In which of the following arrangements the order is NOT according to the property indicated against it? (1) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$ — Increasing ionic size (2) $\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}$ — Increasing first ionization enthalpy (3) $\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}$ — Increasing electron gain enthalpy (with negative sign) (4) $\mathrm{Li}<\mathrm{Na}<\mathrm{K}<\mathrm{Rb}$ — Increasing metallic radius

(B) $\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}$ — Increasing first ionization enthalpy

(A) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$ — Increasing ionic size

(C) $\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}$ — Increasing electron gain enthalpy (with negative sign)

(D) $\mathrm{Li}<\mathrm{Na}<\mathrm{K}<\mathrm{Rb}$ — Increasing metallic radius

JEE Main · 2005 · 08 May 2005 · Offline · Classification of Elements and Periodicity

In which of the following arrangements the order is NOT according to the property indicated against it?
(1)

Al^{3 +} < Mg^{2 +} < Na^{+} < F^{-}

— Increasing ionic size
(2)

B < C < N < O

— Increasing first ionization enthalpy
(3)

I < Br < F < Cl

— Increasing electron gain enthalpy (with negative sign)
(4)

Li < Na < K < Rb

— Increasing metallic radius

(A) $Al^{3 +} < Mg^{2 +} < Na^{+} < F^{-}$ — Increasing ionic size
(B) $B < C < N < O$ — Increasing first ionization enthalpy✓ Correct
(C) $I < Br < F < Cl$ — Increasing electron gain enthalpy (with negative sign)
(D) $Li < Na < K < Rb$ — Increasing metallic radius

Answer: (B)

Solution

This asks which stated trend is the odd one out. Isoelectronic ionic size does grow as nuclear charge falls,

Al X^{3 +} < Mg X^{2 +} < Na X^{+} < F X^{-}

; the electron gain enthalpy does become more negative in the order

I < Br < F < Cl

, with fluorine's small size making it the anomaly; and metallic radius does rise down the alkali metals. The ionisation-enthalpy claim is the faulty one: the true order is

B < C < O < N

, because nitrogen's half-filled

2 p^{3}

shell is especially stable and resists losing an electron, so nitrogen sits above oxygen rather than below it.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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JEE Main · 2005 · 08 May 2005 · Offline · Classification of Elements and Periodicity

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