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• Chapter 8 Financial Statement Analysis
• Chapter 9 Ratio Analysis
• Chapter 10 Computerised Accounting System-Tally

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 3 p-Block Elements – II Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

12th Chemistry Guide p-Block Elements – II Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. In which of the following, NH₃ is not used?
a) Nessler’s reagent
b) Reagent for the analysis of IV group basic radical
c) Reagent for the analysis of III group basic radical
d) Tollen’s reagent
Answer:
a) Nessler’s reagent

2. Which is true regarding nitrogen?
a) least electronegative element
b) has low ionisation enthalpy than oxygen
c) d – orbitals available
d) ability to form pπ -pπ bonds with itself
Answer:
d) ability to form pπ -pπ bonds with itself

3. An element belongs to group 15 and 3rd period of the periodic table, its electronic configuration would be
a) Is² 2s² 2p⁴
b) Is² 2s² 2p³
c) Is² 2s² 2p⁶ 3s2² 3p²
d) Is² 2s² 2p⁶ 3s² 3p³
Answer:
d) Is² 2s² 2p⁶ 3s² 3p³

4. Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously burn in air giving smoky rings. A and B are respectively
a) P₄(red) and PH₃
b) P₄ (white) and PH₃
c) S₈ and H₂S
d) P₄(white) and H₂S
Answer:
b) P₄ (white) and Ph₃

5. On hydrolysis, PCl₃ gives
a) H₃PO₃
b) PH₃
c) H₃PO₄
d) POCl₃
Answer:
a) H₃PO₃

6. P₄O₆ reacts with cold water to give
a) H₃PO₃
b) H₄P₂O₇
c) HPO₃
d) H₃PO₄
Answer:
a) H₃PO₃

7. The basicity of pyrophosphorous acid (H₄P₂O₃) is
a) 4
b) 2
c) 3
d) 5
Answer:
b) 2

8. The molarity of given orthophosphoric acid solution is 2M. Its normality is
a) 6N
b) 4N
c) 2N
d) none of these
Answer:
a) 6N
Normality = M x basicity =2×3 = 6

9. Assertion : bond dissociation energy of fluorine is greater than chlorine gas
Reason : chlorine has more electronic repulsion than flourine
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true bu t reason is not the correct explanation of assertion
c) Assertion is true bu t reason is false
d) Both assertion and reason are false
Answer:
d) Both assertion and reason are false The converse is true

10. Among the following, which is the strongest oxidizing agent?
a) Cl₂
b) F₂
C) Br₂
d) l₂
Answer:
b) F₂

11. The correct order of the thermal stability of hydrogen halide is (PTA – 4)
a) Hl > HBr > HCl > HF
b) HF > HCl > HBr > HI
c) HCl > HF > HBr > HI
d) HI > HCl > HF > HBr
Answer:
b) HF > HCl > HBr > HI

12. Which one of the following compounds is not formed?
a) XeOF₄
b) XeO₃
c) XeF₂
d) NeF₂
Answer:
d) NeF₂

13. Most easily liquefiable gas is
a) Ar
b) Ne
c) He
d) Kr
Answer:
c) He

14. XeFg on complete hydrolysis produces
a) XeOF₄
b) XeO₂F₂
c) XeO₃
d) XeO₂
Answer:
c) XeO₃

15. Which of the following is strongest acid among all?
a) HI
b) HF
c) HBr
d) HCl
Answer:
a) HI
Reason : H-I bond is weakest.

16. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (NEET)
a) Br₂ > I₂ > F₂ > Cl₂
b) F₂ > Cl₂ > Br₂ > I₂
c) I₂ > Br₂ > Cl₂ > F₂
d) Cl₂ > Br₂ > F₂ > I₂
Answer:
d) Cl₂ > Br₂ > F₂ > I₂

17. Among the following the correct order of acidity is (NEET)
a) HClO₂ < HClO < HClO₃ < HClO₄
b) HClO₄ < HClO₂ < HClO < HClO₃
c) HClO₃ < HClO₄ < HClO₂ < HClO
d) HClO < HClO₂ < HClO₃ < HClO₄
Answer:
d) HClO < HClO₂ < HClO₃ < HClO₄

18. When copper is heated with cone HNOs it produces
a) Cu (NO₃)₂, NO and NO₂
b) Cu (NO₃)₂ and N₂O
c) Cu (NO₃)₂ and NO₂
d) Cu (NO₃)₂ and NO
Answer:
c) Cu (NO₃)₂ and NO₂

II. Answer the following questions.

Question 1.
What is the inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belong to p-block.
Answer:
Give reason.

• Chalcogens belong to p-block elements.
• Because their outer electronic configuration is ns² np⁴.
• In these elements the last electron enters np orbital.
• Hence they belong to p-block elements.

Question 3.
Explain why fluorine always exhibits an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have a d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibits an oxidation state of-1 and others in the halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following
Answer:
a) OF₂
b) O₂F₂
c) Cl₂O₃
d) I₂O₄

Halogen	Oxidation State
OF2	-1
O2F2	-1
Cl2O3	+3
I2O4	+4

Question 5.
What are interhalogen compounds? Give examples
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F₂O) and difluorine dioxide (F₂O₂).

Question 6.
Why fluorine is more reactive than other halogens? (PTA – 1, 3)
Answer:
Fluorine is the most reactive element among halogens. This is due to the low value of F – F bond dissociation energy.

Question 7.
Give the uses of helium. (PTA – 2)
Answer:

1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide an inert atmosphere in the electric arc welding of metals
3. Helium has the lowest boiling point hence used in cryogenics (low-temperature science).
4. It is much less denser than air and hence used for filling air balloons.

8. What is the hybridisation of iodine in IF₇? Give its structure. (PTA – 5)
Answer:

Inter halogen	Hybridisation	Structure
IF7	Sp3d3	Pentagonal bipyramidal

9. Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH
Answer:
Chlorine reacts with cold NaOH to give sodium hypochlorite

Chlorine reacts with hot NaOH to give sodium chlorate

10. How will you prepare chlorine in the laboratory? (PTA – 2)
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO₂ + 4HCl → PbCl₂ + 2H₂O + Cl₂
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + 2MnCl + 8H₂O + 5Cl₂
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

11. Give the uses of sulphuric acid.
Answer:
Sulphuric acid is used

• In the manufacture of fertilisers, ammonium sulphate and superphosphates.
• In the manufacture of other chemicals such as hydrochloric acid, nitric acid etc.,
• as a drying agent.
• in the preparation of pigments, explosives, etc.,

12. Give a reason to support that sulphuric acid is a dehydrating agent. (PTA – 1)
Answer:

• Sulphuric acid is highly soluble in water.
• It has a strong affinity towards water.
• Hence it can be used as a dehydrating agent.
• When dissolved in water, it forms mono (H₂SO₄.H₂O) and dihydrates (H₂SO₄.2H₂O) and the reaction is exothermic.
  ex
  C₁₂H₂₂O₁₁ + H₂SO₄ → 12C + H₂SO₄.11H₂O
  HCOOH + H₂SO₄ → CO + H₂SO₄.H₂O

13. Write the reason for the anomalous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electronegativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

14. Write the molecular formula and structural formula for the following molecules
Answer:
a) Nitric add
b) dinitrogen pentoxide
c) phosphoric acid
d) phosphine

15. Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

16. Write the valence shell electronic configuration of group -15 elements.
Answer:
Valence shell electronic configuration of group 15 elements is ns²np³

Elements	Valence Shell Electronic configuration
N	2s2p3
P	3s2p3
S	4s24p3
Sb	5s25p3
Bi	6s2p3

17. Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
Basic Nature :
Phosphine is weakly basic and forms phosphonium salts.
PH₃ +HI → PH₄ I
PH₄I + H₂O \(\underrightarrow { \triangle } \) PH₃ + H₃O⁺ + I^–
It react with halogen to give phosphorous penta halidae
PH₃ +4Cl₂ → PCl₅ + 3HCl

Combustion:
When phosphine is heated with air or oxygen it burns to give metaphosphoric acid

18. Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid rects with a basic oxide to form salt and water.
3 FeO + 10HNO₃ → Fe (NO₃)₃ + NO + 5H₂O

19. What happens when PCl₅ is heated?
Answer:
On heating Phoshorous pentachloride decomposes into phosphorus trichloride and chlorine

20. Suggest a reason why HF is a weak acid, whereas binary acids of all other halogens are strong acids.
Answer:

• HF is only slightly ionised, hence it is a weak acid
• Other halogen acids are almost completely ionised, hence they are strong acids.
• Among halogen acids, the electronegativity difference is maximum (1.9) in HF acid.
• Hence the bond between H and F is stronger and the acid HF is weaker.

21. Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF = x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

22. What type of hybridisation occur in (PTA – 5)
a) BrF₅
b) BrF₃

23. Complete the following reactions
1. NaCl+ MnO₂ + H₂SO₄ →
2. NaNO₂ + HCl →
3. P₄ + Na0H + H₂O →
4. AgNO₃ + PH₃ →
5. Mg + HNO₃ →

8. Sb + Cl₂ →
9. HBr + H₂SO₄ →
10. XeF₆ + H₂O →
11. XeO₆^4- + Mn²⁺ + H⁺ →
12. XeOF₄ + SiO₂ →

Answer:

III. Evaluate yourself

1. Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.

12th Chemistry Guide p-Block Elements – II Additional Questions and Answers

Part II – Additional Questions

I. Match the following

1.

Compound	The oxidation state of Nitrogen
i) NH3	+2
ii) N22	+5
ii) N2O	+4
iv) NO	+3
v) HNO2	0
vi) NO2	+1
vii) HNO3	-3

Answer:

Compound	Oxidation state of Nitrogen
i) NH3	-3
ii) N22	0
ii) N2O	+1
iv) NO	+2
v) HNO2	+3
vi) NO2	+4
vii) HNO3	+5

2.

Name	Molecular Formula
i) Sulpurous acid	H2S2O8
ii) Thiosulphuric acid	H2S2O7
iii) Pyrosulphuric acid	H2S2O6
iv) Marshall’s acid	H2SO3
v) Dithionic acid	H2S2O3

Answer:

Name	Molecular Formula
i) Sulpurous acid	H2SO3
ii) Thiosulphuric acid	H2S2O3
iii) Pyrosulphuric acid	H2S2O7
iv) Marshall’s acid	H2S2O8
v) Dithionic acid	H2S2O6

3.

Inter Halogen compound	Hybridisation
i) IF5	Sp3d3
ii) BeF3	Sp3
iii) IF7	Sp3d2
iv) ClF	Sp3d

Answer:

Inter Halogen compound	Hybridisation
i) IF5	Sp3d2
ii) BeF3	Sp3d
iii) IF7	Sp3d3
iv) ClF	Sp3

4.

Compound	Structure
XeOF2	Linear
XeO3	Square planar
XeF	Pyramidal
XeOF4 I	Distorted octahedron
XeF4T	T shaped
XeP6	Square pyramidal

Answer:

Compound	Structure
XeOF2	T shaped
XeO3	Pyramidal
XeF	Linear
XeOF4	Square pyramidal
XeF4T	Square planar
XeP6	Distorted octahedron

II. Assertion and Reason

i) Both A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) Both A and R are correct, but R does not explain A.

1. Assertion (A) : Aqueous solution of potash Alum is acidic
Reason (R) : Aluminium sulphate undergo hydrolysis. (PTA – 2)
Answer:
i) Both A and R are correct, R is explanation of A

2. Assertion (A) : Elements belonging to group 16 are called chalcogens
Reason (R) : Group 16 elements are saltforming elements
Answer:
ii) A is correct, R is wrong
Correct Reason : Group 16 elements are ore forming elements

3. Assertion (A) : Among halogen acids, HF has low melting and boiling points
Reason (R) : In HF hydrogen bond is present.
Answer:
iii) A is wrong, R is correct.
Correct A : Among halogen acids HF has high melting and boiling points.

4. Assertion (A) : A small piece of Zinc dissolved in dilute nitric acid but hydrogen gas is not evolved. (PTA – 3)
Reason (R) : HNO₃ is an oxidising agent and this oxidizes hydrogen.
Answer:
ii) A is correct but R is wrong.

III. Pick out the Correct statement

1. i) Oxygen is diamagnetic
ii) Oxygen forms hydrogen bonds
iii) Oxygen exists in two allotropic forms
iv) Oxygen exists as a triatomic gas
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement: (i) Oxygen is paramagnetic (iv) Oxygen exists as a diatomic gas

2. i) Sulphur exists in crystalline as well as an amorphous form
ii) Rhombic sulphur has a characteristic yellow colour and composed of Sg molecules.
iii) When heated slowly above % C monoclinic sulphur is converted into Rhombic sulphur
iv) At around 140°C Rhombic sulphur melts to form mobile pale yellow liquid called X sulphur.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (iii) When heated slowly above 96°C, Rhombic sulphur is converted into monoclinic sulphur
(iv) At around 140°C the monoclinic sulphur melts to form mobile pale yellow liquid called X sulphur

3. i) H₂SO₄ is a dibasic acid
ii) H₃PO₃ is a tribasic acid
iii) H₃PO₄ is a dibasic acid
iv) H₃PO₂ is a monobasic acid
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement:
(ii) H₃PO₃ is a dibasic acid
(iii) H₃PO₄ is a tribasic acid

4. i) Krypton is used in cryogenics.
ii) Neon is used in high-speed electronic flashbulbs used by photographers.
iii) Helium is used to provide an inert atmosphere in electric arc welding of metals.
iv) Radon is used as a source of gamma rays.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement: (i) Helium is used in cryogenics
(ii) Xenon is used in high speed electronic flash bulbs used by photographers.

IV. Pick out the incorrect statement

1. i) In inter halogen compounds the central atom will be the smaller halogen
ii) Interhalogen compounds can be formed only between two halogen atoms.
iii) Flourine can act as a central atom.
iv) Interhalogens are strong oxidising agents
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
b) (i) & (iii)
Correct statements:
(i) In interhalogen compounds the central atom will be the larger halogen
(iii) Flourine cannot act as a central atom.

2. i) Nitrogen reacts with group 2 metals to form ionic nitrides.
ii) Ammonia is less soluble in water.
iii) Liquid nitrogen is used in biological preservation.
iv) In the conversion of metal oxides to metal ammonia acts as an oxidising agent.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (ii) & (iv)
Answer:
d) (ii) & (iv)
Correct statements:
(ii) Ammonia is extremely soluble in water.
(iv) In the conversion of metal oxides to metal ammonia acts as a reducing agent

3. i) When reacted with metals, nitric acid liberates hydrogen
ii) Chromium when reacted with nitric acid becomes passive due to the formation of nitrate on its surface.
iii) In most of the reactions nitric acid acts as an oxidising agent
iv) Fuming nitric acid contains oxides of nitrogen
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (i) When reacted with metals, nitric acid does not liberate hydrogen
(ii) Chromium when reacted with nitric acid becomes passive due to the formation of oxide on its surface

4. The rate of decomposition of ozone increases sharply in alkaline solution.
ii) In acidic solution ozone exceeds the oxidising power of fluorine and atomic oxygen
iii) Considerable amount of ozone is formed in the upper atmosphere by the action of UV light
iv) The shape of the Ozone molecule is linear
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement : (i) The rate of decomposition of ozone drops sharply in alkaline solution.
(iv) The shape of the ozone molecule is bent.

V. Pick out the odd man out

1. w.r.t oxidation number pick the odd man out.
a) HPO₃
b) H₃PO₃
c) H₃PO₄
d) H₄P₂O₇
Answer:
b) H₃PO₃ – O.N is +3 while in others the O.N of phosphorous is +5

2. w.r.t the reaction with sulphuric acid pick the odd man out
a) Gold
b) Silver
c) Platinum
d) Copper
Answer:
d) Copper – copper reacts with sulphuric acid while others do not

3. w.r.t reactivity pick the odd man out
a) F₂
b) Cl₂
c) Br₂
d) I₂
Answer:
a) F₂ – F₂ is more reactive than other halogens

4. w.r.t the ability to form oxoacids pick the odd man out
a) fluorine
b) chlorine
c) bromine
d) iodine
Answer:
a) Flour – Fluorine forms only one oxoacid where as other halogens form more than one oxoacid.

