Chapter 22, Problem 22.49QP

(a)

Interpretation Introduction

Interpretation: The charges of the $\text{Mn}$ ions in each of the given mineral are to be calculated. The compound in which there could be high spin and low spin $\text{Mn}$ ions is to be identified.

Concept introduction: Strong ligands produce high crystal field splitting and weak ligands produce small crystal field splitting. High spin complex ions have low crystal field splitting.

To determine: The charges of the $\text{Mn}$ ions in each of the given mineral.

Answer to Problem 22.49QP

Solution

The charge on $\text{Mn}$ in ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ is $\underset{\_}{+3}$ and the charge on $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $\underset{\_}{+4}$.

Explanation of Solution

Explanation

The manganese minerals pyrolusite, ${\text{MnO}}_{\text{2}}$ and hausmannite, ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ contain manganese ions surrounded by oxide ions.

The charge on $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is calculated as,

$\begin{array}{c}\text{Charge}\text{on}\text{compound}=\left(\text{Number}\text{of}\text{Mn}\text{atoms}\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Charge}\text{on}\text{O}\text{atom}\right)\end{array}$

There is $1$ $\text{Mn}$ atom and $2$ $\text{O}$ atoms present in ${\text{MnO}}_{\text{2}}$. Charge on oxygen atom is $-2$. Charge on the compound is $0$.

Substitute the value of number of atoms of manganese and oxygen atoms and charge on the compound in the above equation.

$\begin{array}{c}\text{0}=\left(\text{1}\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\left(\text{2}\times \text{-2}\right)\\ \text{Charge}\text{on}\text{Mn}\text{atom}=\underset{\_}{+4}\end{array}$

Therefore, the charge on $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $\underset{\_}{+4}$.

Manganese is present in two oxidation states $+2$ and $+3$. Formula of ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ is sometimes written as $\text{MnO}{\text{.Mn}}_2{\text{O}}_3$.

The charge on $\text{Mn}$ in $\text{MnO}$ is calculated as,

$\begin{array}{c}\text{Charge}\text{on}\text{compound}=\left(\text{Number}\text{of}\text{Mn}\text{atoms}\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Charge}\text{on}\text{O}\text{atom}\right)\end{array}$

There is $1$ $\text{Mn}$ atom and $1$ $\text{O}$ atom present in $\text{MnO}$. Charge on oxygen atom is $-2$. Charge on the compound is $0$.

Substitute the value of number of atoms of manganese and oxygen atoms and charge on the compound in the above equation.

$\begin{array}{c}\text{0}=\left(1\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\left(1\times \text{-2}\right)\\ \text{Charge}\text{on}\text{Mn}\text{atom}=+2\end{array}$

The charge on $\text{Mn}$ in ${\text{Mn}}_2{\text{O}}_3$ is calculated as,

$\begin{array}{c}\text{Charge}\text{on}\text{compound}=\left(\text{Number}\text{of}\text{Mn}\text{atoms}\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\\ \left(\text{Number}\text{of}\text{O}\text{atoms}\times \text{Charge}\text{on}\text{O}\text{atom}\right)\end{array}$

There are $2$ $\text{Mn}$ atoms and $3$ $\text{O}$ atoms present in ${\text{Mn}}_2{\text{O}}_3$. Charge on oxygen atom is $-2$. Charge on the compound is $0$.

Substitute the value of number of atoms of manganese and oxygen atoms and charge on the compound in the above equation.

$\begin{array}{c}\text{0}=\left(2\times \text{Charge}\text{on}\text{Mn}\text{atom}\right)+\left(3\times \text{-2}\right)\\ \text{Charge}\text{on}\text{Mn}\text{atom}=+3\end{array}$

Therefore, charge on $\text{Mn}$ in ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ is $\underset{\_}{+3}$.

(b)

Interpretation Introduction

To determine: The compound in which there could be high spin and low spin $\text{Mn}$ ions.

Answer to Problem 22.49QP

Solution

The compound in which there could be high spin and low spin $\text{Mn}$ ions is ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$.

Explanation of Solution

Explanation

The charge on $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $+4$.

The electronic configuration of $\text{Mn}$ is $\left[\text{Ar}\right]{\text{3d}}^5{\text{4s}}^{\text{2}}$. With the loss of four electrons, it becomes ${\text{Mn}}^{\text{4+}}$. The electronic configuration of ${\text{Mn}}^{\text{4+}}$ is $\left[\text{Ar}\right]{\text{3d}}^3$. There are three electrons present in the d orbital. Therefore, all the three electrons will occupy the lower energy orbital ( ${\text{t}}_{\text{2}}\text{g}$). The distribution of electrons is as follows:

Figure 1

There is only high spin state in ${\text{MnO}}_{\text{2}}$.

The charge on $\text{Mn}$ in ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ is $+2,+3$.

The electronic configuration of $\text{Mn}$ is $\left[\text{Ar}\right]{\text{3d}}^5{\text{4s}}^{\text{2}}$. With the loss of two electrons, it becomes ${\text{Mn}}^{2+}$. The electronic configuration of ${\text{Mn}}^{2+}$ is $\left[\text{Ar}\right]{\text{3d}}^5$. There are five electrons present in the d orbital. In case of high spin state, the distribution of electrons is as follows:

Figure 2

Three electrons will occupy the lower energy orbital ( ${\text{t}}_{\text{2}}\text{g}$) and two electrons will occupy higher energy orbital ( $\text{eg}$).

In case of low spin state, pairing of electrons takes place. The distribution of electrons is as follows:

Figure 3

All electrons will occupy the lower energy orbital ( ${\text{t}}_{\text{2}}\text{g}$) in low spin state.

Therefore, the compound in which there could be high spin and low spin $\text{Mn}$ ions is ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$.

Conclusion

The compound in which there could be high spin and low spin $\text{Mn}$ ions is ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$.