Calculate the hydrogen and hydroxyl ion concentrations in (i) 0.01M HNO3, (ii) 0.001M KOH solution at 298 K.
Answer:
(i) HNO3 is a strong acid. It ionizes as
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 3
(ii) KOH is a strong base. It ionizes as
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 4

Question 8.
Assuming complete dissociation, calculate the pH of the following solutions:
(i) 0.003M HCl,
(ii) 0.005M NaOH,
(iii) 0.002M HBr,
(iv) 0.002M KOH.
Answer:
(i) HCl → H+ + Cl
[H+] = [HCl] = 0.003M
pH = – log [H+]
= – log [0.003] = 2.52

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

(ii) NaOH → Na+ + OH
[OH] = [NaOH] = 0.005M
pOH = – log[OH]
= – log [0.005] = 2.301
pH = 14 – pOH
= 14 – 2.301 = 11.699

(iii) HBr → H+ + Br
[H+] = [HBr]
= 0.002M
pH = – log [H+] (or)
pH = – log [0.002] = 2.6989

(iv) KOH → K+ + OH
[OH] = [KOH]
= 0.002M
pOH = – log [OH ]
pH = – log (0.002) = 2.6929
pH = 14 – pOH (or)
= 14 – 2.6929 = 11.3011

Question 9.
Calculate the pH of the resulting mixtures: 10 ml of 0.02M Ca(OH)2 + 25 ml of 0.1M HCl.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 5
Number of moles of OH ion = 2 x 0.002 = 0.004
Number of moles of H+ ion = 0.0025
Number of moles of OH remaining after neutralisation = 0.004 – 0.0023 = 0.0015
Molarity of OH ions
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 6

Question 10.
The concentration of H+ ions in 0.1 M solution of a weak acid is 1.0 x 10-5 mol L-1. Calculate the dissociation constant of the acid.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 7
[HA] can be taken as 0.1 as 1 x 10-5 is very small. Hence,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 8

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 11.
Find whether the resulting solution is acidic, neutral, or basic when
(i) a strong acid is mixed with a strong base
(ii) a strong acid is mixed with a weak base
(iii) a weak acid is mixed with a strong base
(iv) a weak acid is mixed with a weak base.
Answer:
(i) A strong acid and a strong base give a neutral solution because both are completely ionized and the reaction goes on completion.
eg:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 9
(ii) A strong acid and a weak base give an acidic solution, as the weak acid is hot completely ionized. The reaction does not go for completion and there is an excess of hydrogen ions in the solution.
eg:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 10
The solution is therefore is acidic.
(iii) A weak acid and a strong base give a basic solution, as the weak acid is not completely ionized. The reaction does not go for completion and there is an excess of hydroxyl ions in the solution.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 11
Hence the solution is basic.
(iv) A weak acid and a weak base give an acidic, neutral, or basic depending on the relative strength of acid or base. If both weak acid and weak base have equal strength, the resulting solution is neutral.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 12

Question 12.
State Ostwald’s dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolytes also increases.
The mathematical form is
α = \(\sqrt{\frac{\mathrm{K}_{a}}{\mathrm{C}}}\)
where ‘α’ is the degree of dissociation, Ka is the dissociation constant of the acid and ‘C’ is its concentration.

Question 13.
Explain the term ‘common ion effect’ with an example.
Answer:
The suppression of the dissociation of a weak electrolyte by the addition of an ion in common with that of the weak electrolyte is known as the common ion effect.
Consider the dissociation of acetic acid in the presence of sodium acetate or HCl.
In the absence of any added salt, the dissociation of acetic acid is represented as
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 13
The addition of acetate ions in the form of sodium acetate increases the concentration of acetate ions. This increases the value of Ka. But Ka is a constant at a given temperature. Thus to maintain the Ka value constant, H+ ions combine with acetate ion to form undissociated acetic acid. i.e., The degree of dissociation of acetic acid in the presence of acetate ion is less than in the absence of sodium acetate. The same happens in the presence of HCl.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 14.
What are buffer solutions? Give example.
Answer:

  1. Buffer solutions resist changes in pH by the addition of small quantities of an acid or a base. This ability is known as buffer action.
  2. Buffer solutions consist of a mixture of a weak acid and a salt of a weak acid and a strong base.
    eg: CH3COOH + CH3COONa. This type of buffer is known as an acid buffer.
  3. Buffer solution which consists of a mixture of a weak base and a salt of a weak base and strong acid is known as a basic buffer.
    eg: NH4OH and NH4Cl.

Question 15.
Give the characteristics of a buffer solution.
Answer:

  1. It should have a definite pH. i.e., it has reserve acidity or alkalinity.
  2. Its pH does not change on standing.
  3. Its pH does not change on dilution.
  4. Its pH is only slightly changed by the addition of a small quantity of acid or base.

Question 16.
Give examples of acidic and basic buffers.
Answer:
Acidic buffer:

  1. CH3COOH + CH3COONa
  2. Boric acid and borax.
  3. Pthalic acid and potassium acid pthalate.

Basic buffer:

  1. NH4OH + NH4Cl
  2. Glycine and glycine hydrochloride.
  3. Aniline and aniline hydrochloride.

Question 17.
Explain buffer action with a suitable example.
Answer:
The buffer action in a solution containing CH3COOH and CH3COONa is explained as follows:
The dissociation of the buffer components occurs as below.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 14
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 15
If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO to form the undissociated weak acid i.e., the increase in the concentration of H+ does not reduce the pH significantly.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 16
If a base is added, it will be neutralized by H3O+, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 17

Question 18.
Explain the terms ‘buffer capacity” and ‘buffer index”.
Answer:
The buffer capacity is the ability of the buffer to resist changes in pH by the addition of small quantities of acid or base. It is measured in terms of buffer capacity. The quantitative measure of buffer capacity is the buffer index. It is defined as the number of equivalents of acid or base to one liter of the buffer solution to change its pH by unity.
β = \(\frac{d \mathrm{~B}}{d(\mathrm{pH})}\)
where P = buffer index, dB = number of gram equivalent of acid/base added to one liter of buffer solution, d(pH) = The change in pH after the addition of an acid or base.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 19.
Derive Henderson equation for the determination of pH of an acid buffer.
Answer:
An acid buffer consists of a mixture of a weak acid and salt with strong acid. The ionization of weak acid, HA is given by
HA ⇌ H+ + A
The dissociation constant of the weak acid is given by,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 18
It can be assumed that the concentration of [A] ions from the complete ionization of the salt compared to the ionization of the weak acid. Hence, [A] concentration may be considered as the concentration of the salt.
So,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 19
Taking logarithms on both sides and reversing the sign,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 20
This is known as Henderson’s equation.

Question 20.
Write Henderson’s equation for a basic buffer.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 21

Question 21.
Suppose it is required to make a buffer solution of pH = 4. Using acetic acid and sodium acetate. How much sodium acetate to be added to 1 liter of N/10 acetic acid? The dissociation constant of acetic acid is 1.8 × 10-5.
Answer:
Applying Henderson’s equation,
The molecular mass of CH3COONa = 82
Amount of salt = 0.018 × 82 = 1.476 g.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 22

Question 22.
Calculate the pH of a buffer solution containing 0.15 moles of NH4OH and 0.25 moles of NH4Cl. (Kb for NH4OH = 1.8 × 10-5)
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 23
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 24

Question 23.
Calculate the pH of 0.1M ammonia solution. Calculate the pH after 50 ml of this solution is treated with 25 ml of 0.1M HCl. The dissociation constant of ammonia, Kb = 1.77 × 10-5.
Answer:
For NH4OH
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 25
When NH3 is added, neutralization occurs
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 26
(MNH4Cl) number of mole of NH4Cl = 0.0025
(MNH4OH) Remaining mole of NH4OH = 0.005 – 0.0025 = 0.0025
According to Henderson’s equation,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 27

Question 24.
Define hydrolysis.
Answer:
The reaction between cation or anion of the salt with water to produce the acid and base from which the salt formed is known as hydrolysis.
salt + H2O = acid + base

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 25.
Define the term degree of hydrolysis.
Answer:
It is the fraction of one mole of the salt that undergoes hydrolysis is known as the degree of hydrolysis.

Question 26.
Define the term hydrolysis constant (Kh), with an example.
Answer:
The equilibrium constant for the reaction, salt + H2O ⇌ acid + base, is known as the hydrolysis constant (Kh). Consider the hydrolysis of acetate ion.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 28

Question 27.
The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of the ammonium acetate solution.
Answer:
Ammonium acetate is the salt of a weak acid and weak base.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 29

Question 28.
The ionization constant of nitrous acid is 4.5 × 10-4. Calculate the pH of 0.04 M sodium nitrite solution and its degree of hydrolysis.
Answer:
Sodium nitrite is a salt of a weak acid and a strong base.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 30
Substituting these values in equation (1)
pH = \(\frac{1}{2}\) [14 + 3.346 – 1.3979]
= 7.974
Degree of hydrolysis (h)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 31

Question 29.
Calculate the degree of hydrolysis and pH of a 0.1M sodium acetate solution. The hydrolysis constant of sodium acetate is 5.6 × 10-10.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 32

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 30.
Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation. pKa for formic acid is 3.8 and pKb of ammonia = 4.8.
Answer:
Ammonium formate is a salt of a weak acid and weak base.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 33

Question 31.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Pyridinium hydrochloride is a salt of a weak base and strong acid.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 34

Question 32.
Define solubility product.
Answer:
It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature.

Question 33.
Give expressions for the solubility product of the following: (i) BaSO4, (ii) H2S, (iii) Sb2S3, (iv) AlI3.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 35

Question 34.
Mention the criteria of precipitation of an electrolyte.
Answer:
The ionic product should exceed the value of the solubility product, for precipitation to occur.

Question 35.
How will you decide whether a solution is saturated or unsaturated, from the ionic product values?
Answer:

  1. When the ionic product is less than the solubility product, the solution is unsaturated.
  2. When an ionic product is equal to the solubility product, the solution is saturated.

