Question

Arrange the atoms and/or ions in the following groups in order of decreasing size.

a.$\mathbf{O}\mathbf{,}{\mathbf{O}}^{\mathbf{-}}\mathbf{,}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}$

b.${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}$

c. ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{\mathbf{K}}^{\mathbf{+}}\mathbf{,}{\mathbf{Cl}}^{\mathbf{-}}$

Short answer

a)${\mathrm{O}}^{2-}>{\mathrm{O}}^{-}>\mathrm{O}$

b)${\mathrm{Fe}}^{2+}>{\mathrm{Ni}}^{2+}>{\mathrm{Zn}}^{2+}$

c) $C{l}^{-}>K^{+}>C{a}^{2+}$

Step-by-step solution

Consider electron-electron repulsion:

a) The number of protons is same for $\mathrm{O},{\mathrm{O}}^{-},{\mathrm{O}}^{2-}$.all the three species have eight electrons. But they differ in their electron count. O has eight electrons, ${\mathrm{O}}^{-}$has 9 electrons and ${\mathrm{O}}^{-2}$ha 10 electrons. Thus the effective nuclear charge decreases in the order $O^{2-}<O^{-}<O$. As the effective nuclear charge decreases, the attraction between the nucleus and the outermost valence electron decreases. Thus the atomic radii decrease in the following order.

$O^{2-}>O^{-}>O$

Consider trend of conic radii in a period:

b) On moving left to right in a period, effective nuclear charge increases. As a result, ionic radii decrease. So, the following trends are observed.

${\mathrm{Fe}}^{2+}>{\mathrm{Ni}}^{2+}>{\mathrm{Zn}}^{2+}$

Consider the iso-electronic ions:

c) are isolectronic. But the number of protons are 17, 19 and 20 for $C{l}^{-},{\mathrm{K}}^{+}$and ${\mathrm{Ca}}^{+2}$respectively. Thus the effective nuclear charge follows the order ${\mathrm{Cl}}^{-}<{\mathrm{K}}^{+}<{\mathrm{Ca}}^{2+}$.with increase in effective nuclear charge, ionic radii decreases. Hence, the following trends are observed.

${\mathrm{Cl}}^{-}>{\mathrm{K}}^{+}>{\mathrm{Ca}}^{2+}$