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. W and X are, respectively

(C) $\mathrm{O}_{2}$ and $\mathrm{P}_{4} \mathrm{O}_{10}$

(A) $\mathrm{O}_{3}$ and $\mathrm{P}_{4} \mathrm{O}_{6}$

(B) $\mathrm{O}_{2}$ and $\mathrm{P}_{4} \mathrm{O}_{6}$

(D) $\mathrm{O}_{3}$ and $\mathrm{P}_{4} \mathrm{O}_{10}$

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.

W and X are, respectively

Answer: (C)

Heating

KClO_{3}

with

MnO_{2}

catalyst produces oxygen gas:

2 KClO_{3} MnO_{2} 2 KCl + 3 O_{2}

So W =

O_{2}

.

Excess oxygen reacts with white phosphorus (

P_{4}

) to form phosphorus pentoxide:

P_{4} + 5 O_{2} ⟶ P_{4} O_{10}

So X =

P_{4} O_{10}

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