The ionic radii of $\mathrm{K}^{+}, \mathrm{Na}^{+}, \mathrm{Al}^{3+}$ and $\mathrm{Mg}^{2+}$ are in the order:

(C) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{K}^{+}$

(A) $\mathrm{K}^{+}<\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}$

(B) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{K}^{+}<\mathrm{Na}^{+}$

(D) $\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Classification of Elements and Periodicity

The ionic radii of

K^{+}, Na^{+}, Al^{3 +}

and

Mg^{2 +}

are in the order:

(A) $K^{+} < Al^{3 +} < Mg^{2 +} < Na^{+}$
(B) $Al^{3 +} < Mg^{2 +} < K^{+} < Na^{+}$
(C) $Al^{3 +} < Mg^{2 +} < Na^{+} < K^{+}$ ✓ Correct
(D) $Na^{+} < K^{+} < Mg^{2 +} < Al^{3 +}$

Answer: (C)

Solution

N a^{+}

M g^{2 +}

and

A l^{3 +}

all have the ten electron neon configuration. Within an isoelectronic set the same number of electrons is held by a growing nuclear charge, so the radius contracts as the charge on the cation rises, giving

A l^{3 +} < M g^{2 +} < N a^{+}

K^{+}

has the eighteen electron argon configuration, one complete shell more than the other three, and an extra principal shell outweighs the larger nuclear charge, so it is the biggest of the four. The order is

A l^{3 +} < M g^{2 +} < N a^{+} < K^{+}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Classification of Elements and Periodicity

Solution

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