Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 9 Locomotion and Movement Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

11th Bio Zoology Guide Locomotion and Movement Text Book Back Questions and Answers

Part I

Question 1.
Muscles are derived from
a) ectoderm
b) mesoderm
c) endoderm
d) neuro ectoderm
Answer:
b) mesoderm

Question 2.
Muscles are formed by
a) myocytes
b) leucocytes
c) osteocytes
d) lymphocytes
Answer:
a) myocytes

Question 3.
The muscles attached to the bones are called
a) skeletal muscle
b) cardiac muscle
c) involuntary muscle
d) smooth muscles
Answer:
a) skeletal muscle

Question 4.
Skeletal muscles are attached to the bones by
a) tendon
b) ligament
c) pectin
d) fibrin
Answer:
a) tendon

Question 5.
The bundle of muscle fibres is called
a) Myofibrils
b) fascicle
c) sarcomere
d) sarcoplasm
Answer:
b) fascicle

Question 6.
The pigment present in the muscle fibre to store oxygen is
a) myoglobin
b) troponin
c) myosin
d) actin
Answer:
a) myoglobin

Question 7.
The functional unit of a muscle fibre is
a) sarcomere
b) sarcoplasm
c) myosin
d) actin
Answer:
a) sarcomere

Question 8.
The protein present in the thick filament is
a) myosin
b) actin
c) pectin
d) leucin
Answer:
a) myosin

Question 9.
The protein present in the thin filament is
a) myosin
b) actin
c) pectin
d) leucin
Answer:
b) actin

Question 10.
The region between two successive Z-discs is called a
a) sarcomere
b) microtubule
c) myoglobin
d) actin
Answer:
a) sarcomere

Question 11.
Each skeletal muscle is covered by
a) epimysium
b) perimysium
c) endomysium
d) hypomysium
Answer:
a) epimysium

Question 12.
Knee joint is an example of
a) saddle joint
b) hinge joint
c) pivot joint
d) gliding joint
Answer:
b) hinge joint

Question 13.
Name of the joint present between the atlas and axis is
a) synovial joint
b) pivot joint
c) saddle joint
d) hinge joint
Answer:
b) pivot joint

Question 14.
ATPase enzyme needed for muscle contraction is located in
a) actinin
b) troponin
c) myosin
d) actin
Answer:
c) myosin

Question 15.
Synovial fluid is found in
a) Ventricles of the brain
b) Spinal cord
c) immovable joint
d) freely movable joints
Answer:
d) freely movable joints

Question 16.
Inflammation of joints due to accumulation of uric acid crystals is called as
a) Gout
b) myasthenia gravis
c) osteoporosis
d) osteomalacia
Answer:
a) Gout

Question 17.
Acetabulum is located in
a) collar bone
b) hip bone
c) shoulder bone
d) thigh bone
Answer:
b) hip bone

Question 18.
Appendicular skeleton is
a) girdles and their limbs
b) vertebrae
c) skull and vertebral column
d) ribs and sternum
Answer:
a) girdles and their limbs

Question 19.
The type of movement exhibits by the macrophages are
a) flagellar
b) ciliary
c) muscular
d) amoeboid
Answer:
d) amoeboid

Question 20.
The pointed portion of the elbow is
a) acromion process
b) glenoid cavity
c) olecranon process
d) symphysis
Answer:
c) olecranon process

Question 21.
Name the different types of movement
Answer:

• Amoeboid movement
• Ciliary movement
• Flagellar movement
• Muscular movement

Question 22.
Name the filaments present in the Sarcomere.
Answer:
Thick and thin filaments are the two types of filaments present inside the sarcomere.

Question 23.
Name the contractile proteins present in the skeletal muscle.
Answer:

• Myosin – Thick filament
• Actin – Thin filament

Question 24.
When describing a skeletal muscle. What does “Striated mean?
Answer:
Each skeletal muscle fibre has a repeated series of dark and light bands. The dark A-bands and light I-bands give a striated appearance to the muscle.

Question 25.
How does an isotonic contraction takes place?
Answer:
In isotonic contraction, the length of the muscle changes but the tension remains constant.
(eg) lifting dumbbells and weight lifting.

Question 26.
How does an isometric contraction take place?
Answer:
In isotonic contraction, the length of the muscle changes but the tension remains constant. The force produced is unchanged, e.g., lifting dumbbells and weight lifting.

Question 27.
Name the bones of the skull.
Answer:
The cranial bones are 8 in number.

1. Paired parietal
2. Paired temporal
3. frontal
4. Sphenoid
5. Occipital
6. Ethmoid

Question 28.
Which is the only jointless bone in the human body?
Answer:
The jointless bone is the hyoid bone in our throat. The hyoid bone (lingual bone) is a horseshoe.

Question 29.
List the three main parts of the axial skeleton.
Answer:

1. Cranium
2. Hyoid (Lingual)
3. Vertebral column
4. Thoracic cavity.

Question 30.
How is tetany caused?
Answer:
Due to the deficiency of parathyroid hormone the level of calcium decreases in the blood that leads to rapid muscle spasm called tetany.

Question 31.
How does rigor mortis happen?
Answer:
After the death of an individual, the membrane of muscle cells becomes more permeable to calcium ions. This happens due to the partial contraction of skeletal muscles. The contracted muscles are unable to relax. This condition is known as rigor mortis.

Question 32.
What are the different types of rib bones that form the rib cage?
Answer:

• True rib bones (First 7 pairs)
• False rib bones (8,9,10th pairs)
• Floating rib bones (11 and 12th pair)

Question 33.
What are the bones that make the pelvic girdle?
Answer:

• Ilium
• Ischium
• Pubis

Question 34.
List the disorders of the muscular system.
Answer:

• Muscle fatigue
• Atrophy
• Muscle pull
• Muscular dystrophy

Arthritis

• Osteoarthritis
• Rheumatoid arthritis
• Gout

Question 35.
Explain the sliding-filament theory of muscle contraction.
Answer:
Sliding filament theory is an active process. It is proposed by Andrw F. Huxley in 1954 and Rolf Niedergerke.

• Muscle contraction is initiated by a nerve impulse sents by the central nervous system through a motor neuron
• When the nerve impulse reaches neuromuscular junction acetylcholine is released and created action potential.
• This action potential triggers the release of calcium from the sarcoplasmic reticulum
• The released calcium ions bind to troponin on thin filaments.
• The active sites are exposed to the heads of myosin to form a cross bridge. Hence actin and myosin form a protein complex called actomyosin.
• Utilizing the energy released from the hydrolysis of ATP the myosin head rotates until it forms a 90° angle with a long axis of the filament.
• The power stroke begins after the myosin head and hinge region tilt from a 90° angle to a 45° angle.
• The cross-bridge transforms into strong high force bond which allows the myosin head to swells it.
• When the myosin head swells it pulls the attached actin filament towards the centre of the A – band.
• The myosin returns back to its relaxed state and releases ADP and phosphate ions. A Newer ATP molecule binds to the head of myosin and the cross-bridge is broken.
• At the end of each power stroke each myosin head detaches from actin then swivels back and binds to a new actin molecule to start another contraction cycle.
• The power stroke repeats many times and the thin filaments move toward the centre of the sarcomere.
• In this process, there is no change in the lengths of thick or thin filaments.
• The Z – discs attached to the actin filaments are also pulled inwards from both sides causing the shortening of the sarcomere. This process continues.
• When motor impulse stops the calcium ions are purnbed back into the sarcoplasmic reticulum results in the masking of the active sites of the actin filament and the myosin head fails to bind with the actin and causes Z – discs back to their original relaxed position.

Question 36.
What are the Benefits of regular exercise?
Answer:

1. The benefits of regular exercise are:
2. The muscles used in exercise grow larger and stronger.
3. The resting heart rate goes down.
4. More enzymes are synthesized in the muscle fibre.
5. Ligaments and tendons become stronger.
6. Joints become more flexible.
7. Protection from a heart attack.
8. Influences hormonal activity.
9. Improves cognitive functions.
10. Prevents obesity.
11. Promotes confidence, esteem.
12. Aesthetically better with a good physique.
13. Overall well-being with good quality of life.
14. Prevents depression, stress, and anxiety.

Part II.

11th Bio Zoology Guide Locomotion and Movement Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
The pseudopodia of amoeba is formed from
a) Cytoplasm
b) Nucleoplasm
c) Sarcoplasm
d) all the above
Answer:
a) Cytoplasm

Question 2.
Where is ciliary movement taking place?
Answer:
a) Respiratory tract
b) Stomach
c) oesophagus
d) Reproductive tract
Answer:
d) Reproductive tract

Question 3.
Name the movement of spermatozoa
a) amoeboid movement
b) flagellated movement
c) Ciliary movement
d) None of the above
Answer:
b) flagellated movement

Question 4.
Muscles are made up of ………………….
a) Myocytes
b) Lymphocytes
c) adenocytes
d) leucocytes
Answer:
a) Myocytes

Question 5.
The cytoplasm of the muscle fibre is called:
a) Myofibril
b) Sarcomere
c) Sarcoplasm
d) Sarcolemma
Answer:
c) Sarcoplasm

Question 6.
Name the connective tissue which covers the entire muscle?
a) Epimycium
b) Perimycium
c) Endomyciurn
d) Mesornyscium
Answer:
a) Epimycium

Question 7.
Name the membrane which covers each fasiculi
a) Epimycium
b) Perirnycium
c) Endomyciurn
d) Mesomvscium
Answer:
b) Perirnycium

Question 8.
Name the membrane which covers each muscle fibre?
a) Epimycium
b) Perimycium
c) Endomycium
d) Mesomysciurn
Answer:
c) Endomycium

Question 9.
Match and Find the Correct Pair
1. Sarcoplasm – a) Respiratory pigment
2. Myoglobin – b) Glucose giver
3. Glycosome – c) Unit of skeletal muscle
4. Sarcomere – d) cytoplasm

Answer:
a) I- d, II – a, III – b,IV- c

Question 10.
Find the correct and wrong statement and arrange the following statement.
1. The contraction of muscle fibres depends on the actin and myosin protein.
2. The thick muscle fibres depends on Myosin.
3. Each meromyosin molecule will have a globular head with a long arm.
4. The head of the meromyosin bears an actin-binding site and an ATP binding site.

Answer:
a) True True False True

Question 11.
Name the protein that regulates muscle contraction.
a) Tropomyosin and actin
b) Troponin and myosin
c) tropomyosin and troponin
d) None of the above
Answer:
c) tropomyosin and troponin

Question 12.
Who has proposed sliding – filaments hypothesis?
a) Andrew F. Huxley and Rolf Nieder gerke
b) Andrew F. Huxley and Nelson
c) Andrew F. Pluxley and Darwin
d) Andrew F. Huxley and Mendal
Answer:
a) Andrew F. Huxley and Rolf Nieder gerke

Question 13.
Which chemical initiates the opening of multiple gated channels in sarcolemma?
a) Epinephrine
b) Norepinephrine
c) acetylcholine
d) erythromycin
Answer:
c) acetylcholine

Question 14.
Find out the wrong pair
a) Fast – oxidative fibres – have high ATP ase activity
b) Slow – oxidative fibres – low rates of ATP ase activity
c) Oxidative fibres – less number of mitochondria
d) Red muscle fibres – oxidative fibres
Answer:
c) Oxidative fibres – less number of mitochondria

Question 15.
The skeletal system is formed from this layer!
a) Ectoderm
b) Mesoderm
c) Endoderm
d) Neuroderm
Answer:
b) Mesoderm

Question 16.
………………………… a number of bones forms the endoskeleton of man?
a) 210
b) 220
c) 206
d) 209
Answer:
c) 206

Question 17.
Hove many bones are there in the axial and appendicular skeleton?
a) 80 and 126
b) 126 and 80
c) 80 and 120
d) 80 and 118
Answer:
a) 80 and 126

Question 18.
How many bones are there in the facial and cranial bones?
a) 14 and 9
b) 14 and 8
c) 14 and 10
d) 14 and 12
Answer:
b) 14 and 8

Question 19.
Name the opening of the temporal bone
a) External auditory meatus
b) Nasal opening
c) Optic opening
d) Mouth
Answer:
a) External auditory meatus

Question 20.
Name the U-shaped single bone present at the base of the buccal cavity.
a) Palantine bone
b) Hyoid bone
c) Ethmoid bone
d) Sphenoid bone
Answer:
b) Hyoid bone

Question 21.
How many bones from the vertebral column?
a) 33
b) 32
c) 30
d) 36
Answer:
a) 33

Question 22.
Match and find the correct pair
1) Cervical vertebrae – a) – 5
2) Thoraic – b)-l
3) Pelvic bone – 0-7
4) Cocyx bone – d) -12

Answer:
a) I -c, II – d, III – a,IV – b

Question 23.
In which bone ………….. situated?
a) Cervical vertebra
b) thoracic vertebra
c) Pelvic vertebra
d) all the above
Answer:
d) all the above

Question 24.
Name the first vertebra?
a) Atlas
b) Maleus
c) Incus
d) Stapes
Answer:
a) Atlas

Question 25.
Name the second vertebra?
a) Atlas
b) Axis
c) Maleus
d) Stapes
Answer:
b) Axis

Question 26.
How many are the number of ribs?
a) 14 pair
b) 13 pair
c) 12 pair
d) 15 pair
Answer:
c) 12 pair

Question 27.
Match and find the correct answer
1) ribs – a) – 1, 7
2) True ribs – b) -11,12
3) False ribs – c) – Bicephalic
4) Floating ribs – d) – 8,10

Answer:
b) I – c, II – a, III – d,IV -b

Question 28.
How many bones are there appendicular skeleton?
a) 130
b) 140
c) 126
d) 122
Answer:
c) 126

Question 29.
Which forms the appendicular skeleton?
a) Upper and hindlimbs
b) Upper limbs and thoracic bones
c) Hind and vertebral column
d) Hind and cranial bones
Answer:
a) Upper and hindlimbs

Question 30.
Name the bones which joint the axial and appendicular skeleton.
a) Clavicle bone
b) Scapula
c) Acromian process
d) None of the above
Answer:
a) Clavicle bone

Question 31.
Flow many bones are there in the upper arm.
a) 30
b) 32
c) 34
d) 36
Answer:
a) 30

Question 32.
Find the wrong pair.
a) Wrist bone – 8
b) Fore arm bones – 30
c) Facial bone -16
d) Cranial bones – 8
Answer:
c) Facial bone -16

Question 33.
Find the wrong pair
a) Palm bones – 5
b) Phalanges -14
c) Thoracic bone – 12
d) Vertebral column – 33
Answer:
d) Vertebral column – 33

Question 34.
Match and find the correct pair.
1) Cervical vertebra – a) – 12
2) Thoracic vertebra – b) – 1
3) Pelvic vertebra – c) – 7
4) Coccyx – d)-5

Answer:
a) I- c, II – a, III – d, IV – b

Question 35.
The oxidative skeletal muscle fibres are termed as
a) Fatty muscle fibres
b) White muscle fibres
c) red muscle fibres
d) yellow muscle
Answer:
c) red muscle fibres

Question 36.
Whether the following statement is correct or wrong find out the correct sequence.
1) Pelvic bone Ilium, Ischium, and pubis
2) Acetabulam cavity is present in the sacrum
3) The head of the thigh bone fits in the acetabulum cavity
4) The pubic bones articulate anteriorly at the pubic symphysis
Sequence:
a) 1 – True, 2 – False, 3 – True, 4 – True
b) 1 – False, 2 – False, 3 – True, 4 – True
c) 1 – False, 2 – True, 3 – True, 4 – True
d) 1 – True, 2 – True, 3 – False, 4 – True
Answer:
a) 1 – True, 2 – False, 3 – True, 4 – True

Question 37.
Which is the prominent bone of the pelvic bone?
a) Ilium
b) Ischium
c) Pubis
d) all the above
Answer:
a) Ilium

Question 38.
What is the number of hind limbs?
a) 32
b) 37
c) 30
d) 34
Answer:
c) 30

Question 39.
Name the longest bone
a) Femur
b) Humerus bone
c) Tibia
d) all the above

Answer:
a) I -a, II – d, III – c, IV -b

Question 40.
Match and find the correct pair.
1)Patella – a)-14
2) Tarsus – b) -5
3) Metatarsus – c) – 7 bones
4) Phalanges – d) – kneecap
Answer:
3) Metatarsus – c) – 7 bones

Question 41.
Name the membrane which covers the femur bone.
a) Periosteum
b) Endosteum
c) Osteoclast
d) Osteoblast
Answer:
a) Periosteum

Question 42.
Name the joints seen in the cranial region
a) Fibrous joints
b) Cartilagenous joints
c) Diarthroses joints
d) Synovial joints
Answer:
a) Fibrous joints

Question 43.
The Synovial fluid is seen in joints.
a) Cartilagenous joints
b) Diarthroses joints
c) Fibrous joints
d) all the above
Answer:
b) Diarthroses joints

Question 44.
Name the diseases due to the deficiency of acetylcholine.
a) Myasthenia gravis
b) Tetany
c) Duchene muscular dystrophy
d) Muscle pull
Answer:
a) Myasthenia gravis

Question 45.
Name the disease due to the deficiency of ATP
a) Muscle fatigue
b) Muscle pull
c) Muscular dystrophy
d) Tetany
Answer:
a) Muscle fatigue

Question 46.
A tear in the muscle
a) Muscle fatigue
b) Muscle pull
c) Atrophy
d) Muscular dystrophy
Answer:
b) Muscle pull

Question 47.
Name the arthritis due to aging.
a) Osteoarthritis
b) Rheumatoid arthritis
c) Gout
d) Osteoporosis
Answer:
a) Osteoarthritis

Question 48.
Osteoporosis is due to
a) Calcium
b) Sodium
c) Magnesium
d) Potassium
Answer:
a) Calcium

Question 49.
The deposition of urate crystals on the joints is called as
a) Gout
b) Rheumatoid
c) arthritis
d) Osteoporosis
Answer:
a) Gout

Question 50.
Assertion A: Acetylcholine is secreted in the Neuromuscular junction.
Reason B: If acetylcholine is not secreted there won’t be any multiple gated channels in the sarcolemma.
a) A true B wrong
b) A true B this explains the action of A
c) A wrong B true
d) A wrong B wrong
Answer:
b) A true B this explains the action of A

Question 51.
Assertion A: The upper limbs are attached to the pectoral girdles.
Reason B: The pectoral girdles are very light and allow the mobility of the hand
a) A wrong B wrong
b) A True B does not explains the A
c) A True B explains the functions of A
d) A True B wrong
Answer:
c) A True B explains the functions of A

Question 52.
Assertion A: In the pelvis bone a deep socket is present called acetabulum.
Reason B: The head of the femur bone fits in the acetabulum.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explains the structure of B
Answer:
a) A True B explains the functions of A

Question 53.
Assertion A: The pelvis of female is shallow wide and flexible in nature
Cause B: This helps during pregnancy.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explains the structure of A
Answer:
a) A True B explains the functions of A

Question 54.
Assertion A: The lower arm carries the entire weight of the body and is subjected to exceptional forces when we jump or run.
Reason B: To bear the weight of the body it has 46 bones
a) A True B True
b) A wrong B wrong
c) A True B does not explains the functions of A
d) A True B Wrong
Answer:
c) A True B does not explains the functions of A

Question 55.
Assertion A: The cranium belongs to immovable fixed joints.
Reason B: Structures of the flat skull bones are fibrous joints.
a) A True B Wrong
b) A True B explains the structure of A
c) A True B True
d) B does not explains the structure of A
Answer:
b) A True B explains the structure of A

Question 56.
Assertion A: The decreased synthesis of acetylcholine in the neuromuscular junction, causes myasthenia gravis.
Reason B: This leads to muscle fatigue weakness, paralysis.
a) A and B are True
b) A and B are wrong
c) A True B wrong
d) A wrong B True
Answer:
a) A and B are True

Question 57.
Match and arrange the sequence :


Answer:
a) I – D, II – C, III-A,IV – B

Question 58.
Match and arrange the sequence:


Answer:
b) I – C, II – D, III – B, IV -A

2 marks

II. Very Short answers

Question 1.
What is amoeboid movement?
Answer:
The movement of cells by streaming movements of the cytoplasm forming pseudo-podia is known as amoeboid movement, e.g., macrophages.

Question 2.
What are the types of muscles?
Answer:

• Skeletal muscle
• Visceral muscles
• Cardiac muscles

Question 3.
Name the muscle protein?
Answer:

• Actin
• Myosin

Question 4.
Name the regulatory proteins in the thin filaments.
Answer:

• Tropomyosin
• Troponin

Question 5.
Classify the muscles on the basis of their rate of shortening?
Answer:

• Fast contraction fibre
• Slow contraction fibre

Question 6.
On the basis of ATP formation, how are muscles classified?
Answer:

• Oxidative fibres
• Glycolytic fibres

Question 7.
How are muscles classified on the basis of ATP are activity?
Answer:

• Fast oxidative fibres
• Slow oxidative fibres

Question 8.
What is perimysium?
Answer:
The connective tissue covering around each fascicle is the perimysium.

Question 9.
Differentiate the oxidative fibre from the glycolytic fibre.
Answer:

Oxidative fibre	Glycolytic fibre
1. Numerous mitochondria	There are few mitochondria
2. Depends on blood flow	Not depend on blood flow
3. Myoglobin is present	No myoglobin
4. These are known as red muscle fibres	These are called muscle fibres as white muscle fibres

Question 10.
What is sarcoplasm?
Answer:
The cytoplasm of the muscle fibre is called the sarcoplasm.

Question 11.
What is meant by exo skeleton?
Answer:
It is a rigid hard case present outside the body of animals (eg) Cockroach.

Question 12.
What is endo skeleton?
Answer:
It is found inside the body of vertebrates. It is composed of bones and cartilages, (eg) Man.

Question 13.
What are the two types of endo skeletons?
Answer:

1. Axial skeleton.
2. Appendicular skeleton.

Question 14.
What are Glycosomes?
Answer:
Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fiber activity.

Question 15.
What is a brainbox?
Answer:
The cranial bones form the hard protective outer covering of the brain and called the brain box.

Question 16.
Name the ear ossicles?
Answer:

• Malleus
• incus
• stapes

Question 17.
Give notes on Jaw bones?
Answer:

• It is composed of the upper jaw which is formed of the maxilla and the lower jaw is formed of the mandible.
• The upper jaw is fused with the cranium and is immovable.
• The lower jaw is connected to the cranium and is movable.

Question 18.
Name the openings of the skull?
Answer:

• The orbits
• Nasal cavity
• Foramen magnum

Question 19.
What is meant by foramen magnum?
Answer:

• It is a large opening found at the posterior base of the skull.
• Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 20.
What are oxidative fibres?
Answer:
The muscle fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. They are also called red muscle fibres.

Question 21.
Name the first two bones of the vertebral column.
Answer:

1. Atlas
2. Axis

Question 22.
What are the functions of the vertebral column?
Answer:

• It protects the spinal cord.
• Supports the head
• Serves as the point of attachment for the ribs and musculature of the back.

Question 23.
Give short notes on sternum?
Answer:

• The sternum is a flat bone on the mid ventral line of j the thorax.
• It provides space for the attachment of the thoracic ribs and abdominal muscles.

Question 24.
What is a hydrostatic skeleton?
Answer:
The skeleton found in soft-bodied invertebrates is called a hydrostatic skeleton. It is a fluid-filled cavity encircled by muscles, e.g., the earthworm.

Question 25.
What are true ribs?
Answer:
The first seven pairs of ribs are called true ribs. Dorsaily they are attached to the thoracic vertebrae and ventrally connected to the sternum.

Question 26.
What are false ribs?
Answer:
The 8th, 9th, and 10th pairs of ribs do not articulate directly with the sternum but joined with the seventh rib.

Question 27.
What is the sternum?
Answer:
The sternum is a flat bone on the midventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

Question 28.
What is meant by appendicular skeleton?
Answer:
The bones of the upper and lower limbs along with their girdles constitute the appendicular skeleton. It is composed by 126 bones.

Question 29.
White the 3 segments of lower limb.
Answer:

1. The thigh
2. the leg or the shank and
3. the foot.

Question 30.
What is meant by acromian process?
Answer:
The scapula has a slightly elevated ridge called the spine which projects as flat expanded process called the acromion.

Question 31.
What is meant by glenoid cavity?
Answer:
Below the acromian is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

Question 32.
What is meant by olecranon process?
Answer:
The radius and ultra bones present in the fore arm that form the pointed portion of the elbow called olecranon process.

Question 33.
What is meant by carpal tunnel?
Answer:

• There are 8 bones in the wrist arranged in two rows of four each.
• The anterior surface of the wrist has tunnel like appearance. This tunnel is termed as carpal tunnel.

Question 34.
Name the bones which forms the coxal bones.
Answer:

• Ilium
• Ischium
• Pubis

Question 35.
What is meant by pubic symphysis?
Answer:
Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.

Question 36.
Where calcium ion binds with the muscle fibre? Name the molecules which binds with calcium?
Answer:

• The calcium released from sarcoplasma binds with the thin fibre of the muscle.
• The released calcium binds to troponin a thin filaments.

III. Fill Up The Blanks With Suitable Options

1. Scapula – Acromian process
……………. – Bones of upper arm
2. First 7 pair of rib bones – True ribs
11 and 12th pair of ribs – …………….
3. Cervical vertebrae – 7
……………. – Lumbar bones
4. Skull bones – 22
……………. – Skull bones
5. Thick fibres – Myosin
……………. – Thin fibres
6. Amoeboid movement – Macrophage cells
……………. – Sperm cells
Answer:

1. Olecronan process
2. Floating ribs
3. 5
4. 8
5. Actin
6. Flagellated movement

3 marks

IV. Short answers

Question 1.
Give short notes on skeletal muscle and their covering membranes.
Answer:

• Each muscle is made up of bundles of muscle fibres called fascicle. Each muscle fibre contains rod like structures called myofibrils.
• The connective tissue covering the muscle is the epimysium.
• The covering around each fascicle is the perimysium.
• The muscle fibre is surrounded by endomysium.

Question 2.
Give the structure of a skeletel muscle fibre.
Answer:

• Each muscle fibre is thin and elongated.
• Most of them taper at one or both ends.
• Muscle fibres are surrounded by sarcolemma the cytoplasm of the muscle fibre is called the sarcoplasm.
• It contains glycosomes myoglobin and sarcoplasmic reticulum.
• Myoglobin is a red coloured respiratory pigment and glycosomes are reserved glycogen.
• Muscle fibres contain muscle protein actin and myosin.

Question 3.
Give notes on slow oxidative fibres.
Answer:

• These fibres have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP.
• This type of fibres seen in long distance swimmers and long distance runners.

Question 4.
Give notes on fast – oxidative fibres.
Answer:

• These fibres have high myosin ATP ase activity and can make large amounts of ATP.
• They are suited for rapid action.

Question 5.
Give notes on fast glycolytic fibres.
Answer:

• These fibres have myosin ATP ase activity but cannot make as much ATP as oxidative fibres because their source of ATP is glycolysis.
• These fibres are best suited for rapid intense actions such as short sprint at maximum speed.

Question 6.
Name the facial bones.
Answer:
There are 14 facial bones.

1. Pair of maxilla
2. Pair of Zygomatic
3. Pair of Palantine
4. Pair of lacrimal
5.  Pair of Nasal
6. Mandible or lower jaw
7. Vomer

Question 7.
Give notes on fibrous joints.
Answer:

• They are immovable fixed joints in which no movement between the bones is possible.
• Sutures of the flat skull bones are fibrous joints.

Question 8.
Give notes on cartilaginous joints.
Answer:
They are slightly movable joints in which the joint surface are separated by a cartilage and slight movement is only possible.

Question 9.
Give notes on synovial joints.
Answer:
They are freely movable joints the articulating bones are separated by a cavity which is filled with synovial fluid.

Question 10.
Give notes on myasthenia gravis.
Answer:

• It is an auto immune disorder affecting the action of acetylcholine at neuro muscular junction leading to fatique.
• Weakening and paralysis of skeletal muscles.
• Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles.
• When the disease progresses it can make chewing swallowing talking and even breathing difficult.

Question 11.
Give notes on muscle fatique.
Answer:

• It is the inability of a muscle to contract after repeated muscle contraction.
• This is due to lack of ATP and accumulation of lactic acid by anaerobic break down of glucose.

Question 12.
Give notes on Atrophy of muscles.
Answer:

• A decrease in the activity of muscles results in the atrophy of muscles.
• There is a reduction in the size of the muscle and makes the muscle to become weak which occurs with lack of usage as in chronic bed ridden patients.

Question 13.
What is meant by muscle pull?
Answer:

• Muscle pull is actually a muscle tear.
  Atraumatic pulling of the fibres produces a tear known as sprain.
• This can occur due to sudden stretching of muscle beyond the point of elasticity.
• Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.

Question 14.
What is meant by muscular dystrophy?
Answer:

• The group of diseases collectively called the muscular dystrophy are associated with the progressive degeneration of skeletel muscle fibres weakening the muscles and leading to death from lung or heart failure.
• (eg) Duchene muscular dystrophy.

Question 15.
What is meant by skeletel muscle glycogen analysis?
Answer:

• This is used to measure an athlete’s muscle glycogen.
• Muscle glycogen provides the main source of energy during anaerobic exercise.
• A single glycogen molecule may contain 5000 glucose molecules.

Question 16.
Give notes on osteoporosis?
Answer:

• It occurs due to deficiency of vitamin D and hormonal imbalance.
• It causes rickets in children and osteomalasia in adult females.
• The bones become soft and fragile.
• It can be minimized with adequate calcium intake vitamin D intake and regular physical activities.

Question 17.
What is carpal tunnel syndrome?
Answer:

• The narrow passage bounded by bones and ligaments in the wrist get narrowed and pinches the median nerve.
• This syndrome is mostly seen among the clerks, software professional and people who constantly play or text in mobile phones.

5 Marks

V. Give Detailed Answers 

Question 1.
Describe about the different types of movements with examples.
Answer:
Types of Movements :

1. Amoeboid movements
2. Ciliary movements
3. Flagellar movements
4. Muscular movements

1. Amoeboid movements:
Cells such as macrophages exhibit amoeboid movement for engulfing pathogens by pseudopodia formed by the streaming movement of the cytoplasm.

2. Ciliary movements :
This type of movement occurs in the respiratory and genital passages.

3. Flagellar movement:
This type of movement occurs in the cells which are having flagella or whiplike motile organels. (eg) Sperm cells.

4. Muscular movement:
The movement of hands legs jaws are caused by the contraction and relaxation of the muscle.

Question 2.
Describe the structure of sarcomere.
Answer:

• The unit of the skeletal muscle is sarcomere A sarcomere is the region of a myofibril between two successive z – discs.
• It contains an ‘A’ band with a half I band which are perfectly aligned with one another.
• This type of arrangement gives the cell a striated appearance.
• Each dark band has a lighter region in its middle called the M – zone.
• Each H – zone is bisected vertically by a dark line called the M – line.
• The I bands have a darker mid line area called the z – disc.