VI. Choose the best answer.

1. The principal gas present in atmosphere is
a) O₂
b) N₂
C) H₂
d) CO₂
Answer:
b) N₂

2. The basicity of hypophosphorous acid is (PTA – 2)
a) 1
b) 2
c) 3
d) 4
Answer:
a) 1

3. Chile salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) KNO₃
Answer:
b) NaNO₃

4. Indian salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) K.NO₃
Answer:
d) KNO₃

5. Inert character of nitrogen is due to its
a) high electronegativity
b) low electro negativity
c) high bonding energy
d) low bonding energv
Answer:
c) high bonding energy

6. Which of the following is caro’s acid?
a) H₂S₂O₈
b) H₂S₂O₇
c) H₂SO₃
d) H₂SO₃
Answer:
c H₂SO₅

7. Haber’s process is used for the synthesis of
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

8. The substance used in cryosurgery for producing low temperature is
a) liquid oxygen
b) liquid nitrogen
c) liquid hydrogen
d) liquid ammonia
Answer:
b) liquid nitrogen

9. Urea on hydrolysis gives
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

10. The catalyst used in Haber’s process is
a) Ni
b) Fe
c) Co
d) Pt
Answer:
b) Fe

11. The smell of ammonia is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
c) pungent

12. Like water, ammonia is a fairly good ionising solvent, because its dielectric constant is
a) considerably low
b) considerably high
c) equal to zero
d) equal to one
Answer:
b) considerably high

13. The process used for the manufacture of nitric acid is known as
a) Haber’s process
b) Deacon’s process
c) contact process
d) Ostwald’s process
Answer:
d) Ostwald’s process

14. With excess of chlorine, ammonia reacts to give an explosive substance
a) N₂
b) NH₄NO₃
c) NH₄Cl
d) NCl₃
Answer:
d) NCl₃

15. The deep blue colour compound formed when excess of ammonia is added to aqueous solution of copper sulphate is
a) [Cu(NO₃)₂]
b) [Cu(NH₃)₂]²⁺
c) [Cu(NH₃)₄]²⁺
d) [Cu(NH₃)₂]⁺
Answer:
c) [Cu(NH₃)₄]²⁺

16. The shape of ammonia molecule is
a) tetrahedral
b) pyramidal
c) square planar
d) octahedral
Answer:
b) pyramidal

17. The bond angle in ammonia is
a) 104°
b) 104°28′
c) 107°
d) 180°
Answer:
c) 107°

18. The colour of Pure nitric acid is
a) colourless
b) brown
c) pale green
d) green
Answer:
a) colourless

19. Fuming nitric acid contains oxides of
a) sulphur
b) hydrogen
c) nitrogen
d) carbon
Answer:
c) nitrogen

20. Nitric acid can act as
a) an acid
b) an oxidising agent
c) nitrating agent
d) all of the above
Answer:
d) all the above

21. The formula of hyponitrous acid is (MARCH 2020)
a) H₂N₂O₂
b) H₄N₂O₄
c) HOONO
d) HNO₂
Answer:
a) H₂N₂O₂

22. The oxidising power of oxo acids follows the order
a) HOX > HXO₂ > HXO₃ > HXO₄
b) HXO₄ > HXO₃ > HXO₂ > HOX
c) HXO₃ > HXO₄ > HXO₂ > HOX
d) HOX > HXO₄ > HXO₃ > HXO₂
Answer:
a) HOX > HXO₂ > HXO₃ > HXO₄

23. White phosphorous is kept under
a) kerosene
b) water
c) alcohol
d) ether
Answer:
b) water

24. White phosphorous becomes yellow phosphorous due to
a) hydrolysis
b) reduction
c) oxidation
d) displacement
Answer:
c) oxidation

25. In the conversion of yellow phosphorous into phosphine, phosphorous acts as
a) oxidising agent
b) reducing agent
c) catalyst
d) hydrolysing agent
Answer:
b) reducing agent

26. In the conversion of phosphorous into orthophosphoric acid, the catalyst used is
a) Cl₂
b) Br₂
c) I₂
d) F₂
Answer:
c) I₂

27. Which is used in match boxes?
a) White phosphorous
b) Red phosphorous
c) Black phosphorous
d) Scarlet phosphorous
Answer:
b) Red Phosphorous

28. The acid having O-O bond in its structure (PTA – 6)
a) H₂SO₃
b) H₂S₂O₆
c) H₂S₂O₈
d) H₂S₄O₆
Answer:
c) H₂S₂O₈

29. The smell of phosphine is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
b) rotten fish

30. The compounds used in Holme’s signal are
a) CaC₂ & Ca₃P₂
b) AlP & Ca₃P₂
c) CaC₂ & P₄
d) AlP & P₄
Answer:
a) CaC₂ & Ca₃P₂

31. The gases liberated in Holme’s signal are
a) C₂H₂ & CH₄
b) C₂H₂ & Ph₃
c) C₂H₄ & PH₃
d) CH₄ & Ph₃
Answer:
b) C₂H₂ & Ph₃

32. The formula of pyrophosphoric acid is
a) H₄P₂O₆
b) H₄P₂O₇
c) H₃PO₂
d) H₃PO₃
Answer:
b) H₄P₂O₇

33. Thermodynamically stable allotrophic form of sulphur is
a) Rhombic sulphur
b) Monoclinic sulphur
c) Plastic sulphur
d) Colloidal sulphur
Answer:
a) Rhombic sulphur

34. The gas found in volcanic eruptions is
a) NO₂
b) NO
c) SO₂
d) SO₃
Answer:
c) SO₂

35. The hybridisation of sulphur in SO₂ is
a) sp
b) sp²
c) sp³
d) dsp²
Answer:
b) sp²

36. The gas liberated when dilute sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
c) H₂

37. The gas liberated when cone, sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
a) SO₂

38. When sulphuric acid reacts with barium chloride solution, the white precipitate formed is
a) PbSO₄
b) BaSO₄
c) (CH₃COO)₂SO₄
d) PbCl₂
Answer:
b) BaSO₄

39. The halogen which exists as a liquid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
c) bromine

40. The halogen which exists as a solid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
d) iodine

41. Chlorine is manufactured by
a) Haber’s process
b) Deacon’s process
c) Contact process
d) Ostwald’s process
Answer:
b) Deacon’s process

42. The colour of chlorine gas is
a) colourless
b) brown
c) greenish yellow
d) pale green
Answer:
c) greenish yellow

43. Aqua regia is a mixture of cone. HCl and cone. HNO₃ in the ratio
a) 1 : 3
b) 3 : 1
c) 2 : 3
d) 3 : 2
Answer:
b) 3 :1

44. The halogen acid which forms hydrogen bond is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

45. Among halogen acids, the strongest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

46. Among halogen acids, the weakest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

47. Among halogen acids, the strongest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

48. Among halogen acids, the weakest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

49. The correct order of acid strength is
a) HF > HCl > HBr > HI
b) HF < HCl < HBr < HI
c) HF > HCl < HBr > HI
d) HF < HCl > HBr < HI
Answer:
b) HF < HCl < HBr < HI

50. Which is more reactive towards hydrogen?
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
a) flourine

51. The number of bond pair and lone pair of electrons present in the interhalogen compound BrF₃ is
a) 1 & 3
b) 3 & 2
C) 5 & 1
d) 7 & 0
Answer:
b) 3 & 2

52. The oxidation number of oxygen in F₂O is
a) -2
b) -1
c) +2
d) +1
Answer:
c) +2

53. The oxidation number of chlorine in Cl₂O₇ is
a) +1
b) +4
c) +6
d) +7
Answer:
d) +7

54. The strongest oxidising agent among the following is
a) chlorous acid
b) chloricacid
c) hypochlorous acid
d) perchloric acid
Answer:
c) hypochlorous acid

55. The first ionisation energy of noble gases is in the order
a) He < Ne < Ar < Kr
b) He > Ne > Ar > Kr
c) He < Ne > Ar < Kr
d) He > Ne < Ar > Kr
Answer:
b) He > Ne > Ar > Kr

56. Among noble gases, chemical reactivity is shown by
a) He & Ne
b) Ar & Kr
c) Kr & Xe
d) Xe & Rn
Answer:
c) Kr & Xe

57. Which among the following is used in cryogenics?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

58. Which is used for filling air balloons?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

59. Which is used in advertisement sign boards?
a) He
b) Ne
c) Ar
d) Kr
Answer:
b) Ne

60. Lamps used in airports as approaching lights is filled with
a) He
b) Ne
c) Ar
d) Kr
Answer:
d) Kr

VII. Two Mark Questions

Question 1.
How is pure nitrogen gas prepared?
Answer:
Pure nitrogen gas is prepared by the thermal decomposition of sodium azide at about 575 K
2NaN₃ \(\underrightarrow { 573K } \) 2Na + 3N₂

Question 2.
Nitrogen does not form any penta halides like phosphorus, why?
Answer:
Nitrogen does not form pentahalide although it exhibit +5 oxidation state. Due to the absence of d-orbitals.
It cannot undergo sp3d hybridization and hence cannot form pentahalides.

Question 3.
What is Haber’s process?
Answer:
The synthesis of ammonia from nitrogen and hydrogen at high pressure and optimum temperature in presence of iron catalyst is known as Haber’s process.

∆H_f = -46.2 Kjmol^-1

Question 4.
Write the uses of nitrogen
Answer:
Nitrogen is used
In the manufacture of ammonia, nitric acid, and calcium cyanamide etc.
Liquid nitrogen is used for producing low temperature required in cryosurgery and so used in biological preservation.

Question 5.
How is ammonia prepared in the laboratory?
Answer:
Ammonia is prepared in the laboratory by heating an ammonium salt with a base
2NH₄Cl + CaO → CaCl₂ + 2NH₃ + H₂O

Question 6.
Write about the reducing property of ammonia.
Answer:
When passed over heated metallic oxides Ammonia reduces metal oxides into metal
3PbO + 2NH₃ → 3Pb + N₂ + 3H₂O

Question 7.
What happens when ammonia reacts with excess of chlorine?
Answer:
With excess of chlorine ammonia reacts to give an explosive substance nitrogen trichloride
2NH₃ + 6Cl₂ → 2NO₃ + 6HO

Question 8.
On standing nitric acid becomes yellow in colour why?
Answer:

• Pure nitric acid is colourless
• Fuming nitric acid contains oxides of nitrogen
• It decomposes on exposure to sunlight or on being heated into nitrogendioxide, water and oxygen.
  4HNO₃ → 4NO₂ + 2H₂O + O₂
• Due to this reaction, pure nitric acid or its concentrated solution becomes yellow on standing

Question 9.
Prove that nitric acid is an oxidising agent.
Answer:
Non metals like carbon, sulphur are oxidised by nitric acid.
C + 4HNO₃ → CO₂ + 4NO₂ + 2H₂O
S + 2HNO₃ → H₂SO₄ + 2NO

Question 10.
Prove that nitric acid is a nitrating agent.
Answer:
Nitric acid replaces hydrogen atom from organic compounds with nitronium ion NO₂⁺. This is called nitration.

Question 11.
Write the uses of nitric acid is used
Answer:

• as an oxidising agent.
• in the preparation of aqua regia.
• Salts of nitric acid are used in photography (AgNO₃) and gunpowder for fire arms (NaNO₃)

Question 12.
How is nitrous oxide prepared?
Answer:
By Heating ammonium nitrate nitrous oxide is prepared
NH₄NO₃ → N₂O + 2H₂O

Question 13.
How is nitrous acid prepared?
Answer:
By treating nitrite salt with acids, nitrous acid is prepared
Ba(NO₂)₂ + H₂SO₄ → 2HNO₂ + BaSO₄

Question 14.
What is phosphorescence?
Answer:
White phosphorous undergoes spontaneous slow oxidation in air giving a greenish yellow glow which is visible in the dark. This is known as phosphorescence. The main product of this slow oxidation is P₂O₃.

Question 15.
How is phosphine prepared?
Answer:
Phosphine is prepared by the action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide

Question 16.
How is orthophosphoric acid prepared in ‘ the laboratory?
When phosphorous is treated with cone, nitric acid in the presence of iodine catalyst, it is oxidised to orthophosphoric acid.

Question 17.
Write the uses of phosphorous Phosphorous is used
Answer:

• in match boxes
• For the production of certain alloys such as phosphor bronze.

Question 18.
What happens when phosphine is heated in the absence of air?
Answer:
Phosphine decomposes into its elements when heated in the absence of air at 317 K
4PH₃ \(\underrightarrow { 317K } \) P₄ + 6H₂

Question 19.
Write about the reducing property of phosphine?
Answer:
Phosphine reduces silver nitrate into silver
PH₃ + 6AgNO₃ + 3H₂O → 6Ag + 6HNO₃ + H₃PO₃

Question 20.
How is phosphorous trichloride prepared?
Answer:

• When a slow stream of chlorine is passed over white phosphorous, PCl₃ is obtained.
• It is also prepared by treating white phosphorous with thionyl chloride.
  P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

Question 21.
Ozone (O₃) acts as a powerful oxidizing agent why? (PTA – 5)
Answer:
Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen.
Nascent oxygen, being a free radical, is very reactive
O₃ \(\underrightarrow { \triangle } \) O₂ + [O]

Question 22.
Write the uses of oxygen
Answer:

• Oxygen is one of the essential components for the survival of living organisms.
• Oxygen is used in oxyacetylene welding.
• Liquid oxygen is used as a rocket fuel.

Question 23.
How is sulphur dioxide prepared in the laboratory?
Answer:
SO₂ is prepared in the laboratory by treating a metal or metal sulphite with sulphuric acid
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
SO₃^2- + 2H⁺ → SO₂ + H₂O

Question 24.
Illustrate the oxidising property of SO₂.
Answer:
SO₂ oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
2H₂S + SO₂ → 3S + 2H₂O
2Mg + SO₂ → 2MgO + S

Question 25.
Write about contact process.
Answer:

• In contact process SO₂ is oxidised to SO₃
• It is used in the manufacture of sulphuric acid.
  

Question 26.
Write the uses of sulphurdioxide.
Answer:

• SO₂ is used in bleaching hair, silk, wool etc.
• SO₂ is used for disinfecting crops and plants in agriculture

Question 27.
Write about the structure of sulphr dioxide.
Answer:

• Sulphur undergoes sp² hybridisation.
• A double bond arises between S and O due to pπ – dπ overlapping

Question 28.
Illustrate the dehydrating property of sulphuric acid.
Answer:

Question 29.
Show that sulphuric acid is a dibasic acid.
Answer:
H₂SO₄ forms two types of salts with NaOH Hence it is dibasic.

Question 30.
How is chlorine is manufactured by Deacon’s process?
Answer:

• A mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, containing pumice stones soaked in cuprous chloride.
• Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
  
• Chlorine obtained is dilute and used for the manufacture of bleaching powder

Question 31.
Write about the bleaching action of chlorine.
Answer:
Chlorine is a strong oxidising and bleaching agent since it produces nascent oxygen.
H₂O + Cl₂ → HCl + HOCl (Hypochlorous acid)
HOCl → HCl +[0]
Colouring matter + Nascent oxygen → Colourless oxidation product.

The bleaching of chlorine is permanent.

Question 32.
How is bleaching powder prepared? (MARCH 2020)
Answer:
Bleaching powder is prepared by passing chlorine gas through dry slaked lime (calcium hydroxide)
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 33.
Write the uses of chlorine Chlorine is used
Answer:

• In the purification of drinking water.
• In the bleaching of cotton textiles, paper, and rayon.
• In the extraction of gold and platinum.

Question 34.
How is hydrochloric and prepared in the laboratory?
Answer:
Hydrochloric add is prepared by the action of sodium chloride and cone, sulphuric acid
NaCl + H₂SO₄ → NaHSO₄. + HCl
NaHSO₄ + NaCl → Na₂SO₄. + HCl
Dry hydrochloric acid is obtained by passing the gas through cone, sulphuric acid

Question 35.
How is xenon trioxide prepared?
Answer:
2XeF₆ + SiO₂ \(\underrightarrow { 50°C } \) 2XeOF₄ + SiF₄
2XeOF₄ + SiO₂ → 2XeO₂F₂ + SiF₄
2XeO₂F₂ + SiO₂ → 2XeO₃ + SiF₄
(or)
XeF₆ + 3H₂ → XeO₃ + 6HF

Question 36.
How is sodium per xenate obtained?
Answer:
When XeF₆ reacts with 2.5 M NaOH, sodium per xenate is obtained.
2XeF₆ + 16NaOH → Na₄XeO₆ + Xe + O₂ + 12NaF + 8H₂O

Question 37.
Show that sodium per xenate is a strong oxidising agent
Answer:
Sodium per xenate oxidises manganese (II) ion into permanganate ion even in the
absence of a catalyst
5XeO^4-₆ + 2Mn²⁺ + 14H⁺ → 2MnO^–₄ + 5XeO₃ + 7H₂O

Question 38.
Give reason: ICl is more reactive than l₂ (PTA – 3)
Answer:

• This is because inter-halogen compounds are in general more reactive than halogens due to w eaker inter-halogen X-X bond than X-X bond.
• So, I₂ is more stable and less reactive than ICl.

VIII. Three Mark Questions

Question 1.
Write about the structure of ammonia.
Answer:

• Ammonia molecule is pyramidal in shape.
• Hybridisation of nitrogen is sp³.
• The shape must be tetrahedral but in one of the tetrahedral positions a lone pair of electrons from the nitrogen atom is present, hence it is pyramidal.
• The N – H bond distance is 1.016 A0.
• The H – H bond distance is 1.645 A°.
• The bond angle is 107°
  

Question 2.
How does red phosphorous react with oxygen?
Answer:
Red phosphorous reacts with oxygen on heating to give phosphorous trioxide and phosphorous pentoxide.
P₄ + 3O₂ → P₄O₆
P₄ + 5O₂ → P₄O₁₀

Question 3.
How is pure phosphine prepared?
Answer:
Pure phosphine is prepared by heating phosphorous acid

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 4.
What happens when phosphine is heated with air?
Answer:
When phosphine is heated with air or oxygen, it undergoes combustion to give meta phosphoric acid.

Question 5.
Write about Holmes signal
Answer:

• In a ship during distress, a container with calcium carbide and calcium phosphide mixture is pierced and thrown into the sea.
• The mixture reacts with seawater liberating acetylene and phosphine gases.
• The liberated phosphine catches fire and ignites acetylene.
• These burning gases with lot of smoke serves as a signal to the approaching ships.
• This is known as Holme’s signal.

Question 6.
Write about the structure of phosphine
Answer:

• Phosphorous shows sp³ hybridisation.
• Three orbitals are occupied by bond pair electrons.
• Fourth orbital is occupied by lone pair of electrons.
• Hence instead of tetrahedral, PH₃ has a pyramidal shape.
• Bond angle is reduced to 94°
  

Question 7.
How is oxygen prepared in the laboratory?
Answer:
Oxygen is prepared in the laboratory by the decomposition of hydrogen peroxide in presence of Mn02 catalyst or by the oxidation of potassium permanganate.

5H2O₂ + 2MnO₄^– + 6H⁺ → 5O₂ + 8H₂O + 2Mn²⁺

Question 8.
Write about ozone
Answer:

• Ozone is an allotropic form of oxygen
• Ozone is triatomic gas
• Although negligible amounts of ozone occurs at sea level, it is formed in the upper atmosphere by the action of UV light.
• In the laboratory ozone is prepared by passing electrical discharge through oxygen.
• At a potential of 20,000 V about 10% of oxygen is converted into ozone, it gives a mixture known as ozonised oxygen.
• Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.
  
• Ozone molecule has a bent shape and symmetrical with delocalised bonding between the oxygen atoms.

Question 9.
Write about the reducing property of sulphur dioxide
Answer:

SO₂ reduces chlorine into hydrochloric acid
SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl

SO₂ reduces potassium permanganate into manganese sulphate (Mn²⁺).
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

SO₂ reduces potassium dichromate into chromic sulphate (Cr³⁺)
K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 10.
Write about the bleaching action of sulphur dioxide.
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property

When the bleached product (Colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour.

Hence bleaching action of sulphurdioxide is temporary

Question 11.
Explain the manufacture of sulphuric acid
Answer:

• Sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen / air
  S + O₂ → SO₂
  4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
• SO₂ is oxidised to SO₃ by air in presence of V₂O₅ or platinised absestos
  
• SO₃ is absorbed in cone H₂SO₄ and oleum is produced
  SO₃ + H₂SO₄ H₂S₂O₇
• Oleum is converted into sulphuric acid by diluting it with water.
  H₂S₂O₇ + H₂O → 2H₂SO₄
• To maximize the yield the plant is operated at 2 bar pressure and 720 K.
  96% pure H₂SO₄ is obtained.

Question 12.
Show that sulphuric acid is an oxidising agent
Answer:
Sulphuric acid is an oxidising agent as it produces nascent oxygen
H₂SO₄ → H₂O + SO₂ + [O] (Nascert oxygen)

Sulphuric acid oxidises carbon into carbon dioxide
C + 2H₂SO₄ → 2SO₂ + 2H₂O + CO₂

Sulphuric acid oxidises phosphorous into orthophosphoric acid
P₄ + 10H₂SO₄ → 4H₃PO₄ + 10SO₂ + 4H₂0

Sulphuric acid oxidises iodide into iodine.
H₂SO₄ + 2HI → 2SO₂ + 2H₂O + I₂

Question 13.
What is the action of sulphuric acid on metals?
Answer:
Dilute sulphuric acid reacts with metals liberating hydrogen gas.
Zn + H₂SO₄ ZnSO₄ + H₂ ↑

Hot cone. Sulphuric acid reacts with metals to give sulphates and sulphur dioxide
Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂ ↑

Sulphuric acid does not react with noble metals like gold, silver and platinum.

