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Potassium Permanganate

X+ K 2 CO 3 +...

Question

$(X) + K_{2} {C O}_{3} + A i r h e a t - - \to (Y)$
$(Y) + {C l}_{2} ⟶ (Z) P u r p l e$
Which of the following is correct?

A

X = b l a c k, M n O_{2}; Y = b l u e, K_{2} C r O_{4}; Z = K M n O_{4}

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B

X = g r e e n, {C r}_{2} O_{3}; Y = y e l l o w, K_{2} C r O_{4}; Z = K_{2} {C r}_{2} O_{7}

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C

X = b l a c k, M n O_{2}; Y = g r e e n, K_{2} M n O_{4}; Z = K M n O_{4}

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D

X = b l a c k, {B i}_{2} O_{3}; Y = c o l o u r l e s s, {K B i O}_{2}; Z = K B i O_{3}

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Solution

The correct option is C $X = b l a c k, M n O_{2}; Y = g r e e n, K_{2} M n O_{4}; Z = K M n O_{4}$
When black coloured manganese dioxide is fused with potassium carbonate in presence of air, green coloured potassium manganate is obtained.
$2 M n O_{2} b l a c k, X + 2 K_{2} {C O}_{3} + O_{2} h e a t - - \to 2 K_{2} M n O_{4} g r e e n, Y + 2 C O_{2}$
Chlorine oxidizes green coloured potassium manganate to pink coloured potassium permanganate.
$2 K_{2} M n O_{4} Y g r e e n, + {C l}_{2} ⟶ 2 K M n O_{4} Z, p u r p l e$

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