• Inside the sarcomere two types of filaments are present namely the thick filaments and thin filaments.
• The thick filaments extend the entire length of the A band, the thin filaments extend across the I band and partly into the A – band.
• The invagination of the sarcolemma forms transverse (T- tubules) tubules and they penetrate into the junction between the A and I bands.

Question 3.
Describe the structure of muscle protein.
Answer:
Contraction of the muscle depends on the presence of contractile proteins such as actin and myosin.
Myosin fibre:

• The thick filaments are composed of the protein myosin.
• Each myosin molecule is made up of monomer called meromyosin.
• The meromyosin have a globular head with a j short arm and a tail.
• The short arm have heavy meromyosin and the tail portion have the light meromyosin.
• The head bears an actin – binding site and an ATP binding site
• It also contains ATP ase enzyme that split ATP to generate energy for the contraction of muscle.

Actin filament:

• Actin has polypeptide subunits called globular actin or G – actin and filamentous form F – actin.
• Each thin filament is made of two F – actins helically wound to each other.
• Each F – actin is a polymer of monomeric G – actins, It also contains a binding site for myosin.
• The thin filament contain several regulatory protein like tropomyosin, troponin, which help in regulating the contraction of muscles along with actin and myosin.

Thick filament:
Each thick filament consists of many myosin molecules whose heads produce at opposite ends of the filament Portion of a thick filament

Thin filament:
A thin filament consists of two strands of actin subunits twisted into a helix plus two types of regulatory proteins (troponin and tropomyosin) Portion of a thin filament.

Question 4.
Give the schematic representation of muscle contraction.
Answer:

Question 5.
Give the four important features of skeletal muscles.
Answer:

1. Excitability :
   Based on the chemical and electrical excitation the muscles contracts.
2. Contractility:
   It is the ability of the muscle which gives movement to the attached organs.
3. Conductivity:
   The excitation at one part of the muscle is connected to the other part of the muscle.
4. Elasticity :
   The ability of the muscle to returns to its original position after the extension of the muscles.

Question 6.
Explain the bones that form the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial, bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3.

These bones are joined by sutures which are immovable. They are paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid. The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull.

A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only bone without any joint. Each middle ear contains three tiny bones- malleus, incus, and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible.

The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. The foramen magnum is a large opening found at the posterior base of the skull. Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 7.
Give an account of vertebral column.
Answer:

• It consists of 33 serially arranged vertebrae which are interconnected by a cartilage known as inter vertebral disc.
• The vertebral column extends from the base of the skull to the pelvis and forms the main framework of the trunk.

It has five major regions, they are

1. Cervical – 7
2. Thoracic vertebrae -12
3. Lumbar vertebrae – 5
4. Sacrum – 5 sacral vertebrae found in the infant which are fused to form one bone in the adult.
5. Coccyx – 1 – 4 Coccy geal vertebrae found in the infant which are fused to form one bone in the adult.

Question 8.
Give an account of the ribcage.
Answer:

• There are 12 pairs of ribs.
• Each rib bone is connected dorsally to the vertebral column and ventrally to the sternum.
• It has two articulation surfaces on its dorsal end called bicephalic.
• The first 7 pairs of ribs are called true ribsorvertebro – sternal ribs.
• Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.
• The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous part of the seventh rib.
• These are called false ribs or vertebro – chondral ribs.
• The last 11th and 12th pairs of ribs are not connected ventrally.

• They are called as floating ribs or vertebral ribs.
• Thoracic vertebrae ribs and sternum form the rib cage.

Question 9.
Give an account of the pectoral girdle?
Answer:

• The upper limbs are attached to the pectoral girdles.
• These are very light and allow the upper limbs a degree of mobility not seen any where else in the body.
• The girdle is formed of two halves.
• Each pectoral girdle consists of a clavicle or collar bone and a scapula.
• The scapula is a large triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs.
• It has a elevated expanded process called the acromion.

• The clavicle articulates this process.
• Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.
• Each clavicle is a long slender bone with two curvatures which lie horizontally and connect the axial skeleton with the appendicular skeleton.

Question 10.
Describe the structure of the upper limb.
Answer:

• The upper limb consists of 30 separate bones and is specialized for mobility.
• The region between the shoulder and elbow is the humerus.
• The head of humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint.
• The distel end of humerus articulates with the two fore arm bones the radius and ulna
• Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow.
• The hand consists of carpals metacarpals and phalanges.
• Carpals the wrist bones 8 in number are arranged in two rows of four each and form a tunnel termed as carpal tunnel.
• Meta carpals the palm bones are 5 in number and phalanges the digit bones are 14 in number.

Question 11.
Give an account of pelvic girdle.
Answer:

• The pelvic girdle is a heavy structure specialised for weight-bearing.
• It is composed of two hib bones called coxal bones that secure the lower limbs to the axial skeleton.
• Together with the sacrum and coccyx the hib bones form the basin like bonypelvis.
• Each coxal bone consists of three fused bones ilium, ischium, and pubis.
• At the point of fusion of these three bones forms a deep hemispherical socket called the acetabulum present on the lateral surface of the pelvis.
• It receives the head of the femur at hip joint and helps in the articulation of the femur.
• Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.
• The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly.
• The ischium is a curved bar of bone. The ‘V’ shaped pubic bones articulate anteriorly at the pubic symphysis.
• The pelvis of male is deep and narrow with larger heavier bones and the female is shallow wide and flexible in nature and this helps during pregnancy which is influenced by female hormones.

Question 12.
Give an account of the lower limb.
Answer:

• The lower limb consists of 30 bones which carries the entire weight of the erect body and is subjected to exceptional forces when we jump or run.
• The bones of the lower limbs are thicker and stronger than the upper limbs.
• Each lower limb consists of the thigh, the leg or the shank and the foot.
• The femur is the strongest and longest bone of the body.
• The head of femur articulates with the acetabulum of the pelvis to form the hip joint.
• The tibia and fibula form the skeleton of the shank.
• A thick triangular patella forms the knee cap which protects the knee joint arteriorly and improves the leverages of thigh muscles acting across the knee.
• The foot includes the bones of ankle the tarsus (7) the metatarsus (5) and the phalanges or toe ebones. (14)
• The foot supports our body weight and acts as a lever to propel the body forward while walking and running.
• The phalanges of the foot are smaller than those of the fingers.

Question 13.
Give an account of a structure of a typical long bone.
Answer:

• The typical long bone has a diaphysis, epiphyses and membranes.
• A tubular diaphysis or shaft forms the long axis of the bone and has a central medullary cavity.
• The epiphyses are the bone ends.
• Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow.
• The region where the diaphysis and epiphysis meet is called metaphysis.
• The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum.
• The outer fibrous layer is dense irregular connective tissue.
• The inner osteogenic layer consists of osteoblasts cell. ( bone-forming cells) and osteoclasts cells (E bone – destroying cells)
• There are primitive stem cells osteogenic cells that give rise to the osteoblasts.
• The periosteum is richly supplied with nerve fibres lymphatic vessels and blood vessels.
• Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum It also contains osteoblasts and osteoclasts cells.
• Between the epiphysis and diaphysis growthplate or epiphyseal plate is present.

Question 14.
Write a short note on Rib cage?
Answer:
There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic.

The first seven pairs of ribs are called ‘true ribs or vertebro-stemal ribs. Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous (hyaline cartilage) part of the seventh rib. These are called ‘false ribs’ or vertebro-chondral ribs.

The last 11th and 12th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs. Thoracic vertebrae, ribs and sternum together form the ribcage. Rib cage protects the lungs, heart, liver and also plays a role in breathing.

Question 15.
Tabulate the differentiate of joints man
Answer:

Pivot joint	between atlas and axis
Gliding joint	between the carpals
Saddle joint	between the carpal and meta carpal
Ball and socket joint	between humerus and pectoral gridle
Hinge joint	Knee joint
Condyloid or Angular or Ellipsoid	Joint between radius joint and carpal

Question 16.
Draw the diagram of different types of fracture and arrange them.
Answer:

Question 17.
Bones of the skeletal system. Table: 1 Bones of skeletal system
Answer:

Notes:

1. The strongest muscle in the human: Massetter in cheeks
2. The smallest muscle in the human: Middle ear in stapedius
3. Well moving muscle: Tongue
4. The largest muscle in the human: Buttock in Glutens Maximus
5. The longest muscle in the human : Hip to knee (sartorius)
6. Total number of bones is adults = 206

Question 18.
Explain the basic categories of exercise and physical activity?
Answer:
Exercise and physical activity fall into four basic categories. Endurance, strength, balance, and flexibility. Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness.

Strength exercises make the muscles stronger. They help to stay independent and cany out everyday activities such as climbing stairs and carrying bags.

Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improve balance.

Flexibility exercises help to stretch body muscles for more freedom of joint movements.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 8 Excretion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

11th Bio Zoology Guide Excretion Text Book Back Questions and Answers

Part – I

Question 1.
Arrange the following structures in the order that a drop of water entering the nephron would encounter them.
a) Afferent arteriole
b) Bowman’s Capsule
c) Collecting duct
d) Distal tubule
e) Glomerulus
f) Loop of Henle
g) Proximal tubule
h) Renal Pelvis
Answer:
(a), (e), (b), (g), (f), (d), (c), (h)

Question 2.
Name the three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule. What components of the blood are usually excluded by these layers.
Answer:
The three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule are,

1. Glomerular capillary endothelium
2. Basal lamina or basement membrane
3. Epithelium of Bowman’s capsule.
4. Blood corpuscles and plasma protein are excluded by these layers.

Question 3.
What forces promote glomerular filtration? What forces oppose them? What is meant by net filtration pressure?
Answer:

• Glomerulus hydrostatic pressure
• Glomerulus pressure
• Opposing pressure: Colloidal osmotic pressure, Capsular hydrostatic pressure
• Net filtration pressure = Glomerular hydrostatic pressure – (Colloidal osmotic pressure + capsular hydrostatic pressure.

Question 4.
Identify the following structures and explain their significance in renal physiology?
Answer:
a) Juxtaglomerular apparatus It is a specialized tissue in the afferent arteriole of the nephron that consists of macula densa and granular cells.

• Action: The macula densa cells sense distal tubular flow and aspect afferent arteriole diameter.
• Actions of Glandular cells: When there is a fall in glomerular blood flow it can activate Juxta glomerulus cells to release renin from the liver which converts angiotensinogen to angiotensin I.
• Angiotensin converting enzyme converts angiotensin I to angiotensin II.
• It stimulates Na+ reabsorption in the proximal convoluted tubule by vasoconstriction of the blood vessel and increases the glomerular blood pressure.

b) Podocytes:

• The visceral layer of Bowman’s capsule is made of epithelial cells called podocytes.
• The podocytes and the foot processes and the basement membrane of glomerulus act as a filtering membrane.
• Podocytes act in glomerulus filtration.

c) Sphincters in the bladder:

• The sphincters of the bladder regulate urination.
• When the bladder is full the stretching of the urinary bladder stimulates the CNS through the sensory neurons of the parasympathetic nervous system and brings about contraction of the bladder.
• Somatic motor neurons induce the sphincters to close.
• Smooth muscles contract results in the opening of the internal sphincters and relaxing the external sphincter.
• The sphincter opens and the urine is expelled out.

Question 5.
In which segment of the nephron most of the re-absorption of substances takes place?
Answer:
In the proximal convoluted tubule.

Question 6.
When a molecule or ion is reabsorbed from the lumen of the nephron, Where does it go?
Answer:
When a molecule or ion is reabsorbed from the lumen of the nephron, it goes to the bloodstream through an efferent arteriole which carries blood away from the glomerulus. If a solute is filtered and not reabsorbed from the tubule, it goes along with urine.

Question 7.
Which segment is the site of secretion and regulated reabsorption of ions and pH homeostasis?
Answer:

• 1. Distal tubule; 2. Collecting duct
• In the renal tubules H⁺ and NH⁴ are secreted and liberated into the tubules and excreted through
  urine thus maintaining acid base balance.
• For each H⁺ ions from the liberated filtrate one Na⁺ is reabsorbed in the tubules.
• The secreted HCO₃, PO₃ and NH₃ combines to form carbonic acid and phosphoric acids.
• Thus the H⁺ are maintained and their reabsorption is prevented.

Question 8.
What solute is normally present in the body to estimate GFR in humans?
Answer:

• The glomerulus filtrate consists of water glucose amino acids creatinine protein salts and urea.
• These solutes decide the glomerular filtration rate.

Question 9.
Which part of the autonomic nervous system is involved in the micturition process?
Answer:
Parasympathetic nervous system.

Question 10.
If the afferent arteriole of the nephron constricts, what happens to the GFR in that nephron? If the efferent arteriole constricts what happens to the GFR in that nephron? Assume that no autoregulation takes place.
Answer:

• Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
• Because afferent arteriole is wider than efferent arteriole.
• The constriction of this does not affect the rate of filtration.

Question 11.
The concentration of urine depends upon which part of the nephron.
Answer:
a) Bowman’s capsule
b) Length of Henle’s loop
c) PCT
d) network of capillaries arising from the glomerulus.
Answer:
b) Length of Henle’s loop

Question 12.
If Henle’s loop were absent from mammalian nephron, which one of the following is to be expected?
a) There will be no urine formation
b) There will be hardly any change in the quality and quantity of urine formed.
c) The urine will be more concentrated
d) The urine will be more dilute
Answer:
d) The urine will be more dilute

Question 13.
What will happen if the stretch receptors of the urinary bladder wall are totally removed?
a) Micturition will continue
b) Urine will continue to collect normally in the bladder
c) There will be micturition
d) Urine will not collect in the bladder
Answer:
a) Micturition will continue

Question 14.
The end product of the Ornithine cycle is
a) Carbon dioxide
b) Uric acid
c) Urea
d) Ammonia
Answer:
c) Urea

Question 15.
Identify the wrong match.

a) Bowman’s capsule	Glomerular filtration
b) DCT	Absorption of glucose
c) Henle’s loop	Concentration of urine
d) PCT	Absorption of Na+ and K+ ions

Answer:
b) DCT – Absorption of glucose

Question 16.
Podocytes are the cells present on the
a) Outer wall of Bowman’s capsule
b) Inner wall of Bowman’s capsule
c) Neck of the nephron
d) Wall glomerular capillaries
Answer:
b) Inner wall of Bowman’s capsule

Question 17.
Glomerular filtrate contains
a) Blood without blood cells and proteins
b) Plasma without sugar
c) Blood with proteins but without cells
d) Blood without urea
Answer:
c) Blood with proteins but without cells

Question 18.
Kidney stones are produced due to the deposition of uric acid and Silicates
b) Minerals
c) Calcium Carbonate
d) Calcium oxalate
Answer:
d) Calcium oxalate

Question 19.
Animal requiring the minimum amount of water to produce urine is
a) Ureotelic
b) Ammonotelic
c) Uricotelic
d) Chemotelic
Answer:
c) Uricotelic

Question 20.
Aldosterone acts at the distal convoluted tubule and collecting duct resulting in the absorption of water through
a) Aquaporins
b) Spectrins
c) GLUT
d) Chloride channels
Answer:
a) Aquaporins

Question 21.
The hormone which helps in the reabsorption of water in kidney tubules is
a) Cholecystokinin
b) Angiotensin Il
c) Antidiuretic hormone
d) Pancreozymin
Answer:
c) Antidiuretic hormone

Question 22.
Malpighian tubules remove excretory products from
a) Mouth
b) Oesophagus
c) Haemolymph
d) Alimentary canal
Answer:
c) Haemolymph

Question 23.
Identify the biological term excretion, glomerulus, urinary bladder, glomerular filtrate, ureters, urine, Bowman’s capsule, urinary system, reabsorption, micturition, osmosis, proteins.
Answer:

• A liquid which gathers in the bladder – Urine
• Produced when blood is filtered in a Bowman’s capsule – Glomerular filtrate
• The temporary storage of urine – Urinary bladder
• A ball of intertwined capillaries – Glomerulus
• Removal of unwanted substances from the body – Excretion
• Each contains a glomerulus – Bowman’s Capsule
• Carry urine from the kidneys to the bladder – Ureter
• The scientific term for urination – Micturition
• Regulation of water and dissolved substances in the blood and tissue fluid – Osmoregulation
• Consists of the kidneys ureters and bladder Excretory system
• Removal of useful substances from glomerular filtrate reabsorption
• What solute the do blood contain that is not present in the glomerular filtrate? – Plasma Protein

Question 24.
With regards to toxicity and the need for dilution in water how different are ureotelic and ureocotelic excretions? Give examples of animals that use these types of excretions?
Answer:

• The type of nitrogenous end product (Urea or Uric acid or ammonia) of an animal excretion depends upon the habitat of the animal.
• For the Excretion of Ammonia, more water is needed.
• Animals that excrete most of their nitrogen in the form of ammonia are called ammonites.
• Fishes Amphibians. Uric acid can be eliminated with a minimum loss of water and is called Uricoteles.
• Reptiles, birds insects, landrails. Mammals and terrestrial amphibians excrete urea and are called ureoteles.
  Nitrogenous

Question 25.
Differentiate protonephridia from metanephridia.
Answer:

Protonephridia	Metanephridia
1. Proto – first it is a network of dead-end tubules lacking internal opening. (E.g) Platyhelminthes	Meta – after They have a tubular internal opening called nephrostome. (E.g) Earthworm
2. Flame cells are excretory structures.	The excretory products are filters and selectively reabsorbed.
3. Excretory product excretes through nephridiopore	The excretory product excretes through the nephridiopore.
4. Excretory structure are usually osmo regulators	Excretory structures are osmo regulators and helps in excretion.

Question 26.
What is the nitrogenous waste produced by amphibian larvae and by adult animals?
Answer:
The adult amphibian’s excretory product is urea. The larval form of amphibian’s excretory product is ammonia.

Question 27.
How is urea formed in the human body?
Answer:
Urine formation involves three main processes namely Glomerular Alteration, tubular reabsorption and tubular secretion.

I. Glomerular filtration:

• Blood enters the kidney from the renal artery into the glomerulus.
• Blood is composed of water colloidal proteins, sugars and nitrogenous end product.
• The filteration is started in the glomerulus. The fluid that leaves the glomerular capillaries and enters the Bowman’s capsule is called the glomerular filtrate. It formed 170-180 litre within 24 hours.
• Glomerular membrane has large surface area. Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
• This is because afferent arteriole is wider than efferent arteriole and the glomerular hydrostatic pressure is around 55 mm Hg.
• Molecules larger than 5mm are barred from entering the tubule.

The two opposing forces are contributed by the plasma proteins in the capillaries.

1. Colloidal osmotic pressure – 30 mm Hg
2. Hydrostatic pressure -15 mm Hg

Both pressures combine. 30 mm Hg + 15mm Hg = 45mm Hg

• The net filtration pressure of 10mm Hg is responsible for the renal filtration.
• Net filtration pressure – Glomerular
• Hydrostatic Pressure – Colloidal Osmotic pressure + Capsular hydrostatic pressure.

Net filteration pressure = 55 mm Hg – (30mm Hg +15mm Hg) = 10mm Hg

Substance	Concentration in blood Plasm / g dm-3	Concentration in glomerular filtrate / g dm-3
Water	900	900
Proteins	80.	0.05
Aminoacids	0.5	0.5
Glucose	1.0	1.0

Substance	Concentration in blood PIasm /gdm-3	Concentration glomerular filtrate/gdm-3
Urea	0.3	0.3
Uric Acid	0.04	0.04
Creatinine	0.01	0.01
Inorganic ions (mainly (Na+, K+and Cl–)	7.2	7.2

II. Tubular reabsorption:

• The volume of filtrate formed per day is around 170 – 180l.
• Urine released is around 1.51 per day. 99% of the glomerular filtrate is reabsorbed by the renal tubules.

Reabsorption in proximal convoluted tubule:

• Glucose lactate amino acids Na in the filtrate are reabsorbed in the PCT.
• Sodium is reabsorbed by active transport through a sodium-potassium pump in the PCT.
• Small amounts of urea and uric acid are reabsorbed.

Reabsorption in Henle Loop:

• Descending limb of Henle’s loop is permeable to water due to the presence of aqua porlns but not permeable to salts.
• Water is lost in the descending limb. Hence Na and Cl get concentrated in the filtrate.
• Ascending limb of Henle’s loop is impermeable to water but permeable to Na⁺, Cl^– andK⁺.

Distal Convoluted Tubule:

• It recovers water and secretes potassium into tubule.
• Na⁺, Cl^-, and water remain in the DCT.
• Reabsorption of HCO₃ takes place to regulate the blood pH.
• Collecting the tubule is permeable to water potassium ions are actively transported into the tubule and Na⁺ to produce concentrated urine.

Tubular Secretion:

• Once it enters the collecting duct water is absorbed and concentrated hypertonic urine is formed.
• For every H secreted into the tubular filtrate, a Na⁺ is absorbed by the tubular cell.
• The H⁺ secreted combines with HCO⁺₃, HPO^–_3, and NH⁺ and gets fixed as H2CO₃, H2PO_3, and NH4⁺.
• Since H⁺ gets fixed in the fluid reabsorption of H⁺is prevented.

Question 28.
Differentiate cortical from medullary nephrons.
Answer:
In the renal tubules proximal convoluted tubule and distal convoluted tubule are situated in the cortical region of the kidney and the loop of Henle is in the medullary region.

Cortical Nephron	Medullary Nephron
1. The loop of Henle is too short and extends only very little into the medulla.	Nephrons have a long loop of Henle that run deep in to the medulla.
2. There is no vasa recta.	Vasa recta is present.

Question 29.
What vessels carry blood to the kidneys? Is this blood arterial or venous?
Answer:
Renal artery right and left artery. The blood is arterial.

Question 30.
Which vessels drain filtered blood from the kidneys?
Answer:
Renal vein. The filtered blood is taken from the kidney to inferior vena cava through renal vein.

Question 31.
What is tubular secretion? Name the substances secreted through the renal tubules?
Answer:

• In the distal convoluted tubule the urine becomes hypotonic.
• It contains urea and salts pass from peri tubular blood in to the cells of DCT.
• For every H⁺ secreted a Na⁺ is absorbed by the tubular cell.
  The H⁺ secreted combines HCO₃, HPO₃ and NH₃ and becomes carbonic acid phosphoric acid and ammonium.
• Since H⁺ gets fixed in the fluid reabsorption of H⁺ is prevented.

Question 32.
How are the kidneys involved in controlling blood volume? How is the volume of blood in the body related to arterial pressure?
Answer:

• When the volume of blood decreases the flow pressure, decreases.
• This can be sensed by hypothalamus and osmo receptors are stimulated and anti diuretic hormone is secreted from the neuro hypophysis.
• The aquaporins in the proximal convoluted tubules and collecting tubule reabsorb water. Hence the blood volume increases and the blood pressure increases.

Question 33.
Name the three main hormones that are involved in the regulation of the renal function?
Answer:

• Renin
• Angio Tensin I
• AngioTensin II

Question 34.
What is the function of anti diuretic hormone? Where is it produced and What stimuli increases or decreases its secretion?
Answer:

• ADH which is also called as vasopressin or Anti diuretic hormone is secreted from neuro hypophysis.
• Fluid loss or if blood pressure increases the osmoreceptors of hypothalamus is stimulated and hence neuro hypophysis is stimulated and secretes ADH.
• When the fluid level and pressure is maintained due to the negative feed back mechanism ADH secretion stops.

Question 35.
What is the effect of aldosterone on kidneys and where is it produced?
Answer:

• Due to the stimulation of Angiotensin II the adrenal cortex secretes aldosterone. That causes reabsorption of Na⁺, K⁺ excretion and absorption of water from distal convoluted tubule and collecting tubule.
• This mechanism is known as Renin – Angio tensin Aldosterone system.

Question 36.
Explain the heart’s role in secreting a hormone that regulates renal function? What hormone is this?
Answer:

• Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or factor (ANF) travels to the kidney where H increases Na+ excretion and increases the blood flow to the glomerulus,
• Acting on the afferent glomerular arterioles as a vasodilator or an efferent arterioles as a vasoconstrictor.
• The first identical natri uretic hormone is atrial natriuretic horome of heart.

Part – II.

11th Bio Zoology Guide Excretion Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
Name the organ that regulates the acid base and water regulation?
a) Kidney
b) Liver
c) Skin
d) Lungs
Answer:
a) Kidney

Question 2.
Whether the following statements are true or false find the correct sequence. Find the correct sequence.
I) According to the environmental changes organism change their osmotic concentration and are called osmo confirmers.
II) Osmo regulators maintain their internal osmotic concentration irrespective of their external osmotic environment.
III) The Euryhaline animals are able to tolerate only narrow fluctuations in the salt concentration?
IV) The steno haline animals can tolerate wide fluctuations in the salt concentration.
Series:
a) I – False, II – True, III – False, IV – True
b) I – True, II – True, III – False, IV- False
c) I – False, II – True, III – False, IV – True
d) I – False, II – False, III – True, IV – True
Answer:
b) I – True, II – True, III – False, IV- False

Question 3.
Name the toxic formed due to the degeneration of amino acid.
a) Urea
b) Uric Acid
c) Ammonia
d) Carbonic Acid
Answer:
c) Ammonia

Question 4.
Find out the odd one out.
a) Ammonoteles – Aquatic Amphibians
b) Urico teles – Birds
c) Ureoteles – Land amphibians
d) Ureoteles – Earthworm (When it is in water)
Answer:
d) Ureoteles – Earthworm (When it is in water)

Question 5.
What is the functional unit of kidney?
a) Nephron
b) Neuron
c) Synapsis
d) Glomerulus
Answer:
a) Nephron

Question 6.
What type of Urine is formed in organism with long Henle’s loop?
a) Isotonic
b) less concentrated
c) Concentrated
d) None of the above
Answer:
c) Concentrated

Question 7.
In the marine organism the kidney with glomerulus produce ………………. type of concentrated urine.
a) Concentrated than body fluid
b) Equal to the concentration of body
c) Less than the concentration of body fluids
d) None of the above
Answer:
b) Equal to the concentration of body

Question 8.
The average weight of human kidney,
a) 100-120g
b)150-200g
c) 120-170g
d) 120-150g
Answer:
c) 120-170g

Question 9.
What is the outer covering of the kidney?
a) Renal fascia perirenal fat capsule pleura
b) Renal fascia perirenal fat capsule peri cardial membrane
c) Renal fascia, Peri renal fat capsule meninges
d) Renal fascia perirenal fat capsule fibrous capsule
Answer:
d) Renal fascia perirenal fat capsule fibrous capsule

Question 10.
What is meant by renal corpuscle?
a) Glomerulus and Bowman’s Capsule
b) Glomerulus and Malpighian capsule
c) Glomerulus and nephron
d) Glomerulus and Henle’s loop
Answer:
d) Glomerulus and Henle’s loop

Question 11.
Regarding renal tubules find the correct sequences.
a) Glomerulus Bowman’s capsule Malpighian capsule uriniferous tubules.
b) Proximal convoluted tubules thick descending loop thin ascending limb distal convoluted tubule.
c) Proximal convoluted tubule, Henle’s loop. Distal convoluted tubule.
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.
Answer:
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.

Question 12.
Urea is synthesisted through this cycle.
a) Citric cycle
b) Carboxylic Acid cycle
c) Ornithinecycle
d) Arginine cycle
Answer:
c) Ornithinecycle

Question 13.
What is the pressure in the afferent arthery of Glomerulus?
a) 55 mm Hg
b) 50 mm Hg
c) 57 mm Hg
d) 58 mm Hg
Answer:
a) 55 mm Hg

Question 14.
What is the amount of filteration of the glomerulus in 24 hours?
a) 200l
b) 180l
c)190l
d) 170l
Answer:
b) 180l

Question 15.
What should be the size of the molecule that can pass through the tubules from the plasma filtrate?
a) 7 nm
b) 4 nm
c) 5 nm
d) 6 nm
Answer:
c) 5 nm

Question 16.
What is the pressure in glomerulus?
a) 50 mm Hg
b) 60 mm Hg
c) 55 mm Hg
d) 62 mm Hg
Answer:
c) 55 mm Hg

Question 17.
The glomerular pressure encounter ………….. of colloidal osmotic pressure and …………… of hydrostatic pressure.
a) 30mmHg;15mmHg
b) 15mmHg;30mmHg
c) 40 mm Hg; 30 mm Hg
d) 30mmHg;40mmHg
Answer:
a) 30mmHg;15mmHg

Question 18.
What is the net filteration pressure?
a) 55mmHg – (15mmHg + 30mmHg) = 10mm Hg
b) 55 mm Hg – (30 mm Hg +15 mm Hg) =10 mm Hg
c) 55 mm Hg – (35 mm Hg + 5 mm Hg) = 15 mm Hg
d) 55 mm Hg – (5 mm Hg + 35 mm Hg) = 15 mm Hg
Answer:
b) 55 mm Hg – (30 mm Hg + 15 mm Hg) =10 mm Hg

Question 19.
What is the amount of filtrate formed in one minute?
a) 100 ml -125 ml
b) 100 ml -150 ml
c) 120 ml -125 ml
d) 130 ml -140 ml
Answer:
c) 120 ml -125 ml

Question 20.
What is the amount of filtrate formed in a day?
a) 170 ml-200 ml
b) 170 ml-190 ml
c) 170 ml-180 ml
d) 170 ml-185 ml
Answer:
c) 170 ml-180 ml

Question 21.
What is the amount of urine excreted in a day?
a) 1.5l
b) 1l
c) 2l
d) 2.5l
Answer:
a) 1.5l

Question 22.
Which of the following hormones are secreted by the kidney?
a) Renin
b) Gastrin
c) Calcitriol
d) secretin
a) (ii) and (iii)
b) (i) and (iii)
c) (ii) and (iv)
d) (i) and (ii)
Answer:
b) (i) and (iii)

Question 23.
Where is aquaporins present?
a) Proximal convoluted tubule
b) Distal convoluted tubule and ascending limb of Henle
c) The descending limb of Henle and distal convoluted tubule
d) All the above
Answer:
b) Distal convoluted tubule and ascending limb of Henle

Question 24.
Name the hormone that helpsin the reabsorption of water in distal convoluted tubule.
a) Vasopressin
b) Serotonin
c) Oxytocin
d) All the above
Answer:
a) Vasopressin

Question 25.
Where is Renin synthesized in Kidney?
a) Afferent arteriole
b) Efferent arteriole
c) Vasarecta
d) Collecting duct
Answer:
a) Afferent arteriole

Question 26.
Name the hormone that helpsin the conversion of plasma protein angioten- sinogen into angiotensin I
a) Renin
b) Vasopressin
c) ADH
d) STH
Answer:
a) Renin

Question 27.
What is micturition?
a) Excretion of urine from the urinary bladder
b) The formation of urine in the glomerulus
c) Urine formation in distal convoluted tubule
d) the absorption of urine in Bowman’s capsule
Answer:
a) Excretion of urine from the urinary bladder

Question 28.
According to the food that maneats the pH of urine ……………… to ……………. can be altered.
a) pH 4.8-7.5
b) PH 4.9-7.9
c) pH 4.5-8.0
d) pH 4.4 -7.4
Answer:
c) pH 4.5-8.0

Question 29.
What is the reason for the yellow colour of the urine?
a) Urochrome
b) Haemoerythrin
c) Haemocyanin
d) None
Answer:
a) Urochrome

Question 30.
What is the normal pH of urine?
a) pH 6.0
b) pH 5.3
c) pH 6.4
d) pH 5.9
Answer:
a) pH 6.0

Question 31.
What is the amount of CO₂ released from lungs in a day?
a) 20l
b) 19l
c)18l
d)119l
Answer:
c)18l

Question 32.
Confirm
Statement A: The important function of sweat gland is to cool the body.
Statement B: The sweat glands excrete sodi¬um chloride urea and lactic acid.
a) Statement A-True B-True
b) Statement A-True B-False
c) Statement A-FalseB-True
d) Statement A – False B – False
Answer:
a) Statement A-True B-True

Question 33.
Confirmation:
Statement S: When kidney fails suddenly there is more chance of recovery.
Statement T: In the chronic kidney failure there may not be any chance of recovery.
a) Statement S-True T-True
b) Statement S-True T-False
c) Statement S-True T-True
d) Statement S – True T – False
Answer:
c) Statement S-True T-True

Question 34.
Confirmation:
Statement A: The kidney infection leads to inflammation of bladder and kidney
Statement B: Urination with pain, urinary urgency bloodytinged urine.
a) Statement A – True Statement B explain the symptom of A.
b) A – True – The statement B does not explains the statement A
c) A and B are false
d) A False B – True
Answer:
a) Statement A – True Statement B explain the symptom of A.

