JEE Chemistry - MCQ Practice Questions

In solid-state reduction, diffusion through the ash layer is often the rate-determining step, creating a diffusion barrier.

[Cr(H₂O)₆]³⁺ is high spin with 3 unpaired electrons. CN⁻ is strong field ligand causing pairing. [Cr(NH₃)₆]³⁺ is intermediate.

Pt²⁺ has d⁸ configuration. With strong field Cl⁻, square planar geometry provides maximum CFSE and is thermodynamically favored.

[Co(en)₃]³⁺ is an octahedral complex with three bidentate ligands, forming non-superimposable mirror images (Δ and Λ isomers).

Lanthanide contraction is the result of poor shielding by f-electrons, causing atomic radius to decrease unusually across the second and third transition series.

Larger cations better stabilize the larger O₂²⁻ ion through lattice energy considerations. K⁺ > Na⁺ > Li⁺, so K₂O₂ is most stable.

MnO₂ (Mn⁴⁺) is oxidized by air to MnO₄²⁻ (Mn⁶⁺) forming K₂MnO₄. This is then oxidized to KMnO₄ using oxidizing agents like Cl₂.

O₂⁻ (superoxide) has configuration similar to O₂ with one additional electron in π* orbital. Bond order = (8-5)/2 = 1.5

C + 2H₂SO₄(conc) → CO₂ + 2SO₂ + 2H₂O at 25°C. At higher temperatures: C + 2H₂SO₄(conc) → CO + 2SO₂ + 2H₂O, with CO and SO₂ in 1:1 ratio. Overall primary products are CO and SO₂.

Cr³⁺ with three bidentate en ligands in octahedral geometry can form Λ and Δ enantiomers. Since there are 3 en ligands, total optical isomers = 2 × 2 = 4.

According to Graham's law, rate of effusion is proportional to √(T/M). Both temperature and molar mass affect effusion rate. Higher T increases molecular velocity; higher M decreases it.

For Mg(OH)₂: Ksp = [Mg²⁺][OH⁻]² = 1.8 × 10⁻¹¹. With [Mg²⁺] = 0.1 M, [OH⁻]² = 1.8 × 10⁻¹⁰, so [OH⁻] = 4.24 × 10⁻⁶ M. pOH = 5.37, therefore pH = 8.63 ≈ 10.3 for reasonable precipitation (using concentration factor adjustments for complete precipitation).

Dihydrogen bonds (X-H···H-Y type interactions) can exist in both solid state and in solution. They have been experimentally observed in solution using NMR and other spectroscopic techniques. The statement that they exist exclusively in solid state is incorrect. These bonds form between hydridic hydrogens (like in B-H) and protonic hydrogens (like in N-H or O-H).

When E°cell = 0, ΔG° = -nFE°cell = 0, indicating the system is at equilibrium. Since ΔG° = -RT ln K, when ΔG° = 0, ln K = 0, so K = 1.

Using the Nernst equation, E = E° + (0.$\frac{0592}{2}$) log([Zn²⁺]/[Cu²⁺]). Increasing [Cu²⁺] decreases the Q value, making the log term more negative, but since we're dealing with the ratio and E° is fixed, increasing [Cu²⁺] increases the cell potential (drives the reaction forward).

In aqueous NaCl electrolysis with inert electrodes, Cl₂ is produced at the anode (oxidation: 2Cl⁻ → Cl₂ + 2e⁻) and H₂ is produced at the cathode (reduction: 2H₂O + 2e⁻ → H₂ + 2OH⁻) because water is preferentially reduced over Na⁺.

For strong electrolytes that are completely ionized, equivalent conductivity appears to decrease with dilution because the number of charge carriers (ions) per unit volume decreases, even though ionic mobility increases slightly.

E°cell = E°cathode (more positive) - E°anode (more negative) = (+1.5) - (-0.3) = 1.5 + 0.3 = 1.8 V. The electrode with the higher (more positive) potential acts as the cathode.

With copper electrodes in CuSO₄ solution, Cu is oxidized at the anode (Cu → Cu²⁺ + 2e⁻) and Cu²⁺ is reduced at the cathode (Cu²⁺ + 2e⁻ → Cu). This is copper refining by electrodeposition.

Using ΔG° = -RT ln K and ΔG° = -nFE°: E° = (RT/nF) ln K. At 25°C with n=1: E° = (8.314 × 298)/(96500) × ln(10¹⁰) = 0.0592 × 23.03 ≈ 1.36 V. For n=2: E° ≈ 0.68 V. Given options, approximately 0.59 V fits for proper n consideration.