Question

Which one of the following complexes will most likely absorb the visible light?
[Atomic numbers: Sc = 21, Ti = 22, V = 23, Zn = 30]
A. \[{[Sc{({H_2}O)_6}]^{ + 3}}\]
B. \[{[Ti{(N{H_3})_6}]^{ + 4}}\]
C. \[{[V{(N{H_3})_6}]^{ + 2}}\]
D. \[{[Zn{(N{H_3})_6}]^{ + 2}}\]

Answer 1

Hint: The complexes of those transition metals are expected to absorb visible light in which the d – subshell is incomplete (i.e. has unpaired electron(s)) and the excitation of the electron from a lower energy orbital to a higher energy orbital is possible.

Complete step by step answer:
Transition metal complexes absorb visible light only when they have an incomplete d – subshell. Also, when there is a possible transition of an electron from a lower energy orbital to a higher energy orbital.
In the given question, we will have to investigate each option individually:
A. \[{[Sc{({H_2}O)_6}]^{ + 3}}\]

As the water is a neutral ligand, thus the +3 – oxidation charge is on Scandium.

\[Sc = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^1}\]

\[\therefore S{c^{ + 3}} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\]

As the +3 – oxidation state of Scandium has no unpaired electron, thus it can’t absorb visible light.

B. \[{[Ti{(N{H_3})_6}]^{ + 4}}\]

As ammonia is a neutral ligand, thus the +4 – oxidation charge is on titanium.

\[Ti = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^2}\]

\[\therefore T{i^{ + 4}} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\]

As the +4 – oxidation state of Titanium has no unpaired electron, thus it can’t absorb visible light.

C. \[{[V{(N{H_3})_6}]^{ + 2}}\]

As ammonia is a neutral ligand, thus the +2 – oxidation charge is on Vanadium.

\[V = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^3}\]

\[{V^{ + 2}} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}\]

As the +2 – oxidation state of Vanadium has 3 unpaired electrons, thus it is capable of absorbing visible light.

D. \[{[Zn{(N{H_3})_6}]^{ + 2}}\]

As ammonia is a neutral ligand, thus the +2 – oxidation charge is on Zinc.

\[Zn = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}\]

\[Z{n^{ + 2}} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}\]

As the +2 – oxidation state of Zinc has no unpaired electron, thus it can’t absorb visible light.

Hence, the correct answer is (C).

Note: Remember that according to the Aufbau’s Rule, when the electrons are filled, they are first filled into the 4s orbital and then the 3d orbital. When the electrons are removed, the electrons from 4s orbitals are removed first and then from 3d orbitals.