Question 35.
What is the normal urea level in the blood
a) 17-30mg/100ml
b) 30-35mg/ 100ml
c) 10-15 mg/100 ml
d) 5-10 mg/100 ml
Answer:
a) 17-30mg/100ml

Question 36.
Find the wrong pair.
a) Renal stone – nephrolithiasis
b) Urine – Urochrome
c) Deficiency of ADH – Urine out put decreases
d) Skin- Lactic acid excretion
Answer:
c) Deficiency of ADH – Urine out put decreases

Question 37.
Confirmation:
Statement A: Bright’s disease is due to the infection of streptococcus in children
statement B: There is inflammation of glomerulus.
a) Statement A and Bare false
b) Statement A True B explain the A
c) A True B False
d) A False B – True
Answer:
b) Statement A True B explain the A

Question 38.
Find the wrong pair.
a) Heporin – Anticoagulating factor
b) Glomerulo nephritis – Accumulation of water in the body
c) Primary kidney – Meso nephridia
d) Uremia – Increase in the blood urea level
Answer:
c) Primary kidney – Meso nephridia

Question 39.
How much urine can be stored up in the bladder?
a) 300-600 ml
b) 200-300 ml
c) 400-700 ml
d) 500-800 ml
Answer:
a) 300-600 ml

Question 40.
For how much time urine can be held in the bladder?
a) 6 hours
b) 2 hours
c) 5 hours
d) 3 hours
Answer:
c) 5 hours

Question 41.
The urinary bladder is made up this muscle ……………………….
a) Detrusor muscle
b) Striated muscle
c) Sphincter muscle
d) None of the above
Answer:
a) Detrusor muscle

Question 42.
Match the following:

1. Steno haline	I Shark
2. Eurihaline	II Otter
3. Osmo regulators	III Goldfish
4. Osrno confirmers	IV Salmon

a) 1-IV 2-III 3-II 4-1
b) 1 -II 2-II 3-IV 4-1
c) 1 -III 2-IV 3-II 4-1
d) 1-1 2-II 3-IV 4-III
Answer:
c) 1 -III 2-IV 3-II 4-1

Question 43.
Find out the less toxic waste among the following:
a) Urea
b) Uric acid
c) Ammonia
d) Creatinine
Answer:
a) Urea

Question 44.
Find out the ammonotelic organisms?
a) Reptiles
b) Birds
c) Aquatic amphibians
d) Frog
Answer:
c) Aquatic amphibians

Question 45.
Find out the wrong pair.
a) Rennette cells – Annelida
b) Molluscs – Metanephridia
c) Amphioxus – Mesonephridia
d) Tapeworm – Flame cells
Answer:
c) Amphioxus – Mesonephridia

Question 46.
Find the excretory structure of prawn?
a) Malpighian tubules
b) Green glands
c) Rennette cells
d) Peyer gland
Answer:
b) Green glands

Question 47.
Kidney with noglomerulus form this types of urine?
a) Hypertonic
b) Isotonic
c) Hypotonic
d) Neutrotonic
Answer:
c) Hypotonic

Question 48.
Match the following:

1. Conical tissue masses	a) Columns of Bertini
2. Extension in renal	b) Renal pelvis
3. Broad part of Hilum	c) Calyces
4. Projection of pelvis	d) Medullary pyramids

a) 1-d 2-a 3-b 4-c
b) 1-a 2-b 3-c 4-d
c) 1-b 2-a 3-c 4-d
d) 1-a 2-c 3-b 4-d
Answer:
a) 1-d 2-a 3-b 4-c

Question 49.
How much urine can be held in the urinary bladder?
a) 300-800ml
b)300-600ml
c) 100-400ml
d)200-300ml
Answer:
b)300-600ml

Question 50.
The muscles of urinary bladder is called by this name?
a) sphincter muscle
b) Detrusor muscle
c) constricting muscle
d) Peristaltic muscle
Answer:
b) Detrusor muscle

Question 51.
Name the nephron where the Henle’s loop is short?
a) Medullary nephron
b) Cortical nephron
c) Juxta medullary nephron
d) Medulla nephron
Answer:
b) Cortical nephron

Question 52.
The vessel which runs parallel to the loop of Henle is called by this name.
a) Efferent artery
b) Afferent artery
c) Vasarecta
d) Vasanervosa
Answer:
c) Vasarecta

Question 53.
What is the process of urea formation?
a) Ross cycle
b) Ornithine cycle
c) Urea cycle
d) b and c
Answer:
b) Ornithine cycle

Question 54.
What is the pressure in the afferent arteriole?
a) 35 mm Hg
b) 55 mm Hg
c) 20 mm Hg
d) 40 mm Hg
Answer:
b) 55 mm Hg

Question 55.
What is the filtrate entered into the Bowman’s capsule from glomerulus?
a) Secondary urine
b) Primary urine
c) Tertiary urine
d) Quarternary urine
Answer:
b) Primary urine

Question 56.
Find the correct option
Assertion: The glomerular filtrate resembles the blood
Reason: The glomerulus filtered the blood received from efferent artery
a) Assertion True Reason True
b) Assertion False Reason False
c) Assertion True The Reason does not explains the statement
d) Assertion True Reason explains A
Answer:
c) Assertion True The Reason does not explains the statement

Question 57.
What is the fate of water in the descending limb of Henle?
a) The concentrations of Na reduces
b) The concentration of Cl^– ions reduces
c) Na and Cl^– gets concentrated in the filtrate.
d) Formation of Hypotonic solution.
Answer:
c) Na and Cl^– gets concentrated in the filtrate.

Question 58.
How can the pH of blood be regulated?
a) Due to glucose reabsorption
b) Due to the reabsorption of HCO^–₃
c) Due to the reabsorption of Na⁺
d) Due to the reabsorption of Cl^–
a) 1-2 II-4 III-3 IV-1
b) 1-1 II-2 III-3 IV-4
c) 1-4 II-3 III-2 IV-1
d) 1-2 II-l III-3 IV-4
Answer:
b) 1-1 II-2 III-3 IV-4

Question 59.
Match the following
Answer:

I. Potassium	1. Descending limb of Renie
II. Water	2. Active transport
III. Glucose	3. Proximal convoluted tubule
IV. Na, cl, lc	4. Aqua porin

Question 60.
How is osma regulation in medulla maintained?
a) Juxtamedullary
b) Counter current
c) Positive current exchanger
d) Negative control
Answer:
b) Counter current

Question 61.
Find the wrong pair.
a) Vasa recta – Proximal convoluted tubule
b) ADH – Neuro hypophysis
c) Specialized nephron tissue – Juxtaglomerular apparatus
d) Renin – Glanularcell
Answer:
a) Vasa recta – Proximal convoluted tubule

Question 62.
This synthesizes angio tenginogen.
a) Kidney
b) Liver
c) Malpighian tubule
d) Glomerular
Answer:
b) Liver

Question 63.
This increases the absorption of sodium ions.
a) Renin
b) Angiotensin I
c) Angio tensin II
d) Angio tensinogen
Answer:
c) Angio tensin II

Question 64.
Name the hormone that decreases the blood pressure
a) Angiotensin I
b) Atrial natriuretic peptide
c) Vasopressin
d) ADH
Answer:
b) Atrial natriuretic peptide

Question 65.
This decreases the excretion of Renin?
a) Aldosterone
b) vasopressin
c) Atrial natriuretic peptide
d) Ventrical natriuretic peptide
Answer:
c) Atrial natriuretic peptide

Question 66.
What is the pH of Urine?
a) 6.5
b) 7.0
c) 6.0
d) 8
Answer:
c) 6.0

Question 67.
What is the reason for the yellow colour of urine?
a) Uro chrome
b) Cytochrome
c) Chlorochrome
d) Phyto chrome
Answer:
a) Uro chrome

Question 68.
The increase in the level of urea
a) Uremia
b) ureomia
c) Uricmia
d) Ketosis
Answer:
a) Uremia

Question 69.
The accumulation of salt in the blood.
a) Poly urea
b) Oligo urea
c) Polydipsia
d) Polyphagia
Answer:
b) Oligo urea

Two marks

II. Very short answer 

Question 1.
What is meant by stenohaline organism?
Answer:
The stenohaline animals can tolerate only narrow fluctuations in the salts concentration. (eg) Goldfish.

Question 2.
What is meant by Eury haline animals.
Answer:
Eury haline animals are able to tolerate wide fluctuations in the salt concentrations. (eg) Salmons Tilapia.

Question 3.
What is meant by Osmoregulation?
Answer:
It is the control of tissue osmotic pressure which acts as a driving force for movement of water across biological membrances.

Question 4.
What is meant by Ionic regulation?
Answer:
It is the control of ionic composition of body fluids.

Question 5.
What are the nitrogenous waste formed due to the degen ration of aminoacid?
Answer:

• Ammonia
• Urea
• Uric acid

Question 6.
Write is kidney situated in the body?
Answer:

• The kidney lie in the superior lymbar region between the levels of the last thoracic and third lumber vertebra close to the dorsal inner wall of the abdominal cavity.
• The right kidney is placed lower than the left kidney.

Question 7.
What is meant by renal hilum?
Answer:
The centre of the inner concave surface of the kidney has a notch called the renal hilum through which ureter blood vessels and nerves innervate.

Question 8.
What is meant by malpighian capsule or renal corpuscle?
Answer:
The Bowman’s capsule and the glomerulus together constitute the renal corpuscle.

Question 9.
What are podocytes?
Answer:

• The visceral wall of gromerulus is made of epithelial cells called podocytes.
• The podocytes end in foot processes which cling to the basement membrance of the glomerulus.

Question 10.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little into the medulla and are called cortical nephrons.

Question 11.
What is meant by Juxta medullary nephrons?
Answer:
Some nephrons have very long loop of Henle that run deep in to the medulla and are called Juxta medullary nephrons.

Question 12.
What is vasa recta?
Answer:
The efferent arteriole serving the juxta medullary nephron forms bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Question 13.
List the three important process of urine formation?
Answer:

1. Glomerulus filtration
2. Tubular reabsorption
3. Tubular secretion

Question 14.
What is meant by Glomerulus filtrate? What is its composition?
Answer:
The blood comes to the glomerulus are filtered and enters the Bowman’s capsule is called glomerular filtrate
Composition Water glucose aminoacids and nitrogenous wastes.

Question 15.
What is meant by Net filtration pressure?
Answer:
The two opposing forces against the glomerular blood pressure

1. Collodial osmotic pressure is 30 mm Hg
2. Capsular hydrostatic pressure is 15mm Hg

Net filtration pressure- Glomerular
Hydrostatic pressure – Colloidal
Osmotic pressure + Capsular hydrostatic pressure = 55 mm Hg – 30 mm Hg + 15 mm Hg = 10 mm Hg

Question 16.
What is meant by Glomerular filtration rate?
Answer:
It is the volume of filtrate formed in a minute in all nephrons of both the kidneys.

Question 17.
What is the amount of glomerular filtrate?
Answer:
In adults the GFR is 120 -125 ml per minute.

Question 18.
What is meant by primary filtrate?
Answer:
The filtrate enters from glomerulus to the Bowman’s capsule is called as primary filtrate.

Question 19.
Why glomercular filtrate resembles blood plasma?
Answer:
In the glomerular filtrate all the contents that present in the blood except the plasma protein is present.

Question 20.
How much filtrate is formed in one day?
Answer:
The amount of filtrate formed in a day is 170 to 180l.

Question 21.
What is meant by selective permeability?
Answer:

• Some substances present in the glomerular filtrate is essential for our body.
• Hence these molecules are reabsorbed in a tubules. This process is called as selective reabsorption.

Question 22.
Name the process in which selective reabsorption is taking place?
Answer:

• Passive transport
• Active transport
• Diffusion
• Osmosis

Question 23.
What are aqua porins?
Answer:
Aquaporins are membrane transport proteins that allow water to move across the epithelial cells.

Question 24.
What are the molecules that move into the filtrate of nephron?
Answer:
Nitrogen potassium ammonia creatinine and organic acids.

Question 25.
What is meant by Isotonic solution?
Answer:
Isotonic condition of a solution indicates no passage of water across the membrane separating two such solution.

Question 26.
What is meant by Hypotonic solution?
Answer:
The solution in which there is a loss of water then that solution is a hypotonic solution.

Question 27.
What is meant by hypertonic solution?
Answer:
When two solutions A and B are separated by a semi permeable membrane when water move from solution A to B across the membrane then the B solution is hypertonic and the solution A where the water loses is known as hypotonic solution.

Question 28.
Name the organ that the angiotensin II acton?
Answer:

• Heart
• Kidney
• Brain
• Adrenal cortex

Question 29.
What are the symptoms of diabetes mellitus?
Answer:

• Excess glucose and ketone bodies in the urine
• Poly dipsia – Excessive drinking of water
• Polyurea – Excretion of large quantities of urea
• Polyphagia – Excessive appetite

Question 30.
Name the organ that excrete nitrogen other than kidney?
Answer:

• Lungs
• Liver
• Skin

Question 31.
What are the significance of sweat glands?
Answer:

• Sweat produced by the sweat glands helps to cool the body.
• It excretes Na⁺ and Cl^– small quantities of urea and lactate.

Question 32.
What is meant by nephrolithiasis?
Answer:
It is the formation of hard stone like masses in the renal tubules of renal pelvis.

Question 33.
What is meant by phleothotomy or lithotripsy?
Answer:
Renal stones can be removed by the technique pyleothotomy or lithotripsy.

Question 34.
What is meant by Bright’s disease?
Answer:
The inflammation of the glomeruli of both kidney of children due to the streptococcal infection is called as Bright’s disease.

Question 35.
What is meant by haemodialysis?
Answer:
The process of removing toxic urea from the blood of renal failure patients is known as haemodialysis.

Question 36.
Why females are prone to urinary tract infection than men?
Answer:
Females are prone to recurring urinary tract infection as they have shorter urethra.

Question 37.
Why men are finding difficult to urinate in their old age?
Answer:
With age prostate in males may enlarge which forces urethra to tighten restricting a normal urinary flow.

Question 38.
What is the change in Urine formation when there is a deficiency of ADH?
Answer:
When there is a deficiency of ADG the reabsorption of water from the proximal convoluted tubule decreases leads to dilute urine formation.

Question 39.
Why there is a increase in the body fluid when we drink large volume of water with out eating anything salty?
Answer:

• When we drink or eat salty products the Na+ enters into the body fluids.
• The sodium ions helps in the reabsorption of water
• But when we drink only water as there is no sodium ions the tubules cannot reabsorb water.
• Hence there is a increase in the urine output.

Question 40.
Consider how different foods affect water and salt balance and how the excretory system must respond to maintain homeostasis.
Answer:

• When we eat different foods the salt present in them goes to renal tubules.
• If sufficient salts are present in the body fluids this salt may not be reabsorbed when it goes through nephron and excreted through urine.
• If there is a decrease in the volume of water in body fluids sodium ions are reabsorbed. Hence water is reabsorbed from filtrate and water volume is increased.

Question 41.
Name the different process that maintains water level? When there is a severe loss of water in the body?
Answer:

• The blood vessels supplies to skin constricts and thus there is a decrease in the secretion of sweat prevents loss of water.
• There is a reduction in the glomerular blood pressure and the rate of filtration decreases.
• The reaborption of water in the proximal distal convoluted tubules increase.
• There is absorption of water from the small intestine and large intestine and thus increases the water content in the blood.

Question 42.
What is meant by Osmolarity?
Answer:

• The solute concentration of a solution of water is known as osmolarity,
• The unit is millosmoles / litre (mOsm /l)

Question 43.
What is meant by aquaporins? What are its functions?
Answer:
Aqua porins are water permeable channels.
Functions
It helps in allowing water to move across the epithelial cells in relation to the osmotic difference from the lumen to the interstitial fluid.

Question 44.
How can we measure that there is a efficient glomerular filtration?
Answer:
If the renal clearance is equal to the glomerular filtration rate with little reabsoption and secretion. Then we know the kidney is functioning efficiently.

Question 45.
What is the Significance of having long and short Henle’s loop of Nephrons?
Answer:

• The main function of Henle’s loop is to reabsorb water from filtrate.
• If the length of the loop is longer then there is more reabsorption of water and if the lengh of Henle’s loop is shorter then the reabsorption of water is less.

Question 46.
Give notes on Capillary to capsule.
Answer:
Bloodcells and most blood proteins are too big to cross the capsular membrane into the capsule space. But the membrane’s slits and pores allow through water mineral salts polypeptides and other small molecules including waste such as urea ammonia and creatinine.

Question 47.
Give short notes on Blood enters the glomerulus.
Answer:
Blood flows from renal arteriole into the knot of capillaries. It enters at pressure which will force water and other out of the capillaries into the capsular space.

Question 48.
Give notes on filteration in proximal convoluted tubule?
Answer:
Proximal tubule is nearer to the Bowman’s capsule. This region allows much water to be reabsorbed into the capillaries and surronding fluids as well as glucose mineral salts and other useful substances.

Question 49.
Give notes on filteration in peritubular capillaries.
Answer:
It is also called the vasarecta this network reabsorbs upto 99 percent of the water in the tubule as well as various other substance using active pumps it also moves sodium from the blood to the tubule.

Question 50.
Give notes on filteration in Henle’s loop (Ascending)
Answer:

• As the loop of the Henle dipsin to the renal medcula more water moves from the tubule into the blood as well as small amounts of salts and some urea and creatinine.
• Some acids and amines may move into the tubule in which ammonia cango in both the direction.

Question 51.
Give notes on filteration in distal tubule?
Answer:

• Distal tubule is far from capsule. This region may see water go in or out of the tubule depending on the concentration of water already in the tubule/ while hydrogen and potassium ions move to regulate both blood and urine pH.
• Acids amines and ammonia compounds may also transported into the tubule.

Question 52.
Give notes on filteration in collecing duct.
Answer:

• Fine adjustment of urine composition continues into the collecting duct system.
• About 5 percent of all the water and sodium being reabsorbed into the blood is recovered here.

Question 53.
Give notes on venousflow?
Answer:

• Blood flowing away from the nephrons carries 99 % of its orginal water.
• 98% of its sodium calcium and cholrides and about 40% of its urea.

Question 54.
We are not consuming urea. But in our body area is produced. Why?
Answer:
Through Arnithine cycle in the lives the nitrogenous waste formed due to the breakdown of amino acid creates urea.

Question 55.
What is meant by Ionic regulation?
Answer:
It is the control of the ionic composition of body fluids.

Question 56.
What are stenohaline animals?
Answer:
They can tolerate only narrow fluctuations in the salt concentration. Ex: Goldfish

Question 57.
What are Euryhaline animals?
Answer:

• They are able to tolerate wide fluctuations in the salt concentrations. Ex: Artemia Salmons.
• Acids amines and ammonia compounds may also transported into the tubule.

Question 58.
List the nitrogenous wastes
Answer:
Ammonia urea uricacid.

Question 59.
What are the other nitrogenous wastes of protein metabolism?
Answer:
Alantonin Alantoic acid, Ornithuriacid creatinine creatine purines pyramidines and pterines.

Question 60.
Which are called ammoneteles?
Answer:
Animals that excreate most of its nitrogen inthe form of ammonia are called ammonoteles.

Question 61.
What are called as uricoteles?
Answer:
Animals that excrete uric acid crystals with minimum loss of water are called uricoteles.

Question 62.
What are ureoteles?
Answer:
Mammals and terrestrial amphibians mainly excrete urea are called ureoteles.

Question 63.
How is Reptiles produced lesshypotonic urine?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henles loop and hence produce very little hypotonic urine.

Question 64.
How is mammals produced concentrated urine?
Answer:
Mammalian kidneys produce concentrated urine due to the presence of Henle’s loop.

Question 65.
What are the three coverings of kidney?
Answer:

• Renal facia
• Perirenal fat capsule
• Fibrous capsule

Question 66.
What are medullary pyramids?
Answer:
The medulla is divided into a few conical tissue masses called medullary pyramids.

Question 67.
What is meant by renal columns of Bertini?
Answer:
The part of cortex that extends in between the medullary pyramids is the renal columns of Bertini.

Question 68.
What is meant by renal pelvis?
Answer:
A broad funnel shaped space inner to the hilum is called renal pelvis.

Question 69.
What is calyces?
Answer:
The projection in the pelvis is called calyces.

Question 70.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little in to the medulla and are called cortical nephrons.

Question 71.
What is meant by juxta medullary nephron?
Answer:
Some nephrons have very long loop of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 72.
What is the functions of aquaporins?
Answer:
This helps in allow water to move across the epithelial cells in relation to the osmotic difference from the human to the interstitial fluid.

Question 73.
What is meant by juxta glomerular apparatus?
Answer:
It is a specialized tissue in the afferent arteriole of the nephron that consists of maculadensa and granular cells.

Question 74.
What is meant by micturition?
Answer:
The process of release of urine from the bladder is called micturition.

Question 75.
What are the symptoms of diabetes mellitus?
Answer:
Presence of glucose and ketone bodies in the urine.

Question 76.
How is lung acting as a excretory organ?
Answer:
Lungs remove large quantities of carbondixide 181 / day and significant quantities of water every day

Question 77.
What is meant by renal clearnace?
Answer:
The amount of solute passing from the urine in a given period of time is renal clearance.

Question 78.
How can we estimate the efficiency of kidney?
Answer:

• The renal clearance is equal to the glomerular filtrate then there is efficient filtration with little reabsorption and secretion.
• Thus we can estimate the clearance is equal.

Three Marks

Short Answer –

Question 1.
a) List the ma j or nitrogenous wastes.
b) What are other wastes formed during protein metabolism?
Answer:
a) 1. Major nitrogenous wastes.

• ammonia
• Urea
• Uricacid

b) Other Wastes:

• Trimethy lamine oxide TMO
• Quanine
• Allantonin
• Creatinine
• Creatine
• Purines

Question 2.
Give notes on ammonoteles uricoteles and ureoteles.
Answer:
Ammonoteles:

• Animals that excrete most of its nitrogen in the form of ammonia are called ammonoteles.
  (e.g) fishes Amphibians aquatic insects.
• In bony fishes ammonia diffuses out acrossthe body surface.

Uricoteles:

• Animals which excrete uricacid crystals with a minimum loss of water is called.
• Uricoteles (e.g) Reptiles Birds land snails and insects.

Ureoteles:

• Animals which excrete urea as a nitrogenous wastes are called ureoteles
• (e.g) Mammals, terrestrial amphibians.

Question 3.
Nephrons are the functional and structural unit of kidney’s. What is the relationship between glomerulus, Henle’s loop and urine formation?
Answer:

• Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop and produce hypotonic urine (dilute)
• Mammalian kidneys produce concentrated urine due to the presence of long Henle’s loop.
• Aglomerular kidneys of marine fishes produce little urine that is iso osmotic to the body fluid.
• Amphibians and fresh water fishlack Henle’s loop hence produce dilute urine.

Question 4.
Differentiate the cortical nephron fron juxta medullary nephron
Answer:

Question 5.
Give Short notes on capillary bed of the nephron:
Answer:
The first capillary bed of the nephron is the glomerulus.
The other is peritubular capillaries.
1. Glomerulus:

Blood enters into the glomerulus through afferent arteriole and drained by the efferent arteriole.

2. Peritubular capillaries:

• The efferent arteriole forms a fine cappillary network around the renal tubule called the peritubular capillaries.
• The efferent arteriole serving the juxta medullary nephrons forms bundles of long straight vessel called vasa recta.
• Vasa recta is absent in cortical nephrons.

Question 6.
What happens to the filtrate that comes to the proximal convoluted tubule? (or) Explain about reabsorption?
Answer:

• In the proximal convoluted tubule glucose lacticacid aminoacid sodium ions are reabsorbed.
• Sodium is reabsorbed – potassium pump in the proximal convoluted tubule.
• Small amounts of urea and uric acid are also reabsorbed.

Question 7.
What happen to the filtrate that comes to the Henle’s loop? (or) Explain the reabsorptionin the Henle’s loop?
Answer:

Descending Loop:
The aquaporin present in the descending limb of Henle permeable to water but not permeable to salts.
Hence Na⁺ and cl^– gets concentrated in the filtrate.

Ascending Limb:
It is impermeable to water but permeable to solutes like Na⁺cl^– and K⁺.

Question 8.
Give an account of tubular reabsorption?
Answer:
The volume of filterate formed perday is 170-180 is and the urine released in a day is 1.5l

• Nearly 99 % of the glomerular filtrate is reabsorbed by the renal tubules. It is called selective reabsorption.
• Reabsorption is taken place by the tubular epithelial cells in different segments of the nephron by active transport or passive transport diffusion and osmosis.

Question 9.
What is happenning to the filtrate in distal convoluted tubule (or) Reabsorption taking place here?
Answer:

• Depending on the body’s need the reabsorption taking place here and is regulated by hormones.
• Reabsorption of bicarbonate HCO₃^–  takes place to regulate the blood pH.
• Homestasis of K⁺ and Na⁺ in the blood is also regulated in this region.

Question 10.
Name the structures that regulate the functioning of kidney?
Answer:

• Hypothalamus
• Juxta glomerular apparatus
• Heart

Question 11.
What is meant by diabetes incipidus?
Answer:
If there is deficiency or absence of ADH that leads to dilute urine called diabetes incipidus.
Symptoms.

• Excessive thirst
• Excretion of large quantities of dilute urine.
• FaIl in blood pressure.

Question 12.
Give notes on juxta glomerular apparatus?
Answer:

• Specialized tissue in the afferent arteriole of ncphron is the juxta glomerular apparatus.
• It consists of macula densa and granular cells.
• The macula densa cells sense distal tubular flow and affect afferent alteriole diameter.
• The granular cells secrete renin.

Question 13.
Why female are prone to urinary tract infections? (Urethritis)
Answer:

• Female’s urethra is very short and its external opening is close to the analopening.
• Hence improper toilet habits can easily carry faecal bacteria into the urethra.
• The urethral mucusa is continuous with the urianary tract and the inflammation of the urethra is called urethriti’s.

Question 14.
What is cystitis?
Answer:
The urinary tract infection leads to inflammation of bladder called cystiti’s.
Symptoms:

• Painful urination
• Urinary Urgency
• Cloudy or bloodtingedurine
• Back pain head ache offen occurs

Question 15.
What is meant by renal failure? What are its types?
Answer:
When the kidney fails to excrete wastes may lead to accumulation of urea with marked reduction in the out put called renal failure.
Types
Acute renal failure
Chronic renal failure

Question 16.
Why the chronic renal failure is dangerous than acute renal failure?
Answer:

• Though the kidney stops its function abruptly there are chances for recovery of kidney function in acute renal failure.
• But in chronic failure there is a progressive loss of function of the nephron which gradually decreases the function of kidney.

Question 17.
What is meant by glomerulo nephritis or Bright’s disease? What are its symptoms?
Answer:
Inflammation of the glomerulus of both the kidneys due to the strepto coccal infection in children is called as Bright’s disease Symptoms

• Haematuria
• Proteinuria
• Salt and water retention – Oligouria (Low urine out put)
• Hypertension and Pulmonaryoedema

Question 18.
a) What is meant by kidney transplantations.
b) Where is graft kidney received from?
c) What are the steps to be taken to avoid graft rejection?
Answer:

• Transfer of healthy kidney from one person (donor) to another person with kidney failure is called kidney transplantation.
• The donated kidney may be taken from a healthy person who is declared brain death or from sibling or close relatives.
• Immuno supressive drugs are administered to the patient to avoid tissue rejection.

Question 19.
a) Name the hormone that the dipsticks contain which tests urine?
b) Which colour indicates the presence of glucose in the urine?
Answer:

• Glucose oxidase and peroxidase.
• Brown coloured compound is produced.

Question 20.
What are Osmo confirmers?
Answer:
Osmo confirmers are able to change their internal osmotic concentration with change in external environments asin marine and sharks. molluscs

Question 21.
What are Osmo regulators?
Answer:
They maintain their internal osmotic concentration irrespective of their external osmotic environment, (eg) Otters.

Question 22.
List the excretory structures of different organisms.
Answer:

• Protonephridia
• Meta nephridia
• Flame cells – Platy helminthes
• Rennette cells – Nematodes
• Malpighian tubules – Insects
• Greenglands – Prawns

Question 23.
What is meant by filtration slits?
Answer:

• The visceral layer of glomerulus is made of epithelial cells called podocytes and ends in foot processes which cling to the basement membrane of the glomerulus.
• The openings between the foot processes are called filtration slits.

Question 24.
Draw the diagram of ornithine cycle?
Answer:

Question 25.
Why there is a pressure reduction when the blood goes through efferent arteriole?
Answer:

• Blood enters the glomerulus faster with greater force through afferent arteriole.
• Because the afferent arteriole is broader than efferent arteriole that is why the pressure reduces when it goes through the efferent arteriole.

Question 26.
What are the changes taking place in our body when there is a fluid loss?
Answer:

• The osmo receptors in the hypothalamus is stimulated.
• The neurohypophysis is stimulated and anti diuretic hormone is liberated.
• The aquaporins in the tubuler are increased and water is reabsorbed and enters into the interstitial cell and the water loss is rectified.

Question 27.
How is skin acted as a excretory organ?
Answer:

• Skin excretes Na⁺ and Cl^– small quantities of urea and lactate.
• Sebaceous glands eliminated certain substances like steroids, hydrocarbons and waxes.

Question 28.
What is meant by urethritis?
Answer:
The urethral mucosa is continuous with the urinary tract. The infection in the urethra is called urethritis.

Question 29.
What is meant by Cystitis?
Answer:
The infection in the urethra can ascend the tract to cause bladder inflammation called cystitis.