Question 36.
Obtain a relationship between solubility and solubility product of the following: (i) AgCl, (ii) Ag2CO3, (iii) CaF2, (iv) Cr(OH)3.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 36
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 37

Question 37.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to the precipitation of copper iodate? Ks for cupric iodate = 7.4 x 10-8.
Answer:
Given, [NaIO3] = 0.002 M
[Cu(ClO3)2] = 0.002 M
∴ [IO3] = 0.002 M
[Cu+2] = 0.002 M
When equal volumes of these solutions are mixed, their concentrations will be halved.
i.e., [IO3] = 0.001 M
[Cu+2] = 0.001 M
Cu(IO3)2 ⇌ Cu+2 + 2IO3
Ks = [Cu+2][IO3]2
= 0.001 x (0.001)2 = 10-9
Since IP is less than solubility products precipitation will not occur.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

Question 38.
What is the minimum volume of water required to dissolve 1 gm of calcium sulfate at 298 K. The Ks for calcium sulfate is 9.1 x 10-6.
Answer:
Let the solubility of CaSO4 at 298 K be ‘s’ mol L-1.
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 39
Minimum volume of water required to dissolve
1 g of CaSO4 = \(\frac{1000}{0.41}\)
= 2439 mL = 2.439 L

Question 39.
The solubility product of BaS04 is 1.5 x 10-9 Find its solubility (i) in pure water and (ii) in 0.1M BaCl2, solution.
Answer:
(i)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 40
Let ‘s’ be the solubility in mol L-1. Then
Ks = s2
(or) 1.5 x 107 = s2
s = \(\sqrt{1.5 \times 10^{-9}}\)
= 3.87 x 10-5 mol L7

(ii) Let (s’) be the solubility of BaSO4 in 0.1M BaCl2 solution.
Total [Ba2+] = [Ba2+] from BaSO4 from [Ba2+] from BaCl2.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 41

Question 40.
Calculate the pH at which Mg(OH)2 begins to precipitate from a solution containing 0.10 Mg2+ ions.
Answer:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 42

Choose the correct answer:

1. What is the conjugate base of OH?
(a) O2
(b) H2O
(c) O
(d) O2-
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 43

2. C2H5ONa is …………. of C2H5OH.
(a) strong acid
(b) weak acid
(c) strong base
(d) weak base
Answer:
(c)
Hint: C2H5OH is a strong base. a weak acid, C2H5O is a strong base

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

3. Among the following, the one which can act as Bronsted acid as well as Bronsted base is:
(a) H3PO4
(b) AlCl3
(c) CH3COO
(d) H2O
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 44

4. At 25° C, the dissociation constant of a base, BOH is 1.0 x 10-12. The concentration of hydroxyl ions in 0.01M aqueous solution of the base would be:
(a) 1.0 x 10-6 mol lit-1
(b) 1.0 x 10-7 mol lit-1
(c) 2.0 x 10-6 mol lit-1
(d) 1.0 x 10-5 mol lit-1
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 45

5. Which of the following is the correct statement?
(a) HCO3 is the conjugate base of CO32-
(b) NH2 is the conjugate acid of NH3
(c) H2SO4 is the conjugate acid of HSO4
(d) NH3 is the conjugate base of NH2
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 46

6. Three reactions involving H2PO4 are given below:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 47
In which of the above H2PO4 act as an acid?
(a) (iii) only
(b) (i) only
(c) (ii) only
(d) (i) and (ii)
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

7. Which one of the following will decrease the pH of 50 ml of 0.01M hydrochloric acid?
(a) Addition of 50 ml of 0.01 M HCl
(b) Addition of 50 ml of 0.002 M HCl
(c) Addition of 150 ml of 0.002 M HCl
(d) Addition of 5 ml of 1 M HCl
Answer:
(d)
Hint: For 0.001 M HCl, [H+] = 10-2M; pH = 2
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 48
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 49

8. If pKa for fluoride ion at 25° C, is 10.83, the ionisation constant of hydrofluoric acid at this temperature is:
(a) 1.74 x 10-5
(b) 3.52 x 10-3
(c) 6.75 x 10-4
(d) 5.38 x 10-2
Answer:
(c)
Hint:
pKa + pKb = 14
pKa = 14 – 10.83 = 3.17
pKa= – log Ka = – log 3.17
log Ka= -3.17 = 4.83
(or) Ka= 6.76 x 10-4

9. The pH of a solution obtained by mixing 100 ml of a solution of pH = 3 with 400 ml of a solution of pH = 4 is:
(a) 3 – log 2.8
(b) 1 – log 2.8
(c) 4 – log 2.8
(d) 5 – log 2.8
Answer:
(c)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 50

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

10. The pH of 0.1M aqueous solution of a weak acid, HA is 3. What is its degree of dissociation?
(a) 1%
(b) 10%
(c) 50%
(d) 25%
Answer:
(a)
Hint: HA ⇌ H+ + A
[H+] = Cα
0.1 x α = 10-3
α = 10-2 i.e., 1%

11. The pKa of acetic acid is 4.74. The concentration of acetic acid is 0.01 M. The pH of acetic acid is:
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 51

12. How many times 1M CH3COOH solution should be diluted so that pH of the solution is doubled?
(a) 20 times
(b) 200 times
(c) 5.55 x 102 times
(d) 5.55 x 104 times
Answer:
(d)
Hint: pH of a weak acid
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 52

13. 40 ml of 0.1M ammonia is mixed with 20 ml 0.1M HCl. What is the pH of the mixture? (pKb for ammonia is 4.74).
(a) 4.74
(b) 2.26
(c) 9.26
(d) 5.00
Answer:
(c)
Hint: 40 ml of 0.1M NH3 solution
= 40 x 0.1 milli = 4 millimol
20 ml of 0.1 M HCl = 20 x 0.1 = 2 millimol
2 millimol of HCl neutralise 2 millimol of NH4OH, to form 2 millimol of NH4OH.
NH4OH left = 2 millimol
Total volume = 60 ml
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 53

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

14. The pH of a solution formed on mixing 20 ml of 0.05M H2SO4 with 5 ml of 0.45 M NaOH at 298 K is:
(a) 6
(b) 2
(c) 12
(d) 1
Answer:
(c)
Hint:
2NaOH + H2SO4 → Na2SO4 + 2H2O
20 ml of 0.05M H2SO4
= 20 ml of 0.05 millimol
= 1.0 millimol

5 ml of 0.45M NaOH
= 5 x 0.45 millimol
= 2.25 millimol

2 millimol of NaOH will react with 1 millimol of H2SO4.
∴ NaOH left in solution = 0.5 millimol Volume of the solution = 25 ml
∴ [OH] = \(\frac{0.5}{25}\) = 0.01 M = 10-2 M
[H+] = 10-12;
pH = 12

15. Which of the following salts will have the highest pH in water?
(a) KCl
(b) NaCl
(c) Na2CO3
(d) CuSO4
Answer:
(c)
Hint: KCl and NaCl, being salts of strong acid and strong base do not get hydrolysed. Their aqueous solutions are neutral and hence their pH = 7.
Na2CO3 being a salt of weak acid and a strong base, gives NaOH as a product of hydrolysis. Hence, its aqueous solution is alkaline and has pH >7.
CuSO4 being a salt of strong acid and weak base, gives H2SO4, as a product of hydrolysis H2SO4 is a strong acid and its pH < 7.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

16. The H3O+ ion concentration of a solution of pH 6.58 is:
(a) antilog (- 6.58)
(c) antilog (- 5.58)
Answer:
(a)
Hint: pH = – log [H3O+];
log [H3O+] = -pH = – 6.58
[H3O+] = antilog (-6.58)
= 2.63 x 10-7 g ion/lit

17. 30 CC of \(\frac{M}{4}\) HCl, 20 CC of \(\frac{M}{2}\) HNO3, and 40 CC of \(\frac{M}{4}\) NaOH solutions are mixed and the volume is made upto 1 dm3. The pH of the resulting solution is:
(a) 2
(b) 1
(c) 3
(d) 8
Answer:
(a)
Hint:
Total millimoles of H
= (30 x \(\frac{1}{3}\)) + (20 x \(\frac{1}{2}\))
= 10 + 10 = 20
Total millimol of OH
= 40 x \(\frac{1}{4}\) = 10
∴ H+ ions left after the neutralisation = 10 millimoles Volume of the solution
= 1 dm3
∴ Molarity of H+ ions
= \(\frac{10}{1000}\) = 10-2M
pH = – log [H+]
= – log [10-2] = 2

18. The pH of 0.1 M solutions of the following salts increases in the order:
(a) NaCl < NH4Cl < NaCN < HCl
(b) HCl < NH4Cl < NaCl < NaCN
(c) NaCN < NH4Cl < NaCl < HCl
(d) HCl < NaCl < NaCN < NH4Cl
Answer:
(b)
Hint: NaCl = neutral (pH = 7); NH4Cl = slightly acidic (pH < 7); NaCN = basic (pH > 7); HCl = strongly acidic (pH << 7).

19. A weak acid HX has the dissociation constant 1 x 10-5. It forms NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is:
(a) 0.0001%
(b) 0.01 %
(c) 0.1%
(d) 0.15 %
Answer:
(b)
Hint: Hydrolysis reaction is
X + H2O ⇌ HX + OH.
For a salt of weak acid with strong base,
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 54

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

20. Among the following hydroxides, the one which has the lowest value of Ks (solubility product) at ordinary temperature is:
(a) Mg(OH)2
(b) Ca(OH)2
(c) Ba(OH)2
(d) Be(OH)2
Answer:
(d)
Hint: The solubility increases down the group due to an increase in the size of the ion and a decrease in lattice energy. Lower the solubility, lower is the Ks.

21. On adding 0.1M solution each of Ag+, Ba2+, Ca2+ ions in a Na2SO4 solution, the species first precipitated is (Ks(CaSO4) = 10-6, Ks(BaSO4) = 10-11, Ks(Ag2SO4) = 10-5:
(a) Ag2SO4
(b) BaSO4
(c) CaSO4
(d) All of these
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 55
The minimum [SO42-] concentration required for precipitation in for BaSO4.

22. The Ks of Ag2CrO4, AgCl, AgBr, and AgI are respectively 1.1 x 10-12, 1.8 x 10-6, 5.0 x 10-13 and 8.3 x 10-17. Which of the following salts will precipitate last of AgNO3 solution containing equal moles of NaCl, NaBr, Nal and Na2CrO4.
(a) AgBr
(b) Ag2CrO4
(c) AgI
(d) AgCl
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 56
As the solubility of Ag2CrO4 is highest, it will be precipitated last at all.

