Chapter 21, Problem 21.1P

Interpretation Introduction

Interpretation:

The electronic configuration of $\text{Mn}$, and the number of unpaired electrons in ${\text{MnO}}_{\text{2}}$ is to be determined.

Concept introduction:

The filling of the electrons in the orbital can be done by the three sets of the principle are as follows:

(a). The Afbau principle:- This principle is stated that the lower energy of the orbital would be assigned electrons before the higher energy of the orbital.

(b). The Hund’s rule:-According to this rule, the pairing of the electrons can be done in the subshell after each of the subshells has a single electron.

(c). The Pauli’s exclusion principle:-This principle is stated that both electrons are present in either positive half-spin or negative half spin.

Answer to Problem 21.1P

The electronic configuration of ${\text{Mn}}^{4+}$ is shown below:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$

The number of unpaired electrons is present in the compound ${\text{MnO}}_{\text{2}}$ is $3$ that is in $3d$ orbital.

Explanation of Solution

The sum of the oxidation number of all the atoms that are present in the neutral compound is equal to zero.

Consider the oxidation number of $\text{Mn}$ will be $x$. When oxygen is reacted with metal than the oxidation number of oxygen will be $\left(-2\right)$

The oxidation number of $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is calculated as follows:

$\left(\text{Oxidation number of Mn}\right)+2\left(\text{Oxidation number of O}\right)=0$

The value of the oxidation number of $\text{Mn}$ is $x$.

The oxidation number of oxygen is $-2$.

Substitute the value in the above expression.

$\left(x\right)+2\left(-2\right)=0$

Rearrange the above expression for $x$.

$\left(x\right)=+4$

Thus, the oxidation number of $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $+4$.

The atomic number of $\text{Mn}$ is $25$. The atomic number of the atom is equal to number electrons or number of the proton. Thus, the number of electrons is present in the manganese atom is $25$. The electronic configuration of neutral $\text{Mn}$ is represented below:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$

In the compound ${\text{MnO}}_{\text{2}}$, the oxidation number of $\text{Mn}$ is $+4$. This shows that $\text{Mn}$ loses four electrons to attain ${\text{Mn}}^{4+}$. The number of electrons are present in ${\text{Mn}}^{4+}$ is $21$. The two electrons are removed from $s$ subshell and the other two electrons are removed from $d$ subshell. The electronic configuration of ${\text{Mn}}^{4+}$ is represented below:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$

The above electronic configuration of ${\text{Mn}}^{4+}$ can be written in the form noble gas configuration is shown below:

$\left[\text{Ar}\right]3d^3$

The number of unpaired electrons is present in the compound ${\text{MnO}}_{\text{2}}$ is calculated as shown below:

The electronic configuration of ${\text{Mn}}^{4+}$ is represented below:

$\left[\text{Ar}\right]3d^3$

The structure of $3d^3$ orbital can be shown below:

$3d^3\to \underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\underset{\_}{0}\underset{\_}{0}$

From the above structure of $3d^3$, there is three unpaired electrons are present.

Therefore, the number of unpaired electrons present in the compound ${\text{MnO}}_{\text{2}}$ is $3$.

Conclusion

The electronic configuration of ${\text{Mn}}^{4+}$ is represented below:

$\left[\text{Ar}\right]3d^3$

The number of unpaired electrons present in the compound ${\text{MnO}}_{\text{2}}$ is $3$.