Question 30.
What is meant by Phelitis or Pyelone phritis?
Answer:
The bladder infection ascend to the renal inflammation called pyelitis or pyelonephritis.

Question 31.
What are the two types of renal failure?
Answer:

1. Acute renal failure
2. Chronic renal failure

1. Acute renal failure

• In acute renal failure the kidney stops its function abruptly.
• There are chances for recovery of kidney function.

2. Chronic renal failure
In chronic renal failure there is a progressive loss of function of the nephron which gradually decreases the function of kidneys.

Question 32.
What is meant by Uremia?
Answer:
Uremia is characterized by increase in Urea uric acid and creatinine in blood.

Question 33.
What is meant by Nephrolithiasis?
Answer:
It is a formation of hard stone like masses in the renal tubules of renal pelvis.

Question 34.
How is water excess taken through drinking too much fruit juice regulated?
Answer:

• When we drink much fruit juice the osmo receptors in hypothalamus is not stimulated and hence the secretion of vaso pressin from neuro hypophysis is reduced.
• The aquaporin escapes from collecting duct to cytoplasm and hence water reabsorption is prevented and formed dilute urine.

Question 35.
What is uremia?
Answer:
Uremia is a condition in which there is a increase level of urea uric acid and creatinine in blood.

Question 36.
What is the amount of Urea present in the blood?
Answer:

• The level of urea in the blood is 17 -30 mg /100ml.
• In chronic kidney failure there is 10 times increase in urea level.

Five marks

IV. Detailed Answers –

Question 1.
Name the different excretory structure and different organisms.
Answer:
1. Invertebrate – Protonephridia / Meta nephridia
2. Platvhelminthes – Flamecells
3.  Amphioxues – Solenocytes
4 Nematodes – Rennette cells
5. Annelida – Metanephridia
6. Insects – Malpighian tubules
7. Prawn / Crustaceans – Green glands / Antenna! glands

Question 2.
The concentration of urine depends on the structure of nephron? Explain?
Answer:

1. In the reptiles the glomerulus is reduced or there may be no glomerulus and 1-lenle’s loop and hence produces dilute urine (chypotonic).
2. In the mammals the long Henle’s loop produces concentrated urine (hypertonic)
3. A glomerular kidneys of marine fishes produce little urine that is isoosmotic to the body fluid.
4. Amphibians and fresh water fish lack Henle’s loop hence produce dilute urine.

Question 3.
Draw the following diagram and mark the following parts.
Answer:
A) Aorta
B) Renal vein
C) Ureter
D) Urinary bladder

Question 4.
a. What is the weight of kidneys? What are its outer coverings?
b. Draw the L.S of kidney and name the parts.
c. Explain the internal structure of kidney
Answer:
a. Each kidney weighs an average of 120 – 170 gms.
The outer layer of the kidney is covered by three layers of supportive tissue namely renal fascia perirenal fat capsule fibrous capsule.

b. Draw the LS of kidney and name the parts.

c. Internal Structure of kidney

• The longitudinal section of kidney shows an outer cortex inner medulla and pelvis.
• The inner concave surface of the kidney is renal hilum through which ureter blood vessels and nerves enter.
• Inner to the hium is a funnel shaped renal pelvis with projection called calyces.
• The calyces collect the urine and empties in to the ureter.
• The medulla consists of conical tissues called medullary pyramids or renal pyramids.

Question 5.
a) What is the structural and functional unit of kidney.
b)Draw the diagram of nephron and name the parts.
c) Give short notes on Malpighian body or Renal Corpuscle.
Answer:
a) Hie structural and functional unit of kidney is nephron. It is composed of Malpighian body or renal corpuscle and Urine ferous tubule.
b) Structure of nephron

c. Malpighian body/Renal Corpuscle.

• The Bowman’s capsule and the glomerulus together constitutes Malpighian corpuscle.
• Bowman’s capsule is made up of two layers. It contains blood vessels called glomerules.
• The endothelial of glomerulus has many pores. The viscral layers of glomerulus is made of epithelial cells called podocytes.
• Tire podocytes end in foot processes which cling to the basement membrance of the glomerulus.
• The openings between the foot processes are called filtration slits.

Question 6.
a) Give the different regions of renal tubule.
b) Where is renal tubule present in the kidney?
c) How is renal tubules differentiated depending on the Henle’s loop?
Answer:
a)
1. proximal convoluted tubule.
2. Henle’s loop
a. Thindescending limb of Henle’s loop.
b. Thick ascending limb.

2. Distal convoluted tubules
The distal convoluced tubules opens in to acollecting duct.
Several collecting ducts fuse to form papillary duct that delivers urine in to the calvces which opens into the renal pelvis.
b) The PCT and DCT are situated in the cortical region of the kidney.
The loop of Henle is in the medulla region.
c) The loop of Henle is too short and extends only very little into the medulla and are called cortical nephron.
Some nephrons have very long loops of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 7.
a) Give notes on capillaries of nephron
b) Give an account of blood vessels of glomerulus.
c) What is vasa recta? Where is it seen?
Answer:
Capillaries of nephron
1. Glomerulus capillary bed
2. Peritubular capillaries

1. Glomerular capillarybed

• It consists of afferent and efferent arteriole.
• The afferent arteriole is broader than efferent arteriole.
• The efferent arteriole that comes out of the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries.

Vasa recta
The efferent arteriole serving the juxta medullary nephrons form bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Question 8.
a) Describe the three important process in the urine formation?
b) Give an account on glomerular filtration
Answer:
a. Main Processes

• Clornerular Filtration
• Tubular reabsorption
• Tubular secretion

b. Glomerular filtration

• Blood enters the kidney from the renal artery.
• Blood enters the glomerulus faster with greater force through the afferent arterioles because the afferent arteriole is wider than efferent arteriole.
• The glomerular hydrostatic pressure is around 55 mm Hg.
• The colloidal osmotic pressure is 30 mm Hg. Capsular hydrostatic pressure is 15 mm Hg. The net pressure of 10 mm Hg is responsible for the renal filtration.
• Next filtration pressure 55 mm Hg – (30 mm Hg +15 mm Hg) 10 mm Hg.
• The kidneys produce 1801 of glomerular filtrate 24 hours and 120 -125 ml/min.
• This filtrate will enter into the Bowman’s capsule and called primary urine. It contains more water colloidal protein, glucose salts and nitrogenous waste.

Question 9.
a) Why glomerular filtrate resembles blood plasma?
b) Tabulate the concentration of substances in the blood plasma and in the glomerular filtrate.
Answer:
As the glomerular filtrate forms it contain all the substances except plasma protein of blood. Hence it resembles blood.

Substance	Concentration in blood Plasma / g dm-3	Concentration in glomerular filtrate / g dm-3
Water	900	900
Proteins	80.0	0.05
Aminoacids	0.5	0.5
Glucose	1.0	1.0
Urea	0.3	0.3
Uric Acid	0.04	0.04
Creatinine	0.01	0.01
Inorganic ions (mainly (Na+, K+ and Cl– )	7.2	7.2

Question 10.
a) What is meant by Tubular Secretion?
b) Give an account of tubular secretion of nephron.
Answer:
Tubular Secretion:

• The collecting tubule of nephron secrete H⁺ NH⁴ , Creatinine and Organic acid and liberated into the tubules and excreted through urine.
• Most of the water is absorbed in the proximal convoluted tubule.
• Na^– is exchanged for water in the loop of Henle.
• The hypotonic fluid enters the distal convol uted tubule.
• Substances such as urea and salts pass from peritubular blood into the cells as distal convoluted tubule and then to collecting duct.
• Water is absorbed and concentrated hypertonic urine is form ed.
• For every H⁺ secreted into the tubular filtrate a Na⁺ is absorbed by the tubular cell.
• The H secreted combines with HCO₃, HPO₃ and NH₃ and gets fixed as carbonic acid CH2CO₃ and Phosphoric acid CH2PO4
• Since H⁺ gets fixed in the fluid reabsorption of H⁺ is prevented.

Question 11.
a) In which process concentrated urine is formed?
b) How is concentrated urine formed in the Hen le’s loop.
Answer:
a) The major function of Henle’s loop is to concentrate Na⁺ and Cl^–.

• There is low osmolarity near the cortex and high osmolarity towards the medulla.
• This osmolarity in the medulla is due to the presence of the solutes transporters and is maintained by
  the arrangement of the loop of Henle collecting duct and vasa recta.
• The osmolarity of interstitial fluid is 300 m Osm.
• The Henle’s loop create a countercurrent multiplier.

• As the fluid enters the descending limb water moves from the lumen into the inter stitial fluid the osmolarity reduces.
• To counteract this dilution the region of the ascending limb actively pumps solutes from the lumen into the interstitial fluid and the osmolarity increases to about 1200 m OSM in medula.

b) The vasa recta maintains the medullary osmotic gradient via counter current exchanger.

• The counter current exchanger of vasa recta preserves the medullary gradient while removing reabsorbed water and solutes.
• The vasa recta leaves the kidney at the junction between the cortex and medulla.
• When the blood leaves the efferent arteriole and enters vasa recta the osmolarity in the medulla increases (1200 rnOsm) and result in passive up take of solutes and loss of water.
• As the blood enters the cortex the osmolarity in the blood decreases and the blood loses solutes and gain water to form concentrated urine.

Question 12.
How is vasa recta helps in producing concentrated urine?
Answer:
Vasa recta maintains the medullary osmotic gradient via counter current exchanger.

• Vasa recta preserves the medullary gradient while removing reabsorbed water and solutes through counter current exchanges.
• The vasa recta leave the kidney at the junction between the cortex and medulla.
• The interstitial fluid at this point is iso – osmotic to blood.
• When the blood leaves the efferent arteriole and enters the vasa recta the osmalarity in the medulla increases (1200 mOsm) and results in passive up take of solutes and loss of water.
• As the blood enters the cortex the osmolarity in the blood decreases (300mOsm) and the blood loses solutes and gains water to form concentrated urine.
• Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Question 13.
a) What are the structures that regulate kidney functioning?
b) What is the role of ADH in Urine formation.
c) What are the symptoms of diabetes insipidus.
Answer:
a) The structures that regulate kidney functioning

• Hypothalamus
• Juxta glomerular apparatus
• Heart

b) The functions of ADH

• When there is excessive loss of fluid from the body or when there is an increase in the blood pressure the osmo receptors of the hypothalamus is stimulated.
• The osmo receptors stimulate the neurohypophysis to secrete antidiuretic hormone (AOH) or vasopressin.
• ADH facilitates reabsorption of water by increasing the number of aquaporins on the cell surface of the distal convoluted tubule and collecting duct and prevents excessloss of water.

Question 14.
a) Name the cell that secretes the enzyme renin.
b) Where is granular cell present?
c) What is the role of renin in Osmoregulation.
Answer:
a) Renin is secreted by granular cells.
b) Granular cells are present in the afferent arteriole.
c) The role of renin

• A fall in glomerular blood flow blood pressure and filtration rate can activate granular cells of juxta glomerular cells to release renin.
• Renin converts the plasma protein angiotensinogen into angiotensis I and angiotensin II.
• Angiotensis II stimulates Na⁺ reabsorption in the proximal convoluted tubule by vaso constriction of the blood vessels and increases the glomerular blood pressure.
  Angiotensis II stimulates adrenal cortex to secrete aldo sterone that causes reabsorption of Na⁺ ,K⁺ excretion and absorption of water.
• This increases the glomerular blood pressure and glomerular filtration rate.
• Hence renin regulates the osmoregulation.

Question 15.
a) Where is atrial natriuretic peptide liberated from?
b) Write its significance in short.
Answer:
a) This is liberated from atrium of heart.
b) Use of Atrial natriuretic peptide

• It increases Na excretion and increases the blood flow to the glomerulus.
• It acts on the afferent glomerular arteriole as a vaso dilator or an efferent arteriole as a vaso constrictor.
• It reduces aldosterone from adrenal cortex and renin secretion.
• Thus decreases the angiotensin II.
• The atrial natri uretic factor acts antagonistically to renin angiotensin system aldosterone and vasopressin.

Question 16.
a) What is micturition?
b) How is central nervous system regulates urination?
Answer:
a) The process of release of Urine from the bladder is called micturition or urination.
b) Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till it receives a signal from the central nervous system.

• The stretch receptors present inthe urinary bladder are stimulated when it gets filled with urine.
• At the same time the internal sphincters opens and relaxing the external sphincter.
• The sphincter opens and the urine is expelled out.

Question 17.
Answer for the following questions.
a) What is the average excretion of an adult human?
b) What is the pH of Urine
c) How much pH is differed due to the food we eat?
d) What is the reason for the yellow colour of urine?
e) How much urea is excreted in a day?
f) If there is more glucose, and ketone what does it indicates?
Answer:

• 1.5l
• pH = 6
• pH = 4.5-8
• Uro chrome
• 25 – 30 g
• Diabetes melitus

Question 18.
Answer for the following question in excretion by other organs.
a) What are the other structures that excrete nitrogenous wastes rather than kidney?
b) How much CO₂ is excreted through lungs?
c) What are the wastes excreted by digestive systems? ’
d) What are the glands that excrete waste through skin?
e) What is the main function of sweat glands?
f) What is the second important function of sweat gland?
g) Name the substance excreted by sebaceous glands.
h) Name the substances excreted by sebaceous glands.
i) Name the waste excreted through saliva.
Answer:

• Lungs, Liver and skin
• 18 Litre, much Water
• Bilirubin, Biliverdin, Cholesterol, Vitamins and drugs.
• Sweat glands and sebaceous glands
• To cool the body
• Na⁺ and Cl^–, small quantities of Urea and Lactate excretion.
• Sebum
• Steroid Hydrocarbon and wax.
• Nitrogenous wastes

Question 19.
Draw the schematic representations of renin hormone in the regulation of body fluid concentration.
Answer:

Question 20.
a) What is meant by Haemodialysis?
b) Why is it called as artifical kidney?
c) Give an account of Haemodialysis.
Answer:
a) in the patients of kidney failure toxic urea can be removed from the blood by a process called haemodialysis.
b) The dialyzing machine is a artificial kidney.
c) Hameodialysis

• A dialyzing machine consists of a long cellulose tube surrounded by the dialysing fluid in a water bath.
• A patient’s blood is drawn from a convenient artery and pumped into the dialysing unit after adding an anti coagulant like heparin.
• The tiny poresin the dialysis tube allows small molecules such as glucose salts and urea to enter into the water bath.
• Where as blood cells and protein molecules do not enter these pores the cleared blood is then pumped back tothe body through a vein.

Question 21.
Give notes on kidney transplantation?
Answer:

• This involves transfer of healthy kidney from one person (donor) to another person who is with kidney failure.
• The donated kidney may be taken from a healthy person who is declared brain dead or from sibling or close relatives to minimize the chances of rejection by the immune system of the host.
• Immuno suppressive drugs are usually administered to the patient to avoid tissue rejection.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 11 Fundamentals of Organic Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
Select the molecule which has only one π bond.
a) CH₃ – CH = CH – CH₃
b) CH₃ – CH = CH – CHO
c) CH₃ – CH = CH – COOH
d) All of these
Answer:
a) CH₃ – CH = CH – CH₃

Question 2.
In the hydrocarbon the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence.
a) sp, sp, sp³, sp², sp³
b) sp², sp, sp³, sp², sp³
c) sp, sp, sp², sp, sp³
d) none of these
Answer:
a) sp, sp, sp³, sp², sp³

Question 3.
The general formula for alkadiene is
a) C_nH_2n
b) C_nH_{2n – 1}
c) C_nH_{2n – 2}
d) C_nH_{n – 2}
Answer:
c) C_nH_{2n – 2}

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept – 2 – ene is

Answer:
a)

Question 5.
The IUPAC name of the compound is

a) 2, 3 – Dimethylheptane
b) 3 – methyl – 4 – ethyloctane
c) 5 – ethyl – 6- methyloctane
d) 4 – Ethyl – 3 methyloctane
Answer:
d) 4 – Ethyl – 3 methyloctane

Question 6.
Which one of the following names does not fit a real name?
a) 3 – Methyl – 3- hexanone
b) 4 – Methyl – 3 – hexanone
c) 3 – Methyl – 3 hexanol
d) 2 – Methyl cyclo hexanone
Answer:
a) 3 – Methyl – 3- hexanone

Question 7.
The IUPAC name of the compound CH₃ – CH = CH – C ≡ CH is
a) Pent – 4- yn – 2 – ene
b) Pent – 3- en – 1- yne
c) Pent – 2 – en – 4 – yne
d) Pent – 1 yn – 3 – ene
Answer:
b) Pent – 3- en – 1- yne

Question 8.
IUPAC name of  is
a) 3, 4, 4 – Trimethylheptane
b) 2 – Ethyl – 3, 3, – dimethyl heptane
c) 3, 4, 4 – Trimethyloctane
d) 2 – Butyl – 2 – methyl – 3 ethyl – butane
Answer:
c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of
is
a) 2,4,4 – Trimethylpent – 2 – ene
b) 2,4,4 – Trimethylpent – 3 – ene
c) 2,2,4 – Trimethylpent – 3 – ene
d) 2,2,4 – Trirnethylpent – 2 – ene
Answer:
c) 2,2,4 – Trimethylpent – 3 – ene

Question 10.
The IUPAC name of the compound is
a) 3 – Ethyl – 2 – hexene
b) 3 – Propyl – 3 – hexene
c) 4 – Ethyl – 4 – hexene
d) 3 – Propyl – 2 – hexene
Answer:
a) 3 – Ethyl – 2 – hexene

Question 11.
The IUPAC name of the compound
is
a) 2 – Hydroxypropionic acid
b) 2 – Hydroxy Propanoic acid
c) Propan – 2 – ol – 1 – oic acid
d) 1 – Carboxyethanol
Answer:
b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of is
a) 2 – Bromo – 3- methyl butanoic acid
b) 2 – methyl – 3 bromo butanoic acid
c) 3 – Bromo – 2 – methylbutanoic acid
d) 3 – Bromo – 2, 3 – dimethyl propanoic acid
Answer:
c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is
a) CH₃ – CH₂ – CH₂ – CH₂

b)

c)

d)
Answer:
c)

Question 14.
The number of stereoisomers of 1, 2 – dihydroxy cyclopentane
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

Question 15.
Which of the following is optically active?
a) 3 – Chloropentane
b) 2- Chloro propane
c) Meso – tartaric acid
d) Glucose
Answer:
d) Glucose

Question 16.
The isomer of ethanol is
a) acetaldehyde
b) dimethyl ether
c) acetone
d) methyl carbinol
Answer:
b) dimethyl ether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C₃H₆O?
a) 4
b) 5
c) 9
d) 10
Answer:
c) 9

Question 18.
Which one of the following shows functional isomerism?
a) ethylene
b) Propane
c) ethanol
d) CH₂Cl₂
Answer:
c) ethanol

Question 19.
are
a) resonating structure
b) tautomers
c) Optical isomers
d) Conformers
Answer:
b) tautomers

Question 20.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
a) Fe₃[Fe(CN)₆]₂
b) Fe₄ [ Fe(CN)₆]₃
c) Fe₄[Fe(CN)₆]₂
d) Fe₃[Fe(CN)₆]₃
Answer:
b) Fe₄ [ Fe(CN)₆]₃

Question 21.
Lassaigne’s test for the detection of nitrogen fails in
a) H₂N – CO – NH.NH₂.HCl
b) NH₂ – NH₂.HCl
c) C₆H₅ – NH – NH₂.HCl
d) C₆H₅CONH₂
Answer:
c) C₆H₅ – NH – NH₂.HCl

Question 22.
Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
a) NH₂ NH₂ HCl and ClCH₂ – CHO
b) NH₂ CS NH₂ and CH₃ – CH₂Cl
c) NH₂ CH₂ COOH and NH₂ CONH₂
d) C₆H₅NH₂ and ClCH₂ – CHO
Answer:
d) C₆H₅NH₂ and ClCH₂ – CHO

Question 23.
Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of
a) [Fe (CN)₅ NO]^3-
b) [Fe (NO)₅ CN]⁺
c) [Fe (CN)₅ NOS]^4-
d) [Fe (CN)₅ NOS]^3-
Answer:
c) [Fe (CN)₅ NOS]^4-

Question 24.
An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to
a) 46 %
b) 34 %
c) 3.4 %
d) 4.6 %
Answer:
b) 34 %

Question 25.
A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50mL of 0.5 M H₂SO₄ The remaining acid after neutralization by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is.
a) 14 %
b) 28 %
c) 42 %
d) 56 %
Answer:
b) 28 %

Question 26.
In an organic compound, phosphorus is estimated as
a) Mg₂P₂O₇
b) Mg₃(PO₄)₂
c) H₃PO₄
d) P₂O ₅
Answer:
a) Mg₂P₂O₇

Question 27.
Ortho and para – nitro phenol can be separated by
a) azeotropic distillation
b) destructive distillation
c) steam distillation
d) cannot be separated
Answer:
c) steam distillation

Question 28.
The purity of an organic – compound is determined by
a) Chromatography
b) Crystallization
c) melting or boiling point
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by
a) distillation at atmospheric pressure
b) distillation under reduced pressure
c) fractional distillation
d) steam distillation
Answer:
b) distillation under reduced pressure

Question 30.
Assertion:

Reason:
The principal functional group gets lowest number followed by double bond (or) triple bond.
a) both the assertion and reason are true and the reason is the correct explanation of assertion.
b) both assertion and reason are true and the reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both the assertion and reason are false
Answer:
a) both the assertion and reason are true and the reason is the correct explanation of assertion.

II. Write brief answers to the following questions:

Question 31.
Give the general characteristics of organic compounds.
Answer:
All organic compounds have the following characteristic properties.

1. They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc…
2. Many of the organic compounds are inflammable (except CCl₄). They possess low boiling and melting points due to their covalent nature.
3. Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.

Question 32.
Describe the classification of organic compounds based on their structure.
Answer:

Question 33.
Write a note on homologous series.
Answer:
Homologous series:
A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH₂ group is called homologous series.
Example:
Alkanes:
Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈) etc..
Alcohols:
Methanol (CH₃OH), Ethanol (C₂H₅OH) Propanol (C₃H₇OH) etc…

Question 34.
What is meant by a functional group?
Identify the functional group in the following compounds.
a) acetaldehyde
b) oxalic acid
e) dimethyl ether
d) methylamine
Answer:
Functional group:
An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties.
a) acetaldehyde → – CHO
b) oxalic acid → – COOH
c) di methyl ether → – O –
d) methyiamine → – NH₂

Question 35.
Give the general formula for the following classes of organic compounds
a) Aliphatic monohydric alcohol
b) Aliphatic ketones
c) Aliphatic amines
Answer:
a) Aliphatic monohydric alcohol → C_nH_{2n + 2}O
b) Aliphatic ketones → C_nH_2nO
c) Aliphatic amines → C_nH_{3n + 2}N

Question 36.
Write the molecular formula of the first six members of homologous series of nitre alkanes.
Answer:

• CH₃NO₂
• C₂H₅NO₂
• C₃H₇NO₂
• C₄H₉NO₂
• C₅H₁₁NO₂
• C₆H₁₃NO₂

Question 37.
Write the molecular formula and possible structural formula of the first four members of homologous series of carboxylic acids.
Answer:

Question 38.
Give the IUPAC names of the following compounds.
i) (CH₃)₂CH – CH₂ – CH(CH₃) – CH(CH₃)₂

ii)

iii) CH₃ – O – CH₃

iv)

v) CH₂ = CH – CH – CH₂

vi)

vii)

viii)

ix)

x)

xi)

xii)

xiii)

xiv)

xv)

xvi)
Answer:
(i) 2, 3, 5 – trimethyihexane
(ii) 2 – bromo – 3 – methylbutane
(iii) methoxymethane
(iv) 2 – hydroxybutanal
(y) buta – 1 ,3 – diene
(vi) 4 – chloropent – 2 – yne
(vii) 1 – bromobut – 2 – ene
(viii) 5 – oxohexanoic acid
(ix) 3 – ethyl – 4 – ethenylheptane
(x) 2, 4, 4 – trimethylpent – 2 – ene
(xi) 2 – methyl -1 – phenyipropan – 1 -amine
(xii) 2, 2 – dimethyl – 4oxopentanenitrile
(xiii) 2 – ethoxypropane
(xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – bromo – 2 – methylpentanal
(xvi) Acetophenone

Question 39.
Give the structure for the following compound
(i) 3 – ethyl – 2 methyl – 1 – pentene
(ii) 1, 3, 5 – Trimethyl cyclohex – 1 – ene
(iii) tetry butyl iodide
(iv) 3 – Chlorobutanal
(V) 3 – Chlorobutanol
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2 – dimethyl – 1 – chloropropane
(viii) 3 – methylbut -1- ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1, 3 – diene
(xi) 1, 3 – Dimethylcyclohexane
(xii) 3 – Chlorobut – 1 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde
Answer:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
This method involves the conversion of covalently bonded N, S or halogen present in the organic compounds to corresponding water soluble ions in the form of sodium salts For this purpose a freshly cut piece of Na of the size of a paper, dried it by pressing between the folds of a filter paper is taken in a fusion tube in an iron tand clamping it just near the upper end and it is gently heated.

When it melts to a shiñing globule, put a pinch of the organic compound on it. Heat the tube with the tip of the flame till all reaction ceases and it becomes red hot. Now plunges it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striding against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for qualitative analysis of nitrogen, sulfur and halogens present in organic compounds.

Test for Nitrogen:
If nitrogen is present it gets converted to sodium cyanide which on reaction with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl gives a Prussian blue color or green color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the excess of NaOH on FeSO₄ which would otherwise mark the Prussian blue precipitate. The following reaction takes part in the formation of Prussian blue.

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Answer:
Estimation of halogens (Carius method):
A known mass of the organic compound is heated with fuming HNO₃ along with AgNO₃. C, H & S gets oxidized to CO₂, H₂O, SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated.

A known mass of the substance is taken along with fuming HNO₃ and AgNO₃ is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made.

Calculation:
Weight of the organic compound: w g
Weight of AgCl precipitate = a g
143. 5 g of AgCl contains 35.5 g of Cl
∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a
w g Organic compound gives a g AgCl
Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\)
Let Weight of silver Bromide be ‘b’ g
188 g of AgBr contains 80 g of Br
∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br
w g Organic compound gives b g AgBr
Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
Let Weight of silver Iodide be ‘c’ g
235 g of AgI contains 127 g of I
∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I
w g Organic compound gives c g AgI
percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)

Question 42.
Give a brief description of the principles of
(i) Fractional distillation
(ii) Column Chromatography
Answer:
(i) Fractional distillation:
This is one method to purify and separate liquids present in the mixture having their boiling point close to each other. In the fractional distillation, a fractionating column is fitted with distillation flask and a condenser. A thermometer is fitted in the fractionating column near the mouth of the condenser. This will enable to record the temperature of vapour passing over the condenser.

The process of separation of the components in a liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. The process of fractional distillation is repeated, if necessary. This method finds a remarkable application in distillation of petroleum, coal-tar and crude oil.

(ii) Column Chromatography:
This is the simplest chromatographic method carried out in long glass column having a stop cock near the lower end. This method involves separation of a mixture over a column of adsorbent (Stationery phase) packed in a column. In the column a plug of cotton or glass wool is placed at the lower end of the column to support the adsorbent powder. The tube is uniformly packed with suitable adsorbent constitute the stationary phase. (Activated aluminum oxides (alumina), Magnesium oxide, starch are also used as adsorbents).

The mixture to be separated is placed on the top of the adsorbent column. Eluent which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Different components depending upon the degree to which the components are adsorbed and complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of ‘ paper acts as an adsorbent. This method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.

A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase. The solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots . on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Structural isomerism:
This type of isomers have same molecular formula but differ in their bonding sequence.

(a) Chain or nuclear or skeletal isomerism:
These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched).

(b) Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton, but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.

(c) Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
Molecular formula C₃H₆O
CH₃ – CH₂ – CHO
propanal (aldehyde group)

Question 45.
Describe optical isomerism with suitable example.
Answer:
Compounds having same physical and chemical property but differ only in the rotation of plane of the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Example:
Some organic compounds such as glucose have the ability to rotate the plane of the plane polarized light and there called are said to be optically active compounds and this property of a compound is called optical activity. The optical isomer, which rotates the plane of the plane polarised light to the right or in clockwise direction is said to be dextrorotary (dexter means right) denoted by the sign (+), whereas the compound which rotates to the left or anticlockwise is said to be leavorotatory (leavues means left) denoted by sign (-). Dextrorotatory compounds are represented as ‘d’ or by sign (+) and leavorotatory compounds are represented as ‘l’ or by sign (-).

Question 46.
Briefly explain geametrical isomerism in alkene by considering 2-butene as an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.

In alkenes, the carbon-carbon double bond is sp² hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The π bond is formed by the head on overlap of sp² hybrid orbitals. The n bond is formed by the side-wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible. This restriction of rotation about C – C double bond is responsible for geometrical isomerism in alkenes.

These two compounds are termed as geometrical isomers and are distinguished from each other by the terms c is and trans. The c is isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis- trans isomerism.

