Chemistry - Complete d-block Question with Solution | TestHub

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 ${\mathrm{K}}_2{\mathrm{MnO}}_4$ which disproportionates in a neutral or acidic solution to give permanganate.

$2{\mathrm{MnO}}_2+4\mathrm{KOH}+{\mathrm{O}}_2\to 2{\mathrm{K}}_2{\mathrm{MnO}}_4+2{\mathrm{H}}_2\mathrm{O}$

$3{{\mathrm{MnO}}_4}^{2-}+4{\mathrm{H}}^{-}\to 2{{\mathrm{MnO}}_4}^{-}+{\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}$

The oxidation state of $\mathrm{Mn} \mathrm{in} {\mathrm{MnO}}_2 \mathrm{is} \mathrm{less} \left({\mathrm{Mn}}^{4+}\right) \mathrm{so}, \mathrm{it} \mathrm{is} \mathrm{B}$

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

Where, $\mathrm{n}=3$

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

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

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

$\Rightarrow \sqrt{15}$

$\Rightarrow 3.87\approx 4 \mathrm{B}.\mathrm{M}.$

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