Question

In which of the following arrangement the order is not according to the property indicated against it?

• A. \(\mathrm{Li}< \mathrm{Na}< \mathrm{K}< \mathrm{Rb}\) (increasing metallic radius)
• B. \(\mathrm{I}< \mathrm{Br}< \mathrm{F}< \mathrm{Cl}\) (increasing electron gain enthalpy, with negative sign)
• C. \(\mathrm{B}< \mathrm{C}< \mathrm{N}< \mathrm{O}\) (increasing first ionisation enthalpy)
• D. \(\mathrm{Al}^{3+}< \mathrm{Mg}^{2+}< \mathrm{Na}^{+}< \mathrm{F}^{-}\) (increasing ionic size)

Hint:

Remember: N has higher ionisation enthalpy than O due to stable half-filled p-subshell.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ionisation enthalpy generally increases across a period, but N has higher IE than O due to half-filled stability.

Step 2: Detailed Explanation:
(A) Correct: Atomic radius increases down Group 1.
(B) Correct: Electron gain enthalpy becomes more negative from I to Cl; F is an exception.
(C) Incorrect: Correct order is B<C<O<N because N (half-filled 2p\(^3\)) has higher IE than O (2p\(^4\)).
(D) Correct: For isoelectronic species, size decreases as nuclear charge increases.

Step 3: Final Answer:
Option (C) is not according to the property.