Category: Class 12 Chemistry Objective Questions

Chemistry MCQs for IIT JEE Exam on “Biomolecules – Proteins – 2”.

1. Proteins are formed primarily from ______ bonds.
a) glycosidic
b) peptide
c) phosphodiester
d) disulphide
Answer: b
Clarification: When many alpha amino acid units arranged themselves in a chain (or any other suitable structure), a protein is formed. These units are linked together by peptide bonds between NH₂ groups and COOH groups.

2. Identify the correct statement.
a) Peptide bond is formed by the loss of water molecule
b) A protein is made of only one type of amino acid
c) Dipeptides consists of different amino acids
d) Glycylalanine is a tripeptide
Answer: a
Clarification: The reaction between two molecules of same of different amino acids proceeds through the linking of COOH group of one and the NH₂ group of the other along with the loss of H₂O and formation of a peptide bond. The product is called a dipeptide.

3. How many peptide linkages does a hexapeptide have?
a) 4
b) 5
c) 6
d) 7
Answer: b
Clarification: A hexapeptide is a compound formed by the combination of six same or different amino acids with the help of the amino and carboxyl groups. The six amino acids are connected by five peptide bonds.

4. Proteins are _______
a) dipeptides
b) tripeptides
c) tetrapeptides
d) polypeptides
Answer: d
Clarification: When the number of amino acids in a peptide is more than ten, it is a polypeptide. But when a polypeptide has more than a 100 amino acid residues, with molecular mass higher than 10000u, it is called a protein. However, this is not true in all cases (like insulin).

5. Alanylglycyl phenylalanine is an example of a ______
a) dipeptide
b) tripeptide
c) tetrapeptide
d) polypeptide
Answer: b
Clarification: It is a tripeptide made from alanine, glycine and phenylalanine. It is abbreviated as Ala-Gly-Phe. The carboxyl group of alanine and the amino group of glycine combine to form one peptide bond. The second peptide linkage is formed between COOH of glycine and NH₂ of phenylalanine.

6. Which of the following bonds in not found in fibrous proteins?
a) Phosphodiester
b) Peptide
c) Hydrogen bonds
d) Disulphide
Answer: a
Clarification: Fibrous proteins are linear polypeptide chains that lie parallel to each other. Peptide binds are prevalent in the individual chains, whereas the threads are held together by hydrogen and disulphide bonds, to form a fibre-like structure.

7. Which of the following is not a fibrous protein?
a) Keratin
b) Myosin
c) Collagen
d) Albumin
Answer: d
Clarification: Keratin is a fibrous protein found in skin, nails, hair and wool. Myosin is present in muscles and collagen in tendons. Albumin is a globular protein.

8. The sequence in which amino acids are arranged in a protein is called ______ structure.
a) primary
b) secondary
c) fibrous
d) sheet
Answer: a
Clarification: Proteins have one or more polypeptide chains, in which each chain consists of a specific sequence of amino acids linked with each other. This is called the primary structure and is the most basic level. Any change in primary structure creates a different protein.

9. Which type of bonds govern the secondary structure of proteins?
a) Covalent
b) Hydrogen
c) Electrostatic
d) Peptide
Answer: b
Clarification: The secondary structure refers to the shape in which the polypeptide chain exists. There are two possible structures which arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between CO and NH groups of peptide bond.

10. Which of the following is soluble in water?
a) Insulin
b) Elastin
c) Fibroin
d) Collagen
Answer: a
Clarification: Fibroin (silk), collagen (tendons) and elastin (skin) are fibrous proteins which are insoluble in water. Insulin is an example of a globular protein which is water soluble.

11. The structure in which all peptide chains are stretched out to full extension and laid side by side through intermolecular hydrogen bonds is called ______
a) α-helix
b) β-pleated sheet
c) tertiary structure
d) quaternary structure
Answer: b
Clarification: β-pleated sheet is one of the secondary structures of proteins formed due to intermolecular hydrogen bonding. The structure resembles the pleated folds of drapery and hence the name.

12. Fibrous and globular proteins are classified on the basis of ______ structure.
a) primary
b) secondary
c) tertiary
d) quaternary

Answer: c
Clarification: Tertiary structure represents the overall folding of the polypeptide chains or the further folding of secondary structures. It gives rise to two major molecular shapes, i.e., fibrous and globular.

13. A protein ‘X’ was found in a biological system with a unique 3-D structure and biological activity. ‘X’ is known as _______
a) tertiary structure
b) quaternary structure
c) native protein
d) globular protein
Answer: c
Clarification: Native state of a protein is its most energetically stable state. When a native protein is subjected to physical change, the hydrogen bonds are disturbed, and the globules unfold.

