Question

Fusion of $Mn{{O}_{2}}$, with $KOH$ in presence of ${{O}_{2}}$ produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt X. The manganese containing ions present in W and X, respectively are Y and Z. Correct statement(s) is (are)
This question has multiple correct options
(A) In both Y and Z, π-bonding occurs between p-orbitals of oxygen and d-orbitals of manganese
(B) Both Y and Z are coloured and have tetrahedral shape
(C) In aqueous acidic solution, Y undergoes disproportionation reaction to give Z and $Mn{{O}_{2}}$
(D) Y is diamagnetic in nature while Z is paramagnetic

Answer 1

In the given question W is potassium manganate and X is potassium permanganate. The Potassium manganate is a salt of manganic acid and the permanganate is that of permanganic acid. In the permanganate salt (X) the oxidation state of manganese is +7 and in manganate salt it is +6.

Complete step by step solution:
-The reaction $Mn{{O}_{2}}$, with KOH in presence of ${{O}_{2}}$ produces can be represented as follows
$Mn{{O}_{2}}\xrightarrow[{{O}_{2}}]{KOH}MnO_{4}^{2-}$
That is when manganese dioxide ($Mn{{O}_{2}}$) is treated with potassium hydroxide in presence of oxygen, manganate salt (W) is being formed. The ion present in this W is $M{{n}^{+6}}$ .That is the ion is in +6 oxidation state.
- When an Alkaline solution of W is subjected to electrolytic oxidation the following reaction occurs
$MnO_{4}^{2-}\xrightarrow[Oxidation]{Electrolytic}MnO_{4}^{-}$
That is when manganate salt (W) undergoes electrolytic oxidation, permanganate salt is being formed and the ion present in X is $M{{n}^{+7}}$. That is the ion is in +7 oxidation state. As we know, in both Y and Z, π-bonding occurs between p-orbitals of oxygen and d-orbitals of manganese and hence option (A) is correct.
- Both the $MnO_{4}^{2-}$ and $MnO_{4}^{-}$ are coloured.The permanganate ion is purple coloured and the manganate ion is green in colour. They both have tetrahedral geometry. Thus option (B) is also correct.
- In acidic medium, $MnO_{4}^{2-}$ undergoes disproportionation reaction and the reaction can be written as follows
$MnO_{4}^{2-}\xrightarrow[medium]{Acidic}Mn{{O}_{2}}+MnO_{4}^{-}$
As we can see the products formed are Z ($MnO_{4}^{-}$) and$Mn{{O}_{2}}$.Hence, option (C) is also correct.
- In $MnO_{4}^{2-}$ an unpaired electron is present and hence it (Y) is paramagnetic in nature and in $MnO_{4}^{-}$ the electrons are paired in the outermost configuration and hence Z is diamagnetic. Thus option (D) is incorrect.

Therefore the correct options are (A), (B) and (C).

Note: It should be noted that, both the permanganates and manganates are powerful oxidizing agents. As we mentioned, permanganate in the metal is in higher oxidation state than in the manganate. Therefore the permanganate is expected to be a more powerful oxidizing agent than the manganate.