Which one of the following complexes will most likely absorb visible light? (At nos. $\mathrm{Sc}=21,…

Which one of the following complexes will most likely absorb visible light? (At nos. $\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Zn}=30$ )
  1. $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
  2. $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{4+}$
  3. $\left[\mathrm{V}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
  4. $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$

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 $\mathrm{Sc}^{3+}(18)=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^0 4 s^0-$ no unpaired $\mathrm{e}^{-}$. $\mathrm{Ti}^{4+}(18)=1 s^2 2 \mathrm{~s}^2 2 p^6 3 s^2 3 p^6 3 d^0 4 s^0-$ no unpaired $\mathrm{e}^{-}$. $\mathrm{V}^{3+}(20)=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^2 4 s^0$ - Two unpaired $\mathrm{e}^{-}$. $\mathrm{Zn}^{2+}(28)=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^0-$ no unpaired $\mathrm{e}^{-}$. hence $\left[\mathrm{V}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ will most likely absorb visible light.

Asked in: JEE Main 2014 (12 Apr Online)

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