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Consider the following reaction
\(\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O}\).
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 \(\qquad\) BM. (nearest integer) (Given atomic number of \(\mathrm{Mn}\) is 25)
\(\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O}\).
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 \(\qquad\) BM. (nearest integer) (Given atomic number of \(\mathrm{Mn}\) is 25)
Solution:
1926 Upvotes
Verified Answer
The correct answer is:
4
$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O}$
(A)
$\begin{aligned}
& \mathrm{K}_2 \mathrm{MnO}_4 \xrightarrow{\text { Neutral/acidic solution }} \mathrm{KMnO}_4+\mathrm{MnO}_2 \\
& \mathrm{Mn}^{+4}:-[\mathrm{Ar}] 3 \mathrm{~d}^3 \\
& \mathrm{n}=3, \mu=\sqrt{3(3+2)}=3.87 \text { B.M. }
\end{aligned}$
Nearest integer is (4)
(A)
$\begin{aligned}
& \mathrm{K}_2 \mathrm{MnO}_4 \xrightarrow{\text { Neutral/acidic solution }} \mathrm{KMnO}_4+\mathrm{MnO}_2 \\
& \mathrm{Mn}^{+4}:-[\mathrm{Ar}] 3 \mathrm{~d}^3 \\
& \mathrm{n}=3, \mu=\sqrt{3(3+2)}=3.87 \text { B.M. }
\end{aligned}$
Nearest integer is (4)
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