Consider the following reaction
$$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O} \text {. }$$
Product '$\mathrm{A}$' in neutral or acidic medium disproportionate to give products '$\mathrm{B}$' and '$\mathrm{C}$' along with water. The sum of spin-only magnetic moment values of $\mathrm{B}$ and $\mathrm{C}$ is ________ BM. (nearest integer) (Given atomic number of $\mathrm{Mn}$ is 25)
Answer (integer)
4
Solution
<p>$\mathrm{A}$ is $\mathrm{K}_2 \mathrm{MnO}_4$</p>
<p>$\mathrm{B}$ and $\mathrm{C}$ are $\mathrm{KMnO}_4$ and $\mathrm{MnO}_2$</p>
<p>$\mathrm{KMnO}_4(\mu=0)$</p>
<p>$\mathrm{MnO}_2(\mathrm{Mn}^{4+})(\mu=3.87)$</p>
<p>Sum $=3.87=4$ (Nearest integer)</p>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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