Consider the following reaction MnO2 + KOH + O2 → A + H2O.

Question

Consider the following reaction

\(MnO_2 + KOH + O_2 \rightarrow A + H_2O.\)

Product ‘A’ in neutral or acidic medium disproportionate 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

Correct answer is : 4

\(\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \underset{(A)}{K_2MnO_4} +H_2O\)

\(\mathrm{K}_2 \mathrm{MnO}_4 \xrightarrow{\text { Neutral/acidic solution }} \mathrm{KMnO}_4+\mathrm{MnO}_2\)

\(\mathrm{Mn}^{+4}:-[\mathrm{Ar}] 3 \mathrm{~d}^3\)

\(\mathrm{n}=3, \mu=\sqrt{3(3+2)}=3.87 \text { B.M. }\)

Nearest integer is (4)