Fusion of MnO₂ with KOH in presence of O₂ produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt X. The Managanese containing ions present in W and X, respectively, are Y and Z. Correct statement (s) is ( are)

$\begin{array}{l}Mn{O}_2+2KOH+\frac{1}{2}{O}_2\stackrel{\Delta }{\to }{K}_{2}Mn{O}_4+H_{2}O\\ \left(W\right)potassium\\ mangate\end{array}$

$\begin{array}{l}{K}_{2}Mn{O}_4\left(aq\right)\stackrel{ }{\rightleftarrows }2K^{+}\left(aq\right)+Mn{O}_4{}^{2-}\left(aq\right)\\ \left(W\right)\left(Y\right)\end{array}$

$Mn{O}_4^{2-}$ion has one unpaired electrons, therefore it gives d-d transition to form green colour. Y complex has paramagnetic nature due to presence of one unpaired electron.In aqueous solution,

$\underset{\left(W\right)}{{\mathrm{K}}_2\mathrm{MnO}}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\stackrel{\text{Electrolytic oxidation}}{\to }{\mathrm{H}}_2+\mathrm{KOH}+\underset{\left(X\right)}{{\mathrm{KMnO}}_4}$${\mathrm{KMnO}}_4\left(aq\right)\stackrel{\mathrm{\Delta }}{⟶}{K}^{+}+\underset{\left(Z\right)}{{\mathrm{MnO}}_4^{-}}$

$Mn{O}_4^{-}$ions gives charge transfer spectrum in which a fraction of electronic charge is transferred between the molecular entities.

$\because Mn{O}_4^{2-}\underset{oxidation}{\overset{Electrolytic}{\to }}Mn{O}_4^{-}+e^{-}$

In acidic medium, Y undergoes disproportionation reaction

$\begin{array}{rl}\underset{\left(Y\right)}{3{\mathrm{MnO}}_4^{2-}}& \left(aq\right)+4{\mathrm{H}}^{+}\to \underset{\left(Z\right)}{2{\mathrm{Mno}}_4^{-}}+{\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}\end{array}$

$\underset{\left(Y\right)}{{\mathrm{MnO}}_4^{2-}} and \underset{\left(z\right)}{{\mathrm{MnO}}_4^{-} }$  both ions form $\pi$-bonding between p-orbitals of oxygen and d-orbitals of manganese. Thus, options (a,c,d) are correct.