Question 14.
Give the test for sulphate / sulphuric acid
Dilute solution of sulphuric acid / Sulphates react with barium chloride or lead acetate solution to give a white precipitate

Question 15.
How is chlorine manufactured by the electrolytic process?
Answer:
When brine solution (NaCl) is electrolyzed, Na⁺ and Cl^– ions are formed.
Na⁺ ions react with OH^– ions of water forming sodium hydroxide.
Hydrogen and chlorine are liberated as gases.
NaCl → Na+ +Cl^–
H₂O → H+ +OH^–
Na⁺ + OH^– → NaOH
At cathode : H⁺ + e^– → H
H + H → H₂
At anode : Cl^– → Cl + e^–
Cl + Cl → Cl₂

Question 16.
What is aqua regia? What is its action on gold?
Answer:
Aqua regia is a mixture of three parts of cone, hydrochloric acid and one part of cone, nitric acid.
This is used for dissolving gold, platinum etc.
AU + 4H⁺ + NO₃^– + 4Cl^– → AuCl₄^– +NO + 2H₂O

Question 17.
HF acid is a weaker acid at low concentration, but becomes stronger as the concentration increases why?
Answer:
0.1 M Solution HF is 10% ionised, hence it is a weak acid.
But 5 M, 15 M solution of HF is stronger due to the equilibrium.

At high concentration, the equilibrium involves the removal of flouride ions and increases the hydrogen ion concentration, HF becomes stronger acid as the concentration increases.

Question 18.
HF acid is not stored in glass bottles why? (MARCH 2020)
Answer:
HF attacks silica and silicates present in glass bottles.

Hence HF is not stored in glass bottles. But HF is stored in Teflon bottles.
SiO₂ + 4HF → SiF₄ + 2H₂O
Na₂SiO₃ + 6HF → Na₂SiF₆ + 3H₂O

Question 19.
Mention the characteristic of interhalogen compounds (PTA – 2)
Answer:

• The central atom will be the larger halogen.
• It can be formed only between two halogens and not more than two halogens.
• Fluorine can’t act as a central atom because it is the smallest among halogens and highly electronegative.
• They are strong oxidising agents and undergo auto ionization.

Question 20.
Give the preparation of xenon fluorides
Answer:

Question 21.
What is the hybridisation in XeOF₂? Give its structure. (PTA – 1)
Answer:

Element	hybridisation	structure
XeOF2	Sp3d	T-shaped

IX. Five Mark Questions

Question 1.
How is nitric acid manufactured using Ostwald’s process?

• Ammonia prepared by Haber’s process is mixed about 10 times of air.
• This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze.
• The temperature rises about 1275 K.
• The metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO.
• NO is oxidized to NO₂
  4NH₃ +5O₂ → 4NO + 6H₂O +120 KJ
  2NO + O₂ → 2NO₂
• NO₂ produced is passed through a series of adsorption towers.
• NO₂ reacts with water to give nitric acid.
• Nitric acid formed is bleached by blowing air.
  6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 2.
Explain the action of nitric acid on metals with one example.
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen.
3Cu + 6HNO₃ → 3CU(NO₃)₂ + 6(H)

Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid
6(H) + 3HNO₃ → 3HNO₂ + 3H₂O

Tertiary reaction:
With dilute acid, the secondary products decompose to give final products.
3HNO₂ → HNO₃ + 2NO + H₂O
Hence overall reaction is
3Cu + 8HNO₃ → 3CU(NO₃)₂ + 2NO + 4H₂O

With concentrated acid the secondary products react to give the final products.
HNO₂ + HNO₃ → 2NO₂ + H₂O
Hence overall reaction is
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 3.
Write the preparation of nitrogen oxides.
Answer:

Question 4.
Write the preparation of oxoacids of nitrogen
Answer:

Question 5.
Explain the structure of oxides of phosphorus
Answer:
Phosphorus trioxide :

• In P₄O₆, four phosphorous atoms lie at the corners of a tetrahedron and six oxygen atoms along the edges.
  The P – O bond distance is 165.6 pm which is shorter than the single bond distance of the P-O bond (184 pm)
• This is due to Pπ – dπ bonding
• This results in the considerable double bond character
  

Phosphorous Pentoxide:

• In P₄O₁₀ each P atom forms a single bond with three oxygen atoms and a coordinate bond with one oxygen atom.
• Terminal coordinate P-O bond length is 143 pm
• This is less than the expected single bond distance
• This may be due to lateral overlap of filled
  p – Orbitals of an oxygen atom with empty
  d – Orbital on phosphorous.

Question 6.
Write the structure of and basicity oxoacids of phosphorous. (PTA – 3)

Answer:

Question 7.
Write the preparation of oxoacids of phosphorous
Answer:

Question 8.
Write the structure of oxo acids of sulphur.
Answer:

Question 9.
List any five compounds of Xenon and mention the type of hybridization and structure of the compounds (PTA – 6)
Answer:

Compound	Hybridisation	shape/structure
1. XeP2	sp3d	Linear
2. XeF4	sp3d2	Square planar
3. XeF6	sp3d3	Distorted octahedron
4. XeOF2	sp3d	T-shaped
5. XeOF4	sp3d2	Square pyramidal
6. XeO3	sp3	Pyramidal

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 2 p-Block Elements – I Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

12th Chemistry Guide p-Block Elements – I Text Book Questions and Answers

I. Choose the qorrect answer

1. An aqueous solution of borax is __________ .
a) neutral
b) acidic
c) basic
d) amphoteric
Answer:
c) basic

2. Boric acid is an acid because its molecule (NEET)
a) contains replaceable H⁺ ion
b) gives up a proton
c) combines with proton to form water molecule
d) accepts OH^– from water, releasing proton
Answer:
d) accepts OH^– from water, releasing proton

3. Which among the following is not a borane?
a) B₂H₆
b) B₃H₆
c) B₄H₁₀
d) none of these
Answer:
b) B₃H₆

4. Which of the following metals has the largest abundance in the earth’s crust?
a) Aluminium
b) Calcium
b) Magnesium
d) Sodium
Answer:
a) Aluminium

5. In diborane, the number of electrons that accounts for banana bonds is
a) six
b) two
c) four
d) three
Answer:
c) four

6. The element that does not show catenation among the following p-block elements is
a) Carbon
b) Silicon
c) Lead
d) germanium
Answer:
c) Lead

7. Carbon atoms in fullerene with formula C60 have
a) sp³ hybridised
b) sp hybridised
c) sp² hybridised
d) partially sp² and partially sp³ hybridised
Answer:
c) sp2 hybridised

8. Oxidation state of carbon in its hybrides
a) +4
b) -4
c) +3
d) +2
Answer:
a) +4

9. The basic structural unit of silicates is (NEET) (PTA – 1)
a) (SiO₃)^2-
b) (SiO₄)^2-
c) (SiO)^–
d) (SiO₄)^4-
Answer:
d) (SiO₄)^4-

10. The repeating unit in silicone is

11. Which of these is not a monomer for a high molecular mass silicone polymer?
a) Me₃SiCl
b) PhSiCl₃
c) MeSiCl₃
d) Me₂SiCl₂
Answer:
a) Me₃SiCl

12. Which of the following is not sp² hybridised?
a) Graphite
b) graphene
c) Fullerene
d) dry ice
Answer:
d) dry ice

13. The geometry at which carbon atom in diamond are bonded to each other is
a) Tetrahedral
b) hexagonal
c) Octahedral
d) None of these
Answer:
a) Tetrahedral

14. Which of the following statements is not correct?
a) Beryl is a cylic silicate
b) Mg₂SiO₄ is an orthosilicate
c) SiO₄^4- is the basic structural unit of silicates
d) Feldspar is not aluminosilicate
Answer:
d) Feldspar is not aluminosilicate

15. Match items in Column-I with the items of Column-II and assign the correct code.

Answer:
a) 2 1 4 3

16. Duralumin is an alloy of
a) Cu, Mn
b) Cu, AZ, Mg
c) AZ, Mn
d) AZ, Cu, Mn, Mg
Answer:
d) Al, Cu, Mn, Mg

17. The compound that is used in nuclear reactors as protective shields and control rods is
a) Metal borides
b) Metal oxides
c) Metal carbonates
d) Metal carbide
Answer:
a) Metal borides

18. The stability of +1 oxidation state increases in the sequence
a) AZ < Ga < In < TZ
b) TZ < In < Ga < Al
c) In < TZ < Ga < Al
d) Ga < In < AZ < TZ
Answer:
a) Al< Ga < In < TZ

II. Answer the following questions

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
The following factors are resposible for the anamolous properties of the first elements of p-blick.
1. Small size of the first member
2. High ionisation enthalpy and high electronegativity.
3. Absence of d-orbitals in their valence shell.

First elemenl	Property of First elements	Other elements in the family
B	Mettaloid	Metals
C	1. Non-metal 2. It can form multiple bonds.	1. Metalloids – Si and Ge. 2. Other elements are metals. 3. It can’t form multiple bonds.
N	1. Non metal 2. It can form multiple bonds 3. Diamagnetic	1. Non metal – “P” Metalloids – As. Sb. 2. It cann’t form multiple bonds
O	1. Non metal and diatomic gas 2. It forms H-bonds	1. S, Se – non metals. 2. Te- metalloid and others are metals.
F	1. Non-metals 2. High electro­ negativity 3. Highly reactive.	1. Non-metals 2. Low reactive than ‘F’

Question 2.
Describe briefly allotropiam in p-block elements with specific reference to carbon.
Answer:

• Some elements exist in more than one crystalline or molecular forms in the same physical state.
• This phenomenon is called allotropism.
• The different forms of an element are called allotropes.
• Example: Carbon exists as diamond, graphite, graphene, fullerenes, carbon nanotubes

Question 3.
Give the uses of Borax.
Answer:

1. Borax is used for the identification of coloured metal ions.
2. In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
3. It is also used as a flux in metallurgy and also acts as a good preservative.

Question 4.
What is catenation? Describe briefly the catenation property of carbon. (MARCH 2020)
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to the maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

1. The valency of element is greater than or equal to two.
2. Element should have the ability to bond with itself.
3. The self-bond must be as strong as its bond with other elements.
4. Kinetic inertness of catenated compound towards other molecules.
5. Carbon possesses all the above properties and forms a wide range of compounds with itself.

Question 5.
Write a note on Fisher Tropsch synthesis. Fischer Tropsch synthesis: (PTA – 4)
Answer:
This is a reaction in which carbon monoxide reacts with hydrogen at a pressure less than 50 atm and temperature 500 – 700 K in presence of metal catalysts to give saturated and unsaturated hydrocarbons.
n CO + (2n+l) H₂ → CnH_2n+2 + nH₂O
n CO + 2n H₂ → C_nH_2n + nH₂O

Question 6.
Give the structure of CO and CO₂.
Answer:
Structure of CO:
Structure is linear.

Structure of CO₂:
Structure is linear.

Question 7.
Give the uses of silicones.
Answer:

1. Used for low temperature lubrication.
2. Used in vacuum pumps.
3. Used in high temperature oil baths.
4. Used for making water proof cloths.
5. Used as insulating material in electrical motor and other applicances.
6. Mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals

Question 8.
Describe the structure of diborane. (PTA – 3)
Answer:

• In diborane two BH₂ units are linked by two bridged hydrogens, rherefore it has eight B-H bonds.
• Diborane has only 12 valence electrons anc are- not sufficient to form normal covalen bonds.
• The four terminal B-H bonds are norma covalent bonds. (2c 2e bond) (Totally 8e-s)
• The remaining four electrons have to be used for the bridged bonds, ie two 3 centred B-H-B bonds utilise two electrons each.
• Hence these bonds are 3c – 2e bonds.
  
• The bridging hydrogen atoms are in a plane.
• In diborane, boron is sp³ hybridised.
• Three sp³ hybridised orbitals contain single electron and the fourth orbital is empty.
• Two half filled sp³ hybridised orbitals of each boron overlap with two hydrogens to form four terminal 2C – 2e bonds.
• One empty and one half filled sp³ hybridised orbital on each boron is left.
• Empty sp³ hybridised orbital of one boron, overlaps with half filled sp³ hybridised orbital of the other boron and Is orbital of hydrogen to form two bridged 3C – 2e B-l 1-B bonds.

Question 9.
Write a short note on hydroboration.
Answer:

• Diborane adds on to alkenes and alkynes in ether solvent at room temperature.
• This reaction is known as hydroboration.
• This is used in synthetic organic chemistry especially for anti Markovnikov addition.
  B₂H₆ + 6 RCH = CHR → 2B (CH₂-CH₂ R)₃

Question 10.
Give one example for each of the following:
Answer:
i) icosogens
ii) tetragen
iii) pnictogen
iv) chalcogen

Group Name	Example
i. Icosagens	Boron
ii. Tetragens	Carbon
iii. Pnictogen	Nitrogen
iv. Chalcogens	Oxy gen

Question 11.
Write a note on metallic nature of p-block elements.
Answer:

• The tendency of an element to form a cation by losing electrons is known as electro positive or metallic character.
• This character depends on the ionisation energy.
• Generally on moving down a group ionisation energy decreases and hence the metallic character increases.
• In p-block, the elements present in lower left part are metals, while the elements in the upper right part are non metals.

Group	Non-metals	Metalloids	Metals
13	–	B	Al, Ga, In, Tl
14	C	Si, Ge	Sn, Pb
15	N, P	As, Sb	Bi
16	O, S, Se	Te, Po	–
17	F, Cl, Br, I	–	–
18	He, Ne, Ar, Kr, Xe	–	–

Question 12.
Complete the following reactions:
a) B(OH)₃ + NH₃ →
b) Na₂B₄O₇ + H₂SO₄ + H₂O →
c) B₂H₆ + 2NaOH + 2H₂O →
d) B₂H₆ + CH₃OH →
e) BF₃ + 9H₂O →
f) HCOOH+ H₂SO₄ →
g) SiCl₄ + NH₃ →
h) SiCl₄ + C₂H₅OH →
I) B + NaOH →
j) H₂B₄O₇ \(\underrightarrow { Red\quad hot } \)
Answer:

Question 13.
How will you identify borate radical? (PTA – 5)
Answer:

• When boric acid or borate salt is heated with ethyl alcohol in presence of cone, sulphuric acid, an ester triaikvl borate is formed.
• The vapour of this ester bums with a green edged flame.
• This is ethyl borate test to identify borate radical,
  
  B(OC₂H₅)₃ Ethyl borate (Green edged flame)

Question 14.
Write a note on zeolites. ( PTA – 2)
Answer:

• Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxvgen in their regular three dimensional frame work.
• They are hydrated sodium alumino silicates.
• General formula is
  Na₂O.(Al₂O₃).x(SiO₂).y(H₂O)
  where x = 2 to 10; y = 2 to 6
• Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
• Si and AI atoms are tetrahedrally coordinated with each other through shared oxygen atoms.
• Zeolites are similar to Clay minerals but they differ in their crystalline structure.
• Zeolites have a three dimensional crystalline structure looks like a honey comb consisting of a network of interconnected tunnels and cages.
• Water molecules move freely In and out of these pores but the zeolite frame work remains rigid.
• Another special aspect of this structure is that the pore/channel sizes are nearly uniform, allowing the crystal to act as a molecular sieve.
• Zeolites are used in the removal of permanent hardness of water.

Question 15.
How will you convert boric acid to boron nitride? (PTA – 3)
Answer:
Fusion of urea with boric acid in an atmosphere of ammonia at 800 -1200 K gives boron nitride.

Question 16.
A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducting agent (C). Identify (A), (B) and ( C) (PTA – 1)
Answer:
A hydride of 2nd period alkali metal (A) is LiH

Lithium hydride reacts with compound of boron (B) B₂H₆ to give reducing agent (C) lithium boro hydride.
∴ Compound B is diborane
Compound C is lithium boro hydride.

Question 17.
A double salt which contains fourth period alkali metal (A) on heating at 500 K gives (B). Aqueous solution of (B) gives white precipitate with BaCl₂ and gives a red colour compound with alizarin. Identify (A) and (B).
Answer:

•  A double salt which contains fourth period alkali metal (A) is Potash alum
  K₂SO₄. Al₂ (SO₄)₃.24H₂O
• (A) on heating at 500 K gives
  K₂SO₄.Al₂(SO₄)₃ (B) which is burnt alum.

Question 18.
CO is a reducing agent, justify with an example.
Answer:

• CO is a strong reducing agent.
• It reduces metallic oxides inlo melais.
  Example : 3CO + Fe₂CO₃ → 2Fe + 3CO₂

III. Evaluate Yourself

Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuraton of group 18 elements.
Answer:

• These elements are gases.
• Their outer electronic configuration is ns²np⁶ which is stable completely filled configuration.
• So they are more stable and least reactive.
• Hence they are called inert gases.

