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)
(Given atomic number of Mn is 25)
Correct Answer: 4
Solution and Explanation
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 \( 4 \, \text{BM} \).
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