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 ______ B.M.

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Solution

Q.
Consider the following reaction

${MnO}_{2} + KOH + O_{2} \to A + H_{2} O .$

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 ______ B.M. (nearest integer) (Given atomic number of Mn is 25). [2024]

Ans.
(4)

Potassium permanganate is prepared by fusion of ${MnO}_{2}$ with an alkali metal hydroxide and an oxidising agent. This produces the dark green $K_{2} {MnO}_{4}$ which disproportionates in a neutral or acidic solution to give permanganate.

$2 {MnO}_{2} + 4 KOH + O_{2} \to 2 K_{2} {MnO}_{4} (A) + 2 H_{2} O$

$3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} (B) + {MnO}_{2} (C) + 2 H_{2} O$

Compound Electronic
configuration of
central metal ion Number of
unpaired
electron in
central metal
ion (n) Spin only
magnetic
moment ( $μ$ ) =
$\sqrt{n (n + 2)}$ BM

${MnO}_{4}^{-} (B)$ ${Mn}^{7 +} = {}_{18}{[Ar]} 3 d^{0}$ 0 0 BM

${MnO}_{2} (C)$ ${Mn}^{4 +} = {}_{18}{[Ar]} 3 d^{3}$ 3 $\sqrt{3 (3 + 2)}$ BM
= 3.87 BM

Sum = (0 + 3.87) BM = 3.87 BM

Nearest integer is 4

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