12th Chemistry Guide Chapter 2 p-Block Elements – I Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. The general electronic configuration of p-block elements is
a) ns¹
b) ns²
c) ns² np^1-6
d) (n-1)s² np^1-6
Answer:
c) ns² np^1-6

2. p-block element consists of the groups
a) 1 & 2
b) 3 – 12
c) 13 – 17
d) 13 – 18
Answer:
d) 13 – 18

3. Group 18 elements are inert because of their
a) unstable incompletely filled orbitals
b) stable completely filled orbitals
c) half filled orbitals
d) stable nucleus
Answer:
b) stable completely filled orbitals

4. As we go down the group ionisation energy
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
a) decreases

5. As we go down the group metallic character
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

6. As ionisation energy decreases, the metallic character of elements
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

7. In p-block, metals are placed in
a) upper right part
b) middle part
c) lower left part
d) top of the group
Answer:
c) lower left part

8. In p-block, non-metals are placed in
a) upper right part
b) middle part
c) lower left part
d) bottom of the group
Answer:
a) upper right part

9. Which of the following factor is not responsible for the anamolous behaviour of the first member of each group in p-block elements?
a) small size
b) high ionisation enthalpy
c) outer electronic configuration
d) absence of d-orbitals
Answer:
c) outer electronic configuration

10. The correct order of catenation property in group 14 elements is
a) C << Si < Ge = Sn < Pb
b) C >> Si > Ge = Sn > Pb
c) C >> Si < Ge = Sn < Pb
d) C << Si » Ge = Sn > Pb
Answer:
b) C >> Si > Ge = Sn > Pb

11. The elements N, O, F readily forms hydrogen bonds due to their high
a) ionisation energy
b) electron affinity
c) electro negativity
d) atomic radius
Answer:
c) electro negativity

12. The most electro negative element is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

13. The element with maximum electron affinity is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
b) Chlorine

14. The most reactive element among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

15. The strongest oxidising agent among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

16. The important property shown by p-block elements is
a) complex formation
b) coloured ion formation
c) inert pair effect
d) metallic character
Answer:
c) inert pair effect

17. In 13th group Tl⁺¹ ion is more stable than Tl³⁺ ion due to
a) high electronegatively
b) inert pair effect
c) high ionisation energy
d) stable electronic configuration
Answer:
b) inert pair effect

18. Diamond and graphite are ______ of carbon.
a) Isotopes
b) Isobars
c) Isomers
d) Allotropes
Answer:
d) Allotropes

19. The formula of Borax is
i) Na₂B₄O₇.10H₂O
ii) Na₂[B₄O₅(OH)₄].8H₂O
iii) Na₂[B₄O₅(OH)₄].2H₂O
a) (i) only
b) (i) & (ii) only
c) (i) & (iii) only
d) (iii) only
Answer:
b) (i) & (ii) only

20. Ortho boric acid on dehydration at 373K produces mainly (PTA – 3)
a) metaboric acid
b) boric anhydride
c) Boron metal and Oxygen
d) tetra boric acid
Answer:
a) metaboric acid

21. The formula of colemanite is
a) Na₂B₄O₇
b) Na₂B₄O₇.10H₂O
c) Ca₂B₆O₁₁
d) NaBO₂
Answer:
c) Ca₂B₆O₁₁

22. Which is used as moderator in nuclear reactors?
a) boron nitride
b) boron
c) borax
d) boric acid
Answer:
b) boron

23. The compound used in eye drops and antiseptics is
a) boron nitride
b) boric acid
c) sodium meta borate
d) boron tri oxide
Answer:
b) boric acid

24. The compound used as a flux in metallurgy is
a) boron nitride
b) boric acid
c) borax
d) boron tri oxide
Answer:
c) borax

25. Boric acid on heating at 413 K gives
a) meta boric acid
b) tetra boric acid
c) boric anhydride
d) borax
Answer:
b) tetra boric acid

26. In ethyl borate test the colour of the flame obtained is
a) red
b) yellow
c) blue
d) green
Answer:
d) green

27. On hydrolysis BF3 gives Boric acid and converted to fluroboric acid. The fluoroboric acid contains the species. (PTA – 6)
a) H⁺, F^– & BF₃
b) H⁺ & [BF₄]
c) [H BF₃]⁺ & F^–
d) H⁺, B³⁺ & F^–
Answer:
b) H⁺ & [BF₄]

28. In organic benzene is
a) diborane
b) borazole
c) borax
d) boric acid
Answer:
b) borazole

29. The formula of Inorganic benzene is
a) B₃N₃
b) B₃N₃H₃
c) B₃N₃H₆
d)B₆N₆H₆
Answer:
c) B₃N₃H₆

30. The most stable form of carbon is
a) graphite
b) diamond
c) fullerene
d) carbon nano tubes
Answer:
a) graphite

31. The formula of buckminster fullerene is
a) C₃₂
b) C₅₀
e) C₆₀
d) C₇₀
Answer:
c) C₆₀

32. The number of six membered and five membered rings fused together respectively in buckminster fullerene is
a) 12 & 20
b) 20 & 12
c) 10 & 22
d) 22 & 10
Answer:
b) 20 & 12

33. Water gas is a mixture of
a) CO₂ + H₂
b) CO + H₂O
c) CO + H₂
d) CO + N₂
Answer:
c) CO + H₂

34. Producer gas is a mixture of
a) CO₂ + H₂
b) CO + H₂O
c) CO + H₂
d) CO + N₂
Answer:
d) CO + N₂

35. In the presence of light carbon monoxide reacts with chlorine to form a poisonous gas called
a) mustard gas
b) phosgene
c) phosphine
d) carbylamine
Answer:
b) phosgene

36. Fischer Tropsch synthesis is used for preparing
a) Silicones
b) Boranes
c) Hydrocarbons
d) Carbonyls
Answer:
c) Hydrocarbons

37. In metal carbonyls the oxidation state of metals is
a) 0
b) +1
c) +2
d) +3
Answer:
a) 0

38. The structure of CO molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

39. The structure of CO₂ molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

40. The critical temperature of CO₂ is
a) 21 °C
b) 31°C
c) 12°C
d) 13°C
Answer:
b) 31°C

41. When CO₂ is dissolved in water, the solution is slightly
a) acidic
b) basic
c) amphoteric
d) neutral
Answer:
a) acidic

42. Which among the following is important for photo synthesis?
a) O₂
b) N₂
C) CO
d) CO₂
Answer:
d) CO₂

43. The water repellant property of silicones is due to the presence of
a) -OH group
b) -Si group
c) -R group
d) -Cl group
Answer:
c) -R group

44. The percentage of silicate minerals and silica present in earth’s crust is
a) 75
b) 85
c) 95
d) 100
Answer:
c) 95

45. The basic unit present in silicates is
a) SiO₂
b) [SiO₃]^–
c) [SiO₄]^2-
d) [SiO₄]^4-
Answer:
d) [SiO₄]^4-

46. Talc is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
b) Phyllo silicates

47. Quartz is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
c) Tecto silicates

48. The formula of Spodumene is
a) Sc₂Si₂O₇
b) Li Ai(SiO₃)₂
c) [Be₃ Al₂(SiO₃)₆]
d) Be₂SiO₄
Answer:
b) Li Ai(SiO₃)₂

49. The silicate which is used in the removal of permanent hardness of water is
a) Feldspar
b) Quartz
c) Zeolites
d) Talc
Answer:
c) Zeolites

50. Thermodynamically the most stable form of carbon is (PTA – 4)
a) Diamond
b) Fullerenes
c) graphite
d) Nano tubes
Answer:
c) graphite

II. Pick the odd man out

1. W.r.t. their metallic character pick the odd man out.
a) Ge
b) Ga
c) B
d) As
Answer:
b) Ga – It is a metal while others are metalloids

2. W.r.t. their metallic character pick the odd man out
a) In
b)Pb
c) Cl
d) Bi
Answer:
c) Cl – It is a non metal while others are metals.

3. Pick the odd man out
a) Borax
b) Kernite
c) Colemanite
d) Bauxite
Answer:
d) Bauxite – It is an ore of aluminium others are ores of boron.

4. W.r.t. to hybridisation pick the odd man out.
a) Graphite
b) Diamond
c) Fullerene
d) Graphene
Answer:
b. Diamond – It is sp³ hybridised while others are sp² hybridised.

III. Assertion and Reason

i) Both A and R are correct, R explains A
ii) A is wrong but R is wrong
iii) A is wrong but R is correct
iv) Both A and R are correct but R does not explain A

1. Assertion (A) : Boron shows non metallic character.
Reason (R) : Atomic radius of boron is small and its nuclear charge is high.
Answer:
(i).Both A and R are correct, R explains A

2. Assertion (A) : As we move down Boron group the elements show less tendency to exhibit +1 oxidation state rather than +3. Reason (R) : As we move down Boron group the elements show inert pair effect.
Ans : (iii).A is wrong but R is correct

3. Assertion (A) : Graphite conducts electricity.
Reason (R) : In Graphite, successive carbon sheets are held together by weak Vander Waals force.
Answer:
(iv). Both A and R are corrrect but R does not explain A

4. Assertion (A) : Silicones are used for making water proofing clothes.
Reason (R) : In silicones the organic side groups which surrounds silicon make the molecule looks like an alkane.
Answer:
(i).Both A and R are correct, R explains A

IV. Choose the correct statement

1. i) Some of the p-block elements show negative oxidation states also.
ii) Halogens gain two electrons to give a stable halide ion.
iii) Inert gases have ns²np⁶ configuration and hence more stable.
iv) p-block elements have a general electronic configuration (n-1)s² np^1-6
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correction:
ii) Halogens gain one electron to give a stable halide ion.
iv) p-block elements have a general electronic configuration ns² np^1-6

2. i) Boron compounds are electron rich compounds.
ii) Boron does not react directly with hydrogen.
iii) Borax is sodium salt of metaboric acid.
iv) Boric acid is used as an antiseptic,
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
c) (ii) & (iv)
Correction:
i) Boron compounds are electron deficient compounds.
iii) Borax is sodium salt of tetraboric acid.

3. i) In graphite carbon atoms are sp³ hybridised.
ii) A single planar sheet of graphite is known as graphene.
iii) In diamond each carbon atom is tetrahedrally surrounded by four other carbon atoms.
iv) Carbon nanotubes do not conduct electricity,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correction:
i) In graphite carbon atoms are sp² hybridised.
iv) Carbon nanotubes conduct electricity.

4. i) Silicones are organo silicon polymers.
ii) Hydrolysis of R₂SiCl₂ yields complex cross linked polymer.
iii) Silicones are good thermal and electrical conductors.
iv) All silicones are water repellent,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d)(i) & (iv)
Correction:
ii) Hydrolysis of R₂SiCl₂ yields a straight chain polymer.
iii) Silicones are good thermal and electrical insulators.

V. Choose the wrong statement

i) Boron is a metal.
ii) Nitrogen is a metalloid.
iii) Oxygen is a non metal.
iv) Antimony is a metalloid.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correction:
i) Boron is a metalloid (or) non metal
ii) Nitrogen is a non metal.

2. i) Aluminium chloride is a Lewis acid.
ii) Alum is a double salt of potassium aluminium sulphate.
iii) Aluminium chloride is used as a styptic agent to arrest bleeding.
iv) Alum is used as a catalyst in Friedel Crafts reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correction:
iii) Alum is used as a styptic agent to arrest bleeding.
iv) Anhydrous Aluminium chloride is used as a catalyst in Friedel Crafts reaction.

3. Which of the following statement about H₃BO₃ is not correct? (PTA – 5)
a) It is a strong tribasic acid
b) It is prepared by acidifying an aqueous solution of borax.
c) It is a layer structure in which planer BO₃ units are joined by hydrogen bonds.
d) It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion.
Answer:
a) It is a strong tribasic acid

4. i) Silicates which contain discrete [SiO₄]^4- units are called neso silicates.
ii) Beryl is an example for amphiboles.
iii) Spodumene is an example for phyllo silicates.
iv) Silicates which contain [Si₇O₇]^6- ions are called Soro silicates,
a) (i) & (ii)
b) (ii) & (iv)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
c) (ii) & (iii)
Correction:
ii) Beryl is an example for cyclic silicates.
iii) Spodumene is an example for chain silicates.

VI. Match the following
1.

Group No.		Group Name
i	13	a) Pnictogens
ii	14	b) Chalcogens
iii	15	c) Inert gases
iv	16	d) Halogens
v	17	e) Icosagens
vi	18	f) Tetragens

Answer:

Group No.		Group Name
i	13	e) Icosagens
ii	14	f) Tetragens
iii	15	a) Pnictogens
iv	16	b) Chalcogens
v	17	d) Halogens
vi	18	c) Inert gases

2.

1. Fluorine	i) Identification of coloured metal ions
2. Borax	ii) strong oxidising agent
3. Aluminium	iii) chalgogens present in volcanic ashes
4. Sulphur	iv) Most abundant element

Answer:

1. Fluorine	ii) strong oxidising agent
2. Borax	i) Identification of coloured metal ions
3. Aluminium	iv) Most abundant element
4. Sulphur	iii) chalgogens present in volcanic ashes

3.

Compound	Uses
1. Boron	a) Eye drops
2. Amorphous boron	b) Pyrex glass
3. Boric acid	c) Moderator
4. Boric oxide	d) Rocket fuel igniter

Answer:

Compound	Uses
1. Boron	c) Moderator
2. Amorphous boron	d) Rocket fuel igniter
3. Boric acid	a) Eye drops
4. Boric oxide	b) Pyrex glass

4.

Type of	Example
1. Ortho silicates	a) Quartz
2. Pyro silicates	b) Asbestos
3. Cyclic silicates	c) Mica
4. Chain silicates	d) Thortveitite
5. Amphiboles	e) Spodumene
6. Sheet silicates	f) Phenacite
7. Tecto silicates	g) Beryl

Answer:

Type of	Example
1. Ortho silicates	f) Phenacite
2. Pyro silicates	d) Thortveitite
3. Cyclic silicates	g) Beryl
4. Chain silicates	e) Spodumene
5. Amphiboles	b) Asbestos
6. Sheet silicates	c) Mica
7. Tecto silicates	a) Quartz

VII. 2 Marks questions

Question 1.
What are ‘p’-block elements? Write their general outer electronic configuration.
Answer:
The elements in which their last electron enters the ‘p’ orbital are called ‘p’-block elements.

• They are placed in 13 -18 groups.
• General outer electronic configuration is ns²np^1-6.

Question 2.
How are the p-block elements classified.
Answer:

• Based on the outer electronic configuration they are classified as 13 -18 group elements.
• Based on the nature of the elements they are classified as non metals, metalloids and metals.

Question 3.
Aluminium (III) chloride is stable where as Thallium (III) chloride is unstable. Why? (PTA – 2)
Answer:

• Due to inert pair effect, as we move down the 13th group ns² electrons remain inert and np¹ electron takes part in the reaction.
• So Tl³⁺ ion is less stable and Tl⁺¹ ion is more stable.
• Hence AlCl₃ is stable where as TICl₃ is unstable and decomposes into TlCl.

Question 4.
How is boric acid prepared from borax?
Answer:
Boric acid can be extracted from borax by treating with HCl or H₂SO₄.
Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl
Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + 2Na₂SO₄

Question 5.
How is boric acid prepared from Colemanite?
Answer:
When sulphur dioxide is passed through colemanite solution, boric acid is obtained.
Ca₂B₆O₁₁ + 2SO₂ + 9H₂O → 2CaSO₃ + 6H₃BO₃

Question 6.
What is the action of sodium hydroxide on boric acid?
Answer:
Boric acid reacts with sodium hydroxide to form sodium metaborate and sodium tetra borate.

Question 7.
Write the action of water on diborane.
Answer:
Diborane reacts with water to form boric acid.
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂

Question 8.
What is the action of NaOH on diborane.
Answer:
Diborane reacts with NaOH to form sodium meta borate.
B₂H₆ + 2NaOH + 2H₂O → 2NaBO₂ + 6H₂

Question 9.
What is the action of air on diborane?
Answer:
At room temperature pure diborane does not react with air or oxygen.

But impure diborane reacts with air or oxygen to giveB203 along with large amount of heat.
B₃H₆ + 3O₂ → B₂O₃ + 3H₂O
∆H =-2165 KJ mol^-1

Question 10.
How does diborane react with methyl alcohol?
Answer:
Diborane reacts with methyl alcohol to give trimethyl borate.
B₂H₆ + 6CH₃OH → 2B(OCH₃)₃ + 6H₂

Question 11.
How does diborane react with metal hydrides?
Answer:
When treated with metal hydrides, diborane forms metal boro hydrides.

Question 12.
How does diborane react with ammonia at low temperature?
Answer:
When treated with excess ammonia at low temperature diborane gives diborane di ammonate.
3B₂H₆ + 6NH₃ \(\underrightarrow { -153K } \) 3B₂H₆.2NH₃

Question 13.
How is inorganic benzene prepared? (PTA – 1)
Answer:

• On heating at higher temperatures with ammonia, diborane forms borazole or borazine.
• Borazole or borazine is called as Inorganic benzene
  

Question 14.
BF₃ acts as a Lewis acid. Give example.
Answer:
BF₃ is an electron deficient compound and accepts electron pairs to form coordinate covalent bonds. Hence BF3 acts as a Lewis acid.
BF₃ + NH₃ → F₃B ← NH₃
BF₃ + H₂O → F₃B ← OH₂

Question 15.
Convert BF3 into hydro fluoro boric acid.
Answer:
On hydrolysis BF3 gives boric acid, which is converted into hydro fluoro boric acid.
4BF₃ + 3H₂O → H₃BO₃ + 3HBF₄
3HBF₄ (Hydro fluoro boric acid)

Question 16.
Write about McAfee process of manufacturing AlCl₃.
Answer:
AlCl₃ is obtained by heating a mixture of alumina and coke in a current of chlorine.
Al₂O₃ + 3C + 3Cl₂ → 2AlCl₃ + 3CO

Question 17.
Write the action of NaOH on AlCl₃
Answer:
With excess of NaOH, AlCl₃ gives sodium alumina te.
AlCl₃ + 4NaOH → NaAlO₂ + 2H₂O + 3NaCl

Question 18.
Write the uses of aluminium chloride.
Answer:
1. Anhydrous AlCl₃ is used as a catalyst in Friedel crafts reaction.
2. AlCl₃ is used for the manufacture of petrol by cracking the mineral oils.
3. AlCl₃ is used as a catalyst in the manufacture of dyes, drugs and perfumes.

Question 19.
What are alums? Give examples.
Answer:
1. Alum is a double salt of potassium aluminium sulphate.
2. Now a days the name alum is used for all the double salts with the formula
M’₂ SO₄ M”2 (SO₄)₃.24H₂O
Where M’ is univalent metal ion or NH₄⁺
M” is trivalent metal ion
Example: K₂SO₄.Al₂(SO₄)₃.24H₂O Potash alum
K₂SO₄.Cr₂(SO₄)₃.24H₂O Chrome alum

Question 20.
Aqueous solution of carbon di oxide is acidic. Why?
Answer:
Aqueous solution of carbon di oxide is slightly acidic as it forms carbonic acid which dissociates to give H⁺ ions.

Question 21.
How is silicon tetra choride prepared?
Answer:
SiCl₄ is prepared by passing dry chlorine over an intimate mixture of silica and carbon heating to 1675 K in a porcelain tube.
SiO₂ + 2C + 2Cl₂ → SiCl₄ + 2CO
SiCl₄ is prepared commercially by the reaction of silicon with hydrogen chloride gas above 600 K.
SiO + 4HCl → SiCl₄ + 2H₂

Question 22.
Write the uses of silicon tetra chloride.
Answer:
Silicon tetra chloride is used
i) In the production of semi conducting silicon.
ii) As a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 23.
What is water gas equilibrium? (PTA – 5)
Answer:
Water gas equilibrium
The equilibrium involved in the reaction between carbon di oxide and hydrogen, has many industrial applications and is called water gas equilibrium.

VIII. Three Marks questions

Question 1.
How is borax prepared from colemanite?
Answer:
When colemanite ore solution is boiled with sodium carbonate solution borax is obtained.

Question 2.
Write the uses of boron.
Answer:
1. ₅B¹⁰ absorbs neutrons, hence it is used as a moderator in nuclear reactors.
2. Amorphous boron is used as a rocket fuel igniter.
3. Boron is essential for the cell walls of plants.
4. Boric acid and borax are used in eye drops, antiseptics, washing powders.
5. Boric oxide is used in the manufacture of pyrex glass.

