Question

Give the electronic configuration of the given metal (atom/ions).
$\text{MnO}_{2}$
A) $[\mathrm{Ar}]^{18} 3 \mathrm{s}^{2} 3 \mathrm{p}^{1}$
B) $[\mathrm{Ar}]^{18} 3 \mathrm{d}^{1} 4 \mathrm{s}^{2}$
C) $[\mathrm{Ar}]^{18} 3 \mathrm{d}^{3}$
D) None of these

Answer 1

We know that the total number of unidentate ligands attached to the central metal atom or ion through coordinate bond is named as the coordination number of the metal ion or atom. The number ligands present also vary with the oxidation state of metal atoms.

Complete step by step answer:
We know that the atomic number of manganese is 25. It belongs to the d-block of the periodic table.

The electronic configuration of manganese as $[\mathrm{Ar}]^{18} 3 \mathrm{d}^{5} 4 \mathrm{s}^{2}$. The last electron in accordance with the Aufbau principle the last electron enters in the 3\text{d} orbital, that’s why it is regarded as the d-block element.
The given compound is $\text{MnO}_{2}$. We can assume the oxidation state of manganese to be x. We know that the stable oxidation state of oxygen atom is -2.
Therefore, we can calculate the oxidation state of manganese in this compound as follows.
x+2(-2) = 0
x = +4
Thus, the oxidation state of manganese in $\text{MnO}_{2}$ is +4.
For the formation of $\text{Mn}^{4+}$ ions, we have to remove four electrons from the manganese atom. In accordance with the Pauli Exclusion Principle the energy for the 4s orbital is more than that of the 3d orbital. Therefore, we will first remove the two electrons from 4s orbital and the other two electrons from 3d orbital for the formation of $\text{Mn}^{4+}$ ion.
Therefore, we can conclude that the configuration of $\text{Mn}^{4+}$ ion is $[\mathrm{Ar}]^{18} 3 \mathrm{d}^{3}$.

Hence the correct option is C.

Note: The oxidation number of the central atom is defined as the charge that it carries as calculated by assigning appropriate charges to the ligands and equating sum of the charges on the central atom and the ligands equal to the charge on the coordination sphere.