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

23. The Ks of Ag2CrO4 is 1.1 x 10 12 at 298 K.
The solubility in mol per litre of Ag2CrO4 in 0.1M. AgNO3 solution is:
(a) 1.1 x 10-11
(b) 1.1 x 10-10
(c) 1.1 x 10-12
(d) 1.1 x 10-9
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 57

24. Using Gibb’s free energy change, AG° = +63.3 kJ for the following reaction:
Ag2CO3 (s) ⇌ 2Ag+(aq) + CO32-(aq)
the Ks for Ag2CO3 in water at 25° C is (R = 8.314 JK-1 mol-1)
(a) 3.2 x 10-26
(b) 8.0 x 10-12
(c) 2.9 x 10-3
(d) 7.9 x 10-2
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 58

25. A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20M. If the equilibrium constant, Kb for NH3 equals 1.8 x 10-5, what is the pH of the solution?
(a) 8.73
(b) 9.08
(c) 9.43
(d) 11.72
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 59
= 4.74 – 0.176 = 4.56
pH = 14 – pOH = 14 – 4.56 = 9.44

26. What is the [H+] in mol/L of a solution that is 0.20M in CH3COONa and 0.10M in CH3COOH? Kb for CH3COOH = 1.8 x 10-5:
(a) 9.0 x 106
(b) 3.5 x 10-6
(b) 1.1 x 10-5
(d) 1.8 x 10-5
Answer:
(a)
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 60

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

27. In what volume ratio of NH4Cl and NH4OH solution (each 1M) should be mixed to get a buffer solution of pH, 9.80 (pKb for NH4OH is 4.74):
(a) 1 : 2.5
(b) 2.5 : 1
(c) 1 : 3.5
(d) 3.5 : 1
Answer:
(c)
Hint:
Suppose ‘V1’ ml of 1M NH4Cl is mixed with ‘V2’ ml of 1M NH4OH solution:
V1 ml of 1M NH4Cl = V1 millimole
V2 ml of 1M NH4OH = V2 millimole
pH = 9.80, Hence pOH = 14 – 9.80 = 4.20
TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium 61

TN Board 12th Chemistry Important Questions Chapter 8 Ionic Equilibrium

28. Which one of the following pairs of solutions is not an acidic buffer?
(a) H2CO3 + Na2CO3
(b) H3PO4 + Na3PO4
(c) HClO4 + NaClO4
(d) CH3COOH + CH3COONa
Answer:
(c)
Hint: HClO4 is a strong acid. Hence, it cannot be used to make an acidic buffer.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Students get through the TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Answer the following questions.
Question 1.
Express the rate of the following reaction in terms of disappearance of the reactant and appearance or formation of the product: A → B
Answer:
Rate of disappearance of the reactant = \(-\frac{\Delta[\mathrm{A}]}{\Delta t}\)
where the negative sign indicates the concentration of the reactant decreases with time.
Rate of formation of the product = \(+\frac{d[\mathrm{B}]}{d t}\)
The positive sign indicates that the concentration of the product increases with time.

Question 2.
Mention the unit of rate of reaction.
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 1
For a gaseous reaction, a unit of rate = atm sec-1 since concentration is expressed in partial pressures.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 3.
Explain how will you determine experimentally determine
(i) average rate of reaction
(ii) instantaneous rate of reaction and
(iii) initial rate of reaction.
Answer:
The changes in concentration of the reactants or products are experimentally determined by withdrawing a small portion of the reaction mixture (usually 2 or 5 ml) at suitable intervals of time and then freeze to about 0°C to stop the reaction. The concentration of the reactant or product is measured by a suitable method. A plot of concentration-time is made from the graph, the average rate, instantaneous and initial rates can be determined.
(i) Average rate: Two equivalent points are taken with respect to the time at which the average rate is to be determined. The concentrations corresponding to these points are noted from the graph. The difference in concentration is divided by the time interval between equidistant points. Two equidistant points with respect to T is selected. Let these be ‘t1,’ and ‘t2’ respectively. The corresponding concentrations are x1, and x2 respectively.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 1
(ii) Instantaneous rate of reaction: For the determination of the instantaneous rate of reaction at any time ‘t’, a tangent is the direction to the curve at the point ‘P’ corresponding to that time. The slope of this tangent gives the instantaneous rate of reaction.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 3
Where Q is the angle of a tangent with the time axis. In a similar manner, the rate of formation of the product can be determined by measuring the slope of the tangent at a particular instant in concentration to the time curve.
(iii) Initial rate of reaction:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 4
The rate of reaction is maximum at the time of mixing the concentration. This cannot be determined experimentally. This is determined from the concentration is time curve. The curve is extrapolated to zero time and the slope of this curve at zero time gives the initial rate of the reaction.
\(\frac{\Delta x}{\Delta t}\) = initial rate.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 4.
In the reaction 2A → products the concentration of A decreases from 0.5 mol L1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this period.
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 5

Question 5.
The concentration of reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate using units of time both in minutes and in seconds.
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 6

Question 6.
The decomposition of N2O5 is expressed by the equation.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 7
If during a certain interval of time the rate of decomposition of N2O5 is 1.8 x 10-3 mol liter 1 min-1, what will be the rates of formation of NO2 and O2 during the same interval.
Answer:
The rate expression for the decomposition of N2O5 is
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 8
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 9

Question 7.
For each of the following reactions, express the given rate of change of concentration of the product or reactant in terms of rate of change of concentration of other reactants or products in that reaction.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 10
Answer:
(a)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 11
(b) The equality in this case is,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 12
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 13
(c) The equality in this case is,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 14

Question 8.
From the concentration of R at different times given below, calculate the average rate of the reaction.
Answer:
R → P during different intervals of time
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 15
Average rate of reaction between 0 → 5 sec
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 16
Average-rate of reaction between 5-10 sec
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 17
Similarly, average rate between 10 → 20 and 20 → 30 see can be calculated.
Note: What we understand from the above problem is that the rate of reaction is not a constant quantity. It decreases with time.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 9.
The decomposition of N2O5 is CCl4 solution at 318K has been studied by monitoring the concentration of the N2O5 in the solution. Initially, the concentration of N2O5 in the solution is 2.33 M and after 184 minutes it is reduced to 2.08 M. The reaction takes place according to the equation
2N2O5 → 4NO2 + O2.
Calculate the average rate of the reaction in terms of hours, minutes and seconds. What is the rate of production during this period?
Answer:
Average rate of reaction (mol L-1 min-1)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 18
For the given reaction
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 19

Question 10.
In hydrogenation reaction at 25°C it is observed that hydrogen gas pressure falls from 2 atm to 1.2 atm in 50 min. Calculate the rate of reaction in molarity per sec.
Answer:
R = 0.0821 lit mol-1 deg-1
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 20
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 21

Question 11.
Explain why the average rate cannot be used to produce the rate of reaction at any instant.
Answer:
Because the rate decreases with time as the reaction proceeds. It is not constant.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 12.
Bring out the difference between the rate and rate constant of a reaction?
Answer:

Rate of a reaction The rate constant of a reaction
It represents the speed at which the reactants are converted into products at any instant. It is a proportionality constant.
It is measured as a decrease in the concentration of the reactants or an increase in the concentration of products. It is equal to the rate of reaction when the concentration of each of the reactants in unity.
It depends on the initial concentration of reactants. It does not depend on the initial concentration of reactants.

Question 13.
The decomposition of dimethyl ether leads to the formation of CH4, H2, and Co and the reaction rate in given by
rate = R [CH3OCH3]3/2
The rate is followed by an increase in pressure in a closed vessel so that the rate is expressed in terms of partial pressure of dimethyl ether.
rate = R[PCH3OCH3]3/2
If the pressure is measured in the bar and the time in seconds then what are theTxmts of rate and rate constant?
Answer:
The rate law of the reaction
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 22

Question 14.
State the order with respect to each reactant, overall reaction, and the units of the rate constant in each of the following reactions.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 23
Answer:
(a) Order with respect to NO and Br2 are 2 and 1 respectively, over all order is 3.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 24
(b) Order \(\frac{3}{2}\)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 25
(c) Order with respect to H2 and NO are 1 and 2 respectively.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 26
= mol-2 litre2 sec-1
(d) Order with respect to CO and Cl2 are 2 and \(\frac{1}{2}\) respectively.
Over all order = 2\(\frac{1}{2}\) = \(\frac{5}{2}\)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 27
(e) Order with respect to H2O2 and I are one respectively. Over all order 1 + 1 = 2
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 28

Question 15.
Identify the reaction order from the following rate constants.
(a) k = 3.1 × 10-4 sec-1
(b) k = 1.12 × 10-2
(c) k = 1.35 × 10-2 mol-2 let2 s-1
(d) k = 3.4 × 10-3 mol let s-1
Answer:
(a) Let the reaction be of nth order
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 29
i.e., reaction is of first order.
(b) (atm)1-n × s-1 = atm-1 s-1
1 – n = -1 : or n = 2
i.e., reaction is of second order.
(c)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 30
i.e., reaction is of third order.
(d)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 31
i.e., reaction is of second order.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 16.
Derive an expression for the rate constant for the first-order reaction.
Answer:
A reaction whose fate depends on the reactant concentration raised to the first power is called a first order reaction. Let us consider the following Cl2 first-order reaction,
A → product
Rate law can be expressed as Rate = k [A]1
Where, k is the first order rate constant.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 32
Integrate the above equation between the limits of time t = 0 and time equal to t, while the concentration varies from the initial concentration [A0] to [A] at the later time.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 33
This equation is in a natural logarithm. To convert it into the usual logarithm with base 10, we have to multiply the term by 2.303.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 34

Question 17.
How will you determine the first-order rate constant graphically?
Answer:
For a first order reaction, the rate constant k is given by
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 35
where [A0] is the initial concentration of the reactant and [A] is the concentration of the reactant at time it\
Rearranging this equation in the form . y = mx + c, we get
log [A0] – log [A] = \(\frac{kt}{2.303}\)
2.303 [log [A] – log [A]] = kt
log [A0] – log [A] = \(\frac{k}{2.303}\) t.
Hence a plot of log [A] vs t gives a straight line with a negative slope having the volume of \(\frac{-k}{2.303}\)

Question 18.
Give example for first order reaction.
Answer:
(i) Decomposition of dinitrogen pentoxide
N2O5 → (g) 2NO2 (g) + \(\frac{1}{2}\) O2 (g)
(ii) Decomposition of thionylchloride;
SO2Cl2(l) → SO2 (g) + Cl2 (g)
(iii) Decomposition of the H2O2 in aqueous solution;
H2O2 (aq) → H2O (l) + \(\frac{1}{2}\) O2 (g)
(iv) Isomerisation of cyclopropane to propene.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 19.
Show that acid hydrolysis of an ester is a pseudo first order reaction.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 36
If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., the concentration of water remains almost constant.
Now, we can define k [H2O] = k’ ; Therefore the above rate equation becomes
Rate = k’ [CH3COOCH3]
Thus it follows first-order kinetics.

