Group 14 Ion Nature from Ionisation Enthalpy

 

❓ Question

Group 14 elements A and B have first ionisation enthalpy values:

A → 708 kJ mol⁻¹
B → 715 kJ mol⁻¹

These are the lowest values in the group.

Find the nature of their ions:

$$A^{2+} \quad \text{and} \quad B^{4+}$$

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🖼 Question Image

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✍️ Short Concept

Ionisation energy generally decreases down the group.

Lowest values in Group 14 correspond to the heaviest elements.

So these numbers help identify the elements.

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🔷 Step 1 — Identify the Elements 💯

Group 14 elements:

C → Si → Ge → Sn → Pb

Ionisation energies:

Sn ≈ 708 kJ/mol
Pb ≈ 715 kJ/mol

So:

$$A = Sn$$
$$B = Pb$$

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🔷 Step 2 — Apply Inert Pair Effect

In heavier p-block elements:

ns² electrons become reluctant to participate in bonding.

This is called:

$$\textbf{Inert Pair Effect}$$

Because of this:

Lower oxidation states become more stable.

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🔷 Step 3 — Nature of $A^{2+}$ (Sn²⁺)

Tin:

Stable oxidation states:

$$+2 \quad \text{and} \quad +4$$

Due to inert pair effect:

$$Sn^{2+}$$

is stable and reducing.

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🔷 Step 4 — Nature of $B^{4+}$(Pb⁴⁺)

Lead prefers:

$$+2$$

state.

So:

$$Pb^{4+}$$

is unstable and behaves as a strong oxidising agent.

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✅ Final Answer

$$\boxed{A^{2+} \text{ is reducing, } B^{4+} \text{ is oxidising}}$$

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⭐ Golden JEE Insight

Down the p-block group:

👉 Inert pair effect increases

So trend becomes:

$$+2 \text{ more stable than } +4$$

Especially for Pb.