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Halides of Group 16

A compound X,...

Question

$A$ compound $X$ , of boron reacts with $N H_{3}$ on heating to give another compound $Y$ which is called inorganic benzene. The compound $X$ can be prepared by treating $B F_{3}$ with Lithium aluminium hydride. The compounds $X$ and $Y$ are represented by the formulas.

A

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

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B

B F_{3}, B_{3} N_{3} H_{6}

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C

$B_{2} O_{3}, B_{3} N_{3} H_{6}$

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D

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

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Solution

The correct option is D $B_{2} H_{6}, B_{3} N_{3} H_{6}$
As per the equation,

$X + N H_{3} \to Y$
(Inorganic benzene)

And

$B F_{3} + L i A l H_{4} \to X$

One needs to find $‘ X^{'} a n d ‘ Y^{'}$

Y is inorganic benzene also known as borazine having formula $B_{3} N_{3} H_{6}$ .

It is called inorganic benzene because it has a similar structure like benzene and it is not organic in nature.

So $Y i s B_{3} N_{3} H_{6}$ .

Borazine / inorganic benzene can be produced by reacting $N H_{3}$ with diborane $B_{2} H_{6}$

$3 B_{2} H_{6} + 6 N H_{3} \to 2 B_{3} N_{3} H_{6} + 12 H_{2}$

$X = B_{2} H_{6}$ $Y = B_{3} N_{3} H_{6}$

Also, to confirm $X = B_{2} H_{6}$ , one can use the second reaction

$4 B F_{3} + 4 L i A l H_{4} \to 2 B_{2} H_{6} + 3 L i F + 3 A l F_{3}$

Thus, it is one of the preparation method of diborane by treating boron trifluoride with $L i A l H_{4}$ in diethyl ether.

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, of boron reacts with

N H_{3}

on heating to give another compound

Y

which is called inorganic benzene. The compound

X

can be prepared by treating

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with lithium aluminum hydride. The compounds

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Y

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