A compound of boron X reacts at $200^{\circ} C$ temperature with $N H_{3}$ to give another compound $Y$ which is called as inorganic benzene. The compound $Y$ is a colourless liquid and is highly sensitive. Its melting point is $- 57^{\circ} C$ . The compound $X$ with excess $N H_{3}$ and at a still higher temperature gives boron nitride ${(B N)}_{n}$ . The compounds $X$ and $Y$ respectively are:

B H_{3} and B_{2} H_{6}

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N a B H_{4} and C_{6} H_{6}

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B_{2} H_{6} and B_{3} N_{3} H_{6}

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B_{4} C_{3} and C_{6} H_{6}

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Solution

The correct option is C $B_{2} H_{6} and B_{3} N_{3} H_{6}$
The reactions involved are
$3 B_{2} H_{6} (X) + 6 N H_{3} ⟶ 2 B_{3} N_{3} H_{6} (Y) + 12 H_{2}$
$B_{2} H_{6} + N H_{3} (e x c e s s) ⟶ (B N)_{n} + H_{2}$

Y is borazole, which is isosteric and isoelectronic with benzene and hence called as Inorganic benzene.

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Similar questions

Q. A compound of boron X reacts at

200^{\circ} C

temperature with

N H_{3}

to give another compound

Y

which is called as inorganic benzene. The compound

Y

is a colourless liquid and is highly sensitive. Its melting point is

- 57^{\circ} C

. The compound

X

with excess

N H_{3}

and at a still higher temperature gives boron nitride

{(B N)}_{n}

. The compounds

X

and

Y

respectively are:

A compound of B (Boron), X reacts at $200^{\circ} C$ with $N H_{3}$ to give another compound Y which is called as inorganic benzene. The compound Y is colorless liquid and is highly light sensitive. The compound X with excess of $N H_{3}$ and at still higher temp gives $(B N)_{n}$ . The compound X and Y respectively.