Question

Consider the following reaction: $\text{MnO}_2 + \text{KOH} + \text{O}_2 \rightarrow \text{A} + \text{H}_2\text{O}$. Product ‘A’ in neutral or acidic medium disproportionates to give products ‘B’ and ‘C’ along with water. The sum of spin-only magnetic moment values of B and C is _________ BM (nearest integer).
(Given atomic number of Mn is 25)

Answer 1

Reaction Analysis:

The reaction:

$\text{MnO}_2 + \text{KOH} + \text{O}_2 \to \text{K}_2\text{MnO}_4 + \text{H}_2\text{O}$

where product $A = \text{K}_2\text{MnO}_4$.

Disproportionation of $\text{K}_2\text{MnO}_4$:

In a neutral or acidic medium, $\text{K}_2\text{MnO}_4$ disproportionates as follows:

$\text{K}_2\text{MnO}_4 \to \text{KMnO}_4 + \text{MnO}_2$

where: Product ‘$B$’ is $\text{KMnO}_4$,
Product ‘$C$’ is $\text{MnO}_2$.

Calculating Spin-Only Magnetic Moment:

For $\text{KMnO}_4$: Manganese in $\text{KMnO}_4$ has an oxidation state of $+7$, which has no unpaired electrons. Therefore, the magnetic moment for $\text{KMnO}_4$ is $0 \, \text{BM}$.

For $\text{MnO}_2$: Manganese in $\text{MnO}_2$ has an oxidation state of $+4$, with a $3d^3$ electron configuration.
Number of unpaired electrons $n = 3$. The spin-only magnetic moment $\mu$ is calculated as:

$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM}$

Sum of Magnetic Moments of B and C:

$\text{Magnetic moment of B (KMnO}_4) + \text{C (MnO}_2) = 0 + 3.87 \, \text{BM (nearest integer)}$

Conclusion:

The sum of the spin-only magnetic moment values of $B$ and $C$ is $4 \, \text{BM}$.