\(\underset{Pyrolusite\\ (A)\ Blackish\ brown}{MnO_2 }+ 4KOH + O_2\xrightarrow{Fuse}\underset{Potassium\ magnate\\ (B)-Green\ coloured}{2K_2MnO_4}+2H_2O\)
\(\underset B{2 KMnO_4} + H20 + (0)_2 \xrightarrow[medium]{Alkaline}\underset{Potasium\ permanganate\\ (C)-Purple\ coloured}{2KMnO_4}+2KOH\)
or \(MnO^{-2}_4 \xrightarrow{} MnO_4^- + e^-\)
(b) When acidic solution of green compound (B), i·.e., potassium manganate is allowed to stand for some time, it disproportionates to give permanganate as follows:
\(3 MnO^{2-}_4+ 4H^+\longrightarrow2 MnO^-_4 + MnO_2 + 2 H_2O\)
This reaction is called disproportionation reaction.

\(4 FeO. Cr_2O_3 + O_2\longrightarrow 2 Fe_2O_3 + 4 Cr_2O_3\)
\(\underline{4 Na_2CO_3 + 2 Cr_2O_3 + 3 O_2 \longrightarrow 4 Na_2CrO_4 + 4 CO_2 \times 2}\)_{_________________}
\(4 FeO. Cr_2O_3 + 8 Na_2CO_3 + 7 O_2\longrightarrow\underset{Sodium\ chromate} {\underset{{A}}{8 Na_2CrO_4}} + 2 Fe_2O_3 + 8 CO_2\)
\(2 Na_2CrO_4 + H_2SO_4 \longrightarrow\underset{(B)\\sodium\ choromate} {Na_2Cr_2O_7} + Na_2SO_4 + H_20\)

The electronic configuration of Hg(I), i.e., Hg⁺ is [Xe] 4f¹⁴5d¹⁰ 6s¹ and thus has one electron in the valence 6s-orbital. If this were so, all Hg (I) compounds should be paramagnetic but actually they are diamagnetic. This behaviour can be explained if we assume that the singly filled 6s-orbitals of the two Hg⁺ ions overlap to form a Hg-Hg covalent bond. Thus, Hg+ions exist as dimeric species, i.e., Hg_2²⁺. In contrast, the electronic configuration of Cu(l) ion, i.e., Cu⁺ is [Ar] 3d¹⁰.Therefore, it has no unpaired b electrons to form dimeric species, i.e., Cu₂²⁺ and hence it always exists as Cu⁺ ion.