Question 3.
Aqueous solution of borax is basic. Why?
Answer:
In hot water borax dissociates into boric acid and sodium hydroxide.
Na₂B₄O₇ + 7H₂O → 4H₃BO₃ + 2NaOH
Boric acid is a weak acid, whereas sodium hydroxide is a strong base.

As a result the resulting solution is basic.

Question 4.
What is the action of heat on borax?
Answer:
On heating borax loses its water of crystallisation first and then decomposes into sodium metaborate and boron trioxide.

Boron trioxide appears as transparent glassy beads.

Question 5.
What is the action of heat on boric acid?
Answer:

Temperature	Compound obtained
373 K	Meta boric acid
413 K	Tetra boric acid
Red hot	Boric anhydride (glassy mass)

Question 6.
Describe the structure of boric acid.
Answer:

• Boric acid has a two dimensional structure.
• It consists of [BO₃]^3- unit.
• These unit are linked to each other by hydrogen bonds.

Question 7.
Write the uses of boric acid.
Answer:
Boric acid is
1. Used in the manufacture of pottery glazes, glass, enamels and pigments.
2. Used as an antiseptic.
3. Used as an eye lotion.
4. Used as a food preservative.

Question 8.
How is diborane prepared?
Answer:

• When sodium boro hydride in diglyme is reacted with iodine diborane is obtained.
  2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂
• On heating magnesium boride with Hcl, a mixture of volatile boranes are obtained.
  2Mg₃B₂ + 12HCl → 6MgCl₂ + B₄H₁₀ + H₂
  B₄H₁₀ + H₂ → 2B₂H₆

Question 9.
Write the uses of diborane.
Answer:
Diborane is
1. Used as a high energy fuel for propellant.
2. Used as a reducing agent in organic

Question 10.
How is boron trifluoride prepared from boron trioxide?
Answer:
When boron trioxide is treated with calcium fluroide in presence of conc.sulphruic acid, boron trifluoride is obtained.

When boron trioxide is reacted with carbon and fluorine, boron trifluoride is obtained.
B₂O₃ + 3C + 3F₂ → 2BF₃ + 3CO

Question 11.
How is boron trifluoride prepared in the laboratory?
Answer:
In the laboratory pure BF₃ is prepared by the. thermal decomposition of benzene, diozonium tetrafluro borate.

Question 12.
How is potash alum prepared? (PTA – 4)
Answer:
Potash alum is prepared from alunite or alum stone.
When alunite is treated with excess of sulphuric acid, the aluminium hydroxide present is converted into aluminium sulphate.
A calculated quantity of potassium sulphate is added.
The solution is crystallised to obtain potash alum.
It is purified bv recrystallisation.

Question 13.
Write the uses of alum.
Answer:
Alum is used
i) for the purification of water.
ii) for water proofing and textiles.
iii) in dyeing, paper and leather tanning industries.
iv) as a styptic agent to arrest bleeding.

Question 14.
Write the uses of carbon monoxide.
Answer:
Carbon monoxide is used
i) as a reducing agent and can reduce many metal oxides to metal.
ii) as an important ligand and forms metal carbonyls.
iii) a mixture of CO & H₂ is called as water gas and a mixture of CO & N₂ is called as producer gas. Both are used as important industrial fuels.

Question 15.
Write the uses of carbon dioxide.
Answer:
Carbon dioxide is used

• to produce an inert atmosphere for chemical processing.
• by plants in photosynthesis.
• as fire extinguisher.
• as a propellant gas.
• in the production of carbonated beverages.
• in the production of foam.

Question 16.
Write note on Boron Neutron Capture Therapy (BNCT).
Answer:

• The affinity of Boron-10 for neutrons is the bases of this technique BNCT for treating patients suffering from brain tumours.
• It is based on the nuclear reaction which occurs when Boron-10 is irradiated with low- energy thermal neutrons to give high linear energy a-particles and a Li particle.
• Boron compounds are injected into a brain tumour patient and the compounds collect preferentially in the tumour.
• The tumour area is then irradiated with / thermal neutrons and results in the release of an alpha particle.
• This a-particle damages the tissue in the tumour each time a Boron-10 nucleus captures a neutron.
• In this wav damage can be limited preferentially to the tumour, leaving the normal brain tissue less affected.
• BNCT has been studied as a treatment for several other tumours of the head and neck, the breast the prostate, the bladder and the liver.

IX. Five Marks questions

Question 1.
How is higher boranes obtained from diborane.
Answer:
At high temperatures diborane forms higher boranes liberating hydrogen.

Question 2.
Explain the allotropes of carbon.
Answer:

• Carbon exists in many allotropic forms.
• Graphite and diamond are the most common allotropes.
• Graphene, fullerenes and carbon nano tubes are other important allotropes of carbon.

Graphite:

• It is the most stable allotrope of carbon at normal temperature and pressure.
• It is composed of flat two dimensional hexagonal sheets of sp² hybridised carbon atoms.
• C-C bond length is 1.41 Å which is close to the C-C bond distance in benzene (1.40 Å )
• Each carbon atom forms three sbonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital forms a p-bond.
• These pelectrons are delocalised over the entire sheet, hence graphite conducts electricity.
• Successive carbon sheets at a distance of 3.40 Ao are held together by weak Vander Waals forces, hence graphite is soft, slippery and used as a lubricant.

Diamond:

• Carbon atoms in diamond are sp3 hybridised.
• Each carbon is bonded tetra hedrally with four other carbon atoms by s-bonds with C-C bond length of 1.54 Å. Hence diamond is hard.
• Since all the four valence electrons of carbon are involved in bonding and there is no free electrons, diamond is not a conductor.
• Being the hardest substance, diamond is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Fullerenes:

• There are newly synthesised allotropes of carbon.
• Unlike graphite and diamond these are discrete molecules of carbon like C₃₂, C₅₀, C₆₀, C₇₀, C₇₆ …….
• These have cage like structures.
• Buck‘minster fullerene or bucky ball have a soccer ball like structure with the formula C₆₀.
  It has a fused’ ring structure with 20 six membered rings and 12 five membered rings.
• Each carbon is sp² hybridised and forms three π sbonds and one delocalised pbond giving aromatic character.
• C-C bond distance is 1.44 Å and C=C bond distance is 1.38 Å.

Carbon nano tubes:

• This is another recently discovered allotropes of carbon.
• They have graphite like tubes with fullerene ends.
• Along the axis, carbon nano tubes are stronger than steel and conduct electricity.
• They have many applications in nano scale electronics, Catalysis, polymers and medicine.

Graphene:

• It is a single planar sheet of graphite.
• In this sp² hybridised carbon atoms are densely packed in a honey comb crystal lattice.

Question 3.
Write about the preparation and structure of silicones.
Answer:

• Silicones or poly siloxanes are organo silicon polymers.
• Their general empirical formula is (R₂SiO)
• Since their empirical formula is similar to Ketones (R₂CO) they are called as Silicones.
• They may be linear or cross linked.
• Due to their very high thermal stability they are called high-temperature polymers.

Types of Silicones:
i) Linear Silicones:
They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl
a) Silicone rubbers:
These are bridged together by methylene or similar groups.
b) Silicone resins:
They are obtained by blending silicones with organic resins such as acrylic esters.

ii) Cyclic Silicones:
These are obtained by the hydrolysis of R₂SiCl₂.

iii) Cross linked Silicones:
These are obtained by the hydrolysis of RSiCl₃.

Preparation:
Vapours of RCl or ArCl are passed over silicon at 570 K with copper catalyst gives R₂SiCl₂ (dialkyl dichloro silanes) or Ar₂SiCl₂ (diaryl dichloro silanes)
2RCl + Si \(\underrightarrow { Cu / 570K } \) R₂SiCl₂

Hydrolysis of R₂SiCl₂ gives a straight chain polymer which grows from both sides.

Hydrolysis of mono alkyl trichloro silanes RSiCl₃ gives a very complex cross linked polymer.

Linear silicones can be converted into cyclic or ring silicones when water molecules are removed from the terminal -OH groups.

Question 4.
Explain various types of silicates.
Answer:
The mineral which contains silicon and oxygen in tetrahedral [SiO₄]^4- units linked together in different patterns are called silicates.

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12th Chemistry Guide Metallurgy Text Book Questions and Answers

I. Choose the correct answer

1. Bauxite has the composition
a) Al₂O₃
b) Al₂O₃.nH₂O
c) Fe₂O₃.2H₂O
d) None of these
Answer:
b) Al₂O₃.nH₂O

2. Roasting of sulphide ore gives the gas (A). (A) is a colorless gas. An aqueous solution of (A) is acidic. The gas (A) is
a) CO₂
b) SO₃
c) SO₂
d)H₂S
Answer:
c) SO₂

3. Which one of the following reaction represents calcinations?
a) 2Zn + O₂ → 2ZnO
b) 2ZnS + 3O₂ → 2ZnO + 2SO₂
c) MgCO₃ → MgO + CO₂
d) Both (a) and (c)
Answer:
c) MgCOa → MgO + CO₂

4. The metal oxide which cannot be reduced to metal by carbon is
a) PbO
b) Al₂O₃
c) ZnO
d) FeO
Answer:
b) Al₂O₃

5. Which of the metal is extracted by Hall – Heroult process?
a) Al
b) Ni
c) Cu
d) Zn
Answer:
a) Al

6. Which of the following statements, about the advantage of roasting of sulphide ore before the reduction is not true?
a) ΔG°f of sulphide is greater than those for CS₂ and H₂S
b) ΔG°r is negative for roasting of sulphide ore to oxide
c) Roasting of the sulphide to its oxide is thermodynamically feasible.
d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
Answer:
d) Carbon and hydrogen are suitable reducing agents for metal sulphides.

7. Match items in Column I – with the items of Column – II and assign the correct code.

Answer:
c) (iv) (ii) (iii) (i)

8. Wolframite ore is separated from tinstone by the process of
a) Smelting
b) Calcination
c) Roasting
d) Electromagnetic separation
Answer:
d) Electromagnetic separation

9. Which one of the following is not feasible
a) Zn(S) + Cu²⁺(aq) → Cu(s) + Zn²⁺(aq)
b) Cu(S) + Zn²⁺(aq) → Zn(s) + Cu²⁺(aq)
c) Cu(S) + 2Ag⁺(aq) → Ag(s) + Cu²⁺(aq)
d) Fe(S) + Cu²⁺(aq) → Cu(s) + Fe²⁺(aq)
Answer:
b) Cu(S) + Zn²⁺(aq) → Zn(s) + Cu²⁺(aq)

10. Electrochemical process is used to extract
a) Iron
b) Lead
c) Sodium
d) Silver
Answer:
c) Sodium

11. Flux is a substance which is used to convert
a) Mineral’into silicate
b) Infusible impurities to soluble impurities
c) Soluble impurities to infusible impurities
d) All of these
Answer:
b) Infusible impurities to soluble impurities

12. Which one of the following ores is best concentrated by froth floatation method?
a) Magnetite
b) Heamatite
c) Galena
d) Cassiterite
Answer:
c) Galena

13. In the extraction of aluminium from alumina by electrolysis, cryolite is added to
a) Lower the melting point of alumina
b) Remove impurities from alumina
c) Decrease the electrical conductivity
d) Increase the rate of reduction
Answer:
a) Lower the melting point of alumina

14. Zinc is obtained from ZnO by
a) Carbon reduction
b) Reduction using silver
c) Electrochemical process
d) Acid leaching
Answer:
a) Carbon reduction

15. Extraction of gold and silver involves leaching with cyanide ion. silver is later recovered by (NEET – 2017)
a) Distillation
b) Zone refining
c) Displacement with zinc
d) liquation
Answer:
c) Displacement with zinc

16. Considering the Ellingham diagram, which of the following metals can be used to reduce alumina? (NEET – 2018)
a) Fe
b) Cu
c) Mg
d) Zn
Answer:
c) Mg

17. The following set of reactions are used in refining Zirconium

This method is known as
a) Liquation
b) Van Arkel process
c) Zone refining
d) Mond’s process
Answer:
b) Van Arkel process

18. Which of the following is used for concentrating ore in metallurgy?
a) Leaching
b) Roasting
c) Froth floatation
d) Both (a) and (c)
Answer:
d) Both (a) and (c)

19. The incorrect statement among the following is
a) Nickel is refined by Mond’s process
b) Titanium is refined by Van Arkel’s process
c) Zinc blende is concentrated by froth floatation
d) In the metallurgy of gold, the metal is leached with a dilute sodium chloride solution
Answer:
d) In the metallurgy of gold, the metal is leached with a dilute sodium chloride solution

20. In the electrolytic refining of copper, which one of the following is used as anode?
a) Pure copper
b) Impure copper
c) Carbon rod
d) Platinum electrode
Answer:
b) Impure copper

21. Which of the following plot gives Ellingham diagram
a) ΔS VsT
b) ΔG° VsT
c) ΔG° Vs1/T
d) ΔG° VsT²
Answer:
b) ΔG°VsT

22. In the Ellingham diagram, for the formation of carbon monoxide

Answer:
c) \(\left(\frac{\Delta \mathrm{G}^{0}}{\Delta \mathrm{T}}\right)\) is negative

23. Which of the following reduction is not thermodynamically feasible?
a) Cr₂O₃ + 2Al → Al₂O₃ + 2Cr
b) Al₂O₃ + 2Cr → Cr₂O₃ + 2Al
c) 3TiO₂ + 4Al → 2Al₂O₃ + 3Ti
d) None of these
Answer:
b) Al₂O₃ + 2Cr → Cr₂O₃ + 2Al

24. Which of the following is not true with respect to the Ellingham diagram?
a) Free energy changes follow a straight line. The deviation occurs when there is a phase change.
b) The graph for the formation of CO₂is a straight line almost parallel to the free energy axis.
c) Negative slope of CO shows that it becomes more stable with an increase in temperature.
d) Positive slope of metal oxides shows that their stabilities decrease with an increase in temperature.
Answer:
b) The graph for the formation of CO₂ is a straight line almost parallel to the free energy axis.

II. Answer the following questions

Question 1.
What is the difference between minerals and ores?
Answer:
Minerals:

1. Minerals contain a low percentage of metal.
2. Metal cannot be extracted easily from minerals.
3. Clay Al₂O₃. SiO₂. 2H₂O is the mineral of aluminium.

Ores:

1. Ores contain a large percentage of metal.
2. Ores can be used for the extraction of metals on a large scale readily and economically.
3. Bauxite Al₂O₃. 2H₂O is the ore of aluminium.

Question 2.
What are the various steps involved in the extraction of pure metals from their ores?
Answer:
Steps involved in the extraction of pure metals from their ores are

1. Concentration of the ore
2. Extraction of the crude metal.
3. Refining of the crude metal.

Question 3.
What is the role of Limestone in the extraction of iron from its oxide Fe₂O₃?
Answer:

• Limestone (CaO) is used as a flux in the extraction of iron from its oxide Fe₂O₃.
• Flux is a chemical substance that forms an easily fusible slag with gangue.
• Oxide of iron can be reduced by carbon monoxide as follows
  Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g) ↑
• In this extraction, a basic flux quick lime (or) lime (CaO) reacts with acidic gangue silica to form the slag calcium silicate.
  

Question 4.
Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
Answer:
Sulphide ores can be concentrated by the froth floatation method.
e.g.,

1. Copper pyrites (CuFeS₂H₂)
2. Zinc blende (ZnS)
3. Galena (PbS)

Question 5.
Describe a method for refining nickel Mond process for refining nickel: (PTA – 3)
Answer:

• Impure nickel is heated in a stream of carbon monoxide at around 350K. Nickel reacts with CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left behind.
  Ni_(S) + 4CO_(g) → Ni(Co)_4(g)
• On heating nickel tetra carbonyl around 460K, the complex decomposes to give a pure nickel.
  Ni(CO)_4(g) → Ni_(S)+ 4CO_(g)

Question 6.
Explain the zone refining process with an example. (PTA – 6 MARCH 2020)
Answer:
1. Zone Refining method is based on the principles of fractional crystallisation. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region, i.e. impurities are more soluble in the melt than in the solid-state metal.

2. In this process, the impure metal is taken in the form of a rod. One end of the rod is heated using a mobile induction heater which results in the melting of the metal on that portion of the rod.

3. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone formed due to the movement of the heater. As the heater moves further away, the molten zone containing impurities also moves along with it.

4. The process is repeated several times by moving the heater in the same direction again and again to achieve the desired purity level.

5. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.

6. Elements such as germanium (Ge), silicon (Si) and gallium (Ga) that are used as semiconductors are refined using this process.

Question 7.
Using the Ellingham diagram
(A) Predict the conditions under which
i) Aluminium might be expected to reduce magnesia.
ii) Magnesium could reduce alumina.
B) It is possible to reduce Fe₂O₃ by coke at a temperature around 1200K
Answer:
A) i) Ellingham diagram for the formation of Al₂O₃ and MgO intersects around 1600K. Above this temperature aluminium line lies below the magnesium line. Hence we can use aluminium to reduce magnesia above 1600K.

ii) In Ellingham diagram below 1600K magnesium line lies below aluminium line. Hence below 1600K magnesium can reduce alumina.

B) In Ellingham diagram above 1000K carbon line lies below the iron line. Hence it is possible to reduce Fe₂O₃ by coke at a temperature around 1200K.

Question 8.
Give the uses of zinc. (PTA – 4)
Answer:

• Metallic zinc is used in galvanisation to protect iron and steel structures from rusting and corrosion.
• Zinc is used to produce die – castings in the automobile, electrical and hardware industries.
• Zinc oxide is used in the manufacture of paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles and electrical equipment.
• Zinc sulphide is used in making luminous paints, fluorescent lights and x-ray screens.
• Brass an alloy of zinc which is highly resistant to corrosion is used in water valves and communication equipment.

Question 9.
Explain the electrometallurgy of aluminium.
Answer:
Hall – Heroult Process
Cathode: Iron tanked lined with carbon
Anode: Carbon blocks
Electrolyte: 20% solution of alumina obtained from bauxite + Molten Cryolite +10 % calcium chloride (lowers the melting point of the mixture)
Temperature: Above 1270K
Ionisation of Alumina Al₂O₃ → +2Al³⁺ + 3O^2-
Reaction at cathode: Al³⁺ (melt) + 3e^– → Al_l
Reaction at anode : 2O^2- (melt) → O₂ (melt) + 4e^–
Since carbon acts as anode the following reaction also takes place on it.
C_(s) + O² (melt) → CO + 2e^–
C_(s) + 2O₂ (melt) → CO₂ + 4e^–
During electrolysis, anodes are slowly consumed due to the above two reactions. Pure aluminium is formed at the cathode and settles at the bottom.
Net electrolysis reaction is
4Al³⁺ (melt) + 6O^2- (melt) + 3C_(s) → 4Al_(l) + 3CO_2(g)

Question 10.
Explain the following terms with suitable examples, i) Gangue ii) Slag (PAT – 2)
Answer:
i) Gangue:
The non-metallic impurities, rocky materials and siliceous matter present in the ores are called gangue.
(eg): SiO₂ is the gangue present in the iron ore Fe₂O₃.

ii) Slag:
Slag is a fusible chemical substance formed by the reaction of gangue with a flux.