The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the n bond so that rotation about a bond becomes possible. Upon cooling, the reformation of the n bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butene.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Solution:
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g

Percentage of carbon:
44 g of carbondioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \(\frac{12 \times 0.88}{44}\) g
0.30 g substance contains,
carbon = \(\frac{12 \times 0.88}{44}\) g

100 g substance Contains \(\frac{12 \times 0.88}{44}\) × \(\frac{100}{0.30}\) = 80 g of carbon
Percentage of carbon = 80 %

Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \(\frac{2 \times 0.54}{18}\) g
0.30 g of substance contains hydrogen = \(\frac{2 \times 0.54}{18 \times 0.30}\) g
100 g of substance contains = \(\frac{2 \times 0.54}{18 \times 0.30}\) × 100 g = 20 g of hydrogen
Percentage of hydrogen = 20 %

Question 48.
The ammonia evolved form 0.20 g of an organic compound by Kjeldahl method neutralized 15 ml of N / 20 sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
weight of organic compound = 0.20 g
Normality of acid = \(\frac{\mathrm{N}}{20}\)
Volume of standard acid neitralized by ammonia = 15 ml
1000 ml of N ammonia contains = 14 g of nitrogen
15 ml of ammonia of normality \(\frac{\mathrm{N}}{20}\) contains nitrogen = \(\frac{14 \times 15 \times 1}{1000 \times 20}\)

0.20 g of compound contains nitrogen = \(\frac{14 \times 15}{1000 \times 20}\)

100 g of compound contains nitrogen = \(\frac{14 \times 15 \times 100}{1000 \times 20 \times 0.20}\) = 5.25 g
Percentage of nitrogen = 5.25 %

Question 49.
0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Solution:
Mass of the substance taken = 0.32 g
Mass of BaSO₄ formed = 0.466 g
Molecular mass of BaSO₄ = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of BaSO₄ = \(\frac{0.466 \times 32}{233}\)

Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %

Question 50.
0.24 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Solution:
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143. 5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) × 0.287

Percentage of chlorine = \(\frac{35.5}{143.5} \times \frac{0.287}{0.24}\) × 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Solution:
Volume of N₂ at NTP = \(\frac{V \times P}{t+273} \times \frac{273}{760}\)
= V₀ ml
Substituting the various values in the above equation,
V₀ = \(\frac{20.7 \times 760}{288} \times \frac{273}{760}\) = 19.62 ml

weight of 19.62 ml of Nitrogen = \(\frac{28}{22400}\) × 19.62 g

∴ Percentage of Nitrogen = \(\frac{28}{22400}\) × 19.62 × \(\frac{100}{0.35}\)
= 4.9 %

11th Chemistry Guide Fundamentals of Organic Chemistry Additional Questions and Answers

I. Choose the correct answer:

Question 1.
Organic compounds can be formed by
a) Plants only
b) Animals only
c) Plants and Animals
d) Plants, animals and can be synthesized in laboratory
Answer:
d) Plants, animals and can be synthesized in laboratory

Question 2.
Generally, organic compounds are
a) Amorphous
b) Complexes
c) Covalent
d) Electrovalent
Answer:
c) Covalent

Question 3.
The vital force theory was proposed by
a) Wohler
b) Berthlot
c) Berzelius
d) Kolbe
Answer:
c) Berzelius

Question 4.
The first carbon compound prepared from its elements is
a) Urea
b) Acetic acid
c) Methane
d) benzene
Answer:
b) Acetic acid

Question 5.
The first organic compound was synthesized in laboratory by
a) Wohler
b) Kolbe
c) Berzelius
d) Neil Barthlot
Answer:
a) Wohler

Question 6.
The first organic compound synthesized in the laboratory from an inorganic compound is
a) NH₄NCO
b) NH₂ – CO – NH₂
c) CH₃COOH
d) CH₄
Answer:
b) NH₂ – CO – NH₂

Question 7.
Marsh gas mainly contains
a) C₂H₂
b) C₂H₄
c) CH₄
d) C₂H₆
Answer:
c) CH₄

Question 8.
Hybridization at 2nd carbon in CH₂ = CH – CH₃ is
a) sp
b) sp²
c) sp³
d) sp³d
Answer:
b) sp²

Question 9.
Number of possible position isomers for Dichlorobenzene is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 10.
n – Butane and isobutane are a pair of
a) chain isomers
b) position isomers
c) metamers
d) functional isomers
Answer:
a) chain isomers

Question 11.
Alkanols and Alkoxyalkanes are
a) Functional isomers
b) Keto – enol tautomers
c) Geometrical isomers
d) Not isomers at all
Answer:
a) Functional isomers

Question 12.
n – propyl alcohol and isopropyl alcohol are examples of
a) Position isomerism
b) Chain isomerism
c) Tautomerism
d) Geometrical isomerism
Answer:
a) Position isomerism

Question 13.
The number of structural alcoholic isomers for C₄H₁₀O is
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 14.
Cycloalkanes are isomeric with
a) Alkadienes
b) Alkynes
c) Aromatic compounds
d) Olefins
Answer:
d) Olefins

Question 15.
Number of possible monochloro benzenes is
a) 1
b) 3
c) 5
d) 6
Answer:
a) 1

Question 16.
Diethyl ether and n – propyl methyl ether are
a) Metamers
b) Chain isomers
c) Geometrical isomers
d) Position isomers
Answer:
a) Metamers

Question 17.
The total number of structural isomers for the compound of the formula C₄H₁₀O is
a) 7
b) 6
c) 4
d) 3
Answer:
a) 7

Question 18.
The number of primary alcoholic isomers with the formula C₄H₁₀O is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 19.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) Butan – 1 – ol
c) 2 – Methylpropan – 2 – ol
d) Butanone
Answer:
d) Butanone

Question 20.
The number of isomeric amines possible for the formula C₃H₉N
a) 4
b) 3
c) 5
d) 6
Answer:
a) 4

Question 21.
Which hybrid orbitals are involved in the CH₃ – CH = CH – CH₃ compound
a) sp and sp³
b) sp² and sp³
c) sp and sp²
d) only sp³
Answer:
b) sp² and sp³

Question 22.
Which of the following bonds is strongest?
a)

b) > C = C <

c)

d) – C – C –
Answer:
a)

Question 23.
According to Huckel’s rule a compound, is said to be aromatic if’ it contains
a) 4n bonds
b) 4n atoms
c) (4n + 2) atoms
d) (4n + 2) π electrons
Answer:
d) (4n + 2) π electrons

Question 24.
Which of the following is an aromatic compound
a) Phenol
b) Naphthalene
c) Pyridine
d) All
Answer:
d) All

Question 25.
Which is a saturated compound?
a) alkanes
b) alkenes
c) alkynes
d) cyclo alkenes
Answer:
a) alkanes

Question 26.
Which is an alicyclic compound?
a) benzene
b) cyclohexane
c) pyridine
d) pyrrole
Answer:
b) cyclohexane

Question 27.
Which of the following is not a cyclic compound?
a) Anthracene
b) Pyrrole
c) Phenol
d) Isobutylene
Answer:
d) Isobutylene

Question 28.
Functional group present in amides is
a) – COOH
b) – NH₂
c) – CONH₂
d) – COO –
Answer:
c) – CONH₂

Question 29.
IUPAC name of ester is
a) Alkoxy alkane
b) Alkyl alkanoate
c) Alkanoyl halide
d) Alkanoic anhydride
Answer:
b) Alkyl alkanoate

Question 30.
IUPAC name of methyl cyanide is
a) Cyano methane
b) Ethanenitrile
c) Methane nitrile
d) Methyl – n – butyl amine
Answer:
b) Ethanenitrile

Question 31.
The correct IUPAC name of is
a) 1, 2 – diethyl butene
b) 2 – ethyl – 3- methyl pentene
c) 3, 4 – dimethyl hex – 3 – ene
d) 2, 3 – dimethyl pent – 2 – ene
Answer:
d) 2, 3 – dimethyl pent – 2 – ene

Question 32.
IUPAC name of CH₂OH – CH₂OH is
a) 1, 2 – dihydroxy ethane
b) ethylene glycol
c) ethane – 1, 2 – diol
d) ethane – 1, 2 – dial
Answer:
c) ethane – 1, 2 – diol

Question 33.
IUPAC name of CH ≡ C – CH = CH₂ is
a) but – 3 – ene – 1 – yne
b) but – 1 – ene – 3 – yne
c) but – 1 – yne – 3 – ene
d) but – 3 – yne – 1 – ene
Answer:
b) but – 1 – ene – 3 – yne

Question 34.
The IUPAC name of is
a) 4 – hydroxy – 2 – methyl pentanal
b) 2 – hydroxy – 4 – methyl pentanal
c) 4 – hydroxy – 2 – methyl pentanol
d) 2 – hydroxy – 4 – methyl pentanol
Answer:
a) 4 – hydroxy – 2 – methyl pentanal

Question 35.
3 – methyl penta -1, 3- diene is
a) CH₂ = CH (CH₂)₂ CH₃
b) CH₂ = CHCH (CH₃) CH₂CH₃
c) CH₃CH = C(CH₃)CH = CH₂
d) CH₃ = C = CH (CH₃)₂
Answer:
c) CH₃CH = C(CH₃)CH = CH₂

Question 36.
The structural formula of methyl amino methane is
a) (CH₃)₂ CH NH₂
b) (CH₃)₃ N
c) (CH₃)₂ NH
d) CH₃NH₂
Answer:
c) (CH₃)₂ NH

Question 37.
Which of the following is the functional isomer of methyl acetate?
a) Ethyl acetate
b) Propanoic acid
c) Ethyl formate
d) Propanone
Answer:
b) Propanoic acid

Question 38.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) 1 – Butanol
c) 2 – Methyl – 2 – propanol
d) Butanone
Answer:
d) Butanone

Question 39.
Which of the following pairs of compounds are tautomers?
a) Propanol & propanone
b) Ethanol & vinyl alcohol
c) Ethanol & allyl alcohol
d) Vinyl alcohol & ethanal
Answer:
d) Vinyl alcohol & ethanal

Question 40.
Which of the following compounds does not have any tertiary hydrogen atoms?
a) (CH₃)₃ CCH₂ CH₃
b) (CH₃)₂ CHCH₂ CH₃
c) (CH₃)₂ CHCH (CH₃)₂
d) (CH₃)₃ CH
Answer:
a) (CH₃)₃ CCH₂ CH₃

Question 41.
The IUPAC name of is
a) 2 – Methyl – 2 – butenoic acid
b) 3 – Methyl – 3 – butenoic acid
c) 3 – Methyl – 2 – butenoic acid
d) 2 – Methyl – 3 – butenoic acid
Answer:
c) 3 – Methyl – 2 – butenoic acid

Question 42.
The IUPAC name of Cinnamaldehyde is
a) 3 – Phenyl prop – 2 – enal
b) 1 – Phenyl – prop – 1 – enal
c) 1 – Phenyl – prop – 2 – enal
d) 3 – Phenyl – prop – 1 – enal
Answer:
a) 3 – Phenyl prop – 2 – enal

Question 43.
The IUPAC name of the Compound CH₃ – CH(OH) – COOH is
a) Lactic acid
b) 2 – Hydroxy propanoic acid
c) 3 – Hydroxy propanoic acid
d) Carboxy propanol
Answer:
b) 2 – Hydroxy propanoic acid

Question 44.
The IUPAC name of the compound is
a) 2 – Ethyl – ethyl acetate
b) Ethyl – 3 – methy1butanoate
c) Ethyl – 2 – methyl butanoate
d) 2- methyl butanoic acid
Answer:
c) Ethyl – 2 – methyl butanoate

Question 45.
The IUPAC name of the given compound is
a) 2,2 — Dimethyl butane
b) lsohexane
c) 2, 3 – Dimethyl butane
d) Di isohexane
Answer:
c) 2, 3 – Dimethyl butane

Question 46.
The IUPAC name of the given compound is
a) Octyl cyclopentane
b) 3 – cyclopentyl octane
c) Cyclopentane octane
d) 6 – cyclopentyl octane
Answer:
b) 3 – cyclopentyl octane

Question 47.
The IUPAC name of is
a) but – 2- ene – 2,3 – diol
b) pent – 2- ene – 2,3 – diol
c) 2 – methyl but – 2 – ene – 2,3 – diol
d) hex – 2- ene – 2,3 – diol
Answer:
b) pent – 2- ene – 2,3 – diol

Question 48.
The structure of 3-bromoprop-1-ene is
a)

b) CH₃ – CH = CH – Br

c)

d) Br – CH₂ – CH ≡ CH₂
Answer:
d) Br – CH₂ – CH ≡ CH₂

Question 49.
Neo-heptyl alcohol is correctly represented as
a)

b)

c)

d)
Answer:
c)

Question 50.
Number of dibromo derivatives possible for propane are
a) 2
b) 3
c) 1
d) 4
Answer:
d) 4

Question 51.
The number of aromatic isomers possible for C₇H₈O is
a) 2
b) 3
c) 4
d) 5
Answer:
d) 5

Question 52.
Isomers of propanoic acid are
a) HCOOC₂H₅ and CH₃COOCH₃
b) H – COOC₂H₅ and C₃H₇COOH
c) CH₃COOCH₃ and C₃H₇OH
d) C₃H₇OH and CH₃COCH₃
Answer:
a) HCOOC₂H₅ and CH₃COOCH₃

Question 53.
IUPAC name of CH₃ – CH (OCH₃) – CH₂ – NH₂
a) 2-methoxy propanamine
b) 1-amino – 2-methoxy propane
c) 1-amino – 2-methyl – 2-methoxy ethane
d) 1 – methoxy- 2-amino propane
Answer:
a) 2-methoxy propanamine

Question 54.
IUPAC name of is
a) 3 – cyanopentane – 1, 5 – dinitrile
b) Propane – 1, 2, 3-tri nitrile
c) 1, 2, 3-tri cyano propane
d) Propane 1, 2, 3-tricarbonitrile
Answer:
d) Propane 1, 2, 3-tricarbonitrile

Question 55.
The IUPAC name of the compound is
a) 3 – Carboxylic pentane – 1,5 -dioic acid
b) Propane – 1, 2, 3 – trioic acid
c) 1, 2, 3- tricarboxylic propane
d) Propane – 1,2, 3 – tricarboxylic acid
Answer:
b) Propane – 1, 2, 3 – trioic acid

Question 56.
The IUPAC name of the following compound
CH₃ – C(CH₃)₂ – CH₂ – CH = CH₂ is
a) 2, 2 – Dimethyl – 4 – pentene
b) 4, 4 – Dimethyl – 1 – pentene
c) 1, 1, 1 – trimethyl – 3 – butene
d) 4, 4, 4 – trimethyl – 1 – butene
Answer:
b) 4, 4 – Dimethyl – 1 – pentene

Question 57.
The IUPAC name of is
a) 2, 4 – Dimethyl pentan – 2 – ol
b) 2, 4 – Dimethyl pentan – 4 – ol
c) 2,2 – Dimethyl butan – 2- ol
d) Butan – 2 – ol
Answer:
a) 2, 4 – Dimethyl pentan – 2 – ol

Question 58.
The IUPAC name the compound is
a) Butane – 2, 3, 4 – triol
b) Butane – 1,2, 3 – triol
c) Pentane – 1, 2, 3 – triol
d) 2, 3 dihydroxy butanol
Answer:
b) Butane – 1,2, 3 – triol

Question 59.
The IUPAC name the compound is
a) 3, 3 – Dimethyl – 1 – cyclohexanol
b) 1, 1- Dimethyl – 3 – hydroxy cyclohexane
c) 3, 3 – Dimethyl – 1 – hydroxy cyclohexane
d) 1, 1 – Dimethyl – 3 – cyclohexanol
Answer:
a) 3, 3 – Dimethyl – 1 – cyclohexanol

Question 60.
The IUPAC name of the compound is
a) 2 – Ethylprop – 2 – en – 1 – ol
b) 2 – Hydroxymethyl butan – 1- ol
c) 2 – Methylene butan – 1 – ol
d) 2 – Ethyl -3 hydroxyprop – 1 – ene
Answer:
a) 2 – Ethylprop – 2 – en – 1 – ol

Question 61.
The number of possible alkynes with molecular formula C₅H₈ is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 62.
What is the IUPAC name of the following is
a) 3 – chloro cyclo hexa – 1, 5 – diene
b) 5 – chloro cyclo hexa – 1, 3 – diene
c) 1 – chloro cyclo hexa – 2, 5 – diene
d) 2 – chloro cyclo hexa – 1, 4 – diene
Answer:
b) 5 – chloro cyclo hexa – 1, 3 – diene

Question 63.
What is the IUPAC name of the following is
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al
b) 4 – hydroxy cyclohex – 1 – ene – 3 – al
C) 2 – hydroxy cyclohex – 5 – ene – 1 – al
d) 1 – formyl cyclohex – 5 – ene – 2 – ol
Answer:
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al

Question 64.
What is the IUPAC name of the following?

a) 1 – chloro – 2 – bromo – 4 – nitrobenzene
b) 1 – bromo – 2- chloro – 4 – mtrobenzene
c) 3 – bromo – 4 – chloro – nitrobenzene
d) 2 – bromo – 1 – chloro – 4- nitrobenzene
Answer:
d) 2 – bromo – 1 – chloro – 4- nitrobenzene

Question 65.
What is the IUPAC name of the following?

a) Ethenyl cyclo pentane
b) cyclopentyl ethene
c) cyclopentyl ethylene
d) vinyl cyclopentane
Answer:
b) cyclopentyl ethene

Question 66.
The hybridization of carbon atoms in C – C single bond is HC ≡ C – CH = CH₂ is
a) sp³ – sp³
b) sp² – sp³
c) sp – sp²
d) sp³ – sp
Answer:
c) sp – sp²

Question 67.
The correct IUPAC name of the compound is
a) 1, 4 – Butane dioicacid
b) Ethane – 1, 2 – dicarboxylic acid
c) Succinic acid
d) 1, 2 – Ethane dioic acid
Answer:
a) 1, 4 – Butane dioic acid

Question 68.
Correct statements about CH₃ – CH₂ – CN is
a) common name of the compound is ethylcyanide.
b) IUPAC name of the compound propane – 1 – nitrile.
c) secondary suffix of the compound is nitrile.
d) IUPAC name of the compound is ethane nitrile.
Answer:
a) common name of the compound is ethylcyanide.

Question 69.
Tautomerism is shown by
a) R – C ≡ N
b) R – NO₂
c) R – OH
d) R – COOH
Answer:
b) R – NO₂

Question 70.
Stereo isomers have different
a) Molecular mass
b) Molecular formula
c) Structural formula
d) Configuration
Answer:
d) Configuration

Question 71.
Geometrical isomerism may be exhibited by compounds having atleast.
a) One double bond
b) One triple bond
c) One asymmetric carbon
d) One polar bond
Answer:
a) One double bond

Question 72.
The prefixes syn – and anti – are used to denote
a) structural isomers
b) conformational isomers
c) geometrical isomers
d) optical isomers
Answer:
c) geometrical isomers

Question 73.
d – tartaric acid and l – tartaric acid are
a) geometrical isomers
b) conformers
c) enantiomers
d) diastereomers
Answer:
c) enantiomers

Question 74.
The method of separation of enantiomers from racemic mixture is known as
a) inversion
b) recemisation
c) resolution
d) asymmetric synthesis
Answer:
c) resolution

Question 75.
Racemic mixture is optically inactive due to
a) internal compensation
b) external compensation
c) inversion
d) plane of symmetry
Answer:
c) inversion

Question 76.
Meso isomers are possible when the organic compound contains
a) one asymmetric carbon
b) two or more dissimilar asymmetric carbons
c) similar asymmetric carbons
d) unsaturation
Answer:
c) similar asymmetric carbons

Question 77.
Optical inactivity of meso isomer is due to
a) element of symmetry and element of asymmetry
b) internal compensation
c) due to lack of asymmetric carbon
d) External compensation
Answer:
b) internal compensation

Question 78.
A racemic mixture is a mixture of
a) meso and its isomers
b) d and l isomers of same compound in equimolar proportions
c) d and l isomers of same compound in different proportions
d) mixture of d and meso isomers
Answer:
b) d and 1 isomers of same compound in equimolar proportions

Question 79.
Which of the following is optically active?
a) HOOC – CH₂ – COOH
b) CH₃ – CO – COOH
c) CH₃ – CH(OH) – COOH
d) CH₃ – CH₂ – COOH
Answer:
c) CH₃ – CH(OH) – COOH

Question 80.
Which of the following is optically active?
a) n – propanal
b) 2 – chlorobutane
c) n – butanal
d) 3 – pentanol
Answer:
b) 2 – chlorobutane

Question 81.
The number of optical enantiomers of tartaric acid
a) 3
b) 2
c) 4
d) 1
Answer:
b) 2

Question 82.
Geometrical isomers differ in
a) position of substituents
b) Position of double bond
c) C – C bond length
d) Spatial arrangement of groups
Answer:
d) Spatial arrangement of groups

Question 83.
Which of the following exhibit cis – trans isomerism
a) propene
b) 1 – butene
c) 2 – butene
d) benzene
Answer:
c) 2 – butene

Question 84.
Which of the following show geometrical isomerism?
a) CH₃CH = CHCH₃
b) (CH₃)₂C = CH₂
c) C₂H₅CH = CH₂
d) CH₃CH = CH₂
Answer:
a) CH₃CH = CHCH₃

Question 85.
Which of the following does not show geometrical isomerism?
a) 1, 2 – dichloro – 1 – pentene
b) 1, 3 – dichloro – 2 – pentene
c) 1, 1 – dichloro – 1 – pentene
d) 1, 4 – dichloro – 2 – pentene
Answer:
c) 1, 1 – dichloro – 1 – pentene

Question 86.
Geometrical isomerism is not shown by
a)

b)

c) CH₂ = C(CI) CH₃

d) CH₃ – CH = CH – CH = CH₂
Answer:
c) CH₂ = C(CI) CH₃

Question 87.
Which of the following is optically active?
a) Glycerine
b) Acetaldehyde
c) Glyceraldehyde
d) Acetone
Answer:
c) Glyceraldehyde

Question 88.
The minimum number of C atoms for a hydrocarbon to exhibit optical isomerism
a) 4
b) 5
c) 6
d) 7
Answer:
d) 7

Question 89.
Which of the following can form cis – trans isomers?
a) C₂H₅Br
b) (CH)₂(COOH)₂
c) CH₃CHO
d) (CH₂)₂COOH
Answer:
b) (CH)₂(COOH)₂

Question 90.
No.of geometrical isomers possible for the compound CH₃ – CH = CH – CH = CH – C₂H₅
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 91.
The number of geometrical isomers of CH₃ – CH = CH – CH = CH – CH = CHCl is
a) 2
b) 4
c) 6
d) 8
Answer:
d) 8

Question 92.
Minimum number of C atoms for an alkene hydrocarbon, that shows geometrical & optical isomerism both
a) 5
b) 6
c) 7
d) 8
Answer:
c) 7

Question 93.
Among the following compounds which exhibits optical isomerism?
a) propanol
b) 2 – propanol
c) 1 – butanol
d) 2 – butanol
Answer:
d) 2 – butanol

Question 94.
Which of the mesoisomer?
a) CH₂OHCHOHCHO
b) CH₂OHCHOHCHOHCHO
c) HOOCCHOHCHOHCOOH
d) HOH₂CCHOHCHOHCOOH
Answer:
c) HOOCCHOHCHOHCOOH

Question 95.
d – tartaric acid and l – tartaric acid can be separated by
a) Salt formation
b) Fractional distillation
c) Fractional crystallization
d) Chromatography
Answer:
a) Salt formation

Question 96.
Paper chromatography is
a) Adsorption chromatography
b) partition chromatography
c) Ion exchange chromatography
d) all of these
Answer:
b) partition chromatography

Question 97.
Simple distillation can be used to separate liquids which differ in their boiling points at least by
a) 5°C
b) 10°C
c) 40 – 50°C
d) 100°C
Answer:
c) 40 – 50°C

Question 98.
In adsorption chromatography mobile phase will be
a) Only solid
b) Only liquid
c) Only gas
d) Liquid as well as gas
Answer:
d) Liquid as well as gas

Question 99.
Which of the following can be used as adsorbent in adsorption chromatography?
a) Silica gel
b) Alumina
c) Cellulose powder
d) All of these
Answer:
d) All of these

Question 100.
Two substances when separated out on the basis of their extent of adsorption by one material, the phenomenon is called
a) Chromatography
b) Crystallization
c) Sublimation
d) Steam distillation
Answer:
a) Chromatography

Question 101.
Chromatographic technique is used for the separation of
a) Camphor
b) Alcohol & Water
c) Acetone and Methanol
d) Plant pigments
Answer:
d) Plant pigments

Question 102.
In column chromatography stationary phase is
a) only solid
b) only liquid
c) only gas
d) All of these
Answer:
a) only solid

Question 103.
Which of the following method is used for the purification of solids?
a) Distillation under reduced pressure
b) Distillation
c) Strain distillation
d) Sublimation
Answer:
d) Sublimation

Question 104.
Vacuum distillation is used to purify liquids which
a) are highly volatile
b) are explosive in nature
c) soluble in water
d) decomposes below their B.P’s
Answer:
c) soluble in water

Question 105.
Impure Napthalene is purified by
a) Fractional crystallization
b) Fractional distillation
C) solvent extraction
d) sublimation
Answer:
d) sublimation

Question 106.
A very common adsorbent used in coloun^n chromatography is
a) Powdered charcoal
b) Alumina
c) Chalk
d) Sodium carbonate
Answer:
b) Alumina

Question 107.
Simple distillation of liquids involves simultaneously
a) Vapourisation and condensation
b) Condensation and vapourisation
c) Vapourisation and sublimation
d) Sublimation and condensation
Answer:
a) Vapourisation and condensation

Question 108.
The latest technique for the purification of organic compounds is
a) Fractional distillation
b) Chromatography
c) Vacuum distillation
d) Crystallization
Answer:
b) Chromatography

Question 109.
Fixed melting point of an organic compound informs
a) Purity of an organic compound
b) Conductivity of compound
c) Chemical nature of compound
d) Whether the compound is liquid or gas
Answer:
a) Purity of an organic compound

Question 110.
Lassaigne’s test is used in qualitative analysis to detect
a) Nitrogen
b) Sulphur
c) Chlorine
d) All of these
Answer:
d) All of these

Question 111.
In Lassaigne’s method organic compound is fused with
a) Sodium metal
b) Zinc dust
c) Sodium carbonate and Zinc dust
d) Calcium metal
Answer:
a) Sodium metal

Question 112.
Presence of nitrogen in organic compound in Lassaigne’s extract as
a) Nitrogen gas
b) NH₃
c) NO
d) CN^–
Answer:
d) CN^–

Question 113.
Medium of Sodium extract is
a) Neutral
b) Basic
c) Acidic
d) Depends on organic compound
Answer:
b) Basic

Question 114.
H₂O vapours on passing through anhydrous CuSO₄ turns it to
a) Green
b) Blue
c) Violet
d) White
Answer:
b) Blue

Question 115.
When a nitrogenous organic compound is fused with sodium, the nitrogen present in the compound is converted into
a) Sodium Nitrate
b) Sodium nitrite
c) Sodamide
d) Sodium cyanide
Answer:
d) Sodium cyanide

Question 116.
In the Lassainge’s test the Sulphur present in the organic compound first changes into
a) Na₂SO₃
b) CS₂
c) Na₂SO₄
d) Na₂S
Answer:
d) Na₂S

Question 117.
Which of the following elements in an organic compound cannot be detected by Lassaigne’s test?
a) N
b) S
c) Cl
d) H
Answer:
d) H

Question 118.
A compound which does not give a positive result in the Lassaigne’s test for nitrogen is
a) Urea
b) Hydroxyl amine
c) Glycine
d) Phenylhydrazine
Answer:
b) Hydroxyl amine

Question 119.
Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of
a) N
b) S
c) O
d) Cl
Answer:
b) S

Question 120.
The presence of halogen in an organic compound is detected by
a) Iodoform test
b) Silver nitrate test
c) Beilstein’s test
d) Million’s test
Answer:
c) Beilstein’s test

Question 121.
The Beilstein’s test in a rapid test used for . organic compound^ to^ detect
a) Phosphorous
b) Sulphur
c) Halogens
d) Nitrogen
Answer:
c) Halogens

Question 122.
Liebig’s method is used for the estimation of
a) Nitrogen
b) Sulphur
c) Carbon and hydrogen
d) Halogens
Answer:
c) Carbon and hydrogen

Question 123.
In Kjeldahl’s method of estimation of nitrogen, copper sulphate act as
a) Oxidizing agent
b) reducing agent
c) Catalytic agent
d) Hydrolysing agent
Answer:
c) Catalytic agent

Question 124.
Percentage of carbon in an organic compound is determined by
a) Duma’s method
b) Kjeldahl’s method
c) Carius method
d) Liebig’s method
Answer:
d) Liebig’s method

Question 125.
Halogen can be estimated by
a) Duma’s method
b) Carius method
c) Leibig’s method
d) All of these
Answer:
b) Carius method

Question 126.
In Garius method halogens are estimated
a) X₂
b) BaX₂
c) PbX₂
d) AgX
Answer:
d) AgX

Question 127.
In Duma’s method nitrogen in organic compound is estimated in the form of
a) N₂
b) NO
c) NH₃
d) N₂O₅
Answer:
a) N₂

Question 128.
In Kjeldahl’s method to estimate nitrogen, compound is heated with conc.H₂SO₄ in presence of
a) CaSO₄
b) (NH₄)₂SO₄
c) CuSO₄
d) P₂O₅
Answer:
c) CuSO₄

Question 129.
In organic compounds, Sulphur is estimated as
a) BaSO₄
b) BaCl₂
c) Ba₃(PO₄)₂
d) H₂SO₄
Answer:
a) BaSO₄

Question 130.
In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is the amount of C0„ formed then
a) %C = \(\frac{12 \times x}{16 \times w}\)

b) %C = \(\frac{12}{44} \times \frac{\mathrm{w}}{\mathrm{x}} \times 100\)

c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

d) %C = \(\frac{12}{44} \times \frac{x}{w}\)
Answer:
c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

Question 131.
In Dumas method for estimating nitrogen in organic compound, the gas finally collected is
a) N₂
b) NO
c) NH₃
d) H₂
Answer:
a) N₂

Question 132.
In Dumas method, the gas which is collected in Nitrometer is
a) N₂
b) NO
c) NH₃
d) H₂
Answer:
a) N₂

Question 133.
In Kjeldahl’s method, the nitrogen presence is estimated as
a) N₂
b) NH₃
c) NO₂
d) N₂O₃
Answer:
b) NH₃

Question 134.
In Kjeldahl’s method, nitrogen present in the organic compound is first converted into
a) NH₃
b) (NH₄)₂SO₄
c) N₂
d) NO
Answer:
b) (NH₄)₂SO₄

Question 135.
In Liebig’s method for the estimation of C and H, the combustion tube is passed over
a) CuO pellets
b) Copper turnings
c) Iron fillings
d) Zinc – copper couple
Answer:
a) CuO pellets

Question 136.
Which gas is introduced into the combustion tube in Liebig’s method?
a) Pure and dry CO₂
b) Pure and dry N₂
c) Pure and dry O₂
d) Pure and dry He
Answer:
c) Pure and dry O₂

Question 137.
Chromatographic techniques of purification can be used for
a) Coloured compounds
b) Liquids
c) Solids
d) All of these
Answer:
d) All of these

Question 138.
Two substances when separated on the basis of partition co – efficient between two liquid phase, then the technique is known as
a) column chromatography
b) Paper chromatography
c) GLC
d) TLC
Answer:
b) Paper chromatography

Question 139.
Ortho and para nitro phenols can be separated by
a) crystallization
b) distillation
c) sublimation
d) solvent extraction
Answer:
b) distillation

Question 140.
In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is
a) Equal to atmospheric pressure
b) Less than atmospheric pressure
c) More than atmospheric pressure
d) Exactly half of the atmospheric pressure
Answer:
a) Equal to atmospheric pressure

Question 141.
Organic compound is fused with metallic sodium for testing nitrogen, sulphur, and halogens because
a) To make the solution alkaline
b) To convert into elemental state of nitrogen, sulphur, and halogens
c) To convert covalent compound into ionic compound
d) To decrease fusion temperature
Answer:
c) To convert covalent compound into ionic compound

Question 142.
Sodium extract gives blood red colour when treated with FeCl₃, Formation of blood-red colour confirms the presence of
a) Only nitrogen
b) Only sulphur
c) Only halogens
d) Both Nitrogen and Sulphur
Answer:
d) Both Nitrogen and Sulphur

Question 143.
The compound not formed in the positive test for nitrogen with the Lussaigne’s solution of an organic compound is
a) Fe₄[Fe(CN)₆]₃
b) Na₃[Fe(CN)₆]
c) Fe(CN)₃
d) Na₃[Fe(CN)₅NOS]
a) b, c, d
b) a, b
c) a, b, c
d) a only
Answer:
a) b, c, d

Question 144.
The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of
a) Na₄[Fe(CN)₆]
b) Fe₃[Fe(CN)₆]₂
c) Fe₂[Fe(CN)₆]
d) Fe₄[Fe(CN)₆]₃
Answer:
d) Fe₄[Fe(CN)₆]₃

Question 145.
Which of the following compounds will answer Lassaigne’s test for nitrogen?
a) NH₂NH₂
b) NH₂OH
c) NaCN
d) NaNO₃
Answer:
c) NaCN

Question 146.
In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is
a) 28
b) 38
c) 18
d) 48
Answer:
a) 28

Question 147.
0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be
a) 49.32
b) 59.32
c) 29.32
d) 98.64
Answer:
a) 49.32

Question 148.
In an estimation of S by Carius method 0.217 g of the compound gave 0.5825 g of BaSO₄. Percentage of S is
a) 36.78 %
b) 35.50 %
c) 36.48 %
d) 35.69 %
Answer:
a) 36.78 %

II. Very short question and answers (2 Marks):

Question 1.
What are Acyclic compounds? Give suitable example.
Answer:
These are the compounds in which carbon atoms are linked to form open chain (straight or branched). These compounds may be saturated (all single bonds) or unsaturated (multiple bonds).
Example:

These acyclic compounds are known as open chain or aliphatic compounds.

