Class 12 Chemistry - Half Yearly Practice Paper

📘 Class 12 Chemistry – Half Yearly Practice Paper

Time: 3 Hours | Max Marks: 70

Topics: Solutions + Electrochemistry + Chemical Kinetics + d-block + f-block + Coordination Compounds

SECTION A - Multiple Choice Questions (15 × 1 = 15 marks)

Q1. Which law explains colligative properties?

  • (a) Raoult's law
  • (b) Henry's law
  • (c) Dalton's law
  • (d) Boyle's law
Answer: (a) Raoult's law

Q2. Which of the following shows negative deviation from Raoult's law?

  • (a) Acetone + Benzene
  • (b) HCl + Water
  • (c) Ethanol + Water
  • (d) CCl₄ + Benzene
Answer: (c) Ethanol + Water

Q3. Molar conductivity of weak electrolytes ______ with dilution.

  • (a) decreases
  • (b) increases sharply
  • (c) constant
  • (d) becomes zero
Answer: (b) increases sharply

Q4. The emf of a Daniell cell depends on:

  • (a) temperature
  • (b) concentration of electrolytes
  • (c) nature of electrodes
  • (d) all of the above
Answer: (d) all of the above

Q5. The unit of rate constant for second order reaction is:

  • (a) mol L⁻¹ s⁻¹
  • (b) L mol⁻¹ s⁻¹
  • (c) s⁻¹
  • (d) dimensionless
Answer: (b) L mol⁻¹ s⁻¹

Q6. Which of the following element has highest melting point?

  • (a) Zn
  • (b) Fe
  • (c) W
  • (d) Hg
Answer: (c) W (Tungsten)

Q7. Lanthanoid contraction is due to:

  • (a) poor shielding of 4f electrons
  • (b) high nuclear charge
  • (c) both (a) and (b)
  • (d) large atomic size
Answer: (c) both (a) and (b)

Q8. [Fe(CN)₆]⁴⁻ is:

  • (a) diamagnetic, octahedral
  • (b) paramagnetic, tetrahedral
  • (c) diamagnetic, square planar
  • (d) paramagnetic, octahedral
Answer: (a) diamagnetic, octahedral

Q9. Which complex shows optical isomerism?

  • (a) [Co(NH₃)₆]³⁺
  • (b) [Co(en)₃]³⁺
  • (c) [Cu(NH₃)₄]²⁺
  • (d) [Zn(NH₃)₄]²⁺
Answer: (b) [Co(en)₃]³⁺

Q10. Which actinoid is used in nuclear power generation?

  • (a) Thorium
  • (b) Uranium
  • (c) Neptunium
  • (d) Californium
Answer: (b) Uranium

Q11. In a first-order reaction, the half-life is:

  • (a) proportional to concentration
  • (b) independent of concentration
  • (c) inversely proportional to concentration
  • (d) none of these
Answer: (b) independent of concentration

Q12. Equivalent conductance is given by:

  • (a) κ × 1000 / N
  • (b) κ × 1000 / M
  • (c) Λm / N
  • (d) None
Answer: (a) κ × 1000 / N

Q13. Which is a strong field ligand?

  • (a) H₂O
  • (b) CN⁻
  • (c) F⁻
  • (d) Cl⁻
Answer: (b) CN⁻

Q14. Which reaction follows pseudo-first order kinetics?

  • (a) Hydrolysis of ester in acid
  • (b) Radioactive decay
  • (c) Decomposition of H₂O₂
  • (d) Both (a) and (c)
Answer: (a) Hydrolysis of ester in acid

Q15. Which of the following is not a colligative property?

  • (a) Osmotic pressure
  • (b) Vapour pressure lowering
  • (c) Boiling point elevation
  • (d) Viscosity
Answer: (d) Viscosity

SECTION B - Assertion-Reason Questions (5 × 1 = 5 marks)

Q16. Assertion (A): Molar conductivity of weak electrolytes increases rapidly on dilution.
Reason (R): Degree of ionisation increases with dilution.