Question 20.
Give examples of zero-order reactions.
Answer:
(i) Photochemical reaction between H2 and I2
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 37
(ii) Decomposition of N2O on a hot platinum surface
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 38
(iii) Iodination of acetone in an acid medium is zero-order with respect to iodine.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 39

Question 21.
Give the general rate equation for nth-order reaction involving one reactant A.
Answer:
The general rate equation for an nth order reaction involving one reactant [A],
A → product
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 40
Consider the case in which n ≠ 1, integration of above equation between [A0] and [A] at time t = 0 and t = t respectively gives
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 41

Question 22.
Derive an expression to calculate t1/2 for a zero-order reaction.
Answer:
Let us calculate the half life period for a zero-order reaction.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 42
The half-life of a zero-order reaction is directly proportional to the initial concentration of the reactant.

Question 23.
Give the general expression for t1/2 of the nth order (n ≠ 1) reaction.
Answer:
Half-life for an nth order reaction involving reactant A and n ≠ 1
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 43

Question 24.
Give the characteristics of first order reaction.
Answer:

  1. A change in concentration unit will not change the numerical value of ‘k’
    k = \(\frac{2.303}{t}\) log \(\frac{a}{a-x}\)
    Where ‘a’ and (a – x) are the initial concentration of the reactant and concentration of the reactant at time Y respectively. Thus for the first-order reaction, any quantity which is proportional to the concentration can be used in place of concentration to calculate ‘k’.
  2. The time taken for the completion of the same fraction of change is independent of the initial concentration.
  3. Time required to calculate nth fraction of reaction t1/2 can be calculated as When x = \(\frac{a}{n}\) ; t = t1/n
    Substituting these values in,
    TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 44
    The above equation can be used to calculate t3/4 of a reaction.
    TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 45
  4. In a first-order reaction, the amount of reactant remaining after V half-lives can be calculated as follows:TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 46
    where [A0] – initial concentration of the reactant, [A] = concentration of the reactant remaining after ‘n’ half-lives.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 25.
Calculate the half life of a first order reaction from their rate constant given below:
(a) 200 sec-1 (b) 2 min-1 (c) 4 year-1
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 47
= 3.465 x 10~3 sec
= 0.1732 year.

Question 26.
The rate constant for a first-order reaction is 60 sec. How much time will it take to reduce the initial concentration to its 1/16th value?
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 48

Question 27.
The thermal decomposition of a compound is of the first order. If 50% of the sample is decomposed in 120 minutes, how long will it take for 90% of the compound to decompose?
Answer:
Half life of the reaction = 120 min.
We know that
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 49
If a= 100, x = 90, a-x = 10
= 5.77 x 10-3 min-1
Applying first order rate equation.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 50

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 28.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law. With t1/2 = 3.00 hrs. What fraction of sucrose remains after 8 hours?
Answer:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 51
Therefore fraction of reactant remaining = 0.1576

Question 29.
Explain the pressure change method in determining a first-order reaction.
Answer:
If pressure is given in gaseous reactions then we use the following kinetic equation.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 52
where p0 is the pressure at the initial stage p0-x is the pressure at time ‘t’ values of p0 and x can be calculated using the following examples.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 53
Case I: If total pressure of the reaction mixture is given in place of pressure of reactants.
pt = (p0 – x + x + x + x)
Where pt is the pressure of reaction vessel at time ‘t’.
Case II: If the pressure of the reaction vessel after a long time or infinite time is given
pt = p0 + p0 +p0

Question 30.
The decomposition of Cl2O7 at 400K in the gas phase to Cl2 and O2 is a first order reaction.
(i) After 55 seconds at 400K, the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate the rate constant.
(ii) Calculate the pressure of Cl2O7 after 100 seconds of decomposition at this temperature.
Answer:
(i) As pressure ∝ concentration.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 54
(ii) Applying the first-order rate equation.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 55

Question 31.
For the decomposition of azo isopropanol to hexane and nitrogen at 543K, the following data are obtained.

t(sec) p (mm of Hr)
0 35.0
360 54.0
720 63.0

Calculate the rate constant for the reaction.
Answer:
The reaction is
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 56
Total pressure – p0 – x + x + x
pt = p0 + x
or x = pt – p0
a ∝ p0
∴ (a – x) ∝ p0 – (pt – p0)
= 2p0 – pt
For the first order reaction
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 57

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 32.
Explain Oswald dilution method for determining the order of the reaction.
Answer:
It is applied for these reactions which involve more than one reactant. In this method, the concentration of all reactants are taken in large excess than one. The changes in the concentration of the reactant that is not taken in excess will influence the rate of reaction.
The concentrations of all other reactants practically remain same during the course of reaction. Consider the reaction.
m1A + m2B + m3C → product
Rate = k[A]α [B]β [C]γ
When B and C are taken in excess, the rate law will reduce to
rate = k’ [A]α
where k’ = k[B]β [C]γ
Two sets are taken in which the concentration of A is [A1] and [A2] while B and C are present in large excess,
r1 = [A1]α
r2 = [A2]α
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 58
From this, the value of α, i.e., the order with respect to A is obtained. The process is repeated one by one and order with respect to others is determined.
Total order of reaction = α + β + γ

Question 33.
Show that if the concentration of a reactant is doubled, the rate of the reaction is also doubled for a first-order reaction, increases four times for a second-order reaction, increases by eight times for a third-order reaction, i.e., A → product.
Answer:
(i) If the concentration of A is doubled.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 59
Hence, for the first-order reaction, if the concentration is doubled, the rate is also doubled.
(ii) For a second-order reaction.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 60
i.e., for a second-order reaction, if the concentration of A is doubled, the rate of the reaction increases by 4 times.
(iii) For a third-order reaction,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 61
Hence, for a third-order reaction, if the concentration of the reactant is doubled the rate increases 8 times.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 34.
Compounds A and B react according to the following chemical equation
A (g) +2B(g) → 2C(g)
Answer:
The concentration of either ‘A’ or ‘B’ was changed keeping the concentration of one of the reactants constant and rates were measured as a function of initial concentration following results were obtained. Find the order with respect to A and B and write the rate law for the reaction
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 62
From experiments 1 and 3,
[B] = constant; [A] = doubled
rate is doubled.
Hence rate ∝ [A] i.e., the order with respect to A is 1.
From experiments 1, and 2,
[A] = constant; [B] doubled rate is quadrupled.
Hence, rate ∝ [B]2
i.e., order with respect by B is 2.
Combining rate = k [A] [B]2.

Question 35.
The initial rate of reactions
3A + 2B + C → products at different initial concentrations are given below.

Initial rate ms-1 [A0] M [B0] M [C0] M
1) 5.0 x 10-3 0.010 0.005 0.010
2) 5.0 x 10-3 0.010 0.005 0.015
3) 1.0 x 10-2 0.010 0.010 0.010
4) 1.25 x 10-3 0.005 0.005 0.010

Answer:
(i) Consider equation (1) and (2)
The concentration of A and B are constant. The rate remains the same for different concentrations of C. Hence, the order with respect to C is zero.
(ii) Consider equations (3) and (1)
The concentration of A and C are constant. The concentration of B is doubled, rate increases by 2 times. Hence order with respect to B is 1.
(iii) Consider equations (4) and (1), The concentration of B and C remain constant. When the concentration of A is doubled, the rate increases by 4 times. Hence order with respect to A is 2.
Alternate method:
Suppose order with respect to A, B and C are α, β and γ respectively. Then the rate law at different concentration of A, B and C are
5.0 × 10-3 = k [0.010]α [0.005]β [0.010]γ …(1)
5.0 × 10-3 = k [0.010]α [0.005]β [0.015]γ …(2)
1.0 × 10-3 = k [0.010]α [0.010]β [0.010]γ …(3)
1.25 × 10-3 = k [0.005]α [0.005]β [0.010]γ …(4)
Dividing (1) by (2)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 63
i.e., order with respect to C is zero.
Dividing (3) by (2)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 64
Dividing (1) by (4) ⇒ 4 = 2α or α = 2
Order with respect to A, B, C are 2, 1 and 0.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 36.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected by increasing the concentration of B, three times, keeping the concentration of A constant?
(iii) How is the rate affected by when the concentration of both A and B are doubled?
Answer:
(i) A + B → product
rate = k [A] [B]2
(ii) If the concentration of B is tripled, the rate increases by 9 times i.e.,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 65
(iii) If the concentration of both A and B are doubled, the rate increases by 8 times.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 66

Question 37.
The following rate data were obtained at 303K for the following reaction.
2A + B → C + D.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 67
What is the order with respect to each reactant and the overall order of the reaction? write the rate law. Also, calculate the rate constant for the reaction and the units of the rate constant.
Answer:
From experiments 1 and 4, it is seen that [B] the concentration of B is the same but [A] has been made 4 times, the rate of the reaction has also become 4 times. Hence the order with respect to A is 1. i.e., rate ∝ [A],
From experiments 2 and 3, it may be noted that [A] is kept the same but [B] is doubled. The rate of reaction increases by 4 times, i.e., the order with respect to B is 2 combining both,
rate = k[A] [B]2. This is the rate law for the reaction.
Alternatively, suppose the order with respect to A is α and the order with respect to B is β, the general rate law for the reaction is
rate = k [A]α [B]β
From the data given
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 68
Now,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 69
Unit of rate constant (k)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 70
Calculation of rate constant: Substitute the values of α and β in any of the equations.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 71
Hence k = 6.0 mol-1 L2 min-1.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 38.
In a reaction between A and B, the initial rate of reaction was measured at a different initial concentration of A and B as given below.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 72
What is the order with respect to A and B?
Answer:
The general rate law is rate = k [A]α [B]β where α, β are ordered with respect to A and B respectively.
The given data is
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 73
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 74
Thus order with respect to A is 0.5 and that of B is 0.

Question 39.
The decomposition of ammonium nitrate in an aqueous solution was studied by placing the apparatus in a thermostat maintained at a particular temperature. The volume of nitrogen gas collected at different intervals of time was as follows:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 75
The above data prove that the reaction is of the first order.
Answer:
For a first-order reaction,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 76
In the present case, V = 35.05 cm3. The value of R at each instant of time can be calculated as follows:
Values of ‘k’ come nearly constant and hence the reaction is first order.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 77

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 40.
What do understand by a fraction of effective collisions? Mention its significances.
Answer:
Fraction of effective collisions (f) is given by the following expression
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 78
To understand the magnitude of collision factor (f), Let us calculate the collision factor (f ) for a reaction having an activation energy of 100 kJ mol-1 at 300K.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 79
Thus, out of 1018 collisions, only four collisions are sufficiently energetic to convert reactants to products.

