Question

The increasing order of the ionic radii of the given isoelectronic species is:

• A. $K^{+},S^{2-},C{a}^{2+}\text{and}C{l}^{-}$
• B. $C{a}^{2+},K^{+},C{l}^{-}\text{and}{S}^{2-}$
• C. $S^{2-},C{l}^{-},C{a}^{2+}\text{and}{K}^{+}$
• D. $C{l}^{-}.C{a}^{2+},K^{+}\text{and}{S}^{2-}$

Answer 1

The correct option is B $C{a}^{2+},K^{+},C{l}^{-}\text{and}{S}^{2-}$

For isoelectronic species $\sigma$ is same for all
$\text{Size}\propto \frac{1}{Z_{\text{eff}}}$
$Z_{\text{eff}}=(Z-\sigma )$
$Z$ is atomic number.
Atomic number of the given species is in order $Ca>K>Cl>S$.
Hence, ionic radii is in order, $C{a}^{2+},K^{+},C{l}^{-}\text{and}{S}^{2-}$