Question 2.
What arc Alicyclic Compounds? Give suitable example.
Answer:
These are saturated or unsaturated carbo-cyclic compounds which resemble the corresponding acylic compounds in their properties.

Question 3.
What are Aromatic heterocyclic Compounds? Give example.
Answer:
These are the heterocyclic compounds which possess aromaticity and resemble, the corresponding aromatic compounds in most of their properties. These are also called non-benzenoid aromatic compounds.

Question 4.
What is functional group? Give example.
Answer:
An atom or group of atoms within a molecule that shows a characteristics set of physical and chemical properties.
Example:
(i) – NH₂ – amines
(ii) = NH – Imines
(iii)

Question 5.
Explain the following terms in IUPAC system of nomenclature. of organic compounds.
(i) Root word
(ii) prefix
(iii) suffix
Answer:
(i) Root word:
Root word denotes the number of carbon atoms in the longest continuous chain in molecules.
(ii) prefix:
Prefix denotes the group(s) attached to the main chain which is placed before the root.
(iii) suffix:
Suffix denotes the funtional group and is paced after the root word.

Question 6.
What is Metamerism? With suitable examples.
Answer:
Metamerism:
This type of isomerism is a special kind of structural isomerism arises due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to the either side of the same functional group and having same molecular formula. This isomerism is shown by compounds having functional group such as ethers, ketones, esters and secondary amines between two alkyl groups.
C₄H₁₀O
CH₃ – O – C₃H₇ – Methyl propyl ether 1 – methoxypropane

C₂H₅ – O – C₂H₅ – diethyl ether ethoxyethane

– methyl iso – propyl ether 2 – methoxypropane.

Question 7.
What is meant by stereochemistry?
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This branch of chemistry dealing with the study of three-dimensional nature (spactial arrangement) of molecules is known as stereo chemistry. The metabolic activities in living organisms, natural synthesis and drug synthesis involve various stereoisomers.

Question 8.
What is enantiomerism?
Answer:
An optically active substance may exist in two or more isomeric forms which have same physical & chemical properties but differ in terms of direction of rotation of plane polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenom-enon is known as enantiomerism.
Example: d and l lactic acid.

Question 9.
What are the conditions for an organic compound is said to be optically active?
Answer:
(1) The molecule must contains at least one chiral or Asymmetric carbon atom.
(2) The object molecule should not be super impossable with its own mirror image.

Question 10.
How will you detect phosphorus present in the given organic compound?
Answer:
Test for phosphorous:
A solid compound is strongly heated with a mixture of Na₂CO₃ & KNO₃. Phosphorous present in the compound is oxidized to sodium phosphate. The residue is extracted with water and boiled with Conc. HNO₃.. A solution of ammonium molybdate is added to the above solution. A canary yellow coloration or precipitate shows the presence of phosphorous.

Question 11.
Explain the sublimation process for the purification of organic compounds.
Answer:
Few substances like benzoic acid, naphthalene and camphor when heated pass directly from solid to vapor without melting (ie liquid). On cooling the vapours will give back solids. Such phenomenon is called sublimation. It is a useful technique to separate volatile and non-volatile solid. It has limited application because only a few substance will sublime.
Example:
naphthalene, benzoic acid.

Question 12.
What is the need for purifying an organic compound?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds they must be in the pure state.

III. Short question and answers (3 Marks):

Question 1.
What is Homologous Series? Give suitable example.
Answer:
Homologous series:
It is a series of compounds in which the adjacent members differ by a – CH₂ unit. Individual members of such series are called homologues, and the phenomenon is called as homology. All the members of such a series of alkane have general formula C_nH_{2n + 2}. Few members of this family are
CH₄ – Methane
C₂H₆ – Ethane
C₃H₈ – Propane
C₄H₁₀ – Butane
C₅H₁₂ – Pentane

Question 2.
What are Alicycic heterocyclic Compounds? Give example.
Answer:
These heterocyclic compounds resemble the corresponding aliphatic compounds in most of their properties.
Example:

Question 3.
Write the IUPAC name of the following compounds.
a)

b)

c)

d)
Answer:
a)

b)

c)

d)

Question 4.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 5.
Write the IUPAC name of the following compounds.
a)
b)

c)

d)
Answer:
a)

b)

c)

d)

Question 6.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 7.
Write the IUPAC name of the following compounds.

Answer:
a)

b)

c)

d)

Question 8.
Classify the following compounds based on the structure.
i) CH ≡ C – CH₂ – C ≡ CH

ii) CH₃ – CH₂ – CH₂ – CH₂ – CH₃

iii)

iv)
Answer:
i) CH ≡ C – CH₂ – C ≡ CH is unsaturated open chain compound

ii) CH₃ – CH₂ – CH₂ – CH₂ – CH₃ is saturated open chain compound

iii) is aromatic benzenoid compound

iv) is alicyclic compound

Question 9.
Give two examples for each of the following type of organic compounds.
(i) non-benzenoid aromatic
(ii) aromatic heterocyclic
(iii) alicyclic
(iv) aliphatic open chain
Answer:
(i) non-benzenoid aromatic

(ii) aromatic heterocyclic

(iii) alicyclic

(iv) aliphatic open chain

Question 10.
Explain the copper oxide test for the detection of carbon and hydrogen present in the given organic compound.
Answer:
Copper oxide test:
The organic substance is mixed intimately with about three times its weight of dry copper oxide by grinding. The mixture is then placed in a hard glass test tube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly and the following reaction take place.
C + 2CuO → CO₂ + 2CuO
2H + CuO → H₂O + Cu

Thus if carbon is present, it is oxidized to CO₂ which turns lime water millcy. If hydrogen is also present, it will be oxidized to water which condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water collected in the bulb is separated on anhydrous CuSO₄ which turn blue. This confirms the presence of C and H in the compound. If however, H is not present water droplet is not obtained in the bulb.

Question 11.
In an estimation of sulphur by carius method. 0.2175 g of the substance gave 0.5825 g of BaSO₄ calculate the percentage composition of S in the compound.
Solution:
Weight of organic compound = 0.2175 g
Weight of BaSO₄ = 0.5825 g
233 g of BaSO₄ contains = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175}\right)\)
0.5825 g of BaSO₄ contains
Percentage of S = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100\right)\)
= 36.78 %

Question 12.
0.16 g of an organic compound was heated in a carius tube and H₂SO₄ acid formed
was precipitated with BaCl₂. The mass of BaSO₄ was 0.35 g. Find the percentage of
sulphur [30.04]
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of sulphur = \(\left(\frac{32}{233} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{32}{233} \times \frac{0.35}{0.16} \times 100\right)\)
= 30.04 %

Question 13.
0.284 g of an organic substance gave 0.287 g AgCl in a Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Solution:
Weight of the organic substance = 0.284 g
Weight of AgCl is = 0.287 g
143.5 g of AgCl contains = 35.5 g of chlorine
0.287 g of AgCl contains = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284}\right)\)

% of chlorine is = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100\right)\)
= 24.56 %

Question 14.
0.185 g of an organic compound when treated with Conc. HNO₃ and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
(Ag = 108, Br = 80)
Answer:
Weight of organic substance (w) = 0.185 g
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\left(\frac{80}{188} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{80}{188} \times \frac{0.32}{0.185} \times 100\right)\)
= 73.6%

Question 15.
0.40 g of an lodo – substituted organic compound gave 0.235 g of AgI by carlus method. Calculate the percentage of iodine in the compound. (Ag = 108 I = 127).
Solution:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
Percentage of iodine = \(\left(\frac{127}{235} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{127}{235} \times \frac{0.235}{0.40} \times 100\right)\)
= 31.75%

Question 16.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of phosphorous in the compound.
Solution:
Weight of an organic compound = 0.24 g
Weight of Mg₂P₂O₇ = 0.66 g
222 g of Mg₂P₂O₇ contains = 62 g of P
0.66 g contains = \(\frac{62}{222} \times \frac{0.66}{0.24}\)

Percentage of p = \(\frac{80}{180} \times \frac{0.66}{0.24}\) = 76.80%

Question 17.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound.
Solution:
Weight of organic substance (w) = 0.33 g
Weight of Mg₂P₂O₇ (x) = 0.397 g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{0.397}{0.33} \times 100\) = 33. 59 %

Question 18.
Explain simple distillation process with suitable example.
Answer:
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. This method is called simple distillation. Liquids with large difference in boiling point(about 40k) and do not decompose under ordinary pressure can be purified by simple distillation.
Example:
The mixture of C₆H₅NO₂ (b.p 484K) &
C₆H₆ (354 K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 35 K).

Question 19.
How will you purify an organic compound by differential extraction process?
Answer:
The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction. When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel. The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl₃).

The organic solvent immiscible with water will form a separate layer and the contents are shaken gently. The solute being more soluble in the organic solvent is transfered to it. The solvent layer is then separated by opening the tap of the separating funnel, and the substance is recovered.

Question 20.
Explain the steam distillation process for purifying organic compound.
Answer:
This method is applicable for solids and liquids. If the compound to be steam distilled the compound should not decompose at the steam temperature, should have a fairly high vapour pressure at 373 K, it should be insoluble in water; the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round-bottom flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam. The mixture in the flask is heated and then a current of steam passed in to it. The vapours of the compound mix up with steam and escape into the condenser.

The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers, also in the manufacture of aniline and turpentine oil.

Question 21.
Write a note on azeotropic distillation with suitable example.
Answer:
These are the mixture of liquids that cannot be separated by fractional distillation. The mixtures that can be purified only by azeotropic distillation are called as azeotropes. These azeotropes are constant boiling mixtures, which distil as a single compound at a fixed temperature. Ethanol and water are the most common examples of azeotropic mixture in the ratio of 95.87 : 4.13.

In this method apart from azeotropic mixture a third component like C₆H₆, CCl₄, ether, glycerol, glycol which act as a dehydrating agent depress the partial pressure of one component so that the boiling point of that component is raised sufficiently and thus other component will distill over.

Dehydrating agents like C₆H₆, CCl₄ have low boiling points and reduce the partial vapour pressure of alcohol more than that of water whereas glycerol & glycol, etc. have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 22.
What is Ring chain isomerism? Give an example.
Answer:
In this type of isomerism, compounds having same molecular formula but differ terms of bonding of carbon atom to form open-chain and cyclic structures.

Question 23.
Briefly explain geometrical isomerism in oximes with suitable example.
Ans:
Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here ‘syn’ and ‘anti’ are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and -OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.
For eg:

IV. Long Question and answers (5 Marks):

Question 1.
Write the structure of the following compounds.
a) Cinnamic acid
b) Lactic acid
c) Phthalic acid
d) Tartaric acid
d) Benzoic acid
Answer:
a) Cinnamic acid

b) Lactic acid

c) Phthalic acid

d) Tartaric acid

d) Benzoic acid
C₆H₅COOH

Question 2.
Explain the methods for the representation of structure of organic compounds with suitable example.
Answer:
The structure of an organic compound can he represented using any one of the below mentioned methods.

1. Lewis structure or dot structure,
2. Dash structure or line bond structure,
3. Condensed structure
4. Bond line structure

We know how to draw the Lewis structure for a molecule. The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a Lewis structure. A single line or dash represents single a covalent bond, double line represents double bond (1 σ bond, 2 π bond) and a triple line represents triple bond (1 σ bond, 2 π bond). Lone pair of electrons on hetero atoms may or may not be shown. This represents the complete structural formula.

This structural formula can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.

For further simplification, organic chemists use another way of representing the structures in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are shown in a zigzag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. These representations can be easily understood by the following illustration.

Question 3.
Explain the molecular model method for the representation of structure of organic compounds.
Answer:
Molecular models:

Molecular models are physical devices that are used for a better visualisation and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available.
(i) Frame work model
(ii) Ball and stick model &
(iii) space filling model.

In the frame work model only the bonds connecting the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of the atom.

In the ball and stick model, both the atoms and the bonds are shown. Ball represent atoms and the stick a bond. Compounds containing C = C can be best represented by using springs in place of sticks and this model is termed as ball and spring model.

The space filling model emphasizes the relative size of each atom based on its Vander Waals radius.

Question 4.
Explain briefly
(i) Fisher projection
(ii) sawhorse projection
(iii) Newman projection formula with neat example.
Answer:
(i) Fisher projection formula:
This is a method of representing three dimensional structures in two dimension. In this method, the chiral atom(s) lies in the plane of paper. The horizontal substituents are pointing towards the observer and the vertical substituents are away from the observer. Fisher projection formula for tartaric acid is given below.

Sawhorse projection formula:
Here the bond between two carbon atoms is drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back. The Fischer projection inadequately portrays the spatial relationship between ligands attached to adjacent atoms. The sawhorse projection attempts to clarify the relative location of the groups.

Newman projection formula:
In this method the molecules are viewed from the front along the carbon-carbon bond axis. The two carbon atom forming a bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point whereas the carbon lying further from the eye is represented by the origin of the circle. Therefore, the C-H bonds of the front carbon are depicted from the circle while C-H bonds of the back carbon are drawn from the circumference of the circle with an angle of 120° to each other.

Question 5.
What is Tautomerism? Explain different types of Tautomerism with suitable example.
Answer:
Tautomerism:
It is a special type of functional isomerism in which a single compound exists in two readily inter con-vertible structures that differ markedly in the relative position of atleast one atomic nucleus, generally hydrogen. The two dif¬ferent structures are known as tautomers. There are several types of tautomerism and the two important types are dyad and triad systems.

Dyad system:
In this system hydrogen atom oscillates between two directly linked polyvalent atoms.
Example:

In this example hydrogen atom oscillates between carbon & nitrogen atom.

Triad system:
In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 migration of hydrogen atom from one polyvalent atom to other within the molecule. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are ketoform and enol-form. The polyvalent atoms involved are one oxygen and two carbon atoms. Enolisation is a process in which keto-form is converted to enol form. Both tautomeric forms are not equally stable. The less stable form is known as labile form.

Question 6.
How is sulphur detected by Lassaigne test?
Answer:
a) To a portion of the lassaigne’s extract, add freshly prepared sodium nitro prusside solution. A deep violet colouration is obtained. This test is also used to detect S^2- in inorganic salt analysis.
Na₂S + Na₂ [Fe(CN₅) NO] → Na₄ [Fe (CN₅) NOS]
sodium nitro prusside

b) Acidify another portion of lassaigne’s extract with acetic acid and add lead acetate solution. A black precipitate is obtained.
(CH₃COO)₂Pb + Na₂S → PbS↓ + 2CH₃COONa
(black ppt)

c) Oxidation test:
The organic substances are fused with a mixture of KNO₃ and Na₂CO₃. The sulphur, if present is oxidized to sulphate.
Na₂CO₃ + S + 3O + Na₂SO₄ + CO₂
The fused mass is extracted with water, acidified with HCl and then BaCl₂ solution is added to it. A white precipitate indicates the pressure of sulphur.
BaCl₂ + Na₂SO₄ -> BaSO₄ + 2NaCl

Question 7.
How is halogen detected by using Lassaigne test?
Answer:
To another portion of the Lassaigne’s filtrate add dil HNO₃ warm gently and add AgNO₃
Solution:
a) Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
b) Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
c) Appearance of a yellow precipitate insoluble in ammonia solution indicates the presence of iodine.

Na + NaX ( where X = Cl, Br, I)
from organic compound
NaX + AgNO₃ → AgX + NaNO₃

If N or S is present in the compound along with the halogen, we might obtain NaCN and Na₂S in the solution, which interfere with the detection of the halogen in the AgNO₃ test Therefore we boil the lassaignes extract with HNO₃ which decomposes NaCN and Na₂S as

NaCN + HNO₃ NaNO₃ + HCN T
Na₂S + 2HNO₃ 2NaNO₃ + H2S
NaCN + AgNO₃ AgCN + NaNO₃
white ppt confusing with AgCl
Na₂S + AgNO₃ → Ag₂S ↓ + NaNO₃
black ppt

Question 8.
How will you estimate carbon and hydrogen present in the given organic compound?
Answer:
Both carbon and hydrogen are estimated by the same method. A known weight of the organic substance is burnt in excess of oxygen and the carbon and hydrogen present in it are oxidized to carbon dioxide and water, respectively.
C_XH_Y + O₂ + XCO₂ + \(\frac{\mathrm{y}}{2}\)H₂O

The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the original substance is calculated. The apparatus employed for the purpose consists of three units (1) oxygen supply (2) combustion tube (3) absorption apparatus.

Oxygen supply:
To remove the moisture from oxygen it is allowed to bubble through sulphuric acid contained in a drech gel bottle and then passed through a U-tube charged with soda lime to remove CO₂. The oxygen gas free from moisture and carbon dioxide enters the combustion tube.

Combustion tube:
A hard glass tube open at both end is used for the combustion of the organic substance. It is filled with
(i) a role of oxidized copper gauze to prevent the backward diffusion of the product of combustion
(ii) a porcelain boat containing a known weight of the organic substance
(iii) coarse copper oxide packed in about 2/3 of the entire length of the tube, and kept in position by loose asbestos plugs on either side and
(iv) a roll of oxidized copper gauze placed towards the end of the combustion tube to prevent any vapours of the organic substance having the tube unoxidized. The combustion tube is enclosed in a furnace, and heated by a gas burner.

Absorption Apparatus:
The products of combustion containing moisture and carbon dioxide are then passed through the absorption apparatus which consists of
(i) a weighed U-tube packed with pumice soaked in Conc. H₂SO₄ to absorb water
(ii) a set of bulbs containing strong solution of KOH to absorb CO₂ and finally
(iii) a guard tube filled with anhydrous CaCl₂ to prevent the entry of moisture from atmosphere.

Procedure:
To start with, before loading it with the boat, the combustion tube is detached from the absorption unit. The tube is heated strongly to dry its content and CO₂ present in it is removed by passing a current of pure, dry oxygen through it. It is then cooled slightly and connected to the absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced.

The tube is again heated strongly till the substance in the boat is burnt away. This takes about 2 hours. Finally, a strong current of oxygen is passed through the combustion tube to sweep away any traces of carbon dioxide or moisture which may have been left in it. The U-tube and the potash bulbs are then detached and the increase in weight of each of them is determined.

Calculation:
Weight of the organic substance taken = w g
Increase in weight of H₂O = x g
Increase in weight of CO₂ = y g
18 g of H₂O contain 2 g of hydrogen
∴ x g of H₂O contain = \(\left(\frac{2}{18} \times \frac{x}{w}\right)\)
% of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\) %

44 g of CO₂ contains = 12 g of carbon
∴ y g of CO₂ contain = \(\left(\frac{12}{44} \times \frac{y}{w}\right)\) g of carbon
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\) %

Question 9.
Explain the Carius method for the estimation of sulphur in an organic compound.
Answer:
Estimation of sulphur Carius method:
A known mass of the organic substance is heated strongly with fuming HNO₃, C & H get oxidized to CO₂ & H₂O while sulphur is oxidized to sulphuric acid as per the following reaction.

C CO₂
2H H₂O
S → SO₂  H₂SO₄

The resulting solution is treated with excess of BaCl₂ solution H₂SO₄ present in the solution in quantitatively converted into BaSO₄, from the mass of BaSO₄, the mass of sulphur and hence the percentage of sulphur in the compound can be calculated.

Procedure:
A known mass of the organic compound is taken in clean carius tube and added a few mL of fuming HNO₃. The tube is the sealed. It is then placed in an iron tube and heated for about 5 hours. The tube is allowed to cool to temperature and a small hole is made to allow gases produced inside to escape.

The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker H₂SO₄ acid formed as a result of the reaction is converted to BaSO₄. The precipitate of BaSO₄ is filtered, washed, dried and weighed. From the mass of BaSO₄, percentage of S is found.

Mass of the organic compound = w g
Mass of the BaSO₄ formed = x g
233 g of BaSO₄ contains = 32 g of sulphur

∴ x g of BaSO₄ contain = \(\left(\frac{32}{233} \times \frac{x}{w}\right)\) g of S

Percentage of sulphur = (\(\left(\frac{32}{233} \times \frac{x}{w}\right)\) × 100) %

Question 10.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g
Percentage of hydrogen
18 g of water contain 2 g of hydrogen
0.039 of water contain = \(\left(\frac{2}{18} \times \frac{0.039}{0.26}\right)\) of hydrogen

% of hydrogen = \(\left(\frac{0.039}{0.26} \times \frac{2}{18} \times 100\right)\) = 1.66%
Percentage of Carbon
44 g of CO₂ contain 12 g of C
0.245 g of CO₂ contains \(\left(\frac{12}{44} \times \frac{0.245}{0.26}\right)\) g of C

% of Carbon = \(\left(\frac{12}{44} \times \frac{0.245}{0.26} \times 100\right)\) = 25.69 %

Question 11.
0.2346 of an organic compound yielded C, H & 0 0.2754 g of H₂O and 0.4488 g CO₂. Calculate the % composition.
Solution:
Weight of organic substance w = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO₂ (y), = 0.4488 g
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\)

= \(\left(\frac{12}{44} \times \frac{0.4488}{0.2346} \times 100\right)\) = 52.17 %

Percentage of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{2}{18} \times \frac{0.2754}{0.2346} \times 100\right)\) = 13. 04%
Percentage of oxygen= [100 – (52.17 + 13.04)]
= 100 – 65.21 = 34.79 %

Question 12.
Explain the estimation of phosphours by Carius method.
Answer:
Carius method:
A known mass of the organic compound (w) containing phosphorous is heated with fuming HNO in a sealed tube where C is converted into CO₂ and H to H₂O. phosphorous present in organic compound is oxidized to phosphoric acid which is precipitated, as ammonium phosphomolybdate by heating with Cone. HNO₃ and then adding ammonium molybdate.

H₃PO₄ + 12(NH₄)₂MoO₄ + 21 HNO₃  (NH₄)₃PO₄ + 21 NH₄NO₃ + 12HNO₃

The precipitate of ammonium phosphomolybdate thus formed is filtered washed, dried, and weighed. In an alternative method, the phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture (a mixture containing MgCl₂, NH₄Cl, and ammonia) This ppt is washed, dried, and ignited to get magnesium pyrophosphate which is washed, dried a weighed. The following are the reaction that takes place.

By knowing the mass of the organic compound and the, mass of ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of P is calculated.

Mass of organic compound = w g
Weight of ammonium phosphomolybdate = x g
Weight of magnesium pyrophosphate = y g
Mole mass of (NH₄)₃PO₄. 12MoO₃ is = 1877 g
[3 × (14 + 4) + 31 + 4(16)]+ 12(96 + 3 × 16)
Molar mass of Mg₃P₂O₇ is 222 g
(2 × 24) + (31 × 2) + (7 × 16)
1877 g of (NH₄)₃PO₄. 12 MoO₃ contains 31 g of P
X g of (NH₄)₃PO₄. 12 MoO₃ in w g of organic
compound contains \(\frac{31}{1877} \times \frac{x}{w}\) of phosphorous
Percentage of phosphorous = \(\frac{31}{1877} \times \frac{x}{w}\) (or)
227 of Mg₂P₂O₇ contains 62 g of P
y g of Mg₂P₂O₂ in w g of Organic compound contains = \(\frac{62}{227} \times \frac{y}{w}\) of P

Percentage of phosphorous = \(\frac{62}{227} \times \frac{y}{w}\)%

Question 13.
Explain the estimation of nitrogen by Dumas method.
Answer:
1. Dumas method:
This method is based upon the fact that nitrogenous compound when heated with cupric oxide in an atmosphere of CO₂ yields free nitrogen. Thus
C_x H_y N_z + (2x + y/2) CuO → x CO₂ + y/2 H₂O Z/2 N₂ +(2X + y/2)Cu.

Traces of oxide of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing overheated copper spiral. The apparatus used in Dumas method consists of CO₂ generator, combustion tube, Schiffs nitrometer.

CO₂ generator:
CO₂ needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube or by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after being dried by bubbling through Cone. H₂SO₄ contained in a Drechel bottle.

Combustion Tube:
The combustion tube is heated in a furnace is charged with
a) A roll of oxidized copper gauze to prevent the back diffusion of the products of combustion and to heat the organic substance mixed with CuO by radiation
b) a weighed amount of the organic substance mixed with an excess of CuO,
c) a layer, of course, CuO packed in about 2/3 of the entire length of the tube and kept in position by loose asbestos plug on either side; this oxidizes the organic vapors passing through it, and
d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion to nitrogen.

Schiff’s nitro meter:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO₂ It is estimated by passing nitrometer when CO₂ is absorbed by KOH and the nitrogen gets collected in the upper part of graduated tube.

Procedure:
To start with the tap of nitro meter is left open. CO₂ is passed through the combustion tube to expels the air in it. When the gas bubbles rising through, the potash solution fails to reach the top of it and is completely absorbed it shows that only CO₂ is coming and that all air has been expelled from the combustion tube. The nitrometer is then filled with KOH solution by lowering the reservoir and the tap is closed. The combustion tube is now heated in the furnace and the temperature rises gradually.

The nitrogen set free from the compound collects in the nitro meter. When the combustion is complete a strong current of CO₂ is sent through, the apparatus in order to sweep the last trace of nitrogen from it. The volume of the gas gets collected is noted after adjusting the reservoir so that the solution in it and the graduated tube is the same. The atmospheric pressure and the temperature are also recorded.

Calculation:
Weight of the substance taken = w g
Volume of nitrogen = V₁ L
Room Temperature = T₁ K
Atmospheric Pressure = P mm of Hg
Aqueous tension at room temperature = P’ mm of Hg
Pressure of diy nitrogen = (P – P’) = P’ mm of Hg.
Let P₀V₀ and T₀ be the pressure, volume and temperature respectively of dry nitrogen at STP,
Then,

Calculation of percentage of nitrogen, 22.4 L of N₂ at STP weight 28 g of N2
∴ V₀ L of N₂ at S.T.P weigh = \(\frac{28}{22.4}\) × V₀
Wg Organic compound contain = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) of nitrogen

∴ percentage of nitrogen = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) × 100

Question 14.
0.1688 g when analyzed by the Dumas method yield 31.7 ml of moist nitrogen measured at 14°C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C = 12 mm)
Solution:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V₁) = 31.7 mL
= 31.7 × 10^-3 L
Temperature (T¹) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C =12 mm of Hg
∴ Pressure of dry nitrogen = (P – P¹)
= 758 – 12 = 746 mm of Hg

Question 15.
Explain the estimation of nitrogen by Kjeldahl’s method.
Answer:
This method is carried much more easily than the Dumas method. It is used largely in the analysis of foods and fertilizers. Kjeldahl’s method is based on the fact that when an organic compound containing nitrogen is heated with Conc. H₂SO₄, the nitrogen in it is quantitatively converted to ammonium sulphate. The resultant liquid is then treated with excess of alkali and then liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia (and hence nitrogen) is determined by finding the amount of acid neutralized by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance (0.3 to 0.5 g) is placed in a special long – necked Kjeldahl flask made of pyrex glass. About 25 mL of Cone. H₂SO₄ together with a little K₂SO₄ and CuSO ₄ (catalyst) are added to it the flask is loosely stoppered by a glass bulb and heated gently in an inclined position. The heating is continued till the brown color of the liquid first produced, disappears leaving the contents clear as before. At this point all the nitrogen in the substance is converted to (NH₄)₂SO₄.

The Kjeldahl flask is then cooled and its contents are diluted with same distilled water and then carefully transferred into a 1 lit round bottom flask. An excess NaOH solution is poured down the side of the flask and it is fitted with a Kjeldahl trap and a water condenser. The lower end of the condenser dips in a measured volume of excess the H₂SO₄ solution. The liquid in the round bottom flask is then heated and the liberated ammonia is distilled into sulphuric acid. The Kjeldahl trap serves to retain any alkali splashed up on vigorous boiling.

When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = w g
Volume of H₂SO₄ required for the complete neutralization of evolved NH₃ = V ml.
Strength of H2S04 used to neutralize NH₃ = N
Let the Volume and the strength of NH₃ formed are V₁ and N₁ respectively.
We know that V₁N₁ = VN
The amount of nitrogen present in the = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times \mathrm{w}}\)

Percentage of Nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times \mathrm{w}}\right)\) × 100 %

Question 16.
0.6 g of an organic compound was kjeldalised and NH3 evolved was absorbed into 50 mL of semi – normal solution of H₂SO₄. The residual acid solution was diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac{\mathbf{N}}{20}\) NaOH solution for complete neutralization. Calculate the % of N in the compound.
Solution:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20ml of diluted solution of unreacted sulphuric acid was neutralized by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric aicd = \(\frac{35 \times 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with Organic compound = V₁ mL
Strength of the diluted H₂SO₄ = 0.5 N
Volume of the diluted H₂SO₄ = 150 mL
Strength of the diluted H₂SO₄ = 0.0875 N
V₁ = \(\frac{150 \times 0.0875}{0.5}\) = 26.25 ml
Volume of H₂SO₄ consumed by ammonia = 50 – 26.25 = 23.75 ml
23.75 ml of 0.5 N H₂SO₄ = 23.75 ml of 0.5 N NH₃
The amount of Nitrogen present in the 0.6 g of organic compound = \(\frac{14 g}{1000 m L \times 1 N}\) × 23.75 × 0.5 N = 0.166 g
Percentage of Nitrogen = \(\frac{0.166}{0.6}\) × 100 = 27.66%

Question 17.
Explain the various steps involved in the purification of organic compounds by crystallization process.
Answer:
(i) Selection of solvent:
Most of the organic substances being covalent do not dissolve in polar solvents like water, hence selection of solvent (suitable) becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is just sufficient to dissolve the solute (ie) organic compound. If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with other solvents like benzene, ether, acetone, and alcohol till the most suitably one is sorted out.

