Question

Correct electronic configuration of $C{{o}^{2+}}$ion is:
(a)- $[Ar]4{{s}^{2}}3{{d}^{5}}$
(b)- $[Ar]3{{d}^{7}}$
(c)- $[Ar]4{{s}^{1}}3{{d}^{6}}$
(d)- $[Ar]4{{s}^{2}}3{{d}^{7}}$

Answer 1

Hint: $Co$ means cobalt, and it is an element of d-block and its atomic number is 27. $C{{o}^{2+}}$ ion means two electrons have been removed from the cobalt element, so the electrons will be removed from the outermost shell.

Complete step by step answer:
Co means cobalt, and it is an element of d-block and its atomic number is 27. So, 27 electrons are arranged in the orbitals of the cobalt. Since it belongs to the d-block, the last electron must enter into the d-orbital. So according to Aufbau’s Principle, the electrons are filled according to:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s,…..so on.
So the electronic configuration of cobalt will be:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{7}}$
Or this configuration can also be written as:
$[Ar]4{{s}^{2}}3{{d}^{7}}$
$C{{o}^{2+}}$ ion means two electrons have been removed from the cobalt element, so the electrons will be removed from the outermost shell. So when 2 electrons are removed from the cobalt element, the electrons will be removed from the 4s orbital. The configuration will be:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{7}}$
Or this configuration can be written as:
$[Ar]3{{d}^{7}}$
Although the energy of the 3d orbital is more than the energy of the 4s orbital, the electrons will be first removed from the 4s orbital because the 4s orbital is a new shell and it is the outermost shell.

So the correct answer is option (b)- $[Ar]3{{d}^{7}}$.

Note: You may get confused that option (a)- $[Ar]4{{s}^{2}}3{{d}^{5}}$ must be correct because the 3d is filling after 4s, so the electrons will be first removed from 3d, but the electrons are first removed from the outermost shell, whether its energy is high or low.