Answer: A and R both true, R explains A.

Q17. Assertion (A): Transition metals show variable oxidation states.
Reason (R): Because of small energy difference between (n-1)d and ns orbitals.

Answer: A and R both true, R explains A.

Q18. Assertion (A): First order reactions have constant half-life.
Reason (R): Half-life does not depend on initial concentration.

Answer: A and R both true, R explains A.

Q19. Assertion (A): [Co(NH₃)₆]³⁺ is violet in colour.
Reason (R): Colour of transition metal complexes is due to d-d transition.

Answer: A and R both true, R explains A.

Q20. Assertion (A): Actinoids show greater tendency to form complexes than lanthanoids.
Reason (R): Because of higher charge density and variable oxidation states.

Answer: A and R both true, R explains A.

SECTION C - Short Answer Questions (2 marks × 5 = 10 marks)

Q21. Define Henry's Law. Write one application.

Answer: Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. One application is in the bottling of carbonated drinks where CO₂ is dissolved under high pressure.

Q22. Write cell reaction of Daniell cell and its EMF under standard conditions.

Answer: Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu
Standard EMF: E°cell = 1.10 V

Q23. Define activation energy. Write Arrhenius equation.

Answer: Activation energy is the minimum amount of energy required for a chemical reaction to occur. Arrhenius equation: k = Ae^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

Q24. Write difference between lanthanoid and actinoid contraction.

Answer: Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of lanthanoids due to poor shielding of 4f electrons. Actinoid contraction is the similar decrease in atomic and ionic radii of actinoids due to poor shielding of 5f electrons. The actinoid contraction is greater than lanthanoid contraction due to higher nuclear charge and poorer shielding of 5f electrons.

Q25. Give IUPAC name of [Cr(NH₃)₄Cl₂]Cl and state geometry.

Answer: IUPAC name: Tetraamminedichloridochromium(III) chloride
Geometry: Octahedral

SECTION D - Numerical Problems (3 marks × 5 = 15 marks)

Q26. Calculate molality of solution containing 18 g glucose (M = 180) in 500 g water.

Answer:
Moles of glucose = 18/180 = 0.1 mol
Mass of solvent = 500 g = 0.5 kg
Molality = moles of solute / mass of solvent in kg = 0.1 / 0.5 = 0.2 m

Q27. Resistance of 0.01 M KCl is 200 Ω. Cell constant = 1.25 cm⁻¹. Find molar conductivity.

Answer:
Conductivity (κ) = Cell constant / Resistance = 1.25 / 200 = 0.00625 S cm⁻¹
Molar conductivity (Λm) = (κ × 1000) / Molarity = (0.00625 × 1000) / 0.01 = 625 S cm² mol⁻¹

Q28. A first-order reaction has k = 3 × 10⁻⁵ s⁻¹. Calculate half-life.

Answer:
For first-order reaction: t₁/₂ = 0.693 / k
t₁/₂ = 0.693 / (3 × 10⁻⁵) = 23100 seconds ≈ 6.42 hours

Q29. Calculate EMF of Zn/Zn²⁺ (0.1M) || Cu²⁺ (1M)/Cu, E° = 1.10 V.

Answer:
Using Nernst equation: E = E° - (0.059/2) log([Zn²⁺]/[Cu²⁺])
E = 1.10 - (0.0295) log(0.1/1) = 1.10 - (0.0295)(-1) = 1.10 + 0.0295 = 1.1295 V ≈ 1.13 V

Q30. If Kb of water = 0.512, calculate boiling point of 1 molal NaCl (assume complete dissociation).

Answer:
For NaCl, i = 2 (complete dissociation)
ΔTb = i × Kb × m = 2 × 0.512 × 1 = 1.024°C
Boiling point = 100 + 1.024 = 101.024°C

SECTION E - Long Answer Questions (5 marks × 5 = 25 marks)

Q31. Derive relation between relative lowering of vapour pressure and molar mass of solute.

