Question

Consider the isoelectronic series:
$K^{+},S^{2-},C{l}^{-}$ and $C{a}^{2+}$, the radii of the ions decrease as:

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

Answer 1

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

A series of atoms, ions, and molecules in which each species contains the same number of electrons but the different nuclear charge is called the isoelectronic series.

For example, one isoelectronic series could include $S^{2-},C{l}^{-},K^{+},C{a}^{2+}$. These all have $18$ electrons. The number of protons, though, increases as atomic number increases, so nuclear charge increases. When we consider effective nuclear charge then, the ions of greater nuclear charge attract those ten electrons more strongly and pull them in more tightly.

Therefore, the radii of the ions in an isoelectronic series decrease as nuclear charge (or atomic number) increases.