Question

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

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

Answer 1

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

An isoelectronic series is useful in understanding the effects of gained or loss electrons on atom size.

If the atom or molecule is negatively charged, as an additional electron occupies an outer orbital, there is increased electron-electron repulsion (and hence, increased shielding) which pushes the electrons further apart. Because the electrons now outnumber the protons in the ion, the protons can not pull the extra electrons as tightly toward the nucleus; this results in decreased $Z_{eff}$. Therefore the ionic radii increase whereas it is vice versa when it is positively charged.

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

$C{a}^{2+},K^{+},C{l}^{-},S^{2-}$