Question

Fusion of MnO\(_2\) with KOH in presence of KNO\(_3\) produces a dark green colour compound "X". X disproportionates in acidic solution and gives 'Y', 'Z' and water. The sum of spin only magnetic moment values of 'Y' and 'Z' is:

• A. 3.87
• B. 5.92
• C. 4.90
• D. 2.83

Hint:

Remember the reactions of MnO\(_2\) with KOH and the disproportionation of manganate ions. Also, know how to calculate the spin-only magnetic moment using the number of unpaired electrons.

Answer 1

The Correct Option is A

1. Formation of X:

\( \text{MnO}_2 + 2\text{KOH} + \text{KNO}_3 \rightarrow \text{K}_2\text{MnO}_4 + \text{KNO}_2 + \text{H}_2\text{O} \)

X is \( \text{K}_2\text{MnO}_4 \) (potassium manganate), which is dark green.

2. Disproportionation of X in acidic solution:

\( 3\text{K}_2\text{MnO}_4 + 4\text{HCl} \rightarrow 2\text{KMnO}_4 + \text{MnO}_2 + 2\text{H}_2\text{O} + 4\text{KCl} \)

Y is \( \text{KMnO}_4 \) (potassium permanganate), which is purple.

Z is \( \text{MnO}_2 \) (manganese dioxide), which is brown.

3. Oxidation states and electronic configurations:

In \( \text{KMnO}_4 \) (Y), Mn is in +7 oxidation state. Mn\(^{7+}\) has a configuration of [Ar] 3d\(^0\), so it has 0 unpaired electrons.

In \( \text{MnO}_2 \) (Z), Mn is in +4 oxidation state. Mn\(^{4+}\) has a configuration of [Ar] 3d\(^3\), so it has 3 unpaired electrons.

4. Spin-only magnetic moment:

Spin-only magnetic moment (\( \mu \)) = \( \sqrt{n(n+2)} \) BM, where \( n \) is the number of unpaired electrons.

For Mn\(^{7+}\) (Y), \( n = 0 \), \( \mu = \sqrt{0(0+2)} = 0 \) BM.

For Mn\(^{4+}\) (Z), \( n = 3 \), \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \) BM.

5. Sum of magnetic moments:

Sum = 0 + 3.87 = 3.87 BM

Therefore, the sum of spin-only magnetic moment values of 'Y' and 'Z' is 3.87.

Final Answer:
3.87.