Chemistry Half-Yearly Practice Paper – Set 2
Class 12 – CBSE/JNV Pattern | Time: 3 Hours | Max Marks: 70
Topics: Solutions + Electrochemistry + Chemical Kinetics + d- and f-Block Elements + Coordination Compounds
Section A – Multiple Choice Questions (15 × 1 = 15 marks)
Q1. Henry's law constant for a gas in water is 1.6 × 10⁵ atm at 298 K. What is the solubility of gas in water at 2 atm pressure?
Explanation: According to Henry's law, S = K_H × P = (1/1.6×10⁵) × 2 = 1.25 × 10⁻⁵ mol/L
Q2. Which of the following electrolytes has the highest conductivity at infinite dilution?
Explanation: KCl has the highest molar conductivity at infinite dilution due to high ionic mobility.
Q3. In a first-order reaction, time required to complete 75% reaction is 60 min. The half-life of the reaction is:
Explanation: For first-order reaction, t75% = 2 × t1/2, so t1/2 = 60/2 = 30 min.
Q4. Which transition metal has maximum number of oxidation states?
Explanation: Manganese shows oxidation states from +2 to +7 due to its electronic configuration [Ar] 3d5 4s2.
Q5. Which of the following complexes shows optical isomerism?
Explanation: [Co(en)₃]³⁺ has a chiral structure with no plane of symmetry, making it optically active.
Q6. Raoult's law is applicable to:
Explanation: Raoult's law is strictly applicable only to ideal solutions.
Q7. EMF of a cell reaction is positive when:
Explanation: According to the relation ΔG = -nFE, when E is positive, ΔG is negative.
Q8. Activation energy is equal to:
Explanation: Activation energy is the minimum extra energy required by reactant molecules to form the activated complex.
Q9. The lanthanoid contraction is responsible for:
Explanation: Due to lanthanoid contraction, the atomic sizes of Zr and Hf become almost identical, leading to similar chemical properties.
Q10. The magnetic moment of [Fe(CN)₆]⁴⁻ is:
Explanation: [Fe(CN)₆]⁴⁻ is diamagnetic with no unpaired electrons (low-spin complex).
Q11. In Nernst equation, if concentration of ions increases, the EMF:
Explanation: According to Nernst equation, E = E° - (RT/nF) ln Q, increasing concentration increases Q, thus decreasing E.
Q12. Which rate constant has units of s⁻¹?
Explanation: For a first-order reaction, rate = k[A], so k has units of s⁻¹.
Q13. Which of the following shows highest ionic conductivity at 298 K?
Explanation: HCl is a strong acid and completely dissociates into ions, showing high conductivity.
Q14. Which is a linkage isomer?
Explanation: The NO₂ ligand can coordinate through N (nitro) or O (nitrito), giving linkage isomers.
Q15. Colligative property depends on:
Explanation: Colligative properties depend only on the number of solute particles, not on their nature.
Section B – Assertion-Reason Questions (5 × 2 = 10 marks)
Q16. Assertion (A): Osmotic pressure is a colligative property.
Reason (R): It depends on the number of solute particles present in solution.
Explanation: Osmotic pressure depends on the number of solute particles, making it a colligative property.
Q17. Assertion (A): Mn shows +2 to +7 oxidation states.
Reason (R): Mn has half-filled d-orbitals which favor variable oxidation states.
Explanation: Manganese has the electronic configuration [Ar] 3d5 4s2 with half-filled d-orbitals, allowing it to show various oxidation states.
Q18. Assertion (A): For first-order reaction, half-life is independent of concentration.
Reason (R): Half-life is inversely proportional to initial concentration.
Explanation: For first-order reactions, half-life is constant and independent of initial concentration, making R false.
Q19. Assertion (A): [FeF₆]³⁻ is high-spin while [Fe(CN)₆]³⁻ is low-spin.
Reason (R): F⁻ is weak field ligand, CN⁻ is strong field ligand.
Explanation: F⁻ is a weak field ligand causing high-spin complexes, while CN⁻ is a strong field ligand causing low-spin complexes.
Q20. Assertion (A): Standard EMF of a cell is always positive for a spontaneous reaction.
Reason (R): ΔG = –nFE°, spontaneity requires ΔG < 0.
Explanation: For a spontaneous reaction, ΔG must be negative, which requires E° to be positive.
Section C – Short Answer Questions (5 × 3 = 15 marks)
Q21. Calculate the molality of glucose solution containing 18 g glucose in 100 g water.
Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
Moles of glucose = 18/180 = 0.1 mol
Mass of solvent = 100 g = 0.1 kg
Molality = moles of solute / mass of solvent in kg = 0.1 / 0.1 = 1 m
Q22. Differentiate between order of reaction and molecularity.