Question 41.
Explain the importance of the proper orientation of molecules in the collision theory.
Answer:
Even if the reactant possesses sufficient activation energy, their collisions need not result in the formation of products. The molecules should orient themselves in such a way that their collision becomes fruit fill. The fraction of effective collisions (f) having orientation is given by steric factor (p). Thus, the rate of reaction is given by
rate = p x f x collision rate
(or) rate = pz\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 42.
Define activation energy of a reaction.
Answer:
Activation energy is the extra amount of energy that is supplied from outside so that colloiding molecules must produce effective collisions.

Question 43.
Arrhenius equation is given by k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Based on this equation answer the following questions.
(i) Can reactions have zero activation energy?
(ii) Can a reaction have negative activation energy?
Answer:
(i) k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\), when the expression becomes k = e° = A.
The rate constant k = collision frequency. This implies that every collision is an effective collision and leads to product formation which is not possible. Thus activation energy can not be zero.
(ii) When Ea is negative, the above equation
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 80
Rate constant having a higher value than collision frequency which is not possible.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 44.
Explain the effect of temperature on reaction rate based on Arrhenius’s theory.
Answer:
An increase in temperature increases the fraction of the total number of molecules having the necessary energy of activation which in terms increases the number of effective collisions and hence the rate.

Question 45.
Write Arrhenius equation and explain the terms? What is the significance of frequency factor A in the equation?
Answer:
The Arrhenius equation is k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\) Where ‘k’ is the rate constant for the reaction, A is called the pre-exponential factor, Ea is the energy of activation and R is the gas constant and T is the absolute temperature. The two quantities A and Ea are called the Arrhenius parameters.
The frequency factor gives the frequency of binary collision of the reacting molecules per liter per second. It practically remains constant and does not vary with temperature.

Question 46.
Explain how the energy of activation for a reaction is determined.
Answer:
The rate constant of a reaction is determined at two different temperatures. If R1 is the rate constant at temperature T1 arid R2 is the rate constant at temperature T2 thereby using a logarithmic form of Arrhenius equation
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 80
Ea can be calculated.

Question 47.
Describe how the energy of activation of a reaction is determined graphically.
Answer:
The Arrhenius equation is written in the form,

This equation is of the form y = mx + C i.e., equation for a straight line. Thus, if a plot of log k is \(\frac{1}{T}\) is a straight line, the validity of the equation is confirmed.
Thus the values of k at different temperatures are found and a plot of log k vs \(\frac{1}{T}\) is made.
A straight line with a negative slope is obtained from the value of the slope, Ea can be calculated.
Slope = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\)

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 48.
The rate constant for a reaction is 1.2 x 10-3 sec at 30°C and 2.1 x 10-3 sec-1 at 40°C. Calculate the energy of activation.
Answer:
Given that k1 = 1.2 x 10-3 sec-1
T1 = 30 + 273 = 303K
k2 = 2.1 x 10-3 sec
T2 = 40 + 273 = 313K
Substituting these values in the equation
Solving Ea = 44126.3J = 44.13kJ

Question 49.
The rate of particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation.
Answer:
Given that T1 = 27°C = 273 +21 = 300K
T2 = 37°C = 273 + 37 = 310K
k2 = 2k1
Substituting these values in the equation.
This as solving for Ea
Ea = 53598.6 J mol-1 or 53.6 kJ mol-1

Question 50.
The activation energy of a reaction is 94.14 kJ mol-1 and the value of the rate constant at 313K. is 1.8 x 10-1 sec-1 . Calculate the frequency factor (A).
Answer:
Given that, Ea = 94.14 kJ mol-1 = 9414 J mol-1
T = 313K,
k = 1.8 x 10-1 sec-1 Substituting these values in
= (log 1.8 – 5)+ 15.7082
= (0.2553 – 5)+ 15.708
= 10.963J
A = Anti log (10.963 J)
= 9.194 x 10-10 sec

Question 51.
The first order rate constant for the decomposition of ethyl iodide by the reaction
C2H5I (g) → C2H4 (g) + HI (g)
at 600 K is 1.60 x 10-1 s-1. Its energy of activation is 209 kJ / mol. Calculate the rate constant of the reaction at 700K.
Answer:
Given T1 = 600 K
k1 = 1.60 x 10-5 sec-1
T2 = 700K
k2 = ?
Ea = 299 kJ mol-1 = 299000 J mol-1
Substituting these values in the equation.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

Question 52.
Rate constant ‘k’ of a reaction varies with temperature according to the equation
where Ea is the energy of activation for the reaction. When a graph is plotted for log k vs \(\frac{1}{T}\) a straight line with slope – 6670 K is obtained.
Calculate the energy of activation for the reaction and mention is units.
(R = 8.314 J k-1 mol-1)
Answer:
Slope of the line = \(\frac{-\mathrm{E}_{a}}{2.303 \mathrm{R}}\) = -6670K
Ea = 2.303 x 8.314 x 6670 = 127711.4 J mol-1

Choose the correct answers:

1. Which of the following statements is not correct about the order of reaction?
(a) The order of a reaction can be fractional.
(b) Order of reaction is an experimental quantity.
(c) The order of the reaction is always equal to the sum of the stoichiometric coefficients of reactants in a balanced chemical equation for the reaction.
(d) The order of a reaction is the sum of the powers of molar concentrations of the reactants in the rate law expression.
Answer:
(c)

2. Which of the following statements is correct?
(a) The rate of reaction decreases with the passage of time as the concentration of the reactant decreases.
(b) The rate of reaction is the same at any time during the reaction.
(c) The rate of reaction is independent of temperature change.
(d) The rate of reaction decreases with an increase in the concentration of reactants.
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

3. Which of the following expression is correct for the rate of reaction given below:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 92
Answer:
(c)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 93

4. Time required for 100 percent completion of a zero-order reaction is:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 110
Answer:
(c)

5. For the reaction aA+ bB → cC,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 94
if , then a, b and c respectively are:
(a) 3,1,2
(b) 2,1,3
(c) 1,3,2
(d) 6, 2, 3
Answer:
(c)
Hint: Dividing throughout by 3, we get
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 95
This is so for the reacting A + 3B → 2C

6. The rate of a gaseous reaction is given by the expression k [A] [B]. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to the original rate will be:
(a) 1/10
(b) 1/8
(c) 8
(d) 16
Answer:
(d)
Hint: Rate = k ab. When volume is reduced to 1/4th, concentrations will become = 4 times.
New rate = k (4a) (4b) = 16 kab = 16 times.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

7. In a reaction A → B, the rate of reaction increases two times on increasing the concentration of the reactant four times, then order of reaction is:
(a) 0
(b) 2
(c) 1/2
(d) 4
Answer:
(c)
Hint: (i) r = kaα, (ii) 2r = k(4a)α
Dividing (ii) by (i),
4α = 2 or 2 = 2 or 2α = 1 or α = 1/2.

8. The rate of the reaction 2 NO + Cl2 → 2 NOCl is given by the rate equation: rate = k[NO]2[Cl2]. The value of the rate constant can be increased by:
(a) increasing the temperature
(b) increasing the concentration of NO
(c) increasing the concentration of Cl2
(d) doing all of these
Answer:
(a)
Hint: The rate constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

9. The unit of rate constant for a zero-order reaction is:
(a) mol L-1 s-1
(b) L mol-1 s-1
(c) L2 mol-2 s-1
(d) s-1
Answer:
(a)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 96

10. Rate constant of a reaction (k) is 175 litre2 mol-2 sec-1. What is the order of reaction?
(a) first
(b) second
(c) third
(d) zero
Answer:
(c)
Hint: On the basis of given units of k, the reaction is of 3rd order.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

11. The reaction A → B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?
(a) 1 hour
(b) 0.5 hour
(c) 0.25 hour
(d) 2 hours
Answer:
(a)
Hint: The fraction of A reacted in each case is same (0.2 / 0.8 = 1/4),
(0.9 – 0.675)/ 0.90 = 0.225 / 0.90 = 1/4. Hence, time taken is same.

12. 75% of the first-order reaction was completed in 32 min. 50% of the reaction was completed in:
(a) 24 min
(b) 8 min
(c) 16 min
(d) 4 min
Answer:
(c)
Hint: 75 % of reaction is completed in two half-lives, i.e., 2 x t1/2 = 32 min or t1/2 = 16 min.

13. 1/[A] vs time is a straight line. The order of the reaction is:
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b)
Hint: log [A] vs time is linear for 1st order reactions, 1/[A] vs time is linear for 2nd order reactions; 1/[A]2 vs time is linear for 3rd order reactions.

14. If a graph is plotted between in k and 1/T for the first-order reaction, the slope of the straight line so obtained is given by:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 97
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

15. 10g of a radioactive isotope is reduced to 1.25g in 12 years. Therefore, half-life period of the isotope is:
(a) 24 years
(b) 4 years
(c) 3 years
(d) 8 years
Answer:
(b)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 98

16. The half-life period of a radioactive element is 20 days. What will be the remaining mass of 100 g of it after 60 days?
(a) 25 g
(b) 50 g
(c) 125 g
(d) 20 g
Answer:
(c)
Hint: 60 days = 3 half-lives, i.e, n = 3;
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 99

17. Activation energy of a chemical reaction can be determined by:
(a) determining the rate constant at standard temperature.
(b) determining the rate constants at two temperatures.
(c) determining the probability of collision.
(d) using catalyst.
Answer:
(b)
Hint:
Knowing k1 and k2 at T1 and T2, Ea can be determined. (Arrhenius equation)

18. According to Arrhenius equation, rate constant k is equal to A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). Which of the following options represents the graph of k vs \(\frac{1}{T}\)?
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 100
Answer:
(a)
Hint:
k = A\(e^{-\mathrm{E}_{a} / \mathrm{RT}}\). In k = ln A – \(\frac{\mathrm{E}_{a}}{\mathrm{RT}}\). Hence, graph of In k vs \(\frac{1}{T}\) will be linear with negative slope and intercept = ln A.

19. Which of the following statement is incorrect about the collison theory of chemical reaction?
(a) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(b) Number of effective collisions determines the rate of reaction.
(c) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(d) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer:
(c)
Hint: (c) is incorrect because the formation of the product depends not only on energy but also on proper orientation at the time or collision.

