Question

The electronic configuration of $M{{n}^{+2}}$ is:
(A)- $\left[ Ar \right]3{{d}^{4}}4{{s}^{1}}$
(B)- $\left[ Ar \right]3{{d}^{5}}4{{s}^{0}}$
(C)- $\left[ Ar \right]4{{d}^{5}}4{{s}^{0}}$
(D)- $\left[ Ar \right]3{{d}^{2}}4{{s}^{2}}$

Answer 1

In order to obtain the electronic configuration of manganese ions, we remove two electrons from its outermost shell of Mn atom (in its ground state) such that the obtained configuration is stable.

Complete step by step answer:
The atomic number of manganese is 25. The filling of electrons in the shell follows the Aufbau principle, which states that the electrons are filled first in the lower energy atomic orbitals before the higher levels following the (n + l) rule. Thus, the electronic configuration of the ground state of Mn is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{5}}$ or $\left[ Ar \right]4{{s}^{2}}3{{d}^{5}}$.

In the excited state, when it loses two electrons, it is removed from the 4s orbital, leaving the 3d orbital in a half-filled state which is very stable. Thus, we have excited state electronic configuration of $M{{n}^{+2}}$ to be $\left[ Ar \right]3{{d}^{5}}4{{s}^{0}}$.
Therefore, the electronic configuration of $M{{n}^{+2}}$ is option (B)- $\left[ Ar \right]3{{d}^{5}}4{{s}^{0}}$.
So, the correct answer is “Option B”.

Additional Information:
In the (n + l) rule, where n is the principal quantum number (related to size of orbital) and l is angular momentum quantum number (related to shape of orbital). The (n + l) represents the energy, so the orbitals are filled in increasing order of (n + l). For orbitals with the same value of (n + l), the increasing order of n is taken into account.

Note: During the filling of electrons, the 4s subshell has lower energy than 3d, but it is still more stable. However, during the removal of electrons, the 3d subshell being closer to the nucleus, makes the loss of electrons easier from the 4s subshell which is farther away (outermost) from it.