Question

Which of the following orders is not correct for the given property?

• A. $ \text{Li} < \text{Na} < \text{K} $ -- metallic radius
• B. $ \text{Br} < \text{F} < \text{Cl} $ -- electron gain enthalpy
• C. $ \text{C} < \text{N} < \text{O} $ -- first ionisation enthalpy
• D. $ \text{Mg}^{+2} < \text{Na}^{+} < \text{F}^{-} $ -- ionic radius

Hint:

Always watch out for the "N vs O" and "Be vs B" traps in Ionisation Enthalpy! Half-filled and full-filled configurations jump ahead of the next element in the period.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ionisation Enthalpy (IE) represents the energy required to remove an electron. While it generally increases across a period due to increasing nuclear charge, stable electronic configurations (half-filled or full-filled subshells) create significant exceptions.
Step 2: Detailed Explanation:
Evaluating the options:
(a) Metallic radius increases down a group (Li \( \to \) Na \( \to \) K) due to additional electron shells. Correct.
(b) Electron gain enthalpy for halogens is most negative for Cl. F is smaller but has high inter-electronic repulsion. Cl \(>\) F \(>\) Br in magnitude. Correct.
(c) Across a period, IE generally increases. However, Nitrogen (\( Z=7 \)) has a \( 2p^3 \) configuration (exactly half-filled, very stable). Oxygen (\( Z=8 \)) has a \( 2p^4 \) configuration. It is easier to remove an electron from Oxygen because of electron-electron repulsion in the paired \( p \)-orbital and the fact that Nitrogen's configuration is extra stable. Thus, IE of N \(>\) IE of O. The correct order is \( \text{C}<\text{O}<\text{N} \). Order (c) is incorrect.
(d) These are isoelectronic species (10 electrons each). Radius decreases as nuclear charge (\( Z \)) increases. \( \text{F}^- (Z=9)>\text{Na}^+ (Z=11)>\text{Mg}^{2+} (Z=12) \). Correct.
Step 3: Final Answer:
The incorrect order is (c) \( \text{C}<\text{N}<\text{O} \) for first ionisation enthalpy.