Question

$K^{+},C{l}^{-},C{a}^{2+}$ and $S^{2-}$ ions are isoelectronic. The decreasing order of their size is :

• A. $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$
• B. $C{a}^{2+}>K^{+}>C{l}^{-}>S^{2-}$
• C. $K^{+}>C{l}^{-}>C{a}^{2+}>S^{2-}$
• D. $C{l}^{-}>S^{2-}>C{a}^{2+}>K^{+}$

Answer 1

The correct option is A $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

For isoelectronic species, anions will be larger than cations.

Anions are formed when an electron is gained. As a result there is greater electron-electron repulsion and the effective nuclear charge experienced by an electron decreases. The size becomes bigger to account for these changes. Hence an electron will be larger than a neutral atom.

Similarly, cations are formed when electrons are lost. Thus there is a greater nuclear charge experienced by each electron. The electrons are pulled closer towards the nucleus. Hence the cations are smaller than the neutral atom.

For isoelectronic species,

Anion > Neutral atom > Cation.

In this case, $S^{2-}$ will be the largest and $C{a}^{2+}$ will be the smallest.

The order is $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$.

Hence, option A is the right answer.