JEE Main 2022 — Chemistry Inorganic Chemistry

Potassium permanganate Is prepared by fusion of ${\mathrm{MnO}}_{2}$MnO2​ with an alkali metal hydroxide and an oxidising agent like ${\mathrm{KNO}}_{3}$KNO3​. This produces the dark green ${K}_{2}{\mathrm{MnO}}_{4}$K2​MnO4​ which disproportionates in a neutral or acidic solution to give permanganate.

$$2{\mathrm{MnO}}_{2}+4\mathrm{KOH}+{O}_{2}\rightarrow 2{K}_{2}{\mathrm{MnO}}_{4}+2{H}_{2}O$$2MnO2​+4KOH+O2​→2K2​MnO4​+2H2​O

$$3{{\mathrm{MnO}}_{4}}^{2-}+4{H}^{-}\rightarrow 2{{\mathrm{MnO}}_{4}}^{-}+{\mathrm{MnO}}_{2}+2{H}_{2}O$$3MnO4​2−+4H−→2MnO4​−+MnO2​+2H2​O

The oxidation state of $\mathrm{Mn}\mathrm{in}{\mathrm{MnO}}_{2}\mathrm{is}\mathrm{less}({\mathrm{Mn}}^{4+})\mathrm{so},\mathrm{it}\mathrm{is}B$MninMnO2​isless(Mn4+)so,itisB

$${\mathrm{Mn}}^{4+}=[\mathrm{Ar}]3{d}^{3}$$Mn4+=[Ar]3d3

Where, $n=3$n=3

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

$$\Rightarrow \sqrt{3(3+2)}$$⇒3(3+2)​

$$\Rightarrow \sqrt{3(5)}$$⇒3(5)​

$$\Rightarrow \sqrt{15}$$⇒15​

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

Therefore, the spin only magnetic moment of ${\mathrm{Mn}}^{4+}$Mn4+ ion is approximately $4B.M.$4B.M.