JEE Advanced 2019 · Paper 1 · Chemistry Q6

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ChemistryMultiple correct+4 / -1Question 24 of 54

Fusion of $\ce{MnO2}$ with KOH in presence of $\ce{O2}$ 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)

In aqueous acidic solution, Y undergoes disproportionation reaction to give Z and

\ce{MnO2}

Both Y and Z are coloured and have tetrahedral shape

Y is diamagnetic in nature while Z is paramagnetic

In both Y and Z, π-bonding occurs between

p

-orbitals of oxygen and

d

-orbitals of manganese

Answer and worked solution

Official answer: A, B, D

Identify the ions.

W is $\ce{K2MnO4}$ , from $\ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O}$ , so Y is manganate, $\ce{MnO4^{2-}}$ .
Electrolytic oxidation gives X = $\ce{KMnO4}$ , so Z is permanganate, $\ce{MnO4-}$ .

The statements.

(A) In acid, $\ce{3MnO4^{2-} + 4H+ -> 2MnO4- + MnO2 + 2H2O}$ . True.
(B) Both are tetrahedral; manganate is green and permanganate purple. True.
(C) Manganate is Mn(VI), $d^1$ , so it is paramagnetic. Permanganate is Mn(VII), $d^0$ , so it is diamagnetic. The statement has them the wrong way round. False.
(D) The Mn-O π bonds in both are $p\pi$ - $d\pi$ . True.

(A), (B), (D)

Topics:Important compounds (KMnO4, K2Cr2O7)Balancing redox reactions

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JEE Advanced 2019 · Paper 1, Chemistry question 6