(ii) Preparation of solution:
The quantity to be purified is taken in a conical flash fitted with a reflux condenser. The solvent selected in first stage is also taken along with solute and the quantity of the solvent just enough to dissolve the whole solid on boiling. Small amount of are animal char-coal can be added before boiling to decolorize any colored substance. The heating may be done over a wire gauze or water bath depending upon the nature of liquid (ie) whether the solvent is low boiling or high boiling.

(iii) Filtration of hot solution:
The solution so obtained is filtered through a fluted filter paper placed in a hot water funnel.

(iv) Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper, the pure solid substance separate as crystal. When copious amount of crystal has been obtained, then the crystallization is complete. If the rate of crystallization is slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of the pure compounds to the solution.

(v) Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration. Filtration is done under reduced pressure using a Bucher funnel. When the whole of the mother liquor has been drained into the filtration flask, the crystals are washed with small quantities of the pure cold solvent and then dried.

Question 18.
How will you purify an organic compound by Thin layer chromatography?
Answer:
This method is another type of adsorption chromatography with this method it is possible to separate even minute quantities of mixtures. A sheet of a glass is coated with a thin layer of adsorbent (cellulose, silica gel or alumina). This sheet of glass is called chromoplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluent (solvent). The eluent is drawn up to the adsorbent layer by capillary action.

The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retardation factors (ie) Rf value
Rf = \(\frac{\text { Distance moved by the substance from baseline }(\mathrm{x})}{\text { Distance moved by the solvent from baseline }(\mathrm{y})}\)

The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under UV light or in another method using iodine crystals or by using an appropriate reagent.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 10 Chemical Bonding Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

11th Chemistry Guide Chemical Bonding Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
In which of the following compound does the central atom obey the octet rule?
a) XeF₄
b) AlCl₃
c) SF₆
d) SCl₂
Answer:
d) SCl₂

Question 2.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
a) -1, 0, +1
b) +1, 0, -1
c) -2, 0, +2
d) 0, 0, 0
Answer:
d) 0, 0, 0

Question 3.
Which of the following is electron deficient?
a) PH₃
b) (CH₃)₂
c) BH₃
d) NH₃
Answer:
c) BH₃

Question 4.
Which of the following molecule contain no π bond?
a) SO₂
b) NO₂
c) CO₂
d) H₂O
Answer:
d) H₂O

Question 5.
The ratio of number of sigma (σ) bond and pi (π) bonds in 2 – butynal is
a) 8/3
b) 5/3
c) 8/2
d) 9/2
Answer:
a) 8/3

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluo-ride molecule?
a) 120°, 80°
b) 109°28’
c) 90°
d) 89°, 117°
Answer:
d) 89°, 117°

Question 7.
Assertion:
Oxygen molecule is paramagnetic.
Reason :
It has two unpaired electron in its bonding molecular orbital
a) both assertion and reason are true and reason is the correct explanation of assertion.
b) both assertion and reason are true but reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both assertion and reason are false.
Answer:
c) assertion is true but reason is false.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when
a) fully filled atomic orbitals overlap
b) half filled atomic orbitals overlap
c) non – bonding atomic orbitals overlap
d) empty atomic orbitals overlap
Answer:
b) half filled atomic orbitals overlap

Question 9.
In ClF₃, NF₃ and BF₃ molecules the chlorine, nitrogen and boron atoms are
a) sp³ hybridised
b) sp³, sp³ and sp² respectively
c) sp³ hybridised
d) sp³d, sp³ and sp hybridised respectively
Answer:
d) sp³d, sp³ and sp hybridised respectively

Question 10.
When one s and three p orbitals hybridise,
a) four equivalent orbitals at 90° to each other will be formed
b) four equivalent orbitals at 109°28’ to each other will be formed
c) four equivalent orbitals, that are lying the same plane will be formed
d) none of these
Answer:
b) four equivalent orbitals at 109°28’ to each other will be formed

Question 11.
Which of these represents the correct order of their increasing bond order.
a) C₂⁺ < C₂^2- < O₂^2- < O₂
b) C₂^2- < C₂⁺ < O₂ < O₂^2-
c) O₂^2- < O₂ < C₂^2- < C₂⁺
d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Answer:
d) O₂^2- < C₂⁺ < O₂ < C₂^2-

Question 12.
Hybridisation of central atom in PCl₅ involves the mixing of orbitals.
a) s, P_x, P_y, d_x², d_{x² – y²}
b) s, p_x, p_y, p_xy, d_{x² – y²}
c) s, p_x, p_y, p_z, d_{x² – y²}
d) s, p_x, P_y, d_xy, d_{x² – y²}
Answer:
c) s, p_x, p_y, p_z, d_{x² – y²}

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is
a) H₂O₂ > O₃ > O₂
b) O₂ > O₃ > H₂O₂
c) O₂ > H₂O₂ > O₃
d) O₃ > O₂ > H₂O₂
Answer:
b) O₂ > O₃ > H₂O₂

Question 14.
Which one of the following is diamagnetic?
a) O₂
b) O₂^2-
c) O₂⁺
d) None of these
Answer:
b) O₂^2-

Question 15.
Bond order of a species is 2.5 and the number of electrons in its bonding molecular orbital is formed to be 8. The no. of electrons in its antibonding molecular orbital is
a) three
b) four
c) Zero
d) can not be calculated from the given information
Answer:
a) three

Question 16.
Shape and hybridisation of IF₅ are
a) Trigonal bipyramidal, sp³d²
b) Trigonal bipyramidal, sp³d
c) Square pyramidal, sp³d²
d) Octahedral, sp³d²
Answer:
c) Square pyramidal, sp³d²

Question 17.
Pick out the incorrect statement from the following:
a) sp³ hybrid orbitals are equivalent and are at an angle of 109°28’ with each other
b) dsp² hybrid orbitals are equivalent and bond angle between any two of them is 90°
c) All five sp³d hybrid orbitals are not equivalent out of these five sp³d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three
d) none of these
Answer:
c) All five sp³d hybrid orbitals are not equivalent out of these five sp³d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are
a) SeF₄, XeO₂F₂
b) SF₄, XeF₂
c) XeOF₄, TeF₄
d) SeCl₄, XeF₄
Answer:
a) SeF₄, XeO₂F₂

Question 19.
In which of the following molecules / ions BF₃, NO₂^– H₂0 the central atom is sp² hybridised?
a) NH₂^– and H₂O
b) NO₂^– and H₂O
c) BF₃ and NO₂^–
d) BF₃ and NH₂^–
Answer:
c) BF₃ and NO₂^–

Question 20.
Some of the following properties of two species, NO₃^– and H₃O⁺ are described below. Which one of them is correct?
a) dissimilar in hybridisation for the central atom with different structure
b) isostructural with same hybridisation for the central atom.
c) different hybridisation for the central atom with same structure.
d) none of these
Answer:
a) dissimilar in hybridisation for the central atom with different structure

Question 21.
The types of hybridisation on the five-carbon atom from right to left in the, 2,3 pentadiene.
a) sp³, sp², sp, sp², sp³
b) sp³, sp, sp, sp, sp³
c) sp², sp, sp², sp², sp³
d) sp³, sp³, sp², sp³, sp³
Answer:
a) sp³, sp², sp, sp², sp³

Question 22.
XeF₂ is isostructural with
a) SbCl₂
b) BaCl₂
c) TeF₂
d) ICl₂^–
Answer:
d) ICl₂^–

Question 23.
The percentage of s-character of the hybrid orbitals in methane, ethane, ethene, and ethyne are respectively
a) 25, 25, 33.3, 50
b) 50, 50, 33.3, 25
c) 50, 25, 33.3, 50
d) 50, 25, 25, 50
Answer:
a) 25, 25, 33.3, 50

Question 24.
Of the following molecules, which have shape similar to carbondioxide?
a) SnCl₂
b) NO₂
c) C₂H₂
d) All of these
Answer:
c) C₂H₂

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order
a) 1. p – 1. p > b. p – b. p > 1. p – b. p
b) b. p – b. p > b. p – 1. p > 1. p – b. p
c) 1. p – 1. p > b. p – 1. p > b. p – b. p
d) b. p – b. p > 1. p – 1. p > b. p – 1. p
Answer:
c) 1. p – 1. p > b. p – 1. p > b. p – b. p

Question 26.
Shape of ClF₃ is
a) Planar triangular
b) Pyramidal
c) “T” Shaped
d) none of these
Answer:
c) “T” Shaped

Question 27.
Non – Zero dipole moment is shown by
a) CO₂
b) p – dichlorobenzene
c) carbontetrachloride
d) water
Answer:
d) water

Question 28.
Which of the following conditions is not correct for resonating structures?
a) the contributing structure must have the same number of unpaired electrons
b) the contributing structures should have similar energies
c) the resonance hybrid should have higher energy than any of the contributing structure.
d) none of these
Answer:
c) the resonance hybrid should have higher energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent, and Coordinate linkage is
a) NH₄Cl
b) NH₃
c) NaCl
d) none of these
Answer:
a) NH₄Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is
a) U
b) 2U
c) U /2
d) 4U
Answer:
d) 4U

II. Write brief answer to the following questions:

Question 31.
Define the following:
i) Bond order
ii) Hybridisation
iii) σ – bond
Answer:
i) Bond order:
The number of bonds formed between the two bonded atoms in a molecule is called the bond order.

ii) Hybridisation:
Hybridisation is the process of mixing of atomic orbitals of the same atom with comparable energy to form an equal number of new equivalent orbitals with the same energy.

iii) σ – bond:
When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond.

Question 32.
What is a pi – bond?
Answer:
When two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi (π) bond. When we consider x – axis as molecular axis, the p_y – p_y and p_z – p_z overlaps will result in the formation of a π – bond.

Question 33.
In CH₄, NH₃ and H₂O, the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?
Answer:
According to VSEPR theory, as H₂O has two lone pairs so it repels the bond pairs much more and makes bond angle shorter of 104.5 degrees, and as NH₃ has one lone pair that repels the three bond pairs but not much effectively and strongly as two lone pairs of water repel one bond pair.

So the bond angle between Hydrogen atom of ammonia is 107.5 greater than that of water. Similarly, methane molecule have no lone pair and bond pair repels each other with the equal bond angle between two adjacent hydrogen atoms becomes 109°.28′.

Question 34.
Explain Sp² hybridization in BF₃.
Answer:
Consider boron trifluoride molecule. The valence shell electronic configuration of boron atom is [He]2s² 2p¹.

In the ground state boron has only one unpaired electron in the valence shell. In order to form three covalent bonds with fluorine atoms, three unpaired electrons are required. To achieve this, one of the paired electrons in the 2s orbital is promoted to the 2p_y orbital in the excite state. In boron, the s orbital and two p orbitals (p_x and p_y) in the valence shell hybridises, to generate three equivalent sp² orbitals as shown in the Figure. These three orbitals lie in the same xy plane and the angle between any two orbitals is equal to 120°.

Overlap with 2p_z orbitals of flourine:
The three sp² hybridised orbitals of boron now overlap with the 2p_z orbitals of fluorine (3 atoms). This overlap takes place along the axis as shown below.

Question 35.
Draw the M.O diagram for oxygen molecule calculate its bond order and show that O₂ is paramagnetic.
Answer:

Electronic configuration of O atom:
1s² 2s² 2p⁴

Electronic configuration of O₂ molecule:
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, σ_2px², π_2py² π_2pz² π_2py^*1 π_2pz^*1

Bond order = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-6}{2}\) = 2
Molecule has two unpaired electrons hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:

Bonding in some hetero nuclear di-atomic molecules:
Molecular orbital diagram of Carbon monoxide molecule (CO)
Electronic configuration of C atom: 1s² 2s² 2p²
Electronic configuration of O atom: 1s² 2s² 2p⁴
Electronic configuration of CO molecule :
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, π_2py², π_2pz² σ_2px²

Bond order = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-4}{2}\)
= 3
Molecule has no unpaired electrons hence it is diamagnetic.

Question 37.
What do you understand by Linear combination of atomic orbitals in MO theory?
Answer:
The wave functions for the molecular orbitals can be obtained by solving Schrodinger wave equation for the molecule. Since solving the Schrodinger equation is too complex, approximation methods are used to obtain the wave function for molecular orbitals. The most common method is the linear combination of atomic orbitals (LCAO).

We know that the atomic orbitals are represented by the wave function ψ. Let us consider two atomic orbitals represented by the wave function ψ_A and ψ_B with comparable energy, combines to form two molecular orbitals. One is bonding molecular orbital(ψ_bonding) and the other is antibonding molecular orbital (ψ_antibonding) The wave functions for these two molecular orbitals can be obtained by the linear combination of the atomic orbitals ψ_A and ψ_B as below,

ψ_bonding = ψ_A + ψ_B;
ψ_antibonding = ψ_A – ψ_B

The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitals and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitals. The formation of the two molecular orbitals from two is orbitals is shown below.

Constructive interaction:
The two 1s orbitals are in phase and have the same sign,

Destructive interaction:
The two orbitals are out phase

Question 38.
Discuss the formation of N₂ molecule using MO theory.
Answer:

Molecular orbital diagram of nitrogen molecule (N₂)
Electronic configuration of N atom 1s² 2s² 2p³
Electronic configuration of N₂ molecule:
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, π_2py², π_2pz² σ_2px²
Bondorder = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-4}{2}\) = 3
Molecule has no unpaired electrons hence it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
The polarity of a covalent bond can be measured in terms of dipole moment which is defined as
μ = q × 2d
Where μ is the dipole moment, q is the charge and 2d is the distance between the two charges. The dipole moment is a vector and the direction of the dipole moment vector points from the negative charge to positive charge.

representation of dipole

The unit of dipole moment is coloumb meter (C m). It is usually expressed in Debye unit (d). The conversion factor is 1 Debye = 3.336 × 10^-30 C m.

Question 40.
Linear form of carbondioxide molecule has two polar bonds. Yet the molecule has Zero dipole moment. Why?
Answer:
The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds. In CO₂, the dipole moments of two polar bonds (CO) are equal in magnitude but have opposite direction. Hence, the net dipole moment of the CO₂ is,
μ = μ₁ + μ₂ = μ₁ + (-μ₁) = 0

Question 41.
Draw the Lewis structures for the following species.
(i) NO₃^–
(ii) SO₄^2-
(iii) HNO₃
(iv) O₃
Answer:
(i) NO₃^–

(ii) SO₄^2-

(iii) HNO₃

(iv) O₃

Question 42.
Explain the bond formation if BeCl₂ and MgCl₂.
Answer:
Bond formation of BeCl₂:
Be = 4;
Electronic configuration of Be atom is = 1s² 2s²

Bond formation of MgCl₂:
Mg (Z = 12), 1s² 2s² 2p⁶ 3s² it loses two of its valence electron and became Mg²⁺ with inert gas configuration Neon.
The chlorine accept one electron in its valence shell and because Cl^– ion with Ar electron configuration.
Mg + Cl₂ → Mg²⁺ + 2 Cl^– → MgCl₂
Magnesium cation and two chlorides are attracted by strong electro static force to form MgCl₂ crystals.
Mg = 12,
Electronic configuration: 1s² 2s² 2p⁶ 3s²
Mg⁺²:
Electronic configuration: 1s² 2s² 2p⁶ 3s⁰
Cl = 17, 1s² 2s² 2p⁶ 3s² 3p⁵
Cl^–:
Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶

Question 43.
Which bond is stronger σ or π? Why?
Answer:
σ bond is stronger than π bond. A sigma bond is formed by head on overlapping of orbital is more effective. Hence it is stronger bond. But pi bonds are formed by sidewise overlapping of orbitals. The sidewise overlapping of orbitals is less effective than head on overlapping. Hence it is a weaker bond.

Question 44.
Define bond energy.
Answer:
The bond enthalpy is defined as the minimum amount of energy required to break one mole of a particular bond in molecules in their gaseous state. The unit of bond enthalpy is kJ mol^-1.

Question 45.
Hydrogen gas is diatomic where as inert gases are monoatomic – Explain on the basis of MO theory.
Answer:
The molecular orbital electronic configuration of hydrogen molecule is (σ_1s²). The molecular orbital energy level diagram of H₂ molecule is given in

Here, N₂ = 2, N_a = 0
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-0}{2}\) = 1

He₂: σ_1s² σ_1s^*2
The molecular orbital energy level diagram of He₂ (hypothetical) is given in

Here, N_b = 2 and N_a = 2
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-2}{2}\) = 0

As the bond order for He₂ comes out to formed between two hydrogen atoms. But as the bond order of helium is zero, there is no bond between helium atoms and hence it is mono atomic.

Result:
As the bond order of H₂ molecule is one, it is diatomic and a single bond is formed between two hydrogen atoms. But as the bond order of helium is zero, there is no bond between helium atoms and hence it is mono atomic.

Question 46.
What is the Polar Covalent bond? Explain with example.
Answer:
Polar covalent bond is a chemical bond in which the electrons required to form a bond is unequally shared between two atoms. The atom which is more electronegative attracts more electrons from the bonded pair than the other atom. As a result there is a slight separation of charges in a molecule in which more electronegative atom (comparatively) carries a slight negative charge and less electronegative atom carries a positive charge. The bonds formed between two atoms have a permanent electric dipole.

In water (H₂O) molecule, covalent bond exists between Hydrogen and Oxygen. Being more electronegative Oxygen carries negative charge and Hydrogen carries positive charge. In a water molecule Oxygen carries negative charge (anionic in nature) and hydrogen’s carries positive charge (cationic in nature) which forms a polar covalent bond. The size of an oxygen atom is comparatively higher than a hydrogen atom, hence it polarizes the molecule towards itself i.e., it attracts shared pair of bonding electrons towards itself and makes the bond to be more polarized. Hydrogen which is comparatively a small sized cation makes the bond to be polarized better.

Question 47.
considering x-axis as molecular axis which out of the following will form a sigma bond.
i) 1s and 2p_y
ii) 2p_x and 2p_y
iii) 2p_x and 2p_z
iv) 1s and 2p_z
Answer:
i) 1s and 2p_y: No sigma bond
ii) 2p_x and 2p_y: sigma bond
iii) 2p_x and 2p_z: No sigma bond
iv) 1s and 2p_z: No sigma bond

Question 48.
Explain resonance with reference to carbonate ion.
Answer:
It is evident from the experimental results that all carbon – oxygen bonds in carbonate ions are equivalent. The actual structure of the molecules is said to be resonance hybrid, an average of these three resonance forms. It is important to note that carbonate ion does not change from one structure to another and vice versa. is not possible to picturise the reasonance hybrid by drawing a single Lewis structure. However, the following structure gives a qualitative idea about the correct structure.

(b) Resonance structure of CO₃^2-:

Resonance Hybrid structure of CO₃^2-:
It is found that the energy of the resonance hybrid (structure 4) is lower than that of all possible canonical structures (Structure 1, 2 & 3). The difference in energy between structure 1 or 2 or 3, (most stable canonical structure) and structure 4 (resonance hybrid) is called resonance energy.

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in ethylene:
The bonding in ethylene can be explained using hybridization concept. The molecular formula of ethylene is C₂H₄. The valency of carbon is 4. The electronic configuration of valence shell of carbon in ground state is [He] 2s² 2p_x¹ 2p_y¹ 2p_z⁰. To satisfy the valency of carbon promote an electron from 2s orbital to 2p_z orbital in the excited state.

In ethylene both the carbon atoms undergoes sp² hybridisation involving 2s, 2p_x, and 2p_y, orbitals, resulting in three equivalent sp² hybridised orbitals lying in the xy plane at an angle of 120° to each other. The unhybridised 2p_z orbital lies perpendicular to the xy plane.

Formation of sigma bond:
One of the sp² hybridised orbitals of each carbon lying on the molecular axis (x-axis) linearly overlaps with each other resulting in the formation a C-C sigma bond. Other two sp² hybridised orbitals of both carbons linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C-H sigma bonds on each carbon.

Formation of sigma bond:
The unhybridized 2p_z orbital of both carbon atoms can overlap only sideways as they are not in the molecular axis. This lateral overlap results in the formation of a pi bond between the two carbon atoms as shown in the figure.

Bonding in acetylene :

Similar to ethylene, the bonding in acetylene can also be explained using hybridisation concept. The molecular formula of acetylene is C₂H₂. The electronic configuration of valence shell of carbon in ground state is [He] 2s² 2p_x¹ 2p_y¹ 2p_z⁰. To satisfy the valency of carbon promote an electron from 2s orbital to 2pz orbital in the excited state.

In acetylene molecule, both the carbon atoms are in sp hybridized state. The 2s and 2p_x orbitals, resulting in two equivalent sp hybridized orbitals lying in a straight line along the molecular axis (x-axis). The unhybridized 2p_y and 2p_z orbitals lie perpendicular to the molecular axis.

Formation of sigma bond:
One of the two sp hybridized orbitals of each carbon lineraly overlaps with each other resulting in the formation a C – C sigma bond. The other sp hybridized orbitals of both carbons linearly overlap with the two 1s orbitals of two hydrogen atoms leading to the formation of one C – H sigma bonds on each carbon.

Formation of pi bond :
The unhybridized 2p_y and 2p_z orbitals of each carbon overlap sideways. This lateral overlap results in the formation of two pi bonds (p_y – p_y and p_z – p_z) between the two carbon atoms as shown in the figure.

Question 50.
What type of hybridisations are possible in the following geometries?
a) octahedral b) tetrahedral c) square planar
Answer:
a) octahedral: sp³d²
b) tetrahedral: sp³
c) square planar: dsp²

Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of IF₇ and SF₆.
Answer:
Lewis concept of structure of molecules deals with the relative position of atoms in the molecules and sharing of electron pairs between them. However, we cannot predict the shape of the molecule using Lewis concept. Lewis theory in combination with VSEPR theory will be useful in predicting the shape of molecules.

IF₇:
Iodine has 7 valence electrons in the valence shell. In excited state it has 6 valency electrons by 1st and 2nd excited state & under sp³d², hybridization and combines with 7 Fluorine to get pentagonal bipyramidal shape. There are no lone pairs.
I = ns² np⁵ nd⁰

SF₆:
Sulphar attain six valence electron in 1st and 2nd excited states undergoes sp³d² hybridization and combines with six ‘F’ atoms, as there no lone pair electrons it geometry in octahedral.
S = 16 = ns² np⁴ nd⁰

Question 52.
CO₂ and H₂O both are triatomic molecule but their dipole moment values are different. Why?
Answer:

Sum of the dipole moment are cancelled.

Water is ‘v’ shape sum of the dipole moments are not equal to zero.

Question 53.
Which one of the following has highest bond order?
(i) N₂
(ii) N₂⁺
(iii) N₂^–
Answer:
(i) N₂
N₂ = 14
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px²
Bond order = \(\frac{6}{2}\) = 3

(ii) N₂⁺
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px¹
Bond order = \(\frac{5}{2}\) = 2.5

(iii) N₂^–
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px^*1
Bond order = \(\frac{5}{2}\) = 2.5
The highest bond order is N₂

Question 54.
Explain the covalent character in ionic bond.
Answer:
Like the partial ionic character in covalent compounds, ionic compounds show partial covalent character. For example, the ionic compound, lithium chloride shows covalent character and is soluble in organic solvents such as ethanol.

The partial covalent character in ionic compounds can be explained on the basis of a phenomenon called polarisation. We know that in an ionic compound, there is an electrostatic attractive force between the cation and anion. The positively charged cation attracts the valence electrons of anion while repelling the nucleus.

This causes a distortion in the electron cloud of the anion and its electron density shift towards the cation, which results in some sharing of the valence electrons between these ions. Thus, a partial covalent character is developed between them. This phenomenon is called polarisation.

The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisability.

Question 55.
Describe Fajan’s rule.
Answer:
(i) To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation, greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on the anion, greater is its polarisabiity. Hence, the increase in charge on cation or in anion increases the covalent character.

Let us consider three ionic compounds aluminum chloride, magnesium chloride and sodium chloride. Since the charge of the cation increase in the order Na⁺ < Mg²⁺ <Al³⁺ the covalent character also follows the same order NaCl < MgCl₂ < AlCl₃.

(ii) The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. Lithium chloride is more covalent than sodium chloride. The size of Li⁺ is smaller than Na⁺ and hence the polarizing power of Li⁺ is more. Lithium iodide is more covalent than lithium chloride as the size of I^– is larger than the Cl^–. Hence I^– will be more polarized than Cl^– by the cation, Li⁺.

(iii) Cations having ns² np⁶ nd¹⁰ configuration show greater polarizing power than the cations with ns² np⁶ configuration. Hence, they show greater covalent character. CuCl is more covalent than NaCl. Compared to Na⁺ (1.13 Å). Cu⁺(0.6 Å) is small and have 3s² 3p⁶ 3d¹⁰ configuration.
Electronic configuration of Cu⁺: [Ar] 3s², 3p⁶, 3d¹⁰
Electronic Configuration of Na⁺: [He] 2s², 2p⁶.

11th Chemistry Guide Chemical Bonding Additional Questions and Answers

I. Choose the best answer:

Question 1.
Which among the following elements has the tendency to form covalent compounds?
a) Ba
b) Be
c) Mg
d) Ca
Answer:
b) Be

Question 2.
The valency of C in CO₃^2- is
a) 2
b) 3
c) 4
d) -3
Answer:
c) 4

Question 3.
Two elements X and Y have following electronic configurations
X = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Y = 1s² 2s² 2p⁶ 3s² 3p⁵
The expressed compound formed by combination of X and Y will be expressed as
a) X₂
b) X₅Y₂
c) X₂Y₅
d) XY₅
Answer:
a) X₂

Question 4.
Lattice energy of an ionic compound depends upon:
a) Charge on the ions only
b) Size of the ions only
c) Packing of the ions only
d) Charge and size of the ion
Answer:
d) Charge and size of the ion

Question 5.
If the cyanide ion, the formal negative charge is on
a) C
b) N
c) Both C and N
d) Resonate between C and N
Answer:
b) N

Question 6.
The formal charge of the O – atoms in the ion is
a) -2
b) + 1
c) – 1
d) 0
Answer:
d) 0

Question 7.
A compound with the maximum ionic character is formed from
a) Na and F
b) Cs and F
c) Cs and I
d) Na and Cl
Answer:
b) Cs and F

Question 8.
Which of the following has the highest ionic character?
a) MgCl₂
b) CaCl₂
c) BaCl₂
d) BeCl₂
Answer:
c) BaCl₂

Question 9.
Which of the following compounds has the maximum covalent nature?
a) LiCl
b) NaCl
c) KCl
d) CsCl
Answer:
a) LiCl

Question 10.
Among the following the maximum covalent character is shown by the compound
a) FeCl₂
b) SnCl₂
c) AlCl₃
d) MgCl₂
Answer:
c) AlCl₃

Question 11.
Polarization is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct?
a) maximum polarization is brought about by a cation of high charge
b) Minimum polarization is brought about by a cation of low radius
c) A large cation is likely to bring about a large degree of polarization
d) A small anion is likely to undergo a large degree of polarization
Answer:
a) maximum polarization is brought about by a cation of high charge

Question 12.
Which of the following Lewis structure does not contribute in resonance?

a) I
b) II
c) III
d) IV
Answer:
b) II

Question 13.
Point out incorrect statement about resonance
a) Resonance structures should have equal energy
b) In resonance structures, the constituent atoms should be in the same position
c) In resonance structures, there should not be the same number of electron pairs
d) Resonance structures should differ only in the location of electrons around the constituent atoms
Answer:
c) In resonance structures, there should not be the same number of electron pairs

Question 14.
A diatomic molecule has dipole moment of 1.2 D. If the bond distance is 1 Å. What percentage of electronic charge exists on each atom?
a) 12 % of e
b) 19 % of e
c) 25 % of e
d) 29 % of e
Answer:
c) 25 % of e

Question 15.
The electronegativity difference between two atoms A and B is 2, then percentage of covalent character in the molecule is
a) 54 %
b) 46 %
c) 23 %
d) 72 %
Answer:
a) 54 %

Question 16.
The electronegativity of H and Cl are 2.1 and 3.0 respectively. The correct statement (s) about the nature of HCl is/are:
a) 17 % ionic
b) 83 % ionic
c) 50 % ionic
d) 100 % ionic
Answer:
a) 17 % ionic

Question 17.
For the formation of covalent bond, the difference in the value of electronegativities should be
a) Equal to or less than 1.7
b) More than 1.7
c) 1.7 or more
d) None of these
Answer:
a) Equal to or less than 1.7

Question 18.
Pick out the molecule which has zero dipole moment
a) NH₃
b) H₂O
c) BCl₃
d) SO₂
Answer:
c) BCl₃

Question 19.
The dipole moment of HBr is 1.6 × 10^-30 cm and interatomic spacing is 1 Å. The % ionic character of HBr is
a) 7
b) 10
c) 15
d) 27
Answer:
b) 10

Question 20.
In a polar molecule, the ionic charge is 4.8 × 10^-10 esu. If the inter ionic distance is 1 Å unit, then the dipole moment is
a) 41.8 debye
b) 4.18 debye
c) 4.8 debye
d) 0.48 debye
Answer:
c) 4.8 debye

Question 21.
Of the following molecules, the one, which has permanent dipole moment is:
a) SiF₄
b) BF₃
c) PF₃
d) PF₅
Answer:
c) PF₃

Question 22.
Increasing order of dipole moment is:
a) CF₄ < NH₃ < NF₃ < H₂O
b) CF₄ < NH₃ < H₂O < NF₃
c) CF₄ < NF₃ < H₂O < NH₃
d) CF₄ < NF₃ < NH₃ < H₂O
Answer:
d) CF₄ < NF₃ < NH₃ < H₂O

Question 23.
The correct sequence of dipole moments among the chlorides of methane is
a) CHCl₃ < CH₂Cl₂ > CH₃Cl > CCl₄
b) CH₂Cl₂ > CH₃Cl > CHCl₃ > CCl₄
c) CH₃Cl > CH₂Cl₂ > CHCl₃ > CCl₄
d) CH₂Cl₂ > CHCl₃ > CH₃Cl > CCl₄
Answer:
c) CH₃Cl > CH₂Cl₂ > CHCl₃ > CCl₄

Question 24.
Which of the following has been arranged in order of decreasing dipole moment?
a) CH₃Cl > CH₃F > CH₃Br > CH₃I
b) CH₃F > CH₃Cl > CH₃Br > CH₃I
c) CH₃Cl > CH₃Br > CH₃I > CH₃F
d) CH₃F > CH₃Cl > CH₃I > CH₃Br
Answer:
a) CH₃Cl > CH₃F > CH₃Br > CH₃I

Question 25.
Which of the following has the least dipole moment?
a) NF₃
b) CO₂
c) SO₂
d) NH₃
Answer:
b) CO₂

Question 26.
Among the following compounds, the one that is polar and has central atom with sp² hybridization is
a) H₂CO₃
b) SiF₄
c) BF₃
d) HClO₂
Answer:
a) H₂CO₃

Question 27.
Which of the following will provide the most efficient overlap?
a) s – s
b) s – p
c) sp² – sp²
d) sp – sp
Answer:
d) sp – sp

