Question

Which one of the following complexes will most likely absorb visible light ?
(At nos. $Sc=21,Ti=22,V=23,Zn=30)$

• A. ${\left[Sc{\left(H_{2}O\right)}_6\right]}^{3+}$
• B. ${\left[Ti{\left(N{H}_3\right)}_6\right]}^{4+}$
• C. ${\left[V{\left(N{H}_3\right)}_6\right]}^{3+}$
• D. ${\left[Zn{\left(N{H}_3\right)}_6\right]}^{2+}$

Answer 1

Answer: (C)

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 J-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

$S{c}^{3+}(18)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0$ -no impaired $e^{-}.$

$T{i}^{4+}(18)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0$ - no unpaired $e^{-}.$

$V^{4+}(20)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0$ - Two unpaired $e^{-}.$

$Z{n}^{2+}(28)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0$ - no unpaired $e^{-}.$

hence $[V(N{H}_3{)}_6{]}^{3+}$ will most likely absorb visible light.