Upon heating $\mathrm{KClO}_{3}$ in the presence of catalytic amount of $\mathrm{MnO}_{2}$, a gas $\mathbf{W}$ is formed. Excess amount of $\mathbf{W}$ reacts with white phosphorus to give $\mathbf{X}$. The reaction of $\mathbf{X}$ with pure $\mathrm{HNO}_{3}$ gives Y and Z. Y and Z are, respectively

(B) $\mathrm{N}_{2} \mathrm{O}_{5}$ and $\mathrm{HPO}_{3}$

(A) $\mathrm{N}_{2} \mathrm{O}_{3}$ and $\mathrm{H}_{3} \mathrm{PO}_{4}$

(C) $\mathrm{N}_{2} \mathrm{O}_{4}$ and $\mathrm{HPO}_{3}$

(D) $\mathrm{N}_{2} \mathrm{O}_{4}$ and $\mathrm{H}_{3} \mathrm{PO}_{3}$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · p-Block Elements

Passage

Upon heating

KClO_{3}

in the presence of catalytic amount of

MnO_{2}

, a gas

W

is formed. Excess amount of

W

reacts with white phosphorus to give

X

. The reaction of

X

with pure

HNO_{3}

gives Y and Z.

Upon heating

KClO_{3}

in the presence of catalytic amount of

MnO_{2}

, a gas

W

is formed. Excess amount of

W

reacts with white phosphorus to give

X

. The reaction of

X

with pure

HNO_{3}

gives Y and Z.

Y and Z are, respectively

Answer: (B)

From Q.33, X =

P_{4} O_{10}

.

Reaction of

P_{4} O_{10}

with pure (concentrated)

HNO_{3}

P_{4} O_{10} + 4 HNO_{3} ⟶ 4 HPO_{3} + 2 N_{2} O_{5}

So Y =

N_{2} O_{5}

and Z =

HPO_{3}

(metaphosphoric acid).

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option