The disproportionation of MnO _ 4 ^ 2 - in acidic medium resulted...

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

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

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

The oxidation state of $Mnin{MnO}_{2}isless\left({Mn}^{4 +}\right)so, itisB$

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

Where, $n = 3$

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

$⇒ \sqrt{3\left(3 + 2\right)}$

$⇒ \sqrt{3\left(5\right)}$

$⇒ \sqrt{15}$

$⇒ 3 . 87 \approx 4B . M .$

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