14. The primary structure of protein is unaffected by denaturation.
a) True
b) False
Answer: a
Clarification: Denaturation is the process of altering the physical and biological properties of proteins without affecting chemical composition. It is caused by subjecting the protein to physical changes like temperature, pH, etc. During this, the secondary and tertiary structures are destroyed but the primary structure remains intact.

15. Boiling an egg is an example of reversible denaturation.
a) True
b) False
Answer: b
Clarification: The egg white gets coagulated on boiling. The globular proteins in egg white (albumin) change to a rubber like insoluble mass. This is irreversible denaturation as the protein cannot return to its original state.

Chemistry MCQs for IIT JEE Exam,

Chemistry Multiple Choice Questions on “Classification of Crystalline Solids”.

1. Which of the following consists of either atoms or molecules formed by non polar covalent bonds?
a) Non polar molecular solid
b) Metallic solid
c) Polar molecular solid
d) Ionic solid
Answer: a
Clarification: Non polar molecular solids consists of either of the atoms or molecules formed by non polar covalent bonds. Metallic solids and ionic solids consist of free mobile electrons and ions respectively. Whereas polar molecular solids are formed by polar covalent bonds.

2. Polar molecular solids are _________
a) bad conductors of electricity
b) good conductors of electricity
c) solid at room temperature
d) brittle
Answer: a
Clarification: In polar molecular solids, the molecules are formed by covalent bonds and held together by strong dipole-dipole interaction. Therefore, polar molecular solids are non-conductors of electricity.

3. The molecules in polar molecular solid are held together by __________
a) dipole-dipole interaction
b) london forces
c) ionic bond
d) metallic bond
Answer: a
Clarification: The force responsible for holding together the molecules of polar molecular solids is dipole-dipole force of attraction. Polar molecular solids are non-conductors of electricity.

4. Which of the following tend to be volatile liquids or soft solids at room temperature and pressure?
a) Non polar molecular solids
b) Metallic solids
c) Polar molecular solids
d) Hydrogen bonded molecular solids
Answer: d
Clarification: In hydrogen bonded molecular solid, the intermolecular forces are strong hydrogen bonds. Hence, it tends to be volatile liquids or soft solids at room temperature and pressure.

5. Which type of solids are formed by three-dimensional arrangement of cations and anions bound by strong electrostatic force?
a) Polar molecular solids
b) Ionic solids
c) Covalent solids
d) Metallic solids
View Answer

Answer: b
Clarification: Ionic solids are made up of three dimensional arrangement of cations and anions bound by strong electrostatic force.

6. Which of the following is non-conductor of electricity at solid state but can conduct electricity in the molten state or when dissolved in water?
a) Non polar molecular solids
b) Metallic solids
c) Ionic solids
d) Hydrogen bonded molecular solids
Answer: c
Clarification: Ionic Solids, when dissolved in water, tend to separate cations and the anions which allows the solution to conduct electricity. Also, they have high melting and boiling points.

7. Which of the following is an orderly collection of positive ions surrounded and held together by a sea of electrons?
a) Gas
b) Non-metal
c) Metal
d) Metalloids
Answer: c
Clarification: Metal is said to be an orderly collection of positive ions surrounded and held together by a sea of delocalized electrons. These delocalized electrons are mobile and are responsible for the conduction of electricity.

8. Which of the following are responsible for high electrical and thermal conductivity of a metal?
a) Ions
b) Covalent bonds
c) Free H+ ions
d) Free and mobile electrons
Answer: d
Clarification: Each atom of a metal gives one or more electrons towards the sea of mobile electrons it is surrounded by which ultimately contributes to the electrical and thermal conductivity of the metal.

9. Which of the following is not a characteristic of metals?
a) Lustre
b) Ductile
c) Malleable
d) Brittle
Answer: d
Clarification: Metals are lustrous, ductile and malleable. They cannot be brittle. Non-metals tend to be brittle in their solid state.

10. Which type of crystalline solid is also called as giant molecules?
a) Ionic solids
b) Covalent solids
c) Polar molecular solids
d) Metallic solids
Answer: b
Clarification: Covalent solids consists of a long chain of covalent bonds between the adjacent molecules throughout the crystal. Hence, they are called giant molecules. They are hard and brittle in nature.

Chemistry Multiple Choice Questions on “Abnormal Molar Masses”.

1. Only if the calculated molar mass is higher than the actual molar mass of the solute, the calculated molar mass is considered to be abnormal molar mass.
a) True
b) False
Answer: b
Clarification: When we correlate molar mass with osmotic pressure as a colligative property, we sometimes encounter a situation where the calculated molar mass of the solute is either higher or lower than the actual molar mass. This is called the abnormal molar mass. Abnormal molar mass occurs as a result of dissociation or association of molecules.