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

20. A first-order reaction is 50% completed in 1.26 x 1014 s. How much time would it take for 100% completion?
(a) 1.26 x 1015 s
(b) 2.52 x 1014 s
(c) 2.52 x 1028 s
(d) infinite
Answer:
(d)
Hint: The whole of the substance never reacts because in every half-life, 50% of the substance reacts. Hence, the time taken for 100% completion of a reaction is infinite.

21. Compounds ‘A’ and ‘B’ react according to the following chemical equation:
A (g) + 2 B(g) → 2C (g)
The concentration of either ‘A’ or ‘B’ was changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 101
(a) Rate = k [A]2 [B]
(b) Rate = k [A] [B]2
(c) Rate = k [A] [B]
(d) Rate = k [A]2 [B]0
Answer:
(b)
Hint: From expts.l and 3, [B] = constant, [A] = doubled, rate is doubled. Hence, Rate ∝ [A].
From expts. 1 and 2, [A] = constant, [B] = doubled, rate quadrupled. Hence, Rate ∝ [B]2.
Combining, Rate = k [A][B]2.

22. The rate constant of reaction A → B is 0.6 x 103 mole per liter per second. If the concentration of A is 5 M, then the concentration of B after 20 minutes is:
(a) 0.36 M
(b) 0.72 M
(c) 1.08M
(d) 3.60 M
Answer:
(b)
Hint: The unit mol L-1 s-1 of the rate constant show that it is a reaction of zero order. For a zero order reaction, x = kt. Amount of A reached (x) = (0.6 x 10-3 mol L-1 s-1 ) (20 x 60 s) = 0.72 mol L-1 [B] formed = [A] reacted = 0.72 mol L-1

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

23. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L-1 to 0.04 mg L-1, is:
(a) 414 s
(b) 552 s
(c) 690 s
(d) 276 s
Answer:
(c)
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 102

24. The following data is obtained during the first-order thermal decomposition of A (g) → B (g) + C (s) at constant volume and temperature.

S. No. Time The total pressure in pascals
1. At the end of 10 minutes 300
2. After completion 200

The rate constant in min-1 is
(a) 0.0693
(b) 6.93
(c) 0.00693
(d) 69.3
Answer:
(a)

25. t1/4 can be taken as the time taken for the A concentration of a reactant to drop to its initial value. If the rate constant for a first-order reaction is k, then t1/4 can be written as:
(a) 0.10/k
(b) 029/k
(c) 0.69/k
(d) 0.75/k
Answer:
(b)
Hint:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 103

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

26. A plot of log t1/2, versus log C0 is given in the adjoining fig. The conclusion that can be drawn from this graph is:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 104
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 105
Answer:
(b)

27. In the catalysed decomposition of benzene diazonium chloride,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 106
the half-life period is found to be independent of the initial concentration of the reactant. After 10 minutes, the volume of N2 gas collected is 10 L and after the reaction is complete, it is 50 L. Hence rate constant of the reaction (in min-1) is:
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 107
Answer:
(b)

28. The activation energy of a reaction can be determined from the slope of which of the following graph?
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 108
Answer:
(a)
Hint: From the Arrhenius equation,
TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics 109

29. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:
(a) 60.5 kJ mol-1
(b) 53.6 kJ mol-1
(c) 48.6 kJ mol-1
(d) 58.5 kJ mol-1
Answer:
(b)

30. In the presence of a catalyst, the activation energy of a reaction is lowered by 2 kcal at 27°C. The rate of reaction will increase by:
(a) 2 times
(b) 14 times
(c) 28 times
(d) 20 times
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 7 Chemical Kinetics

31. Which one of the following is not correct?
(a) Every bimolecular collision does not result into a chemical reaction.
(b) Collision theory is not applicable to unimolecular reaction.
(c) According to collision frequency, k = PZAB \(e^{-\mathrm{E}_{a} / \mathrm{RT}}\) where ZAB is collision frequency and P is steric factor.
(d) Collision theory assumes molecules to be hard spheres.
Answer:
(b)
Hint: Collision theory is applicable to unimolecular reactions also.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Students get through the TN Board 12th Chemistry Important Questions Chapter 6 Solid State which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 6 Solid State

Answer the following questions.

Question 1.
Why are solids rigid?
Answer:
In solids, the constituent particles (atoms, molecules, or ions) are not free to move but can only oscillate about their mean position due to strong interatomic or intermolecular forces. This imparts rigidity.

Question 2.
Classify the following as amorphous or crystalline solids.
Answer:
Polyurethane, naphthalene, benzoic acid Teflon, potassium nitrate, cellophane, polyvinyl chloride fiberglass, copper.
Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, and fiberglass. Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, and copper.

Question 3.
Explain why glass is considered an amorphous solid.
Answer:
Like liquids, it has a tendency to flow, though very slowly. It does not have a sharp melting point. It is isotropic.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 4.
Classify the following solids in different categories based on the nature of intermolecular forces operating in potassium sulfate, tin, benzene, urea, ammonia water, zinc sulfide, graphite, rubidium, argon, silicon carbide.
Answer:
Ionic solids: potassium sulfate, zinc sulfide. Covalent solids: graphite, silicon carbide. Molecular solids: benzene, urea, ammonia water, argon.
Metallic solids: rubidium, tin.

Question 5.
Ionic solids conduct electricity in the molten state but not in the solid state. Explain.
Answer:
In ionic solids, free ions are present in a molten state and they move towards the oppositely charged electrode under the influence of electric current. In a solid state, these ions are fixed in their lattice position and cannot move under the influence of electric current.

Question 6.
Briefly outline the properties of covalent solids.
Answer:

  1. They have a high melting point.
  2. They are poor thermal and electrical conductors.

Question 7.
What are covalent solids? Give example.
Answer:
In covalent solids, the constituents (atoms) are bound together in a three-dimensional network entirely by covalent bonds. eg; Diamond, silicon carbide, etc.

Question 8.
What are molecular solids? Give examples.
Answer:
In molecular solids, the constituents are neutral molecules. They are held together by weak Van der Waals forces. Generally, molecular solids are soft and they do not conduct electricity, eg: I2, graphite.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 9.
Explain various types of molecular solids for example.
Answer:
Molecular solids are classified into (i) non-polar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.

  1. In nonpolar molecular solids, the constituent molecules are held together by weak dispersion or London forces, eg: naphthalene, anthracene.
  2. In polar molecular solids, the constituent molecules are formed by polar covalent bonds, eg solid CO2, solid NH3.
  3. In hydrogen-bonded molecular solids, the constituent molecules are held together by hydrogen bonds, eg: ice (H2O), glucose, urea, etc.

Question 10.
Explain the types of the force of attraction that exist (i) nonpolar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.
Answer:

  1. In non-polar molecular solids, the constituent particles are held together by weak dispersion or London forces.
  2. In polar molecular solids, the constituent molecules are held by covalent bonds. They are attracted to each other by relatively strong dipole-dipole attraction.
  3. In hydrogen-bonded molecular solids, the molecules are held together by hydrogen bonds.

Question 11.
What are the characteristics of metallic solids? What type of attractive forces exists among the constituent particles?
Answer:
The constituents of a metallic solid are held together by a metallic bonding. The lattice points are occupied by metals ion and the electron’s charge bond encloses the metal ion.
They are hard, possess a high melting point, They conduct heat and electricity. They possess metallic luster.

Question 12
Give examples for metallic solids.
Answer:
Metal and metal alloys, eg: Cu, Fe, Zn, Ag, Au, Cu, Zn, etc.

Question 13.
Define the following: (i) Lattice point, (ii) Crystal lattice, (iii) Unit cell.
Answer:

  1. Lattice point: Each lattice point represents one constituent particle of a solid. This may be an atom, molecule, or ion.
  2. The crystal lattice is a three-dimensional arrangement of identical points in the space which represents how the constituent particles (atoms, molecules, or ions) are arranged in a crystal.
  3. Unit cell A unit cell is the smallest portion of a space lattice which when repeated in different directions generates the entire lattice.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 14.
What are the characteristics of a unit cell?
Answer:
A unit cell is characterized by the three edge lengths or lattice constants a, b, and c and the angle between the edges α, β, and γ.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 1

Question 15.
Briefly explain how constituent particles are arranged in (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells.
Answer:

  1. In the simple cubic unit cell, each comer is occupied by identical atoms or ions or molecules. And they touch along the edges of the cube, do not touch diagonally. The coordination number of each atom is 6.
  2. In a body-centered cubic unit cell, each comer is occupied by an identical particle, and in addition to that one atom occupies the body center. Those atoms which occupy the comers do not touch each other, however, they all touch the one that occupies the body center. Hence, each atom is surrounded by eight nearest neighbors and the coordination number is 8.
  3. In a face-centered cubic unit cell, identical atoms lie at each comer as well as in the center of each face. Those atoms in the comers touch those in the faces but not each other. The coordination number is 2.

Question 16.
Distinguish between primitive and nonprimitive unit cells.
Answer:

Primitive unit cells Nonprimitive unit cells
A unit cell that contains one lattice point is called a primitive unit cell. In a primitive unit cell, constituent particles are present only at the comers of a unit cell. In the nonprimitive unit cells, the constituent particles are present not only at the comer but also in the centers of the face and body of the unit cell.

Question 17.
Define coordination number? What is the coordination number of each constituent particle present at (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells?
Answer:
The number of constituent particles that surround an atom/ion in a crystal lattice is called the coordination number.

  1. The coordination number of each atom in a simple cubic unit cell is 6.
  2. The coordination number of each atom in a body-centered cubic unit cell is 8.
  3. The coordination number of an atom is a face-centered cubic unit cell is 2.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 18.
Explain how will you calculate the number of particles in a unit cell.
Answer:

  1. A point that lies at the comer of a unit cell is shared among the eight-unit cells and therefore, only one eight (\(\frac{1}{8}\)th) of each point lies within the unit cell.
  2. A point along the edge is shared by 4 unit cells and only \(\frac{1}{4}\)th of it lies within any one cell.
  3. A face-centered point is shared by 2 unit cells and only \(\frac{1}{2}\) of it is present in a given unit cell.
  4. A body-centered point lies entirely within the unit cell and contributes one complete point to the cell.
    TN Board 12th Chemistry Important Questions Chapter 6 Solid State 2
    Total number of particles in an unit cell (z) = (\(\frac{1}{8}\) × occupied by corners) + \(\frac{1}{4}\) × (occupied by edge centres) + \(\frac{1}{2}\) × (occupied by 24 face centres) + one occupied by body centre.