Answer:
According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute:
(P° - P)/P° = X₂
Where X₂ = n₂/(n₁ + n₂) ≈ n₂/n₁ (for dilute solutions)
So, (P° - P)/P° = n₂/n₁ = (W₂/M₂) / (W₁/M₁)
Where W₂ and M₂ are mass and molar mass of solute, W₁ and M₁ are mass and molar mass of solvent.
Thus, M₂ = (W₂ × M₁) / [W₁ × (P° - P)/P°]
This equation can be used to determine the molar mass of the solute.

Q32. Write the Nernst equation and calculate EMF of cell: Zn/Zn²⁺ (0.01 M) || Cu²⁺ (0.1 M)/Cu. E° = 1.10 V.

Answer:
Nernst equation: E = E° - (0.059/n) log(Q)
For the cell: Zn + Cu²⁺ → Zn²⁺ + Cu (n = 2)
Q = [Zn²⁺]/[Cu²⁺] = 0.01/0.1 = 0.1
E = 1.10 - (0.059/2) log(0.1) = 1.10 - (0.0295)(-1) = 1.10 + 0.0295 = 1.1295 V ≈ 1.13 V

Q33. Explain: (a) Optical isomerism in [Co(en)₃]³⁺ (b) Geometrical isomerism in [Pt(NH₃)₂Cl₂].

Answer:
(a) Optical isomerism in [Co(en)₃]³⁺: The complex [Co(en)₃]³⁺ has three bidentate ethylenediamine (en) ligands arranged octahedrally around the central Co³⁺ ion. This creates a chiral complex that is not superimposable on its mirror image, resulting in two enantiomers (d and l forms).
(b) Geometrical isomerism in [Pt(NH₃)₂Cl₂]: This square planar complex exhibits cis-trans isomerism. In the cis isomer, the two Cl ligands are adjacent to each other (90° apart), while in the trans isomer, they are opposite each other (180° apart). Cisplatin (cis form) has significant medicinal properties as an anticancer drug.

Q34. Discuss important trends in d-block: (a) Oxidation states (b) Magnetic properties (c) Catalysis.

Answer:
(a) Oxidation states: Transition elements show variable oxidation states due to similar energies of (n-1)d and ns orbitals. Higher oxidation states are more stable for heavier elements in each group.
(b) Magnetic properties: Transition metals and their compounds exhibit paramagnetism due to unpaired electrons. Some show ferromagnetism (Fe, Co, Ni) and others show antiferromagnetism or diamagnetism.
(c) Catalysis: Many transition metals and their compounds act as catalysts due to their ability to adopt multiple oxidation states, provide active surface area, and form unstable intermediate complexes. Examples: Fe in Haber process, V₂O₅ in Contact process.

Q35. Write note on actinoid elements with examples: (a) Oxidation states (b) Complex formation (c) Uses.

Answer:
(a) Oxidation states: Actinoids show a wide range of oxidation states (typically +3 to +6), with +3 being most common. Some like Uranium can show +6 oxidation state (e.g., in UO₂²⁺).
(b) Complex formation: Actinoids form complexes with ligands like halides, sulfates, and organic ligands. Their complex formation tendency is higher than lanthanoids due to higher charge density.
(c) Uses: - Uranium and Plutonium are used as nuclear fuels in reactors and weapons. - Americium-241 is used in smoke detectors. - Thorium is used in incandescent gas mantles. - Some actinoids are used in medical applications like cancer treatment.
Section
A: MCQs
Questions
15
Marks Each
1
Total
15
Section
B: Assertion-Reason
Questions
5
Marks Each
1
Total
5
Section
C: Short Answer
Questions
5
Marks Each
2
Total
10
Section
D: Numerical
Questions
5
Marks Each
3
Total
15
Section
E: Long Answer
Questions
5
Marks Each
5
Total
25
Grand Total
70

Willer Academy Chemistry Department | Half-Yearly Examination Practice Paper

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