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Compound(X) on reduction with LiAlH₄ gives a hydride(Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify Y

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${BF}_{3}$

$B_{2} H_{6}$

${NaBH}_{4}$

${Na}_{2} B_{4} O_{7}$

answer is B.

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Detailed Solution

$Y + air \overset{}{\to} B_{2} O_{3}$

Y should be $B_{2} H_{6}$

% of $H_{2}$ in $B_{2} H_{6} = \frac{6}{28} \times 100 \approx 21 .72$

$4 {BCl}_{3} + 3 LiAl H_{4} \overset{}{\to} 2 B_{2} H_{6} + 3 LiCl + 3 {AlCl}_{3}$

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