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

Concept: Periodic properties follow specific trends across periods and down groups. However, there are significant exceptions caused by electronic configurations (half-filled and fully-filled stability) and electron-electron repulsions.
• Atomic/Metallic Radius: Increases down a group due to the addition of new shells.
• Electron Gain Enthalpy: Generally becomes more negative across a period. For halogens, $ \text{Cl} > \text{F} $ due to inter-electronic repulsion in the small F atom.
• Ionisation Enthalpy: Generally increases across a period but decreases if an atom has a stable configuration.
• Ionic Radius: In isoelectronic species, radius decreases as nuclear charge ($ Z $) increases.

Step 1: Evaluating Option (a): Metallic Radius Li, Na, and K are Alkali metals (Group 1). As we move down the group from Lithium to Potassium, the number of energy shells increases ($ n=2 $ for Li, $ n=3 $ for Na, $ n=4 $ for K). This outweighs the increase in nuclear charge, resulting in an increase in metallic radius. Order: $ \text{Li} < \text{Na} < \text{K} $. (Correct)

Step 2: Evaluating Option (b): Electron Gain Enthalpy Usually, Electron Gain Enthalpy decreases down a group.
However, Fluorine (F) has a very small size. When an electron is added to its $ 2p $ orbital, it faces significant inter-electronic repulsion. Chlorine (Cl) is larger, allowing the electron to be added more easily with a higher release of energy. Order of magnitude: $ \text{Br} < \text{F} < \text{Cl} $. (Correct)

Step 3: Evaluating Option (c): First Ionisation Enthalpy Across the second period (C, N, O), ionisation enthalpy generally increases. Let's look at the electronic configurations:
• $ \text{C} (Z=6): 1s^2 2s^2 2p^2 $
• $ \text{N} (Z=7): 1s^2 2s^2 2p^3 $ (Exactly half-filled $ p $-subshell)
• $ \text{O} (Z=8): 1s^2 2s^2 2p^4 $ Nitrogen has a half-filled $ 2p $ subshell, which is exceptionally stable. Removing an electron from Nitrogen requires more energy than removing one from Oxygen, where electron-electron repulsion in the paired $ 2p $ orbital makes removal easier. Correct Order: $ \text{C} < \text{O} < \text{N} $. The given order $ \text{C} < \text{N} < \text{O} $ is Incorrect.

Step 4: Evaluating Option (d): Ionic Radius $ \text{Mg}^{+2} $, $ \text{Na}^{+} $, and $ \text{F}^{-} $ all have 10 electrons (isoelectronic).
• $ \text{Mg}^{+2} $: $ Z=12 $ (Strongest nuclear pull)
• $ \text{Na}^{+} $: $ Z=11 $
• $ \text{F}^{-} $: $ Z=9 $ (Weakest nuclear pull) As $ Z $ increases, the radius decreases. Order: $ \text{Mg}^{+2} < \text{Na}^{+} < \text{F}^{-} $. (Correct)