Question

The electronic configuration of the $M{{n}^{4+}}$ ion is:
(A) $\left[ Ar \right]3{{d}^{4}}4{{s}^{0}}$
(B) $\left[ Ar \right]3{{d}^{2}}4{{s}^{1}}$
(C) $\left[ Ar \right]3{{d}^{1}}4{{s}^{2}}$
(D) $\left[ Ar \right]3{{d}^{3}}4{{s}^{2}}$

Answer 1

Hint: To solve this question you must know about the atomic no. of Mn that is manganese with the atomic no. of 25. Now write the electronic configuration of Mn by putting the no of electrons in each subshell.

Complete step by step solution:
From your chemistry lessons, you have learnt about the electronic configuration and the process through which we calculate it.
 To write the electronic configuration you should know how much electrons can be filled in each cell, like s can hold a maximum of 2 electrons only and p of 6 and d of 10.
So, this is written as E.C= ${{s}^{2}}{{p}^{6}}{{d}^{10}}{{f}^{14}}$${{s}^{2}}{{p}^{6}}{{d}^{10}}{{f}^{14}}$
Now in the question, it is given that we have to find the electronic configuration of$M{{n}^{4+}}$,
Firstly we have to find the electronic configuration of Mn with atomic no 25.
..So, the electronic configuration of Mn=$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{5}}$
Or you can write it as$\left[ Ar \right]3{{d}^{5}}4{{s}^{2}}$ because argon a noble gas has an atomic no of 18.
But, here we have to find the electronic configuration of$M{{n}^{4+}}$,
So, the number of electrons in $M{{n}^{4+}}$ is 21.
E.C=$\left[ Ar \right]3{{d}^{1}}4{{s}^{2}}$
Here we have removed the electrons from the Mn configuration and the removal of electrons will be from the outermost orbital i.e, from the d orbitals.

Thus the correct option will be (C).

Note: Electronic configuration should be written carefully. Mn belongs to group seven therefore the outermost orbital of Mn will contain seven electrons only. At the time of electron filling, the no of electrons should be filled in each cell up to their maximum capacity only.