Preparation of potassium permanganate from $\mathrm{MnO}_{2}$ involves two step process in which the $1^{\text {st }}$ step is a reaction with KOH and $\mathrm{KNO}_{3}$ to produce

(D) $\mathrm{K}_{2} \mathrm{MnO}_{4}$

(A) $\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{OH})_{6}\right]$

(B) $\mathrm{K}_{3} \mathrm{MnO}_{4}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · d and f Block Elements

Preparation of potassium permanganate from

MnO_{2}

involves two step process in which the

1^{st}

step is a reaction with KOH and

KNO_{3}

to produce

(A) $K_{4} [Mn (OH)_{6}]$
(B) $K_{3} MnO_{4}$
(C) $KMnO_{4}$
(D) $K_{2} MnO_{4}$ ✓ Correct

Answer: (D)

Solution

Manganese dioxide is converted to potassium permanganate in two stages.

In the first stage,

M n O_{2}

is fused with KOH in the presence of an oxidant such as

K N O_{3}

(or air), oxidising manganese from +4 to +6:

2 M n O_{2} + 4 K O H + O_{2} Δ 2 K_{2} M n O_{4} + 2 H_{2} O

This gives dark green potassium manganate,

K_{2} M n O_{4}

.

In the second stage, this manganate is oxidised further — electrolytically, or chemically with chlorine or ozone — to purple potassium permanganate,

K M n O_{4}

, in which manganese is +7.

The product of the first step is

K_{2} M n O_{4}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · d and f Block Elements

Solution

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