Question 28.
The number and type of bonds between two carbon atoms in CaC₂ are:
a) one sigma (σ) and one pi (π) bonds
b) one sigma (σ) and two pi (π) bonds
c) one sigma (σ) and one half pi (π) bonds
d) one sigma (σ) bond
Answer:
b) one sigma (σ) and two pi (π) bonds

Question 29.
Which is not true according to VBT?
a) A covalent bond is formed by the overlapping of orbitals with unpaired electrons of opposite spins
b) A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins
c) The greater the extent of overlapping the stronger is the bond
d) Overlapping takes place only in the direction of maximum electron density of the orbital
Answer:
b) A covalent bond is formed by the overlapping of orbitals with unpaired electrons of same spins

Question 30.
The hybridization of carbon atoms in C – C single bond of H – C ≡ C = CH = CH₂ is
a) sp³ – sp³
b) sp² – sp
c) sp – sp²
d) sp³ – sp
Answer:
b) sp² – sp

Question 31.
The bond in the formation of fluorine molecule will be
a) Due to s – s overlapping
b) Due to s – p overlapping
c) Due to p – p overlapping
d) Due to hybridization
Answer:
c) Due to p – p overlapping

Question 32.
Which of the following overlaps gives a a bond with x as internuclear axis?
a) P_z and P_z
b) s and p_z
c) s and P_x
d) d_{x² – y²} and d_{x² – y²}
Answer:
c) s and P_x

Question 33.
The strength of bonds by overlapping of atomic orbitals is in the order
a) s – s > s – p > p – p
b) s – s < p – p < s – p
c) s – p < s – s < p – p
d) p – p < s – s < s – p
Answer:
a) s – s > s – p > p – p

Question 34.
Which cannot be explained by VBT?
a) Overlapping
b) Bond formation
c) Paramagnetic nature of oxygen
d) Shapes of molecules
Answer:
c) Paramagnetic nature of oxygen

Question 35.
The structure of IF₇ is
a) square pyramidal
b) trigonal bipyramidal
c) octahedral
d) pentagonal bipyramidal
Answer:
d) pentagonal bipyramidal

Question 36.
The structure of XeOF₄ is
a) tetrahedral
b) square pyramidal
c) square planner
d) octahedral
Answer:
b) square pyramidal

Question 37.
The molecule that has linear structure is :
a) CO₂
b) NO₂
c) SO₂
d) SiO₂
Answer:
a) CO₂

Question 38.
The molecule which has pyramidal shape is:
a) PCl₅
b) SO₃
c) CO₃^2-
d) NO₃^–
Answer:
a) PCl₅

Question 39.
The type of hybrid orbitals used by the chlorine atom in ClO₂^–
a) sp³
b) sp²
c) sp
d) none of these
Answer:
a) sp³

Question 40.
Which one of the following molecule is planar?
a) NF₃
b) NCl₃
c) PH₃
d) BF₃
Answer:
d) BF₃

Question 41.
Which one of the following compounds has sp2 hybridization?
a) CO₂
b) SO₂
c) NO₂⁺
d) CO
Answer:
b) SO₂

Question 42.
Which of the following molecules has trigonal planar geometry?
a) IF₃
b) PCl₃
c) NH₃
d) BF₃
Answer:
d) BF₃

Question 43.
The percentage s – character of the hybrid orbitals in methane, ethane, and ethyne are respectively
a) 25, 33, 50
b) 25, 50, 75
c) 50, 75, 100
d) 10, 20, 40
Answer:
a) 25, 33, 50

Question 44.
Number of lone pair (s) in XeOF₄ is / are
a) 0
b) 1
c) 2
d) 3
Answer:
b) 1

Question 45.
Which of the following molecule contains one pair of non – bonding electrons?
a) CH₄
b) NH₃
c) H₂O
d) HF
Answer:
b) NH₃

Question 46.
If XeF₂, XeF₄ and XeF₆, the number of lone pair of electrons on Xe are respectively.
a) 2, 3, 1
b) 1, 2, 3
c) 4, 1, 2
d) 3, 2, 1
Answer:
d) 3, 2, 1

Question 47.
The shape of XeO₂ F₂ molecule is
a) Trigonal bipyramidal
b) square planar
c) tetrahedral
d) see – saw
Answer:
d) see – saw

Question 48.
According to MO theory,
a) O₂⁺ is paramagnetic and bond order is greater than O₂
b) O₂⁺ is paramagnetic and bond order is less than O₂
c) O₂⁺ is diamagnetic and bond order is less than O₂
d) O₂⁺ is diamagnetic and bond order is more than O₂
Answer:
a) O₂⁺ is paramagnetic and bond order is greater than O₂

Question 49.
Bond order of O₂, O₂⁺, O₂^– and O₂^2- is in order
a) O₂^– < O₂^2-  < O₂ < O₂⁺
b) O₂² < O₂^– < O_2, < O₂⁺
c) O₂⁺ < O₂ < O₂^– < O₂^2-
d) O_2, < O₂⁺ < O₂^– < O₂^2-
Answer:
b) O₂² < O₂^– < O_2, < O₂⁺

Question 50.
Which of the following pairs have Identical values of bond order?
a) N₂⁺ and O₂⁺
b) F₂ and Ne₂
c) O₂ and B₂
d) C₂ and N₂
Answer:
a) N₂⁺ and O₂⁺

Question 51.
Which of the following is paramagnetic?
a) O₂^–
b) CN^–
c) CO
d) NO⁺
Answer:
a) O₂^–

Question 52.
Which of the following compounds is paramagnetic?
a) CO
b) NO
c) O₂^2-
d) O₃
Answer:
b) NO

Question 53.
The number of antibonding electron pairs O₂^2- molecular ion on the basis of molecular orbital theory is
a) 4
b) 3
c) 2
d) 5
Answer:
a) 4

Question 54.
The bond length of the species O₂, O₂⁺ and O₂^– if are in the order of
a) O₂⁺ > O₂ > O
b) O₂⁺ > O₂^– > O₂
c) O₂ > O₂⁺ > O₂^–
d) O₂^– > O₂ > O₂⁺
Answer:
a) O₂⁺ > O₂ > O

Question 55.
Which of the following is a zero overlap which leads to non – bonding?
a)
b)
c)
d) All
Answer:
a)

Question 56.
Identify the least stable ion amongst the following:
a) Li^–
b) Be^–
c) B^–
d) C^–
Answer:
b) Be^–

Question 57.
Which of the following molecular species has unpaired electrons(s)?
a) N₂
b) F₂
c) O₂^–
d) O₂^2-
Answer:
c) O₂^–

Question 58.
Among the following species which has minimum bond length?
a) B₂
b) C₂
c) F₂
d) O₂^–
Answer:
c) F₂

Question 59.
The correct order of bond strength is :
a) O₂^– < O₂ < O₂⁺ < O₂^2-
b) O₂^2- < O₂^– < O₂ < O₂⁺
c) O₂^– < O₂^2- < O₂ < O₂⁺
d) O₂⁺ < O₂ < O₂^– < O₂^2-
Answer:
b) O₂^2- < O₂^– < O₂ < O₂⁺

Question 60.
The species having bond order different from that in CO is
a) NO^–
b) NO⁺
c) CN^–
d) N₂
Answer:
a) NO^–

Question 61.
Which one of the following species is diamagnetic in nature?
a) He₂⁺
b) H₂
c) H₂⁺
d) H₂^–
Answer:
b) H₂

Question 62.
Which of the following species exhibits the diamagnetic behavior?
a) O₂^2-
b) O₂⁺
c) O₂
d) NO
Answer:
a) O₂^2-

Question 63.
Which one of the following pairs of species have the same bond order?
a) CN^– and NO⁺
b) CN^– and CN⁺
c) O₂^– and CN^–
d) NO⁺ and CN⁺
Answer:
b) CN^– and CN⁺

Question 64.
Which one is the the electron deficient compound?
a) ICl
b) NH₃
c) BCl₃
d) PCl₃
Answer:
c) BCl₃

Question 65.
The number of electrons shared by each outermost shell of N₂ is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 66.
Which of the following has zero dipole moment?
a) CH₂Cl₂
b) CH₄
c) NH₃
d) PH₃
Answer:
b) CH₄

Question 67.
Which of the following statement is not correct?
a) Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals
b) sp² hybrid orbitals are formed from two p atomic orbitals and one s atomic orbital.
c) d²sp³ hybrid orbitals are directed towards the corners of a regular octahedron
d) dsp³ hybrid orbitals are all at 90° to one another
Answer:
d) dsp³ hybrid orbitals are all at 90° to one another

Question 68.
Shape of XeF₄ molecule is
a) linear
b) pyramidal
c) tetrahedral
d) square planar
Answer:
d) square planar

Question 69.
Which of the following compounds, the one having a linear structure is
a) NH₂
b) CH₄
c) C₂H₂
d) H₂O
Answer:
c) C₂H₂

Question 70.
The isoelectronic pair is
a) Cl₂, ICl₂^–
b) ICl₂^–, ClO₂
c) IF₂⁺, I₃^–
d) ClO₂^–, ClF₂⁺
Answer:
d) ClO₂^–, ClF₂⁺

Question 71.
The bond order is maximum in
a) O₂
b) O₂^–
c) O₂⁺
d) O₂^2-
Answer:
c) O₂⁺

Question 72.
Which of the following does not exist on the basis of molecular orbital theory?
a) H₂⁺
b) He₂⁺
c) He₂
d) Li₂
Answer:
c) He₂

Question 73.
Which of the following species have maximum number of unpaired electrons?
a) O₂
b) O₂⁺
c) O₂^–
d) O₂^2-
Answer:
a) O₂

Question 74.
In a polar molecule, the ionic charge is 4.8 × 10^-10 esu. If the inter ionic distance is 1 Å unit, then the dipole moment is
a) 41.8 debye
b) 4.18 debye
c) 4.8 debye
d) 0.48 debye
Answer:
c) 4.8 debye

Question 75.
The molecule of CO₂ has 180° bond angle. It can be explained on the basis of
a) sp³ hybridisation
b) sp² hybridisation
c) sp hybridisation
d) d²sp³ hybridisation
Answer:
c) sp hybridisation

Question 76.
Which of the following have both polar and non – polar bonds?
a) C₂H₆
b) NH₄Cl
c) HCl
d) AlCl₃
Answer:
b) NH₄Cl

Question 77.
Blue vitriol has
a) Ionic bond
b) Coordinate bond
c) Hydrogen bond
d) All the above
Answer:
d) All the above

Question 78.
The number of ionic, covalent, and coordinate bond NH₄Cl are respectively
a) 1, 3 and 1
b) 1, 3 and 2
c) 1, 2 and 3
d) 1, 1 and 3
Answer:
b) 1, 3 and 2

Question 79.
Which of the following does not contain a coordinate bond?
a) H₃O⁺
b) BF₄^–
c) HF₂^–
d) NH₄⁺
Answer:
c) HF₂^–

Question 80.
Maximum covalent character is associated with the compound
a) NaI
b) MgI₂
c) AlCl₃
d) AlI₃
Answer:
d) AlI₃

Question 81.
Amongst LiCl, RbCl, BeCl_3, and MgCl₂ the compounds with the greatest and the least ionic character, respectively, are
a) LiCl and RbCl
b) RbCl and BeCl₂
c) RbCl and MgCl₂
d) MgCl₂ and BeCl₂
Answer:
b) RbCl and BeCl₂

Question 82.
LiF is least soluble among the fluorides of alkali metals, because
a) smaller size Li⁺ impart significant covalent character in LiF
b) the hydration energies of Li⁺ and F^– are quite higher
c) lattice energy of LiF is quite higher due to the smaller size of Li⁺ and F^–
d) LiF has strong polymeric network in solid
Answer:
c) lattice energy of LiF is quite higher due to the smaller size of Li⁺ and F^–

Question 83.
The molecule which has zero dipole moment is
a) CH₂Cl₂
b) BF₃
c) NF₃
d) ClO₂
Answer:
b) BF₃

Question 84.
Carbon tetrachloride has no net dipole moment because of
a) Its planar structure
b) Its regular tetrahedral structure
c) similar sizes of carbon and chlorine atoms
d) Similar electron affinities of carbon and chlorine
Answer:
b) Its regular tetrahedral structure

Question 85.
Of the following compounds, which will have a zero dipole moment?
a) 1, 1 dichloroethylene
b) cis – 1,2 dichloroethylene
c) trans – 1,2 dichloroethylene
d) none of these
Answer:
c) trans – 1,2 dichloroethylene

Question 86.
Which contains both polar and non – polar bonds?
a) NH ₄Cl
b) HCN
c) H₂O₂
d) CH₄
Answer:
c) H₂O₂

Question 87.
In which of the following species, is the underlined carbon has sp3 hybridisation:
a) CH₃COOH
b) CH₃CH₂OH
c) CH₃COCH₃
d) CH₂ = CH – CH₃
Answer:
b) CH₃CH₂OH

Question 88.
Ration of a and bonds is maximum in
a) naphthalene
b) tetracyano methane
c) enolic form of urea
d) equal
Answer:
c) enolic form of urea

Question 89.
HCN and HNC molecules have equal number of
a) lone pair of σ bonds
b) σ bonds and π bonds
c) π bonds and lone pairs
d) lone pairs, σ bonds, and π bonds
Answer:
d) lone pairs, σ bonds, and π bonds

Question 90.
Allyl cyanide has
a) 9 sigma bonds and 4 pi bonds
b) 9 sigma bonds, 3 pi bonds, and 1 lone pair
c) 8 sigma bonds and 5 pi bonds
d) 8 sigma bonds, 3 pi bonds
Answer:
b) 9 sigma bonds, 3 pi bonds, and 1 lone pair

Question 91.
Effective overlapping will be shown by:
a) \(\oplus \Theta+\oplus \Theta\)

b) \(\left(\frac{\oplus}{\Theta}\right)+\left(\frac{\Theta}{\oplus}\right)\)

c) \(\oplus \Theta+\Theta \oplus\)
d) All the above
Answer:
c) \(\oplus \Theta+\Theta \oplus\)

Question 92.
In which of the following pairs, the two species are not Isostructural?
a) CO₃^2- and NO₃^–
b) PCl₄⁺; and SiCl₄
c) PF₅ and BrF₅
d) AlF₆^3- and SF₆
Answer:
c) PF₅ and BrF₅

Question 93.
The hybridization of orbitals of N atom in NO₃^–, NO₃⁺ and NH₄⁺ are respectively.
a) sp, sp², sp³
b) sp², sp, sp³
c) sp, sp³, sp²
d) sp², sp³, sp
Answer:
b) sp², sp, sp³

Question 94.
On hybridization of one s and one p -orbital we get:
a) two mutually perpendicular orbitals
b) two orbitals at 180°
c) four orbitals directed tetrahedrally
d) three orbitais in a plane
Answer:
b) two orbitals at 180°

Question 95.
A molecule XY₂ contains two σ bonds, two π bonds, and one lone pair of electrons in the valence shell of X. The arrangement of lone pair, as well as bond pairs, is
a) Square pyramidal
b) Linear
c) Trigonal planar
d) Unpredictable
Answer:
c) Trigonal planar

Question 96.
The maximum number of 90° angles between bond pair of electron is observed in :
a) sp³d² hybridisation
b) sp³d hybridisation
c) dsp² hybridisation
d) dsp³ hybridisation
Answer:
a) sp³ d² hybridisation

Question 97.
Which of the following has a 3 centred 2 electron bond?
a) BF₃
b) NH₃
c) CO₂
d) B₂H₆
Answer:
d) B₂H₆

Question 98.
Which has regular tetrahedral geometry?
a) SF₄
b) BF₄^–
c) XeF₄
d) [Ni(CN)₄]^2-
Answer:
b) BF₄^–

Question 99.
Which of the following statement is incorrect?
a) During N₂⁺ formation, one electron is removed from the bonding molecular orbital of N₂.
b) During O₂⁺ formation, one electron is removed from the anti-bonding molecular orbital of O₂.
c) During O₂^– formation, one electron is added to the bonding molecular orbital of O₂
d) During CN^– formation, one electron is added to the bonding molecular orbital of CN
Answer:
c) During O₂^– formation, one electron is added to the bonding molecular orbital of O₂

Question 100.
Which concept best explains that O – nitrophenol is more volatile than p – nitrophenol?
a) Resonance
b) Hyper conjugation
c) Hydrogen bonding
d) Steric hindrance
Answer:
c) Hydrogen bonding

II. Very short question and answers (2 Marks):

Question 1.
State Octet rule.
Answer:
“The atoms transfer or share electrons so that all atoms involved in chemical bonding obtain 8 electrons in their outer shell (valence shell)”.

Question 2.
What is an electrovalent bond?
Answer:
The complete transfer of electron leads to the formation of a cation and an anion. Both these ions are held together by the electrostatic attractive force which is known as ionic bond.

Question 3.
What is covalent bond?
Answer:
This type of mutual sharing of one or more pairs of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond

Question 4.
What is Co-ordinate bond?
Answer:
One of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation, and these electrons are shared by both the combining atoms. These type of bonds are called coordinate covalent bond or coordinate bond.

Question 5.
Draw the lewis dot structure of the following.
(i) SO₃
(ii) NH₃
(iii) N₂O₅
Answer:

Question 6.
What is bond length?
Answer:
The distance between the nuclei of the two covalently bonded atoms is called bond length.
Example:
Carbon-carbon single bond length (1.54 Å).

Question 7.
What is bond angle?
Answer:
Covalent bonds are directional in nature and are oriented in specific directions in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.
Example :
Molecule = CH₄;
Atoms defining the angle = H-C-H
Bond angle is 109° 28′

Question 8.
What is resonance?
Answer:
When we write Lewis structures for a molecule, more than one valid Lewis structures are possible in certain cases.
They only differ in the position of bonding and lone pair of electrons. Such structures are called structures (canonical structures) and this phenomenon is called resonance.

Question 9.
Draw the resonance structure of CO₃^2- ion?
Answer:

III. Short question and answers (3 Marks):

Question 1.
What is dipole moment of a covalent bond in a polar molecule?
Answer:
Dipole moment of a covalent bond in a polar molecule is defined as the product of the magnitude of the charge present on either of the two atoms and the distance by which the two atoms are separated in the molecule. It is a vector quantity and works in the direction of the line joining the positively charged end to the negatively charged end.
The unit of dipole moment is Debye.
Thus 1D = 1 × 10^-18 e.s.u.cm.

Question 2.
Distinguish between σ – molecular orbital & π – molecular orbital.
Answer:

σ molecular orbital	π molecular orbital
1. It is formed by head on overlapping of atomic orbitals.	It is formed by the sidewise overlapping of atomic orbitals.
2. The overlapping of atomic orbital is maximum.	The overlapping of atomic orbital is less.
3. The orbital is symmetrical to rotation about the intemuclear axis.	The orbital is not symmetrical to rotation about the internuclear axis.
4. The resulting covalent bond is strong.	The resulting covalent bond is weak.

Question 3.
Explain formation of H₂ molecule by MO theory.
Answer:

Molecular orbital diagram of hydrogen molecule (H₂)
Electronic configuration of H atom 1s¹
Electronic configuration of H₂ molecule: σ_1s²
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-0}{2}\) = 1
Molecule has no impaired electrons hence it is dimagnetic.

Question 4.
Explain the formation of Li₂ molecule by MOT.
Answer:

Molecular orbital diagram of Lithium molecule (Li₂)
Electronic configuration of Li atom 1s² 2s¹
Electronic configuration of Li₂ molecule: σ_1s² σ_1s^*2 σ_2s²
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{4-2}{2}\) = 1
Molecule has no impaired electrons hence it is dimagnetic.

Question 5.
Explain the formation of B₂ molecule by MOT.
Answer:

Molecular orbital diagram of Boron molecule (B₂)
Electronic configuration of B atom 1s² 2s² 2P¹
Electronic configuration of B₂ molecule: σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py¹ π_2pz¹
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{6-4}{2}\) = 1
Molecule has two unpaired electrons hence it is paramagnetic.

Question 6.
Explain the formation of C₂ molecule by MOT.
Answer:

Molecular orbital diagram of Carbon molecule (B₂)
Electronic configuration of C atom 1s² 2s² 2p¹
Electronic configuration of C₂ molecule: σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py¹ π_2pz¹
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{8-4}{2}\) = 2
Molecule has two unpaired electrons hence it is diamagnetic.

Question 7.
How is σ and π – bond formed?
Answer:
When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond. This overlap is also called ‘Tread – on overlap” or “axial overlap”. Overlap involves an orbital (s- s and s – p overlaps) will always result in a sigma bond as the s orbital is spherical. Overlap between two p orbitals along the molecular axis will also result in sigma bond formation. When we consider x-axis as molecular axis, the p_x – p_x overlap will result in σ – bond.

When two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi bond. When we consider x – axis as molecular axis, p_y – p_y and p_z – p_z overlap will result in the formation of a π – bond.

Question 8.
Explain Salient features of VB theory?
Answer:
Salient features :

1. When half filled orbitais of two atoms overlap, a covalent bond will be formed between them.
2. The resultant overlapping orbital is occupied by the two electrons with opposite spins. For example, when H₂ is formed, the two is electrons of two hydrogen atoms get paired up and occupy the overlapped orbital.
3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.
4. Each atomic orbital has a specific direction (except s-orbital which is spherical) and hence orbital overlap takes place in the direction that maximizes overlap.
5. Let us explain the covalent bond formation in hydrogen, fluorine, and hydrogen fluoride using VB theory.

Question 9.
How is HF & F₂ molecule formed by using VB theory?
Answer:
Formation of HF molecule:
1. Electronic configuration of hydrogen atom is 1s¹
2. Valence shell electronic configuration of fluorine atom: 2s², 2p_x², 2p_y², 2p_z¹
3. When half filled 1s orbital of hydrogen linearly overlaps with a half filled 2p_z orbital of fluorine, a σ – covalent bond is formed between hydrogen and fluorine.

Formation of fluorine molecule (F₂):
1. Valence shell electronic configuration of fluorine atom: 2s², 2p_x², 2P_y² 2p_z¹
2. When the half filled p_z orbitals of two fluorine overlaps along the z-axis, a σ – covalent bond is formed between them.

IV. Long question and answers (5 Marks):

Question 1.
Explain the Postulates of VSEPR theory.
Answer:

1. The shape of the molecule depends on the number of valence shell electron pair around the central atom.
2. There are two types of electron pairs namely bond pairs and lone pairs. The bond pair of electrons are those shared between two atoms, while the lone pairs are the valence electron pairs that are not involved in bonding.
3. Each pair of valence electrons around the central atom repels each other and hence, they are located as far away as possible in three dimensional space to minimize the repulsion between them.
4. The repulsive interaction between the different types of electron pairs is in the following order.
   1p – 1p > 1p – bp > bp – bp
   1p – lone pair; bp – bond pair
5. The lone pair of electrons are localised only on the central atom and interacts with only one nucleus whereas the bond pairs are shared between two atoms and they interact with two nuclei.
6. Because of this lone pairs occupy more space and have greater repulsive power than the bond pairs in a molecule.

Question 2.
Explain the qualitative treatment of VB theory for the formation of H₂ molecule.
Answer:
A simple qualitative treatment of VB theory for the formation of hydrogen molecule is discussed below.
Consider a situation where in two hydrogen atoms (H_a and H_b) are separated by infinite distance. At this stage there is no interaction between these two atoms and the potential energy of this system is arbitrarily taken as zero. As these two atoms approach each other, in addition to the electrostatic attractive force between the nucleus and its own electron (purple arrows), the following new forces begins to operate.

VB theory for the formation of hydrogen molecule

The new attractive forces (green arrows) arise between
(i) nucleus of H_a and valence electron of H_b
(ii) nucleus of H_b and the valence electron of H_a.

The new repulsive forces (red arrows) arise between
(i) nucleus of H_a and H_b
(ii) valence electrons of H_a and H_b

The attractive forces tend to bring H_a and H_b together whereas the repulsive forces tends to push them apart. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than the repulsive forces and the potential energy decreases. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.

At this stage, there is a maximum overlap between the atomic orbitals of H_a and H_b, and the atoms H_a and H_b are now said to be bonded together by a covalent bond. The internuclear distance at this stage gives the H – H bond length and is equal to 74 pm.

The liberated energy is 436 kJ mol^-1 and is known as bond energy. Since the energy is released during the bond formation, the resultant molecule is more stable. If the distance between the two atoms is decreased further, the repulsive forces dominate the attractive forces and the potential energy of the system sharply increases.

Question 3.
Explain hybridization and geometry of BeCl₂ molecule.
Answer:
Let us consider the bond formation in beryllium chloride. The valence shell electronic configuration of beryllium in the ground state is shown in the figure.

In BeCl₂ both the Be-Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explain this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital and the electronic configuration at the excited state is shown.

Now, the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction as shown in the figure.

Overlap with orbital of chlorine:
Each of the sp hybridized orbitals linearly overlap with p_z orbital of the chlorine to form a covalent bond between Be and Cl as shown in the Figure

sp hybridization of BeCl₂

Question 4.
Distinguish between σ – bond and π – bond.
Answer:

Sigma (σ) Bond	Pi (π) Bond
1. Sigma (σ) bond is formed by axial overlap of atomic orbitals.	Pi (π) bond is formed by the sidewise overlap of atomic orbitals.
2. This bond can be formed by overlap of s-s, s-p or p-p orbitals.	It involves of overlap of p-p orbitals only.
3. The bond is strong because, overlapping can take place to a large extent.	The bond is weak because the overlapping occurs to a small extent.
4. The electron cloud formed by axial overlap is symmetrical about the inter nuclear axis and consists of single charged cloud.	The electron cloud of Pi bond is discontinuous and consists of two changed clouds above and below the plane of atoms.
5. There can be a free rotation of atoms about the σ bond.	Free rotation of atoms around π bond is not possible because it involves breaking of π -bond.
6. The bond may be present between the two atoms either alone or along with a π -bond.	The bond is always present between the two  atoms along with the sigma (σ) bond.
7. The shape of molecule is determined by the sigma framework around the central atom.	The π bonds do not contribute to the shape.

Question 5.
Explain hybridisation & geometry of methane molecule?
Answer:
sp³ hybridisation can be explained by considering methane as an example. In methane molecule the central carbon atom bound to four hydrogen atoms. The ground state valence shell electronic configuration of carbon is [He]2s² 2p_x¹ 2p_y¹ 2P_z⁰.

In order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in the 2s orbital of
carbon is promoted to its 2p_z orbital in the excite state. The one 2s orbital and the three 2p orbitals of carbon mixes to give four equivalent sp³ hybridised orbitals. The angle between any two sp³ hybridised orbitals is 109° 28′.

Overlap with 1s orbitals of hydrogen:
The 1s orbitals of the four hydrogen atoms overlap linearly with the four sp³ hybridised orbitals of carbon to form four C-H σ-bonds in the methane molecule, as shown below.

sp3 hybridization of CH₄

Question 6.
Explain hybridization & geometry of PCl₅ molecule.
Answer:
In the molecules such as PCl₅, the central atom phosphorus is covalently bound to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes sp³d hybridization which involves its one 3s orbital, three 3p orbitals and one vacant 3d orbital (d_z²). The ground state electronic configuration of phosphorous is [Ne] 3s² 3p_x² 3p_y¹ 3p_z¹ as shown below.

One of the paired electrons in the 3s orbital of phosphorous is promoted to one of its vacant 3d orbital (d_z²) in the excite state. One 3s orbital, three 3p orbitals, and one 3d_z². orbital of phosphorus atom mixes to give five equivalent sp³d hybridised orbitals. The orbitals geometry of sp³d hyrbridised orbitals is trigonal bi – pyramidal.

Overlap with 3p_z orbitals of chlorine :
The 3p_z orbitals of the five chlorine atoms linearly overlap along the axis with the five sp³d hybridized orbitals of phosphorous to form the five P – Cl σ – bonds, as shown below.

Question 7.
Explain hybridization & geometry of SF₆ molecule.
Answer:
In sulphur hexafluoride (SF₆) the central atom sulphur extend its octet to undergo sp³d² hybridization to generate six sp³d² hybridized orbitals which accounts for six equivalent S – F bonds. The ground state electronic configuration of sulphur is [Ne] 3s² 3p_x², 3p_y², 3p_z¹

One electron each from 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals (d_z² and d_{x² – y²}) in the excite state. A total of six valence orbitals from sulphur (one 3s orbital, three 3p orbitals, and two 3d orbitals) mixes to give six equivalent sp³d² hybridized orbitals. The orbital geometry is octahedral as shown in the figure.

Overlap with 2p_z orbitals of fluorine:
The six sp³d² hybridized orbitals of sulphur overlaps linearly with 2p_z orbitals of six fluorine atoms to form the six S – F bonds in the sulphur hexafluoride molecule.

Question 8.
Write the postulates of molecular orbital theory.
Answer:

1. When atoms combines to form molecules, their individual atomic orbitals lose their identity and forms new orbitals called molecule orbitals.
2. The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
3. The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy than the corresponding atomic orbital, while the remaining molecular orbitals will have higher energy.
4. The molecular orbital with lower energy is called bonding molecular orbital and the one with higher energy is called anti-bonding molecular orbital. The bonding molecular orbitals are represented as σ (Sigma), π (pi), δ (delta) and the corresponding antibonding orbitals are denoted as σ*, π* and δ*.
5. The electrons in a molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follows Aufbau’s principle, Pauli’s exclusion principle, and Hund’s rule as in the case of filling of electrons in atomic orbitals.
6. Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using the following equation
7. Bond order = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
   Where, N_b = Total number of electrons present in the bonding molecular orbitals.
   N_a = Total number of electrons present in the antibonding molecular orbitals.
   A bond order of zero value indicates that the molecule doesn’t exist.

Question 9.
Distinguish between molecular orbital & anti bonding molecular orbitals.
Answer:

Bonding molecular orbital	Anti-bonding molecular orbital
1. A bonding molecular orbital is formed when the electron waves of combining atoms are in phase, i.e the lobes of atomic orbitals have same sign.	An anti-bonding molecular orbital is formed when the electron waves of the combining atoms are not in phase ie the lobes of atomic orbital have opposite sign.
2. For bonding molecular orbital wave function are summed up.	For anti-bonding molecular orbital, the wave functions are subtracted.
3. The electron density is centered between the nuclei of the combining atoms.	The probability of finding the electron between the nuclei of the combining atom is negligible.
4. The energy of bonding molecular orbital is less than that of the atomic orbitals of the combining atoms.	The energy of an antibonding molecular orbital is higher than the atomic orbitals of combining atoms.
5. Electron present in bonding molecular orbital lead to attraction between the atoms and stabilize the molecule.	Electrons present in anti-bonding molecular orbitals lead to repulsion between the atoms and destabilize the molecule.

Question 10.
Explain the formation of NO by MOT?
Answer:

Molecular orbital diagram of nitrogen molecule (NO)

Electronic configuration of N atom 1s² 2s² 2p³
Electronic configuration of NO molecule
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² σ_2px² π_2py^*1
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{10-5}{2}\) = 2.5
Molecule has one unpaired electrons hence it is paramagnetic.