Question 11.
Give the basic requirement for vapour phase refining.
Answer:

• The metal is treated with a suitable reagent to form a volatile compound.
• Then the volatile compound is decomposed to give the pure metal.

Question 12.
Describe the role of the following in the process mentioned.
i) Silica in the extraction of copper.
ii) Cryolite in the extraction of aluminium.
iii) Iodine in the refining of Zirconium.
iv) Sodium cyanide in froth floatation.
Answer:
i) In the extraction of copper silica acts as an acidic flux to remove FeO as slag FeSiO₃.

ii) As Al₂O₂ is a poor conductor cryolite improves the electrical conductivity.
In addition, crvolite serves as an added impurity and lowers the melting point of the electrolyte.

iii) First Iodine forms a Volatile tetraiodide with impure metal, which decomposes to give pure metal. Impure zirconium metal is heated in an evacuated vessel with iodine to form the volatile zirconium tetraiodide (Zrl₄). The impurities are left behind, as they do not react with iodine.
Zr_(S) + 2I_2(S) → Zrl_4(Vapour)

On passing volatile zirconium tetraiodide vapour over a tungsten filament, it is decomposed to give pure zirconium.
Zrl_4(Vapour) → Zrl_(S) + 2I_2(S)

iv) Sodium cyanide acts as a depressing agent in froth floatation process. It prevent other metal sulphides from coming to the froth.
eg: NaCN depresses the floatation property ZnS present in Galena (PbS) by forming a layer of Zinc complex Na₂[Zn(CN)₄] on the surface of Zinc sulphide.

Question 13.
Explain the principle of electrolytic refining with an example. (PTA – 5)
Answer:

• Crude metal is refined by electrolysis carried out in an electrolytic cell.
• Cathode: Thin strips of pure metal.
  Anode: Impure metal to be refined.
  Electrolyte: Aqueous solution of the salt of the metal with dilute acid.
• As the current is passed, the metal of interest dissolves from the anode and passes into the electrolytic solution.
• At the same time, same amount of metal ions from the electrolytic solution will be deposited at the cathode.
• Less electro positive impurities in the anode settle down as anode mud.
• eg: Electrorefining of silver:
  Cathode: Pure silver Anode: Impure silver rods.
  Electrolyte: Acidified aqueous solution of silver nitrate.
• On passing current the following reactions will take place.
  Reaction at anode: Ag_(s) → Ag⁺_(aq) + e^–
  Reaction at cathode: Ag⁺_(aq) + e^– → Ag_(s)
• At anode silver atoms lose electrons and enter the solution. From the solution silver ions migrate towards the cathode. At cathode silver ions get discharged by gaining electrons and deposited on the cathode.

Question 14.
The selection of reducing agent depends on the thermodynamic factor: Explain with an example.
Answer:

• A suitable reducing agent is selected based on the thermodynamic considerations.
• For a spontaneous reaction AG should be negative.
• Thermodynamically, the reduction of metal oxide with a given reducing agent can occur if AG for the coupled reaction is negative.
• Hence the reducing agent is selected in such a way that it provides a large negative value for the coupled reaction.
• Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal.
• Any metal can reduce the oxides of other metals that are located above it in the diagram.
• Ellingham diagram for the formation of FeO and CO intersects around 1000K. Below this temperature the carbon line lies above the iron line.
• Hence FeO is more stable than CO and the reduction is not thermodynamically feasible.
• However above 1000K carbon line lies below the iron line. Hence at this condition, FeO is less stable than CO and the reduction is thermodynamically feasible. So coke can be used as a reducing agent above this temperature.
• Following free energy calculation also confirm that the reduction is thermo¬dynamically favoured.
• From theEllingham diagram at 1500K
  2Fe_(s) + O_2(g) → 2FeO_(g) = 350 KJmol^-1 …………….. 1
  2C_(s) + 02_2(g) → 2CO_(g) = 480 KJmol^-1 ……………… 2
  Reverse the reaction 1
  2FeO_(s) → 2Fe_(s) + O_2(g) = 350 KJmol^-1 ……………… 3
  Couple the reactions 2 and 3
  2FeO_(s) + 2C_(s) → 2Fe_(s)+ 2CO_(g) = 130 KJmol^-1 ……………… 4
• The standard tree energy change for the reduction of one mole of FeO is = \(\frac{\Delta \mathrm{G}_{3}}{2}\) = -65 KJmol^-1

Question 15.
Give the limitations of Ellingham diagram.
Answer:

• Ellingham diagram is constructed based only on thermodynamic considerations.
• It gives information about the thermodynamic feasibility of a reaction.
• It does not tell anything about the rate of the reaction.
• Moreover, it does not: give an idea about the possibility of other reactions that might be taking place.
• The interpreparation of G is based on the assumption that the reactants are in equilibrium with the product which is not always true.

Question 16.
Write a short note on electrochemical principles of metallurgy.
Answer:

• Reduction of oxides of active metals such as sodium, potassium etc. by carbon is thermodynamically not feasible.
• Such metals are extracted from their ores by using electrochemical methods.
• In this method the metal salts are taken infused form or in solution form.
• The metal ion present can be reduced by treating the solution with a suitable reducing agent or by electrolysis.
• Gibbs free energy change for the electrolysis is ∆G° = nFE°
  n = number of electrons involved in the reduction
  F = Faraday = 96500 coulombs
  E° = electrode potential of the redox couple.
• If E° is positive, is negative and the reduction is spontaneous.
• Hence a redox reaction is planned in such a way that the e.mi of the net redox reaction is positive.
  A more reactive metal displaces a less reactive metal from its salt solution.
  eg;Cu²⁺_(aq) + Zn_(s) → Cu_(s) + Zn²⁺_(aq)
• Zinc is more reactive than copper and displaces copper from its salt solution.

III. Evaluate yourself

Question 1.
Write the equation for the extraction of silver by leaching with sodium cyanide and show that the leaching process is a redox reaction.
Answer:
Ag → Ag⁺ (O.N increases from 0 to +1, hence oxidation)
O₂ OH^– (O.N decreases from 0 to -2, hence reduction)
The leaching of silver is a redox reaction.

Question 2.
Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction
Answer:
MgCO₃\(\underrightarrow { \triangle } \) MgO + CO₂

Question 3.
Using the Ellingham diagram (fig 1.4) indicates the lowest temperature at which ZnO can be reduced to Zinc metal by carbon. Write the overall reduction reaction at this temperature
Answer:
Ellingham diagram for the formation of ZnO and CO intersects around 1200K Below this temperature, Carbon line lies above Zinc line. Hence ZnO is more stable than CO so the reduction is thermodynamically not feasible at this temperature range. However above 1200K carbon line lies below the zinc line, hence carbon can be used as a reducing agent above 1200K.
2Zn + O₂ → 2ZnO ………………….. 1
2C + O₂ → 2CO ………………….. 2
Reversing 1 and adding with equation 2
2ZnO → 2Zn + O₂
2C + O₂ → 2CO
2ZnO +2C → 2Zn + 2CO

Question 4.
Metallic Sodium is extracted by the electrolysis of brine (aq.NaCl). After electrolysis, the electrolytic solution becomes basic in nature. Write the possible electrode reactions.
Answer:
2NaCl_(aq) → 2Na⁺_(aq) + 2Cl_(aq)
Anode: 2Cl_(aq) → Cl_2(g) + 2e
Cathode: 2H₂O_(l) + 2e → H_2(g) + 2OH_(aq)
Nothing happens to sodium ion but it is still important. Na⁺ ions are spectator ions and combine with OH^– ions to form NaOH
Three products are H₂, Cl₂ and NaOH
Over all equation is
2NaCl_(aq) + 2H₂O → H_2(g) + Cl_2(g) + 2NaOH_(aq)
Ionic equation is
2H₂O_(l) + 2Cl^–_(aq) + 2Na^–_(aq) → 2Na⁺_(aq) + 2OH_(aq) + H_2(g) + Cl_2(g)
(or)
2H₂O_(l) + 2Cl^–_(aq) → 2OH_(aq) + H_2(g) + Cl_2(g)

12th Chemistry Guide Metallurgy Additional Questions and Answers

Part – II – Additional Questions one mark

I. Match the following
1.

Ore	Formula
1. Magnetite	a) ZnCO3
2. Cuprite	b) PbCO3
3. Calamine	c) Fe3O4
4. Cerrusite	d) SnO2
5. Cassiterite	e) Cu2O

Answer:

Ore	Formula
1. Magnetite	c) Fe3O4
2. Cuprite	e) Cu2O
3. Calamine	a) ZnCO3
4. Cerrusite	b) PbCO3
5. Cassiterite	d) SnO2

2.

Ore of metal	Name
1. Ore of copper	a) Diaspore
2. Ore of aluminium	b) Chlorargyrite
3. Ore of iron	c) Malachite
4. Ore of lead	d) Limonite
5. Ore of silver	e) Anglesite

Answer:

Ore of metal	Name
1. Ore of copper	c) Malachite
2. Ore of aluminium	a) Diaspore
3. Ore of iron	d) Limonite
4. Ore of lead	e) Anglesite
5. Ore of silver	b) Chlorargyrite

3.

Concentration	Ore
Gravity separation	a) Pyrolusite
Froth floatation	b) Alumina
Cyanide leaching	c) Zinc blende
Alkali leaching	d) Tinstone
Magnetic separation	e) Gold

Answer:

Concentration	Ore
Gravity separation	d) Tinstone
Froth floatation	c) Zinc blende
Cyanide leaching	e) Gold
Alkali leaching	b) Alumina
Magnetic separation	a) Pyrolusite

4.

Purification	Metal
1. Distillation	a) Silicon
2. Liquation	b) Zinc
3. Electrolytic refining	c) Nickel
4. Zone refining	d) Tin
5. Mond process	e) Silver

Answer:

Purification	Metal
1. Distillation	b) Zinc
2. Liquation	d) Tin
3. Electrolytic refining	e) Silver
4. Zone refining	a) Silicon
5. Mond process	c) Nickel

II. Fill in the blanks

1. The metal which shows high resistance to corrosion and used in the design of Chemical reactors is ___________ .
Answer:
Aluminium

2. ___________ are used for increasing the efficiency of the solar cells.
Answer:
Gold nanoparticles

3. The removal of gangue from ores is called as ___________ .
Answer:
Concentration of ores

4. ___________ is the process in which concentrated ore is strongly heated in the absence of air.
Answer:
Calcination

III. Find the odd man out.

1. a) Sphalerite b) Galena c) Azurite d) Iron pyrite
Answer:
c) Azurite. This is a basic carbonate ore others are sulphide ores.

2. a) Malachite b) Limonite c) Siderite d) Haematite
Answer:
a) Malachite. This is the ore of copper, others are ores of iron.

IV. Choose the incorrect pair.

1. a) Malachite, Azurite b) Ruby silver, Horn silver c) Zincite, Cuprite d) Anglesite, Cerrusite
Answer:
c) Zincite, Cuprite. They are ores of Zinc and copper

2. a) Kaolinite, Aluminium b) Stefinite, Silver c) Galena, Lead d) Prousitite, Tin
Answer:
d) Prousitite, Tin. Correct pair is prousitite, silver.

V. Choose the correct pair.

1. Choose the correct pair.
a) Cerrusite, Cassiterite b) Siderite, Limonite c) Anglesite, Zincite d) Azurite, Kaolinite
Answer:
b) Siderite, Limonite. Both are ores of irons.

2. a) Diaspore, Copper b) Galena, Tin c) Stefinite, Silver d) Malachite, Aluminium
Answer:
c) Stefinite, Silver. Stefinite is the ore of silver

VI. Assertion and Reason

1. Assertion (A): Tinstone ore is concentrated by magnetic separation.
Reason (R) : Wolframite impurities are magnetic
i) A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) A and R are correct but R does not explain A.
Answer:
i) A and R are correct, R explains A. 2

2. Correct Assertion (A): Aluminium can be commercially extracted from china clay which is a profitable one
Reason (R): China clay is a mineral of aluminium.
i) A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) A and R are correct but R does not explain A.
Answer:
iii) A is wrong, R is correct
Correct Assertion: Aluminium can be commercially extracted from bauxite which is a profitable ore

3. Assertion (A): Zinc blend can be concentrated by the froth floatation method.
Reason (R) : Metallic ore particles are preferentially wetted by water and settle at the bottom.
A and R are correct, R explains A.
A is correct, R is wrong A is wrong, R is correct A and R are correct but R does not explain A.
Answer:
ii) A is correct, R is wrong
Correct (R) : Metallic particles are preferentially wetted by oil and rise to the surface.

4. Assertion (A) : Cr₂O₃ is reduced into chromium by aluminothermic process.
Reason (R): Aluminium acts as the reducing agent.
i) A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) A and R are correct but R does not explain A.
Answer:
i) A and R are correct, R explains A.

VII. Choose the correct statement

1. a) Metals having more chemical reactivity occur as native elements.
b) Removal of gangue from ores is called refining.
c) Tin stone ore is concentrated by gravity separation.
d) Silver glance is a carbonate ore.
Answer:
c) Tin stone ore is concentrated by gravity separation.

2. a) In froth floatation sodium ethyl xanthate acts as a collector.
b) In leaching the ore is converted into insoluble salt or complex and the gangue remains in the solution.
c) Ammonia leaching is suitable for gold and silver.
d) Bauxite ore is subjected to acid leaching.
Answer:
a) In froth floatation sodium ethyl xanthate acts as a collector.

3. a) Calcination is the process in which concentrated ore is strongly heated in the presence of air.
b) Flux is a chemical substance that forms an easily fusible slag with gangue.
c) In aluminothermic process the ignition mixture used is magnesium peroxide and barium.
d) Any metal can reduce the oxides of other metals that are located below it in Ellingham diagram.
Answer:
b) Flux is a chemical substance that forms an easily fusible slag with gangue.

4. a) In electrorefining pure metal is taken as anode and impure metal is taken as cathode.
b) Distillation is employed for high boiling nonvolatile metals.
c) Zone refining is based on the principle of fractional crystallisation.
d) Mond’s process is used for refining titanium.
Answer:
c) Zone refining is based on the principle of fractional crystallisation.

VIII. Choose the incorrect statement

1. i) In cyanide leaching gold is converted into an insoluble cyanide complex.
ii) In ammonia leaching nickel forms a soluble complex.
iii) In alkali leaching aluminum forms an insoluble complex,
a) i & ii
b) i & iii
c) ii & iii
d) i, ii, iii
Answer:
D) i,ii & iii

2. i) In the Ellingham diagram for most of the metal oxide forming the slope is negative.
ii) Oxygen gas is consumed during the formation of metal oxides resulting in the increase of randomness.
iii) As temperature increases value for the formation of the metal oxide become more negative
a) i & ii
b) i & iii
c) ii & iii
d) i, ii, & iii
Answer:
c) ii & iii

3. i) The reduction of oxides of active metals such as sodium, potassium, etc. by carbon is thermodynamically feasible
ii) When a more reactive metal is added to the solution containing less reactive metal, the less reactive metal will go into the solution.
iii) Copper displaces zinc from zinc salt
solution.
a) i & ii
b) i & iii
c) ii & iii
d) i, ii, iii
Answer:
d) i, ii & iii

4. i) When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the solid region.
ii) Zone refining is carried out in an inert gas atmosphere to prevent the reduction of metals.
iii) Elements such as germanium, silicon, and gallium are refined by zone refining.
a) i & ii
b) i & iii
c) ii & iii
d) i, ii, iii
Answer:
a) i & ii

IX. Choose the best answer.

1. Which of the following is not an oxide ore?
a) Cuprite
b) Siderite
c) Cassiterite
d) Zincite
Answer:
b) Siderite

2. Which of the following is an oxide ore?
a) Sphalerite
b) Calamine
c) Cassiterite
d) Stefinite
Answer:
c) Cassiterite

3. The process of converting hydrated alumina into anhydrous alumina is called
a) Roasting
b) Smelting
c) Auto-reduction
d) Calcination
Answer:
d) Calcination

4. Which of the following is a sulphide ore?
a) Pyrargyrite
b) Malachite
c) Limonite
d) Kaolinite
Answer:
a) Pyrargyrite

5. Which of the following is not a carbonate ore?
a) Siderite
b) Calamine
c) Cerrusite
d) Cassiterite
Answer:
d) Cassiterite

6. Which of the following is a carbonate ore?
a) Limonite
b) Siderite
c) Magnetite
d) Haematite
Answer:
b) Siderite

7. Which of the following is the ore of iron?
a) Limonite
b) Azurite
c) Stefinite
d) Cerrusite
Answer:
a) Limonite

8. Which of the following is not an ore of iron?
a) Haematite
b) Magnetite
c) Siderite
d) Anglesite
Answer:
d) Anglesite

9. Which of the following is an ore of silver?
a) Azurite
b) Prousitite
c) Cerrusite
d) Limonite
Answer:
b) Prousitite

10. Which of the following is a sulphate ore?
a) Galena
b) Zinc blende
c) Cerrusite
d) Anglesite
Answer:
d) Anglesite

11. Non-metallic impurities, rocky materials, and siliceous matter which are associated with ores are called as.
a) Slag
b) Flux
c) Gangue
d) residue
Answer:
c) Gangue

12. Gravity separation is suitable for
a) Oxide ore
b) Sulphide ore
c) Carbonate ore
d) Sulphate ore
Answer:
a) Oxide ore

13. Froth floatation is suitable for
a) Oxide ore
b) Sulphide ore
c) Carbonate ore
d) Sulphate ore
Answer:
b) Sulphide ore

14. In froth floatation, pine oil is used as a
a) Collector
b) depressing agent
c) Frothing agent
d) Flux
Answer:
c) Frothing agent

15. In froth floatation sodium ethyl Xanthate is used as a
a) Collector
b) depressing agent
c) frothing agent
d) Flux
Answer:
a) Collector

16. In froth floatation sodium cyanide is used as a
a) Collector
b) depressing agent
c) frothing agent
d) Flux
Answer:
b) depressing agent

17. The floatation property of the impurity ZnS present in galena is depressed by adding
a) Pure oil
b) Eucalyptus oil
c) Sodium cyanide
d) Sodium ethyl Xanthate
Answer:
c) Sodium cyanide

