Thermal decomposition of a Mn compound $(\mathrm{X})$ at 513 K results in compound $\mathrm{Y}$, $\mathrm{MnO}_{2}$ and a gaseous product. $\mathrm{MnO}_{2}$ reacts with NaCl and concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$ to give a pungent gas Z. $\mathrm{X}$, $\mathrm{Y}$ and Z, respectively.

(D) $\mathrm{KMnO}_{4}$, $\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{Cl}_{2}$

(A) $\mathrm{K}_{2} \mathrm{MnO}_{4}$, $\mathrm{KMnO}_{4}$ and $\mathrm{SO}_{2}$

(B) $\mathrm{K}_{3} \mathrm{MnO}_{4}$, $\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{Cl}_{2}$

(C) $\mathrm{K}_{2} \mathrm{MnO}_{4}$, $\mathrm{KMnO}_{4}$ and $\mathrm{Cl}_{2}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · d and f Block Elements

Thermal decomposition of a Mn compound

(X)

at 513 K results in compound

Y

MnO_{2}

and a gaseous product.

MnO_{2}

reacts with NaCl and concentrated

H_{2} SO_{4}

to give a pungent gas Z.

X

Y

and Z, respectively.

Answer: (D)

Heating potassium permanganate at

513

K gives

2 KMnO X_{4} - > K X_{2} MnO X_{4} + MnO X_{2} + O X_{2}

, so

X

KMnO X_{4}

Y

K X_{2} MnO X_{4}

and the gas is oxygen. The

MnO X_{2}

then reacts with

NaCl

and concentrated

H X_{2} SO X_{4}

to liberate the pungent gas

Cl X_{2}

. So

X

Y

and

Z

are

KMnO X_{4}

K X_{2} MnO X_{4}

and

Cl X_{2}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option