More practice questions

4. Q4. Why is the Cu⁺ ion not stable in aqueous solution? (NCERT Intext Question 4.9.)

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   In water Cu⁺ disproportionates, 2Cu⁺ → Cu²⁺ + Cu. Cu²⁺(aq) is the more stable ion because its hydration enthalpy is far more negative than that of Cu⁺, which more than makes up for copper’s second ionisation enthalpy (NCERT page 100).

5. Q5. Which is the stronger reducing agent, Cr²⁺ or Fe²⁺, and why? (NCERT Intext Question 4.7.)

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   Cr²⁺. On oxidation Cr²⁺ goes from d⁴ to d³, while Fe²⁺ goes from d⁶ to d⁵, and in a medium like water d³ is more stable than d⁵ (NCERT’s answer to Intext Question 4.7, page 117). NCERT’s “see CFSE” points to this: d³, as t_2g³, a half-filled t_2g level (NCERT Example 4.4, page 98), gains crystal field stabilisation energy, while high-spin d⁵ gains none (reasoned). Strictly, d-electron counts alone do not decide such comparisons between different metals (reasoned). The E^⊖ values agree: E^⊖(Cr³⁺/Cr²⁺) is −0.41 V while E^⊖(Fe³⁺/Fe²⁺) is +0.77 V (NCERT Table 4.2, page 94), so Cr²⁺ is far more easily oxidised. Cr²⁺ even liberates hydrogen from dilute acid (NCERT page 101); Fe²⁺ does not (reasoned).

6. Q6. For high-spin ions in the usual octahedral aqueous environment, find the number of unpaired electrons and the spin-only magnetic moment of Ti³⁺ (Z = 22) and of Mn³⁺ (Z = 25).

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   Ti³⁺ is 3d^¹: 1 unpaired electron, μ = √(1 × 3) = √3 = 1.73 BM. Mn³⁺ is 3d^⁴: 4 unpaired electrons, μ = √(4 × 6) = √24 = 4.90 BM.

7. Q7. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? (NCERT Intext Question 4.6.)

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   NCERT’s answer: because of their small size and high electronegativity, oxygen and fluorine can oxidise the metal to its highest oxidation state (page 117). Fluorine stabilises them through a high lattice energy or high bond enthalpies (CoF₃, VF₅), and oxygen does even better because it can form multiple bonds to the metal: manganese reaches +7 in Mn₂O₇ but only +4 in MnF₄ (NCERT pages 99–100). Strictly, “only” is too strong: NCERT’s own Table 4.5 gives titanium its highest state, +4, in all four halides TiX₄ (reasoned from page 99).

8. Q8. Show that the two steps by which iron(III) catalyses the reaction of iodide with persulphate add up to the overall reaction, and say what happens to the catalyst.

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   Adding 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ and 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻, the 2Fe³⁺ and 2Fe²⁺ on both sides cancel, leaving 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻. Iron is reduced to Fe²⁺ in the first step and oxidised back to Fe³⁺ in the second, so it is not used up (NCERT page 104).

9. Q9. Interstitial compounds have formulas such as VH_0.56 and TiH_1.7. Why do such formulas not give a sensible oxidation state for the metal?

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   The small atoms occupy interstitial sites, often with only some sites occupied. They do bond to the metal, with mixed bonding character; a simple model of isolated integer-charge ions is often inadequate. NCERT calls them usually non-stoichiometric, with formulas that do not correspond to any normal oxidation state of the metal (page 104); a nonintegral composition describes average site occupancy, not a literal fractional atom (reasoned).