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 aluminum hydride. The compounds $X$ and $Y$ are represented by formula:

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

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

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

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

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Solution

The correct option is B $B_{2} H_{6}, B_{3} N_{3} H_{6}$
A compound $X$ , of boron, reacts with $N H_{3}$ on heating to give another compound $Y$ which is called inorganic benzene.
$3 B_{2} H_{6} X (D i b o r a n e) + 6 N H_{3} ⟶ 3 {[B H_{2} {(N H_{3})}_{2}]}^{+} {[B H_{4}]}^{-} h e a t - - \to 2 B_{3} N_{3} H_{6} Y (Borazole/Inorganic Benzene) + 12 H_{2}$
$4 B F_{3} + 3 L i A l H_{4} ⟶ 2 B_{2} H_{6} X + 3 L i F + 3 A l F_{3}$

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