18. Which method of purification represented by the equation?
Ti(Impure) + 2I₂ \(\underrightarrow { 550K } \) Til₄ \(\underrightarrow { 1800K } \) Ti(pure) + 2I₂
a) Cupellation
b) Zone refining
c) Van-Arkel method
d) Mond’s process
Answer:
c) Van-Arkel method

19. Concentration of gold ore is done by
a) Cyanide leaching
b) Ammonia leaching
c) Alkali leaching
d) Acid leaching
Answer:
a) Cyanide leaching

20. Ammonia leaching is done for the concentration of the ore of
a) Silver
b) Copper
c) Aluminium
d) Zinc
Answer:
b) Copper

21. During roasting sulphide ores are converted into their
a) Metals
b) Oxides
c) Carbonates
d) nitrates
Answer:
b) Oxides

22. During the calcination of carbonate ore the expelled gas is
a) Carbon monoxide
b) Carbon dioxide
c) Sulphur dioxide
d) Nitrogen dioxide
Answer:
b) Carbon dioxide

23. Sulphite ores of metals are usually concentrated by froth floatation process. Which one of the following sulphide ore offers an exception and is concentrated by chemical leaching.
a) Argentite
b) galena
c) Copper pyrites
d) Sphalerite
Answer:
a) Argentite

24. Cinnabar is converted into mercury by
a) Reduction by metal
b) Reduction by hydrogen
c) Reduction by carbon
d) Auto reduction
Answer:
d) Auto reduction

25. Thermodynamically the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is
a) Positive
b) Negative
c) One
d) Zero
Answer:
b) Negative

26. For the reduction of metal oxide into metal a reducing agent is selected in such a way that for the coupled reaction it provides a
a) Large positive G value
b) Small positive G value
c) Large negative G value
d) Small negative G value
Answer:
c) Large negative G value

27. For the formation of various metal oxides Ellingham diagram is a graphical representation between
a) G° & S
b) G° & H
c) G° & T
d) H & S
Answer:
c) G° & T

28. In the Ellingham diagram, for most of the metal oxide formation the slope is
a) Positive
b) Negative
c) Zero
d) One
Answer:
a) Positive

29. Elements like Silicon and Germanium to be used as a semiconductor is purified by (PTA – 1)
a) heating under Vaccum
b) Van-Arkel method
c) Zone refining
d) Electrolysis
Answer:
c)Zone refining

30. If the e.m.f of the net redox reaction is positive, its G is
a) Positive
b) Negative
c) Zero
d) One
Answer:
b) Negative

31. Which of the following metal is refined by distillation?
a) Tin
b) Lead
c) Zinc
d) Bismuth
Answer:
c) Zinc

32. Which of the following is not refined by zone refining?
a) Germanium
b) Zirconium
c) Silicon
d) Gallium
Answer:
b) Zirconium

33. Which of the following is refined by the Mond process?
a) Silicon
b) Copper
c) Nickel
d) Zinc
Answer:
c) Nickel

34. Which of the following is defined by Van
Arkel method?
a) Gallium
b) Titanium
c) Germanium
d) Silicon
Answer:
b) Titanium

35. Which of the following metal is used in galvanization?
a) Copper
b) Aluminium
c) Zinc
d) Gold
Answer:
c) Zinc

36. Which is used in making luminous paints, fluorescent lights, and x-ray screens?
a) Brass
b) Zinc sulphide
c) Cast iron
d) Gold nanoparticles
Answer:
b) Zinc sulphide

37. Which is used for increasing the efficiency of solar cells?
a) Brass
b) Zinc sulphide
c) Cast iron
d) Gold nanoparticles
Answer:
d) Gold nanoparticles

38. Which is not refined by liquation?
a) Tin
b) Zinc
c) Lead
d) Bismuth
Answer:
b) Zinc. Zinc is refined by distillation.

39. Which is not refined by zone refining?
a) Silicon
b) Gallium
c) Zirconium
d) Germanium
Answer:
c) Zirconium. Zirconium is refined by Van Arkel method

X. Two Mark Questions

Question 1.
What is a mineral?
Answer:
A naturally occurring substance obtained by mining, which contains the metal in a free state or in the form of compounds like oxides, sulphides, etc; is called a mineral.

Question 2.
What is an ore?
Answer:
A mineral which contains high percentage of metal, from which it can be extracted conveniently and economically is called an ore.

Question 3.
What is a concentration of ores?
Answer:
The removal of non metallic impurities, rocky materials and siliceous matter (called as gangue) from the ores is known as concentration of ores.

Question 4.
What is leaching?
Answer:
The process of dissolving metal present in an ore in a suitable solvent to form a soluble metal salt or complex leaving the gangue undissolved is called leaching.

Question 5.
What is the reaction of Ammonia with Iron and copper salts? (PTA – 4)
Answer:
Ammonia reacts with metallic salts to give metal hydroxides (in case of Fe) or forming complexes (in case of Cu)
Fe³⁺ + 3NH⁺₄ → Fe(OH)₃ + 3NH⁺₄
Cu²⁺ + 4NH₃ → [Cu(NH3)4 ]²⁺
Tetra ammine copper (II) ion

Question 6.
What is acid leaching?
Answer:

• Sulphide ores ZnS, PbS can be leached with hot aqueous sulphuric acid.
• In this process, the insoluble sulphide is converted into soluble sulphate and elemental sulphur.
  2ZnS_(s) + 2H₂SO_4(aq) + O_2(g) → 2ZnSO_4(aq) + 2S_(s) + 2H₂O

Question 7.
What is the role of the depressing agent In the froth flotation process? (PTA – 1)
Answer:
When impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the flotation property of ZnS by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of zinc sulphide.

Question 8.
In the extraction of metal, the ore is first converted into metal oxide before reduction into metal, why?
Answer:

• In the concentrated ore, the metal exists in a positive oxidation state and hence it is to be reduced to an elemental state.
• From the principles of thermodynamics, the reduction of oxide is easier compared to the reduction of other compounds of metal.
• Hence before reduction, the ore is first converted into metal oxide.

Question 9.
Write about roasting.
Answer:

• Roasting is applied for the conversion of sulphide ores into their oxides.
• Concentrated ore is oxidised by heating with excess of oxygen below the melting point of the metal in a suitable furnace.
  2PbS + 3O₂ → 2PbO+ 2SO₂ ↑
• Roasting also removes impurities like aresenic, sulphur, phosphorous into their volatile oxides.
  4As + 3O₂ → 2AS₂O₃

Question 10.
Write about the extraction of metal by the process of reduction by carbon.
Answer:

• In this method oxide ore of the metal is mixed with coal (coke) and heated strongly in a blast furnace.
• This method can be applied to metals which do not form carbides with carbon at the reduction temperature.
  ZnO_(s) + C → Zn_(s) + CO_(g) ↑

Question 11.
Write about the extraction of metal by the process of reduction by hydrogen.
Answer:

•  This method can be applied to the oxides of the metals (Fe, Pb, Cu) which are less electropositive than hydrogen.
  Ag₂O_(s) + H_2(g) 2Ag_(s) + H₂O_(l) ↑
• Nickel oxide is reduced to nickel by a mixture of hydrogen and carbon monoxide (water gas)
  2NiO_(s) + CO_(g) + H_2(g) → 2Ni_(s) + CO_2(g) + H₂O_(l) ↑

Question 12.
Write about the extraction of metal by the process of reduction by metal.
Answer:

• In this process a metal oxide is reduced to metal by some active metals, like sodium, potassium and calcium.
  Rb₂O₃ + 3Mg → 2Rb + 3MgO
  TiO₂ + 2Mg → Ti + 2MgO
  ThO₂ + 2Ca \(\underrightarrow { 1260K } \) Th + 2CaO
• Alumino thermite process is also an example of reduction by metal.

Question 13.
How Cr₂O₃ is reduced to Cr by Al powder? (PTA – 6)
Answer:

• In this method a metal oxide is reduced to metal by aluminium.
• It is an exothermic process where heat is liberated.
• Cr₂O₃ is mixed with aluminium powder in a fire clay crucible.
• Ignition mixture is magnesium and barium peroxide.
  BBaO₂ + Mg —> BaO + MgO
• Temperature = 2400°C Heat liberated = 852KJmol^-1 This heat helps the reduction of Cr₂O₃ by Al.
  Cr₂O₃ + 2Al \(\underrightarrow { \triangle } \) 2Cr + Al₂O₃

Question 14.
What is auto reduction of metallic ores?
Answer:

• Simple roasting of some of the metallic ores give the crude metal.
• Use of reducing agent is not necessary.
• (eg) Cinnabar is roasted to give mercury.
  HgS_(s) + O_2(g) → Hg_(l) + SO_2(g) ↑

Question 15.
What is the role of graphite rods in the electrometallurgy of Aluminium? (PTA – 1)
Answer:
Electrolysis is carried in an iron tank lined with carbon which acts as a cathode. The carbon blocks immersed in the electrolyte acts as a anode.

Question 16.
Write about the distillation process of refining a metal?
Answer:
In this method, impure metal is heated to evaporate and the vapours are condensed to get pure metal.

This method is used for low boiling volatile metals like zinc and mercury.

Question 17.
Write about the liquation process of refining a metal?
Answer:

• This method is used to remove the impurities with high melting points from metals having relatively low melting points,
  (eg) Tin, lead, mercury, bismuth.
• Crude metal is heated to form a fusible liquid and allowed to flow on a sloping surface.
• Impure metal is placed on the sloping hearth of a reverberatory furnace.
• Impure metal is heated just above the melting point of the metal in the absence of air.
• The molten pure metal flows down. Impurities are left behind.
• Molten metal is collected and solidified.

Question 18.
Write the applications or uses of copper.
Answer:

• Copper is the first metal used by humans and extended use of its alloy bronze resulted in a new era, ‘Bronze age’.
• Used for making coins and ornaments along with gold and other metals.
• Copper and its alloys are used for making wires, water pipes and other electrical parts.

Question 19.
Write the applications or uses of gold.
Answer:

• Gold is one of the expensive and precious metals.
• Used for coinage and has been used as standard for monetary systems in some countries.
• Extensively used in jewellery in its alloy form with copper.
• Used in electroplating to cover other metals with a thin layer of gold in watches, artificial limb joints, cheap jewelry, dental fillings and electrical connectors.
• Gold nanoparticles are used for increasing the efficiency of solar cells.
• Used as catalyst.

Question 20.
Describe the underlying principle of froth floation process. (PTA – 3)
Answer:
This method is commonly used to concentrate sulphide ores such as galena (pbs) Zinc blende (Zns)

In this method, the metalic ore particles which are preferentially wetted by oil can be separated from gangue.

XI. Three Mark Questions

Question 1.
Write about gravity separation or hydraulic wash?
Answer:

• Ore with high specific gravity is separated from gangue with low specific gravity by simply washing with running water.
• The finely powdered ore is treated with rapidly flowing current of water.
• Lighter gangue particles are washed away by the running water.
• This method is used for concentrating native ore such as gold and oxide ores such as haematite, tinstone.

Question 2.
What is cyanide leaching?
Answer:

• Crushed ore of gold is leached with aerated dilute solution of sodium cyanide.
• Gold is converted into a soluble cyanide complex.
• The gangue alumino silicate remains insoluble.
  4Au_(s) + 8CN^–_(aq) + O_2(g)+ 2H₂O_(l) → 4[Au(CN)₂]^–_(aq) + 4OH^–_(aq)
• Gold can be recovered by reacting the deoxygenated leached solution with Zinc. Gold is reduced to its elemental state (zero oxidation state.)
• This process is called cementation.
  Zn_(s) + 2[Au(CN)₂]^–_(aq) → [Zn(CN)₄]^2-_(aq) + 2Au_(s)

Question 3.
Write about alkali leaching?
Answer:

• In this method, the ore is heated with aqueous alkali to form a soluble complex.
• Bauxite is heated with a solution of sodium hydroxide or sodium carbonate at 470K – 520K and 35 atm to form soluble sodium meta aluminate.
• The impurities iron oxide and titanium oxide are left behind.
  Al₂O_3(s) + 2NaOH_(aq) + 3H₂O_(l) → 2Na [Al(OH)₄]_(aq)
• The hot solution is decanted, cooled and diluted.
• This solution is neutralised by passing CO₂ gas to form a hydrated Al₂O₃ precipitate.
  2Na[AZ(0H)₄]_(aq)4CO_2(g) → Al₂O₃.XH₂O_(s) + 2NaHCO_3(aq)
• The precipitate is filtered off and heated around 1670K to get pure Alumina Al₂O₃.

Question 4.
Write about magnetic separation.
Answer:

• This method is applicable to ferromagnetic ores.
• It is based on the difference in the magnetic properties of the ore and the impurities.
• Non-magnetic tin stone can be separated from the magnetic impurities wolframite.
• Similarly magnetic ores chromite, pyrolusite can be removed from non-magnetic siliceous impurities.
  
• Crushed ore is poured on to an electromagnetic separator with a belt moving over two rollers of which one is magnetic.
• The magnetic part of the ore is attracted towards the magnet and falls as a heap close to the magnetic region.
• The non-magnetic part falls away from it.

Question 5.
Write about calcination. (PTA – 4)
Answer:

• Calcination is the process in which the concentrated ore is strongly heated in the absence of air.
• During this process water of crystallisation present in the hydrated oxide escapes as moisture.
• Any organic matter present also get expelled leaving the ore porous.
• This method can also be carried out with a limited supply of air.
• During the calcination of carbonate ore, carbon dioxide is liberated.
  PbCO₃ \(\underrightarrow { \triangle } \) PbO + CO₂ ↑
  CaCO₃ \(\underrightarrow { \triangle } \) CaO + CO₂ ↑
  Al₂O₃. 2H₂O \(\underrightarrow { \triangle } \) Al₂O_3(s) + 2H₂O_(g) ↑

Question 6.
Write about smelting.
Answer:

• Smelting is a process in which the concentrated ore is mixed with a mixture of a flux and reducing agent in a smelting furnace.
• Flux is a chemical substance which forms an easily fusible slag with gangue.
• Carbon, carbon monoxide, and aluminium are used as reducing agents.
• Iron oxide can be reduced by carbon monoxide.
  Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g) ↑
• Silica gangue present m the ore is acidic, hence a basic flux lime combines with it forming slag calcium silicate.
  

Question 7.
Write about the Ellingham diagram.
Answer:

• The graphical representation of the variation of the standard Gibbs free energy for the formation of various metal oxides with temperature is called the Ellingham diagram.
• Change in Gibbs free energy ∆G is given as ∆G = ∆H – T∆S
  ∆H = Enthalpy change T = Temperature in Kelvin S = Entropy change.
• For an equilibrium, ∆G° can be calculated using the equilibrium constant by the equation.
  ∆G° = -RT In Kp
• By treating the reduction of metal oxides as an equilibrium process Harold Ellingham used the above relationship to calculate ∆G° values at various temperatures.
• Fie Plotted T in the x-axis and ∆G° for the formation of metal oxides in the y axis.
• He obtained a straight line graph with ∆S as slope and ∆H as the y-intercept.

Question 8.
Write the observations from the Ellingham diagram.
Answer:

• For most of the metal oxide forming the slope is positive. This can be explained as follows. Oxygen gas is consumed during the formation of metal oxides resulting in the decrease of randomness. Hence ∆S becomes negative, T∆S is positive in the straight line equation.
• For the formation of carbon monoxide the graph is a straight line with a negative slope. In this case ∆S is positive because 2 moles of CO. gas is formed by consuming 1 mole of oxygen gas. This shows CO is more stable at higher temperature.
• As temperature increases ∆G for the formation of metal oxide becomes less negative and becomes zero at a particular temperature. Below this temperature is negative and the oxide is stable. Above this temperature ∆G is positive and the oxide is less stable. Metal oxides become less stable at higher temperature and their decomposition becomes easier.
• Due to phase transition (melting or evaporation) there is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO.

Question 9.
Write about Van – Arkel method for refining zirconium/titanium?
Answer:

• This method is based on the thermal decomposition of metal compounds to metals.
  (eg) Titanium and Zirconium.
• Impure titanium is heated in an evacuated vessel with iodine at 550K to form volatile titanium tetraiodide.
• The impurities do not react with iodine.
  Ti_(s) + 2I_2(s) → Til_4(vapour)
• Volatile titanium tetraiodide is passed over a tungsten filament at 1800K.
• Titanium tetraiodide is decomposed to pure titanium which is deposited over the filament.
• Iodine is reused. TiI_4(vapour) → Ti_(s) + 2I_2(s)

Question 10.
Write the applications or uses of aluminium.
Answer:

• For making heat exchangers/ sinks.
• For making our day to day vessels.
• For making aluminium foils for packing, food items.
• Alloys of aluminium with copper, manganese, magnesium, silicon are lightweight and strong hence used in design of aeroplanes and other forms of transport.
• Due to its high resistance to corrosion, it is used in the design of chemical reactors, medical equipments, refrigeration units and gas pipelines.
• It is a good electrical conductor and cheap, hence used in electrical overhead cables with a steel core for strength.

Question 11.
Write the applications or uses of iron.
Answer:

• Iron is one the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
• Cast iron is used to make pipes, valves and pumps stoves etc.
• Magnets can be made of iron and its alloys and compounds.
• An important alloy of iron is stainless steel which is very resistant to corrosion.
• It is used in architecture, bearings, cutlery, surgical instruments and jewellery.
• Nickel steel is used for making cables, automobiles, and aeroplane parts.
• Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 12.
Out of coke and CO, which is a better reducing agent for the reduction of ZnO? why? (PTA – 2)
Answer:

• Out of coke and CO, coke is a better reducing agent than CO for the reduction of ZnO.
• Reduction by carbon can be applied to zinc which does not form carbide with carbon at the reduction temperature.
  ZnO_(s) + C → Zn_(s) + CO_(g) ↑
• ZnO lies above CO in the Ellingham diagram meaning that CO is more stable than ZnO. Hence carbon can be used as a reducing agent for the reduction of ZnO. During reduction oxygen from ZnO combines with carbon used for reduction.

XII. Five Mark Questions

Question 1.
Explain the froth floatation method.
Answer:

• This is used to concentrate sulphide ores such as galena (PbS) Zinc blende (ZnS) etc.
• Metallic ore particles preferentially wetted by oil can be separated from gangue.
• Crushed ore is mixed with water and a frothing agent like pine oil or eucalyptus oil.
• A small amount of sodium ethyl xanthate is added as a collector.
• A froth is formed by blowing air through the mixture.
• The collector molecules attach to the ore particles and make them water repellent.
• As a result ore particles wetted by the oil rise to the surface along with the froth.
• The froth is skimmed off and dried to recover the concentrated ore.
• Gangue particles preferentially wetted by water settle at the bottom.
• If the sulphide ore contains other metal sulphides as impurities, they are selectively prevented from coming to the froth by using depressing agents like sodium cyanide, sodium carbonate, etc.
  