Question 19.
Calculate the number of particles present in (i) simple cubic (ii) body-centered and (iii) face-centered unit cells.
Answer:

  1. In simple cubic unit cell particles are present only at the comers of the cube and this is shared by 8 other unit cells. Hence, the contribution of the point at the comers to the given unit cell is
    ∴ Z = 8 × \(\frac{1}{8}\) = 1
    i.e, A simple cubic unit cell has a single constituent (atom/molecule/ion) unit per unit cell.
  2. In the body-centered cubic unit cell, comers, as well as the center of the unit cell, are occupied.
    ∴ Z = 8 × \(\frac{1}{8}\) + 1 = 2
    Hence body-centered cubic (bcc) has 2 constituent units per unit cell.
  3. In face-centered cubic unit cell (fee) comers as well as face centers are occupied.
    ∴ Z = 8 × \(\frac{1}{8}\) + 6 + 2 = 4
    Hence, the face-centered cubic unit cell has four constituent units per unit cell.
    [Note: Remember that the number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence, it helps to predict the formula of the compound.]

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 20.
A compound formed by elements A and B has a cubic structure in which A atoms are at the comers and B atoms are at face centers. Derive the formula of the compound.
Answer:
As ‘A’ atoms are present at the 8 comers of the cube, therefore a number of atoms of A in the unit cell = 8 × \(\frac{1}{8}\) = 1.
As ‘B’ atoms are present at the face centers of the 6 faces of the cube, therefore, the number
of atoms of B in the unit cell = \(\frac{1}{2}\) × 6 = 3.
∴ Ratio of atoms A : B = 1 : 3
∴ The formula of the compound is AB3.

Question 21.
A cubic solid is made up of two elements X and Y. Atoms ‘Y’ are present at the comers of the cube and atoms X at the body center. What is the formula of the compound? What are the coordination numbers of X and Y?
Answer:
As Y atoms are present at 8 comers, of the cube, the number of atoms of Y in the unit cell
= \(\frac{1}{8}\) × 8 = 1
As atoms X are present at the body center, therefore, the number of atoms of X in the unit cell = 1.
∴ Ratio of atoms X : Y = 1 : 1
Hence the formula of the compound is XY.
The coordination number of each X and Y = 8.

Question 22.
A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms are at the comers of the cube and X atoms are at alternate faces. What is the formula of the compound?
Answer:
As there are 8 Y atoms are at the comers of the cube and the contribution of each
= \(\frac{1}{8}\) therefore no. of Y atoms per unit cell
= 8 × \(\frac{1}{8}\) = 1
These can only be 2 × atoms an alternate forces.
As contribution of each of them = \(\frac{1}{2}\).
Therefore number of X atom per unit cell
= 2 × \(\frac{1}{2}\) = 1
Ratio of atom X : Y = 1 : 1
Hence the formula of the compound is 1 : 1

Question 23.
Give the expression to find out the inter planar distance (d) between two successive planes of atoms.
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 3
Where n is the number of planes,
λ is the wavelength of X-ray used,
θ is the angle of diffraction.
Using these values, the edge of the unit cell can be calculated.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 24.
X-rays of wavelength 1.54Å strike a crystal and are observed to be deflected at an angle of 22.5°. Assuming that n = 1, calculate the spacing between the planes of atoms that are responsible for this reflection.
Answer:
Applying Bragg equation,
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 4

Question 25.
Derive an expression to find the density of a crystal from lattice parameters.
Answer:
Using the edge length of a unit cell, we can calculate the density (ρ) of the crystal by considering a cubic unit cell as follows.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 5
substitute (3) in (2)
mass of the unit cell = n × \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) …(4)
For a cubic unit cell, all the edge lengths are equal i.e, a = b = c
Volume of the unit cell = a × a × a = a3 …(5)
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 6
Equation (6) contains four variables namely ρ, n, M, and a. If any three variables are known, the fourth one can be calculated.

Question 26.
The edge length of a unit cell is 408 pm. Its density is 10.6 g cm-3. Predict whether the metal is body-centered, or face-centered, or simple cubic.
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 7
The value of Z indicates that the metal has a fee structure.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 27.
An element crystallizes in bcc structure the edge length of its unit cell is 288 pm. The density of the crystal is 7.2g cm-3. What is the atomic mass of the element?
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 8

Question 28.
A metallic element exists as a body-centered cubic lattice. Each edge of the unit cell is 2.88 pm. The density of the metal is 7.2g cm-3. How many atoms and unit cells are there in 100 g of the atom?
Answer:
Let the atomic mass of the element be M.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 9

Question 29.
The density of a face-centered cubic clement is 6.25 g cm-3. Calculate the length of the unit cell. (Atomic mass of the element = 60.2 AMU).
Answer:
Given Z = 4; M = 60.2 amu; ρ = 625 g cm-3
NA = 6.023 × 1023
Applying the formula:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 10

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 30.
KBr has NaCl type structure. What is the distance between K+ and Br in KBr, if the density is 2.75 g cm-3?
Answer:
Edge length ‘a’ of the unit cell is calculated as
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 11

Question 31.
Derive an expression to calculate the packing efficiency in a simple cubic arrangement.
Answer:
Let us calculate the packing efficiency in simple cubic arrangement,
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 12
Let us consider a cube with an edge length ‘a’ as shown in fig.
Volume of the cube with edge length a is = a × a × a = a3
Let V is the radius of the sphere. From the figure, a = 2r ⇒ r = \(\frac{a}{2}\)
∴ Volume of the sphere with radius ‘r’
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 13
In a simple cubic arrangement, the number of spheres that belongs to a unit cell is equal to one
∴ Total volume occupied by the spheres in sc unit cell = 1 × \(\left(\frac{\pi a^{3}}{6}\right)\) ……(2)
Dividing (2) by (3)
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 14
i.e., only 52.31% of the available volume is occupied by the spheres in simple cubic packing, making inefficient use of available space and hence minimizing the attractive forces.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 32.
Show that the packing efficiency in the face-centered cubic unit cells is 74%.
Answer:
The cubic close packing is based on the face-centered cubic unit cell. Let us calculate the packing efficiency in the fcc unit cell.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 15
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 16
The total number of spheres belongs to a single fcc unit cell is 4
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 17

Question 33.
Give the relationship between the nearest neighbor distance (d) and the edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: d = a
For face centered cubic: d = \(\frac{a}{\sqrt{2}}\) = 0.707a
For body centered cubic:
d = \(\frac{\sqrt{3}}{2} a\)

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 34.
Give the relationship between atomic radius (r), (which is d/2 for crystals of pure substances) and edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: r = \(\frac{a}{2}\)
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 18

Question 35.
Xenon crystallizes in face-centered cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbor distance and what is the radius of the xenon atom?
Answer:
Here, a = 620 pm; d = 2 ; r = ?
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 19

Question 36.
CsCl has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 20
The bcc arrangement of CsCl is shown in fig. Where black circle is Cs+ ion and colored circles are Cl ions. The aim is to find half of the body diagonal AE. If the edge of the unit cell is ‘a’.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 21

Question 37.
If the radius of the atom is 75 pm and the lattice type is body-centered cubic, what is the edge of the unit cell?
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 22

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 38.
The radius of an atom of an element is 500 pm. If it crystallizes as fee lattice, what is the length of the side of the unit cell?
Answer:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 23

Question 39.
A solid AB has a CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of the closest approach between A+ / B ions.
Answer:
The distance of the closest approach is equal to the distance between nearest neighbors (d). As CsCl has a bcc lattice.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 24

Question 40.
What is the radius ratio? Mention its importance.
Answer:
In ionic solids, the ratio of the radius of the cation to the radius of an anion is known as the radius ratio.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 25
The radius ratio of an ionic solid gives the structural arrangement of ionic solids.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 26

Question 41.
What is the meaning of the term imperfection in solids?
Answer:
Imperfection refers to the departure from the perfect periodic arrangement of atoms, ions, or molecules in the structure of crystalline solids.

Question 42.
What are the types of lattice imperfections found in crystals?
Answer:

  1. Stoichiometric defect i.e., Schottky defect and Frenkel defect.
  2. Nonstoichiometric defect viz metal excess, metal deficiency, and impurity defects.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 43.
What are interstitials in a crystal?
Answer:
Atoms or ions that fill the normal vacant interstitial voids in a crystal are known as interstitials.

Question 44.
What is the Schottky defect?
Answer:
If an equal number of cations and anions are missing from their lattice sites, the defect is known as the Schottky defect.

Question 45.
What is Frenkel’s defect?
Answer:
When some ions (usual cation) are missing from the lattice sites and they occupy the interstitial sites. So that, the electro-neutrality, as well as stoichiometry are maintained, it is called Frenkel defect.

Question 46.
Which crystal defect lowers the density of the solid?
Answer:
Schottky defect.

Question 47.
Which crystal defect in crystals does not alter the density of a relevant solid?
Answer:
Frenkel defect.

Question 48.
Which point does defect increase the density of the solid?
Answer:
Schottky defect.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 49.
Which point defect in the crystal increases the density of the solid?
Answer:
Interstitial defect.

Question 50.
Name one solid in which both Frenkel and Schottky defects occur.
Answer:
Silver Bromide.

Question 51.
Why does the Frenkel defect does not change the density of AgCl crystal?
Answer:
Because in Frenkel defect, no ion is missing from the lattice sites. Therefore, no change in density occurs.

Question 52.
What are F. Centres?
Answer:
The free-electron trapped in the anion vacancy is called F.Centres.

Question 53.
What are non-stoichiometric defects?
Answer:
If as a result of imperfections in the crystal, the ratio of the cation to the anions becomes different from that indicated by their ideal chemical formula, then the defect is termed a non-stoichiometric defect.

Question 54.
Why does table salt, NaCl appear yellow in color?
Answer:
Yellow color in sodium chloride is due to metal excess defect due to which unpaired electrons occupy anionic sites. These sites are called F. Centres. These electrons absorb energy from the visible region for excitation which makes the crystal yellow.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 55.
Why is FeO (s) not formed in stoichiometric composition?
Answer:
In the crystal of FeO, some of Fe+2 cations are replaced by Fe+3 cations. Three Fe+2 cations are replaced by two Fe+3 cations to make up for the loss of positive charge. Eventually, there would be fewer atoms of the metal as compared to stoichiometric proportion.

Question 56.
How would you account for the following?

  1. Frenkel defects are not found in alkali metal halides.
  2. Schottky defects lower the density of the solid.
  3. Impurity doped silicon is a semiconductor.

