The correct order of ionic radii for the ions, $\mathrm{P}^{3-}, \mathrm{S}^{2-}, \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}$ is:

(D) $\mathrm{P}^{3-}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$

(A) $\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{P}^{3-}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}$

(B) $\mathrm{P}^{3-}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}$

(C) $\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{P}^{3-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Classification of Elements and Periodicity

The correct order of ionic radii for the ions,

P^{3 -}, S^{2 -}, Ca^{2 +}, K^{+}, Cl^{-}

is:

(A) $K^{+} > Ca^{2 +} > P^{3 -} > S^{2 -} > Cl^{-}$
(B) $P^{3 -} > S^{2 -} > Cl^{-} > Ca^{2 +} > K^{+}$
(C) $Cl^{-} > S^{2 -} > P^{3 -} > Ca^{2 +} > K^{+}$
(D) $P^{3 -} > S^{2 -} > Cl^{-} > K^{+} > Ca^{2 +}$ ✓ Correct

Answer: (D)

Solution

The five ions are isoelectronic, each having

18

electrons, so their sizes are set by nuclear charge: the more protons, the more tightly the electron cloud is drawn in and the smaller the ion. With nuclear charges

15, 16, 17, 19

and

20

for

P^{3 -}, S^{2 -}, C l^{-}, K^{+}

and

C a^{2 +}

, the radii fall in the order

P^{3 -} > S^{2 -} > C l^{-} > K^{+} > C a^{2 +}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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