JEE Main 2014 — Coordination Compounds Question with Solution

The absorption of visible light and hence coloured nature of the transition metal cation is due to the promotion of one or more unpaired - $d$ - electron from a lower to higher level within same $d$-subshell. Hence higher will be the number of unpaired electron higher will be the absorpion in visible light. The electronic configuration of the given elements is ${\mathrm{Sc}}^{3+}(18)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0-$ no unpaired ${\mathrm{e}}^{-}$. ${\mathrm{Ti}}^{4+}(18)=1s^{2}2{\mathrm{s}}^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0-$ no unpaired ${\mathrm{e}}^{-}$. ${\mathrm{V}}^{3+}(20)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}4s^0$ - Two unpaired ${\mathrm{e}}^{-}$. ${\mathrm{Zn}}^{2+}(28)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^0-$ no unpaired ${\mathrm{e}}^{-}$. hence ${\left[\mathrm{V}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ will most likely absorb visible light.