Answer:

  1. This is because alkali metal ions have larger sizes that cannot fit into interstitial sites.
  2. As the number of ions decreases as a result of the Schottky defect, the mass decreases whereas the volume remains the same.
  3. This is due to additional electrons or the creation of holes and doping with impurity. Creation of hole results in p-type semiconductor and creation of electron results in the n-type semiconductor.

Question 57.
Briefly explain the metal excess defect.
Answer:
The metal excess defect arises due to the presence of more metal ions as compared to anions. Alkali metal halides NaCl, KCl show this type of defect. The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electrons present in the interstitial position.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

Question 58.
Briefly explain metal deficiency defects.
Answer:
Metal deficiency defect arises due to the presence of fewer cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.

Question 59.
Briefly explain the impurity defect.
Answer:
A general method of introducing defects in ionic solids is by adding impurity ions. If the impurity ions are in a different valance states from that of the host, vacancies are created in the crystal lattice of the host. For example, the addition of CdCl2 to silver chloride yields solid solutions where the divalent cation Cd2+ occupies the position of Ag+. This will disturb the electrical neutrality of the crystal. In order to maintain the same, a proportional number of Ag+ ions leaves the lattice. This produces a cation vacancy in the lattice, such kinds of crystal defects are called impurity defects.

Choose the correct answer.

1. Which one of the following is a covalent crystal?
(a) Rock salt
(b) Ice
(c) Quartz
(d) Dry ice
Answer: (c)
Hint: Rock salt is an ionic solid, while dry ice and ice are molecular solids.

2. Total volume of atoms present in a face centered cubic unit cell of a metal is:
(r = atomic radius)
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 29
Answer: (b)
Hint: Total no. of atoms in fee unit cell = 4
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 30

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

3. The fraction of the total volume occupied by the atoms present in a simple cube is:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 31
Answer: (b)
In a simple cube, no. of atoms / unit cell = 8 × \(\frac{1}{8}\) = 1
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 32

4. Three elements A, B and C crystallise into a cubic solid lattice. Atoms A occupies the comers, B atoms, the cubic centres, and atoms C, the edges. The formula of the compound is:
(a) ABC
(b) ABC2
(c) ABC3
(d) ABC4
Answer: (c)
Hint:
Atoms A per unit cell = 8 × \(\frac{1}{8}\) = 1
Atom B per unit cell = 1
Atoms C per unit cell = 12 × \(\frac{1}{4}\) = 3
Ratio A : B : C = 1 : 1 : 3.
Hence the formula is ABC3.

5. An alloy of copper, silver and gold is found to have copper forming the simple cubic close packed lattice. If silver atoms occupy the comers and gold atoms are present at the body centres, the formula of the alloy will be:
(a) Cu3 Ag Au
(b) Cu Ag3 Au
(c) Cu4 Ag2 Au
(d) Cu Ag Au.
Answer: (b)
Hint:
Cu atoms per unit cell (present at the comer) = 8 × \(\frac{1}{8}\) = 1
Ag atoms per unit cell (present at the face centres) = 6 × \(\frac{1}{2}\) = 3
Au atoms per unit cell (present at body centre) = 1
Hence the formula is Cu Ag3 Au.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

6. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y) will be:
(a) 275.1 pm
(b) 322.5 pm
(c) 241.5 pm
(d) 165.7 pm
Answer: (c)
Hint: NaCl has a face centered cubic structure. Cl ions and Na+ ions are present in the octahedral voids. Hence for such a solid, radius of the cation is = 0.414 × radius of the anion.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 33

7. The number of atoms in 100g of the fcc crystal with density = 10 g/cm3, and the cell edge equal to 200 pm is equal to:
(a) 5 × 1024
(b) 5 × 1025
(c) 6 × 1023
(d) 2 × 1025
Answer:
(a)
Hint:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 34
Thus 12g of the atom contain = 6.023 × 1023 atom.
∴ 100g of the atom will contain
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 35

8. Which of the following statements is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature.
(b) They may become crystalline or keeping for long time.
(c) Amorphous solids can be moulded by heating.
(d) They are anisotropic in nature.
Answer:
(d)

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

9. The sharp melting point of crystalline solids is due to:
(a) a regular arrangement of constituent particles observed over a short distance in the crystal lattice.
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(c) same arrangement of constituent particles in different directions.
(d) different arrangement of constituent particles in different directions.
Answer:
(b)

10. Iodine molecules are held in the crystal lattice by:
(a) London forces
(b) dipole – dipole interaction
(c) covalent bonds
(d) columbic forces
Answer:
(a)

11. Which of the following is a network solid?
(a) SO2 (solid)
(b) I2
(c) diamond
(d) H2O (ice)
Answer:
(c)

12. Which of the following solids is not an electrical conductor?
(i) Mg(s)
(ii) TiO(s)
(iii) I2 (s)
(iv) H2O (ice)
(a) (i) only
(b) (ii) only
(c) (iii) and (iv)
(d) (ii), (iii) and (iv)
Answer:
(c)

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

13. The lattice sites in a pure crystal cannot be occupied by:
(a) molecule
(b) ion
(c) electron
(d) atom
Answer:
(c)
Hint: Electrons can occupy only in the interstitial sites.

14. Cations are present in interstitial sites in:
(a) Frankel defect
(b) Schottky defect
(c) Valency defect
(d) Metal deficiency defect
Answer:
(a)

15. Schottky defect is obtained in crystal when:
(a) Some cations move from their lattice sites to interstitial sites.
(b) equal number of cations and anions are missing from the lattice.
(c) Some lattice sites are occupied by electrons.
(d) Some impurity is present in the lattice.
Answer:
(b)

16. The total number of tetrahedral void in the face centered unit cell is:
(a) 6
(b) 8
(c) 10
(d) 12
Answer:
(b)
Hint: No. of atoms per unit cell in fee = 4
No. of tetrahedral void: 2 × 4 = 8.

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

17. Which of the following statements is not true about hexagonal close packing?
(a) The coordination number is 12.
(b) It has 74% packing efficiency.
(c) Tetrahedral voids of the second layer are covered by spheres of the third layer.
(d) In this arrangement spheres of the fourth layer are exactly aligned with those of the first.
Answer: (d)
Hint: In hcp arrangement ABAB type and not ABC ABC type. Hence (d) is not true.

18. What is the coordination number in square close packed structure in two dimensions?
(a) 2
(b) 3
(c) 4
(d) 6
Answer:
(c)
Hint: The arrangement is AA AA type. In this arrangement each sphere is touching four other spheres touching the particular sphere, a square is formed. Hence, the coordination number is 4.

19. The correct order to packing efficiency in different type of unit cells is:
(a) fee < bee < simple cubic
(b) fee > bcc > simple cubic
(c) fee < bcc > simple cubic
(d) bcc < fee > simple cubic
Answer:
(b)
Hint: fcc = 74%, bcc = 68%; simple cubic = 52.4%

20. In cubic close packing, the unit cell has:
(a) 4 tetrahedral voids each of which is shared by adjacent unit cells. .
(b) 4 tetrahedral voids within the unit cell.
(c) 8 tetrahedral voids each of which is shared by four adjacent unit cells.
(d) 8 tetrahedral voids within the unit cells.
Answer:
(d)
Hint: ccp = hcp
No. of atoms per unit cell in fcc unit cell = 4.
∴ No. of tetrahedral voids = 2 × 4 = 8

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

21. KCl crystallises in the same type of lattice as does NaCl.
Given that \(r_{\mathrm{Na}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.55 and \(r_{\mathrm{K}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.74, Calculate the ratio of the side of the unit cell of KCl and NaCl.
(a) 1.123
(b) 0.891
(c) 1.414
(d) 0.414
Answer:
(a)
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 36

22. Ice crystallises in a hexagonal lattice having a volume of the unit cell is 132 × 10 cm3. It density of ice at the given temperature is 0.92 g cm-3, then number of H2O molecules per unit cell is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d)
Hint:
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 37

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

24. The edge length of the face-centered cubic unit cell is 508 pm. If the radius of the cation is 110 pm, the radius of the anion will be:
(a) 144 pm
(b) 288 pm
(c) 618 pm
(d) 398 pm
Answer:
(a)
Hint: For the face-centered cubic unit cell (eg: NaCl),
the edge length = 2 × distance between cation and anion
= 2 (r+ + r)
2(r+ + r) = 508 pm
given r+ = 110 pm
2(110 + r) = 508 pm
r = 144 pm

25. Which of the following statements is not correct?
(a) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48
(b) Molecular solids are generally volatile.
(c) The number of carbon atoms in a unit cell of a diamond is 4.
(d) The number of brave lattices in which a crystal can be categorized it.
Answer:
(c)

26. Match the entities with a column I with appropriate entities in column II and choose the correct option using the code given.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 38
(a) (A) – r; (B) – p, r, s; (C) – r, (D) – q
(b) (A) – p,r; (B) – s; (C) – r; (D) – q
(c) (A) – r; (B) – p,s; (C) – q; (D) – r
(d) (A) – r, (B) – s; (C) – r; (D) – p
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

27. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 39
a = cell edge
d = nearest neighbour distance
r = radius
(a) (A) – q; (B) – r; (C) – p; (D) – s
(b) (A) – q; (B) – p; (C) – r; (D) – 5
(c) (A) – q; (B) – s; (C) – q; (D) – p
(d) (A) – q; (B) – s; (C) – r; (D) – p
Answer:
(c)

28. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.
TN Board 12th Chemistry Important Questions Chapter 6 Solid State 40
(a) (A) – s; (B) – q; (C) – r; (D)-p
(b) (A) – q; (B) – p; (C) – r; (D) – s
(c) (A) – r; (B) – s; (C) – q; (D) – p
(d) (A) – s; (B) – r; (C) – p; (D) – q
Answer:
(a)

TN Board 12th Chemistry Important Questions Chapter 6 Solid State

29. Assertion (A): Hexagonal close packing is more closely packed than cubic close packing.
Reason (R): Hexagonal close-packing has a coordination number 12 whereas the cubic close packing has a coordination number 8.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true, and the reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false.
(d) If both assertion and reason are false.
Answer:
(d)
Hint:
Correct assertion: Hexagonal close packing and cubic close packing are equally close-packed with a packing efficiency of 74%.
Correct reason: Both have the coordination number 12.

30. Assertion (A): Frankel defects are shown by silver halides.
Reason (R): Ag+ ion is smaller in size.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true, and reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false.
(d) If both assertion and reason are false.
Answer:
(a)