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. $\left[Ti \left(NH_{3}\right)_{6}\right]^{4+}$
• C. $\left[V\left(NH_{3}\right)_{6}\right]^{3+}$
• D. $\left[Zn\left(NH_{3}\right)_{6}\right]^{2+}$

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

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

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

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

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

hence $[V(NH_3)_6]^{3+}$ will most likely absorb visible light.