The disproportionation of… - JEE Main 2022 Chemistry PYQ

JEE Main 2022 — Chemistry Inorganic Chemistry

2022integermedium

The disproportionation of ${\mathrm{MnO}}_{4}^{2-}$ in acidic medium resulted in the formation of two manganese compounds $A$ and $B$ . If the oxidation state of $\mathrm{Mn}$ in $B$ is smaller than that of $A$ , then the spin-only magnetic moment $(\mu )$ value of $B$ in $\mathrm{BM}$ is____ . (Nearest integer)

Official previous-year question

Held on 28 Jul 2022 · Verified 6 Jul 2026.

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Solution

Potassium permanganate Is prepared by fusion of ${\mathrm{MnO}}_{2}$ with an alkali metal hydroxide and an oxidising agent like ${\mathrm{KNO}}_{3}$ . This produces the dark green ${K}_{2}{\mathrm{MnO}}_{4}$ which disproportionates in a neutral or acidic solution to give permanganate.

$2{\mathrm{MnO}}_{2}+4\mathrm{KOH}+{O}_{2}\rightarrow 2{K}_{2}{\mathrm{MnO}}_{4}+2{H}_{2}O$

$3{{\mathrm{MnO}}_{4}}^{2-}+4{H}^{-}\rightarrow 2{{\mathrm{MnO}}_{4}}^{-}+{\mathrm{MnO}}_{2}+2{H}_{2}O$

The oxidation state of $\mathrm{Mn}\mathrm{in}{\mathrm{MnO}}_{2}\mathrm{is}\mathrm{less}({\mathrm{Mn}}^{4+})\mathrm{so},\mathrm{it}\mathrm{is}B$

${\mathrm{Mn}}^{4+}=[\mathrm{Ar}]3{d}^{3}$

Where, $n=3$

${\mu }_{s}=\sqrt{n(n+2)}$

$\Rightarrow \sqrt{3(3+2)}$

$\Rightarrow \sqrt{3(5)}$

$\Rightarrow \sqrt{15}$

$\Rightarrow 3.87\approx 4B.M.$

Therefore, the spin only magnetic moment of ${\mathrm{Mn}}^{4+}$ ion is approximately $4B.M.$

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Solution

$2{\mathrm{MnO}}_{2}+4\mathrm{KOH}+{O}_{2}\rightarrow 2{K}_{2}{\mathrm{MnO}}_{4}+2{H}_{2}O$

$3{{\mathrm{MnO}}_{4}}^{2-}+4{H}^{-}\rightarrow 2{{\mathrm{MnO}}_{4}}^{-}+{\mathrm{MnO}}_{2}+2{H}_{2}O$

The oxidation state of $\mathrm{Mn}\mathrm{in}{\mathrm{MnO}}_{2}\mathrm{is}\mathrm{less}({\mathrm{Mn}}^{4+})\mathrm{so},\mathrm{it}\mathrm{is}B$

${\mathrm{Mn}}^{4+}=[\mathrm{Ar}]3{d}^{3}$

Where, $n=3$

${\mu }_{s}=\sqrt{n(n+2)}$

$\Rightarrow \sqrt{3(3+2)}$

$\Rightarrow \sqrt{3(5)}$

$\Rightarrow \sqrt{15}$

$\Rightarrow 3.87\approx 4B.M.$

Therefore, the spin only magnetic moment of ${\mathrm{Mn}}^{4+}$ ion is approximately $4B.M.$

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Solution

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JEE Main Inorganic Chemistry in other years

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Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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