2. Which of the following statements is correct?
a) Solutes that dissociate in water have molar mass higher than the molar mass of the solute calculated theoretically
b) Solutes that associate in water have molar mass higher than the molar mass of the solute calculated theoretically
c) Solutes that dissociate in water experience a decrease in colligative properties
d) Colligative properties are independent of the number of particles of the solute in the solution
View Answer

Answer: b
Clarification: On association of solute molecules, the number of particles in a solution is reduced. Since, colligative properties are directly proportional to the number of particles in a solution, the colligative properties reduce. The molecular mass of the solute is inversely proportional to its colligative properties. Therefore, solutes that associate in water have a higher molar mass than its value calculated theoretically.

3. What is the value of the Van’t Hoff factor (i) for solutes that dissociate in water?
a) > 1
b) c) = 0
d) Not defined
Answer: a
Clarification: Van’t Hoff’s factor (i) is defined as the ratio of the observed colligative property to the calculated colligative property. Since, the abnormal mass for solutes that dissociate is lesser than its normal molar mass, the value of Van’t Hoff’s factor will always be > 1.

4. In which of the following solutions will the Van’t Hoff Factor for the solute be lesser than 1?
a) Sodium chloride in water
b) Benzoic acid in benzene
c) Acetic acid in benzene
d) Phenol in benzene
Answer: a
Clarification: The Van’t Hoff factor is less than 1 for solutes that dissociate in solution. In the given list, only sodium chloride dissociates in water whereas the remaining carboxylic acids associate in benzene. Hence, the value of the Van’t Hoff factor is lesser than 1 for a solution of sodium chloride in water.

5. The Van’t Hoff Factor for a solution of glucose in water is equal to 1.
a) True
b) False
Answer: a
Clarification: The Van’t Hoff Factor (i) is equal to the ratio between the normal molar mass of the solute and its abnormal mass. The glucose molecules neither associate nor dissociate. Hence, the abnormal mass is equal to the normal molar mass of the solute and Van’t Hoff factor is equal to 1.

6. The depression of freezing point of a solution of acetic acid in benzene is – 0.2°C. If the molality of acetic acid is 0.1 m, then find the ratio of the normal mass to the abnormal mass. (Assume K_f of acetic acid = 4.0°C m^-1)
a) 1.5
b) 0.8
c) 0.5
d) 0.2
Answer: c
Clarification: Given,
Depression in freezing point(ΔT) = -0.2 °C
Molality of the solution(m) = 0.1 m
Freezing point depression constant (K_f) = 4.0 °C m^-1
Let the Van’t Hoff factor = i
We know that, ΔT = i x K_f x m
0.2 = i x 4 x 0.1
i = 0.2/0.4 = 0.5
Therefore, the ratio between the normal mass and abnormal mass, which is equal to the Van’t Hoff factor is equal to 0.5.

7. What is the Van’t Hoff Factor for 1 mole of BaCl₂, assuming 100% dissociation?
a) 0.33
b) 1
c) 3
d) 2
View Answer

Answer: c
Clarification: Van’t Hoff Factor can also be written as the ratio between the total number of moles of particles after association / dissociation and the total number of moles of particles before association / dissociation. Since BaCl₂ completely dissociates(into one Ba²⁺ and two Cl^– ions), the total number of moles after dissociation is equal to 3. Therefore, the Van’t Hoff Factor for BaCl₂ is 3.

8. Which of the following aqueous solutions should have the least boiling point?
a) 1.0 M KOH
b) 1.0 M (NH₄)₂SO₄
c) 1.0 M K₂CO₃
d) 1.0 M K₂SO₄
Answer: a
Clarification: KOH→K⁺ + OH^–
(NH₄)₂SO₄→2NH₄⁺ + SO4^2-
K₂CO₃→2K⁺ + CO₃^2-
K₂SO₄→2K⁺ + SO₄^2-
Concentration of particles in 1.0 M KOH solution is minimum (= 2 M). Hence, it will minimum elevation in boiling point.

9. The pH of a 2 M solution of a weak monobasic acid (HA) is 4. What is the value of the Van’t Hoff factor?
a) 0.00005
b) 1.005
c) 1.0005
d) 1.00005
Answer: d
Clarification: pH = 4 means [H⁺] = 10^-4 M

HA ⇌ H+ + A–
Initial	C mol L-1	0	0
After dissociation	C – Cα	Cα	Cα

Total = C (1+α)
Thus, [H⁺] = Cα, i.e., 10^-4 = 2 x α or α = 5 x 10^-5 = 0.00005
i = 1 + α = 1 + 0.00005 = 1.00005.