• Sodium cyanide depresses the floatation property of the impurity ZnS present in galena (PbS) by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of ZnS.

Question 2.
How is copper extracted from its ore. (PTA – 5)
Answer:

• Principle ore: Copper pyrites.
• Concentration: Froth floatation Concentrated ore is heated in a reverberatory furnace with an acidic flux silica.
• The basic ferrous oxide formed reacts with silica to form the slag ferrous silicate.
• Mutually soluble metal sulphides Cu2S and FeS known as copper matte is formed.
  
• Matte is removed from the slag and fed to the converting furnace.
• FeS present in the matte is first converted to FeO.
• FeO is removed as slag with silica.
• The remaining copper sulfide is oxidised to cuprous oxide.
• Cuprous oxide and copper sulphide react to form metallic copper. :
  2Cu₂S_(l,s) + 3O_2(g) 2Cu₂O_(l,s) + 2SO_2(g)
  2Cu₂O_(l) 6Cu_(l) + SO_2(g)
• SO₂ is liberated through molten copper and on solidification it has blistered appearance. This copper is called blister copper.

Electrorefining:
Cathode: Thin pure sheet of copper.
Anode: Impure Copper
Electrolyte: CuSO₄ solution + dil H₂SO₄
On passing, current pure copper is deposited at the cathode.

Question 3.
Explain the thermodynamic principle of metallurgy.
Answer:

• Extraction of metals can be carried out by using different reducing agents.
• Consider the reduction of a metal oxide M_x O_y
  \(\frac{2}{Y}\) M_xO_y(s) → \(\frac{2}{Y}\)M_(s) + O_2(g) ……………….. 1
• Above reduction may be carried out with carbon.
• In this case the reducing agent carbon may be oxidised to CO or CO₂
  C + O₂ → CO_2(g) …………….. 2
  2C + O₂ → 2CO_(g) ……………………… 3
• If carbon monoxide is used as a reducing agent, it is oxidised to CO₂
  2CO + O₂ → 2CO₂ ……………………… 4
• A suitable reducing agent is selected based on thermodynamic considerations.
• For a spontaneous reaction, the change in free energy ∆G should be negative.
• Thermodynamically the reduction of metal oxide [equation (1)] with a given reducing agent [equation 2,3, or 4] can occur if the free energy change for the coupled reaction [equation 1&2, 1&3 or 1&4] is negative.
• Hence the reducing agent which gives large negative ∆G value for the coupled reaction is selected.

Question 4.
Write the applications of Eliingham diagram.
Answer:

• Eliingham diagram helps us to select a suitable reducing agent and appropriate temperature range for reduction.
• Reduction of metal oxide to metal is considered as a competition between the element used for reduction and the metal to combine with oxygen.
• If metal oxide is more stable, oxygen remains with the metal.
• If oxide of the element used for reduction is more stable, oxygen from metal oxide combines with the element used for reduction.
• From Eliingham diagram the relative stablility of different metal oxides at a given temperature can be inferred.
• Eliingham diagram for the formation of Ag₂O and HgO is at the upper part and their decomposition temperatures are 600 and 700K respectively. This shows that these oxides are unstable at moderate temperatures and will decompose on heating even in the absence of a reducing agent.
• Eliingham diagram is used to predict the thermodynamic feasibility of reduction of oxides of one metal by another metal.
• For example, in the Eliingham diagram, the line for the formation chromium oxide ties above that of aluminium, meaning that Al₂O₃ is more stable than Cr₂O₃. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide.
• However, aluminium can not be used to reduce the oxides of magnesium and calcium since they occupy lower position than aluminium oxide in the Eliingham diagram.
• Any metal can reduce the oxides of other metals that are located above it in the diagram.
• Carbon line cuts across the lines of many metal oxides and hence it can reduce all these metal oxides at sufficently high temperature.

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Textual Exercises (Text Book Page No. 134)

Read and Understand (Text Book Page No. 134)

A. Read the following statements. Say True or False.

1. Gulliver was the captain of the ship.
Answer:
False

2. One of the Lilliputians gave a ten minutes talk in Gulliver’s language.
Answer:
False

3. Gulliver took the small creatures in his hand and crushed them.
Answer:
False

4. The horses were four and a half inches tall.
Answer:
True

5. The war between the two kingdoms ended in peace.
Answer:
True

B. Identify the speaker/character.

1. He felt something moving along his body almost up to his chin.
Answer:
Gulliver

2. They somehow managed to put him on the platform.
Answer:
The Lilliputians

3. “Don’t let us down now, Gulliver; we need your help”.
Answer:
The Emperor

C. Choose the right answer. (Text Book Page No. 135)

1. Gulliver managed to reach the land as he was
a) A doctor
b) One of the crews
c) a swimmer
d) the captain
Answer:
c) a swimmer

2. Gulliver was set free because the emperor
a) was afraid of him
b) confirmed that he was not harmful.
c) was a kind hearted person.
d) wanted to get something from him.
Answer:
b) confirmed that he was not harmful

3. Gulliver was hailed as a hero because he
a) made the army of Blefuscu giddy
b) fought with the army of Blefuscu.
c) drowned the army of Blefuscu in the water.
d) defeated the emperor of Blefuscu.
Answer:
d) defeated the emperor of Blefuscu

D. Discuss in groups. Retell the story in your own words. Each one should say one sentence.

You can begin like this:
Gulliver was travelling in a ship. One stormy night, the ship was wrecked ………

Gulliver was travelling in a ship. One stormy night, the ship was wrecked. Gulliver was a good swimmer. So, he reached the land by swimming. The land appeared strange and lonely. He saw no human beings, he was hungry and tired. He fell on a patch of grass and slept. When he woke up he found his hands and legs tied down. There were a number of creatures not more than six inches high. He later learned they were Lilliputians. He was taken to the Emperor. He received Gulliver. He set him free with a condition that he would do no harm to them. Once, the neighbouring kingdom, declared war on Lilliput. The Emperor needed Gulliver’s help.

Gulliver walked into a sea and tied with a long rope all the ships of the enemy together he pulled the ships going round and round the whole day. They were not in a position to fight. So, the enemy king begged for peace between the two countries. Gulliver was hailed as a hero.

E. Think and answer.

1. How did Gulliver overcome the adversity?
Answer:
Gulliver’s ship was wrecked. He reached a land where the Lilliputians lived. They were not more than six inches high. Gulliver was taken to the emperor with his hands and legs tied. The emperor received him. Gulliver was given food and drink. He was lodged in an old and amused temple. Gulliver overcame his adversity by helping the Emperor and his people in many ways. Once the neighbouring kingdom declared war on Lilliput. It was Gulliver who defeated the enemy king made peace with the Emperor. Thus Gulliver overcame his adversity by being kind, helping, and wise.

2. How should one react to adversity?
Answer:
When Gulliver faced adversity he remained patient. He was kind to all the Lilliputians and the Emperor. Once the neighbouring country declared war on Lilliput. Then it was Gulliver who saved Lilliput and earned the respect and love of the Emperor and the people.

3. Describe Gulliver’s encounter with the army of Blefuscu?
Answer:
The king of Blefuscu once declared war on Lilliput. The Emperor needed help from Gulliver. Gulliver walked into the sea and tied with a long rope all the hundred ships of the enemy together. He dragged them in the water and pulled the ships, the whole day going round and round. The army of Blefuscu was giddy and was not in a position to fight. The king surrendered to the Emperor and made peace with him. It was Gulliver who brought victory to the Emperor.

Project

F. Look at the traffic signals and write down the traffic rule against each signal.

Answer:

Connecting to Self

G. Look at these pictures. Think of how you must behave when you visit these places. Discuss with your partner and complete the table.

Answer:

Use dust bins	Do not throw rubbish in public places
Put waste into dustbins	Don’t throw down uneaten food.
Use toilet	Don’t spit on the road.
Keep the coast clean	Don’t pass urine on the way.
Keep the environment clear	Don’t walk the meadow.
Dust-free	Don’t pluck flowers.

Step to Success

e.g. 1. Leaves 2. Fruit 3. Seed 4. Flowers 5. Root
a) 2,4,5,1,3
b) 3,5,1,4,2
c) 1,2,3,4,5
d) 5,3,1,2,4
Answer:
b) 3, 5,1, 4, 2

1. 1. Drive 2. Get in 3. Arrive 4. Park 5. Open door
a) 2,1,3,5,4
b) 5,2,1,3,4
c) 3,4,5,1,2
d) 3,5,1,2,4
Answer:
b) 5, 2,1,3, 4

2. 1. Travel 2. Book 3. Plan 4. Confirm 5. Enjoy
a) 3,2,4,1,5
b) 4,5,3,2,1
c) 1,2,3,4,5
d) 5,4,2,3,1
Answer:
a) 3, 2, 4,1, 5

3. 1. Rest 2. Return 3. Supper 4. Go out 5. Visit
a) 1,2,3,4,5
b) 4,5,1,2,3
c) 4,5,2,3,1
d) 5,4,3,2,1
Answer:
c) 4, 5, 2,3,1

4. 1. Check out 2. Pack 3. Pay bill 4. Vacate 5. Drive
a) 3,4,5,2,1
b) 1,2,3,4,5
c) 5,4,3,2,1
d) 2,4,3,1,5
Answer:
d) 2, 4, 3,1, 5

5. 1. Wait 2. Slow 3. Go 4. Stop 5. Get ready
a) 2,4,1,5,3
b) 1,2,3,4,5
c) 5,4,3,2,1
d) 3,4,2,1,5
Answer:
a) 2,4,1, 5,3

Gulliver’s Travel Summary in English

While Gulliver was going, on a long voyage, his ship sank in a violent storm. However, he managed to escape and reached the island by nearing. He was hungry and exhausted so he fell to a patch of grass. When he woke he was tied down by dwarf (six inches) people who called themselves Lilliputians. He was taken to the capital of the Island. The Emperor received him. He was set free on condition. He was happy there. The neighbouring king of Blefuscu declared war on Lilliput. The Emperor needed Gulliver’s help. Gulliver pulled the all shipping the whole day. The king of Blefuicu begged for peace. The Emperor agreed. Gulliver was hailed as a hero by Lilliputians.

Gulliver’s Travel Summary in Tamil

கலிபா நீண்ட பயணத்தில் செல்லும்போது, அவரது கப்பல் கடுமையான புயலில் மூழ்கியது. இருப்பினும், அவர் தப்பித்து அருகிலுள்ள தீவை அடைந்தார். அவர் பசியில் களைப்பும் அடைந்ததார். அதனால், அவர் புல்லின் மீது விழுந்தார். அவர் எழுந்தபோது லில்லிபுத்தியர்கள் என்று அழைத்த குள்ளர்கள் (ஆறு அங்குல) மக்களால் அவர் கட்டப்ட்டார். அவர் தீவின் தலைநகருக்கு கொண்டு செல்லப்ட்டார். சக்கரவர்த்தி அவரை வரவேற்றார். அவர் அங்கு மகிழ்ச்சியாக இருந்தார். பக்கத்து மன்னர் பிலிபியூஸ்க்கு லில்லிபுட்டின்மீது போரை அறிவித்தார். மன்னர் கலிபரின் உதவியை வேண்டினார். கலிபர் அனைத்து கப்பல்களையும் கட்டி நாள்முழுவதும் இழுத்தார். பிலியூயூஸ்க்கு அரசன் சமாதானத்தை கெஞ்சிக் கேட்டார். ஒப்புக்கொண்டார். கலிபர் தலைவர் என புகழப்பட்டார்.

Tamilnadu State Board New Syllabus Samacheer Kalvi 6th English Guide Pdf Term 2 Supplementary Chapter 1 Think to Win Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 6th English Solutions Term 2 Supplementary Chapter 1 Think to Win

6th English Guide Think to Win Text Book Back Questions and Answers

Textual Exercises (Text Book Page No. 105)

A. Choose the correct answer.

1. “Girls you are a good team.”
Which team do the girls belong to?
a) Badminton
b) Hockey
c) Relay
d) Volleyball
Answer:
c) Relay

2. Springfields is the name of a …………………….. .
a) team
b) house
c) company
d) school
Answer:
d) school

3. The inter-school sports meet refers to competitions among the …………………….. .
a) teams of the same school
b) schools in the locality
c) schools in neighborhood
d) schools from other districts
Answer:
b) schools in the locality

4. Seema is Rucha’s …………………….. .
a) friend
b) teammate
c) younger sister
d) opponent
Answer:
c) younger sister

5. Order the names of the members in Team B relay event. Shabnam was followed by
a) Neelam, Aruna, Rucha
b) Arana, Rucha, Neelam
c) Neelam, Rucha, Arana
d) Arana, Neelam, Rucha
Answer:
a) Neelam, Aruna, Rucha

B. Answer the following:

1. How does Rucha differ from her sister?
Answer:
Rucha is overprotective of herself. She never did things rashly. Even in running ‘ and playing, she would be conscious of her movements. But Seema, her younger sister was bold and was ready to take any risks.

2. ‘Springfields has a runner and they call her P.T. Usha.’ Why did they call her so?
Answer:
The runner of Springfields ran as fast as RT. Usha. So she was called P. T. Usha.

3. Describe the qualities of the new P.T. Instructor Mr. Prakash.
Answer:
The new P.T. Instructor was very enthusiastic about sports and drove the children hard – praising them, scolding and correcting them. But most of the time, he encourages and advises them a lot.

4. What words were ringing in Rucha’s ears when she was running in the relay?
Answer:
The words “Think to win” were running in ‘Rucha’s ears when she was running in the relay.

5. What did Rucha finally realise about herself?
Answer:
She realized that she had overcome her hesitations and denials. She could win, whenever she chose to do so.

C. Read the given lines and answer the questions : (Text Book Page No. 106)

1. His voice came from some distance away and,
Rucha realized that he was not holding the bicycle anymore.
“I will fall / will fall” she wailed.
a) What was she afraid of?
Answer:
She was afraid of falling off the bicycle.

b) Was the boy closely following her?
Answer:
No, the boy was not closely following her.

2. For the past two weeks he had been teaching her to ride.
a) Who was teaching whom? ‘
Answer:
Vishnu was teaching Rucha to ride the bicycle.

b) What was he teaching?
Answer:
He was teaching me how to ride a bicycle.

3. Even ‘P. T. Usha’came to shake her hand.
“I thought I was fast, but you were simply superb!”
she shook hands with her.
a) Who does the word ‘you’ refer to?
Answer:
The word ‘you’ refers to Rucha.

b) What quality of the speaker is revealed?
Answer:
The true spirit of a sportsperson is revealed here

D. Think and answer.

Imagine you are Rucha and make a diary entry on your feelings about the day’s happenings and your victory. The start is given. Complete the diary.

Answer:

E. Discuss in class

How did Rucha overcome her self-doubts? How can shyness and fear be overcome?
Answer:
Rucha overcame her self-doubts. She was always thinking of Prakash Sir’s words “Think to A win’. Then he said that he wanted her to have a positive attitude. These words of Prakash sir made Rucha overcome her self-doubts.

We can overcome shyness and fear only by forgetting ourselves while playing the game.

Step to Success (Text Book Page No. 108)

Identify the sport name from the given wuzzles. One is done for you.

Answer:

Find the odd one out:

eg. Weight lifting, Boxing, Silambam, Fencing
Answer:
Weight lifting

1. Hide and Seek, Kho-Kho, Tennikoit, Kabaddi
Answer:
Tennikoit

2. Badminton, Cycling, Tennis, Squash
Answer:
Cycling

3. Trapeze, Throw Ball, Bowling, Goalball
Answer:
Trapeze

4. Snooker, Polo, Five Pins, Carrom Board
Answer:
Polo

5. Cricket, Base Ball, Hockey, Basket Ball
Answer:
Base Ball

Think to Win Summary in English

Rucha was a ten years old girl, she learned to ride a bicycle. When she was riding the bicycle, she was wobbled and fell down. Even in running and playing, she would be conscious of her movements. Inter-house badminton matches will start in her school next week. Rucha was on the senior team. She lost the game by just two points. She became gloomy and despair. She would train with the rest of them for relay matches under the new PT instructor, Prakash Sir. The new P.T instructor Prakash Sir encouraged her during the relay match by saying that only ‘Think to Win’. At last, She won the relay match.

Think to Win Summary in Tamil

ருச்சா ஒரு பத்துவயது பெண். அவள் சைக்கிள் ஓட்டக் கற்றுக்கொண்டாள். அவள் சைக்கிள் ஓட்டும்போது அவள் தள்ளாடி கீழே விழுந்தாள். ஓடுவதிலும் விளையாடுவதிலும் கூட அவள் எச்சரிக்கை விழிப்புணர்வுடன் இருப்பாள். இண்டர்ஹவுஸ் பூப்பந்து போட்டிகள் அவரது பள்ளியில் அடுத்த வாரத்தில் தொடங்கும். ருச்சா இளையோர் அணியில் இருந்தாள். அவன் இரண்டு புள்ளிகளால் ஆட்டத்தை இழந்தாள். அதனால் அவள் விரக்தியடைந்தாள். பிரகாஷ் ஐயா என்ற புதிய பயிற்சியாளரின் அறிவுறுத்தலின் கீழ் தொடர் ஓட்டப்போட்டிக்காக அவள் மற்றவர்களுடன் பயிற்சி பெறுவாள். புதிய பயிற்சியாளர் பிரகாஷ் சார் வெற்றி பெற யோசி மட்டுமே என்று ருச்சாவை தொடர் ஓட்டப் போட்தொடர் ஓட்டப்போட்டியின் போது அவளுக்கு அறிவுரை கூறி ஊக்குவித்தார். கடைசியாக அவள் தொடர் ஓட்டப் போட்டியில் வென்றாள்.

Think to Win About the Author in English

The author of this entry Mrs.Lata Bilaney Kaku, known as Lata Kaku’ is one of the famous freelance writers in India. She was bom in 1969 and completed her schooling in Presentation Convent, Delhi. She has attained the Honours degree for Zoology at the University of Delhi.

Think to Win About the Author in Tamil

லதா காக்கு என அறியப்படுகிற இந்தக் கதையின் ஆசிரியரான லதா பிலானி காக்கு இந்தியாவின் பத்திரிக்கை சாராத புகழ்பெற்ற பொது எழுத்தாளர்களில் ஒருவர் ஆவார். 1969இல் பிறந்த இவர், டெல்லியில் உள்ள ப்ரசண்டேஷன் கான்வென்ட்டில் தன்னுடைய பள்ளிப்படிபை முடித்தார். டெல்லி பல்கலைக் கழகத்தில் விலங்கியல் துறையில் இவர் கௌரவப் பட்டம் பெற்றுள்ளார்.