CBSE 2025, Set 3

Question

Give reasons for the following:

(I) Ce$^{4+}$ in aqueous solution is a good oxidising agent.

(II) The actinoid contraction is greater from element to element than lanthanoid contraction.

(III) $E°_{	ext{Zn}^{2+}/	ext{Zn}}$ value is more negative than expected, whereas $E°_{	ext{Cu}^{2+}/	ext{Cu}}$ is positive.

Edvaya · Accepted Answer

The correct answer is Subjective. **(I) Ce$^{4+}$ is a good oxidising agent:** Ce (Z=58) has configuration [Xe]4f$^1$5d$^1$6s$^2$. Ce$^{4+}$ has configuration [Xe] — an **empty 4f subshell** and a noble gas configuration. Ce$^{4+}$ readily gets reduced to Ce$^{3+}$ ([Xe]4f$^1$), which has an extra stability (one f-electron). The strong tendency to attain Ce$^{3+}$ drives Ce$^{4+}$ to accept electrons — hence it is a **good oxidising agent** in aqueous solution. **(II) Actinoid contraction > Lanthanoid contraction (element to element):** In actinoids (5f elements), the **5f electrons provide even poorer shielding** of nuclear charge than 4f electrons in lanthanoids. The 5f orbitals are more diffuse and spatially more similar to 6d and 7s, meaning they shield each other less effectively. As a result, the effective nuclear charge increases more rapidly across the actinoid series, causing a **greater decrease in atomic/ionic radius** per element than in lanthanoids. **(III) $E°_{	ext{Zn}^{2+}/	ext{Zn}}$ is more negative than expected; $E°_{	ext{Cu}^{2+}/	ext{Cu}}$ is positive:** - **Zn:** Has a completely filled d$^{10}$ configuration. The high stability of the d$^{10}$ configuration makes Zn$^{2+}$ less thermodynamically favourable to form (requires breaking a stable configuration), requiring more energy. Additionally, Zn has **high atomisation enthalpy** relative to the gain from hydration. The sum of ionisation enthalpy and atomisation enthalpy exceeds the hydration energy, making $E°$ more negative than expected. - **Cu:** Has high **atomisation enthalpy** AND high ionisation enthalpy. The energy released by Cu$^{2+}$ hydration ($\Delta_{hyd}H$) is **relatively small** (not enough to compensate). This makes Cu$^{2+}$ formation thermodynamically unfavourable → **positive $E°$** (Cu does not dissolve in dilute acids).

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