Order of reaction:
- Sum of powers of concentration terms in the rate law
- Experimental quantity, can be fractional or zero
- Applicable to elementary and complex reactions
Molecularity:
- Number of reactant molecules colliding in an elementary step
- Theoretical concept, always a whole number
- Applicable only to elementary reactions
Q23. Why is conductivity of electrolytes higher in aqueous solution than in molten state?
The conductivity of electrolytes is higher in aqueous solutions than in molten state because:
1. Water molecules help in dissociation of electrolytes into ions
2. Hydration of ions increases their mobility
3. The dielectric constant of water is high, which reduces interionic attractions
4. In molten state, ions are less mobile due to higher viscosity and stronger interionic forces
Q24. Write IUPAC names of:
(a) [Cr(H₂O)₆]Cl₃
(b) K₄[Fe(CN)₆]
(a) Hexaaquachromium(III) chloride
(b) Potassium hexacyanidoferrate(II)
Q25. Explain the trend in ionization enthalpy of transition metals.
The ionization enthalpy of transition metals shows the following trends:
1. Generally increases across a period due to increasing nuclear charge
2. The increase is not regular due to poor shielding of d-electrons
3. Lower values for Mn and Zn due to stable half-filled and fully-filled configurations
4. Higher values for later transition elements due to stronger nuclear charge
5. The first ionization enthalpy is less than that of s-block elements but more than p-block elements
Section D – Long Answer Questions (3 × 5 = 15 marks)
Q26. The rate constant for first-order decomposition of H₂O₂ is 1.2 × 10⁻³ s⁻¹. Calculate half-life and time required for 75% decomposition.
For a first-order reaction:
Half-life (t1/2) = 0.693/k = 0.693/(1.2 × 10⁻³) = 577.5 seconds
Time for 75% decomposition:
t = (2.303/k) log(a/(a-x))
When 75% decomposed, x = 0.75a, so (a-x) = 0.25a
t = (2.303/(1.2 × 10⁻³)) log(1/0.25)
t = (1919.17) log(4) = 1919.17 × 0.6021 = 1155.5 seconds
Q27. Draw and explain the crystal field splitting in octahedral complexes. Explain why [Ti(H₂O)₆]³⁺ is coloured.
Crystal field splitting in octahedral complexes:
In an octahedral complex, the d-orbitals split into two sets:
- t2g orbitals (dxy, dxz, dyz) at lower energy
- eg orbitals (dx²-y², dz²) at higher energy
The energy difference between these sets is called Δo (octahedral splitting parameter).
Why [Ti(H₂O)₆]³⁺ is colored:
Ti³⁺ has electronic configuration [Ar] 3d1. In an octahedral field, this single electron occupies one of the t2g orbitals. It can absorb visible light (green light, ~500 nm) to get excited to the eg level. The complementary color (purple) is observed, making the complex colored.
Q28. The cell: Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(1 M) | Cu(s). E°cell = 1.10 V. Calculate Ecell.
The cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Using Nernst equation:
Ecell = E°cell - (0.0591/n) log([Zn²⁺]/[Cu²⁺])
n = 2 (number of electrons transferred)
Ecell = 1.10 - (0.0591/2) log(0.1/1)
Ecell = 1.10 - (0.02955) log(0.1)
Ecell = 1.10 - (0.02955)(-1)
Ecell = 1.10 + 0.02955 = 1.12955 V ≈ 1.13 V
Section E – Very Long Answer (1 × 5 = 5 marks)
Q29. Write the preparation, structure, and properties of KMnO₄.
Preparation of KMnO₄:
1. From pyrolusite ore (MnO₂):
- MnO₂ is fused with KOH in presence of air to give potassium manganate (K₂MnO₄)
- 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
- K₂MnO₄ is electrolytically oxidized to KMnO₄
- 2K₂MnO₄ + 2H₂O → 2KMnO₄ + 2KOH + H₂
Structure:
- KMnO₄ has tetrahedral MnO₄⁻ ions
- Manganese is in +7 oxidation state
- The four oxygen atoms are arranged tetrahedrally around the central Mn atom
Properties:
1. Oxidizing agent: Powerful oxidizing agent in acidic, neutral and alkaline media
- Acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
- Neutral/alkaline medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
2. Disproportionation: In neutral or slightly alkaline solution, it disproportionates:
- 3MnO₄²⁻ + 2H₂O → 2MnO₄⁻ + MnO₂ + 4OH⁻
3. Uses: Used as an oxidizing agent in organic synthesis, water treatment, and as a disinfectant
A: MCQs
15
1
15
B: Assertion-Reason
5
2
10
C: Short Answer
5
3
15
D: Long Answer
3
5
15
E: Very Long Answer
1
5
5
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