p-Block Elements-I - Result Question 27
27. Compound $X$ on reduction with $LiAlH _4$ gives a hydride $Y$ containing $21.72 %$ hydrogen alongwith other products. The compound $Y$ reacts with air explosively resulting in boron trioxide. Identify $X$ and $Y$. Give balanced reactions involved in the formation of $Y$ and its reaction with air.
Draw the structure of $Y$.
(2001, 5M)
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Solution:
- Compound $X \xrightarrow{LiAlH _4} Y$ a hydride + other compound. Hydride $Y$ contains $21.72 %$ hydrogen.
$ Y+O _2 \stackrel{\Delta}{\longrightarrow} B _2 O _3+H _2 O $
Therefore, $Y$ is a hydride of boron and it is obtained by reduction of $X$ with $LiAlH _4$. So, $X$ is either $BCl _3$ or $BF _3$.
$ \underset{X}{4 BCl _3}+LiAlH _4 \longrightarrow \underset{Y}{B _2 H _6} +\underbrace{3 AlCl _3+3 LiCl} _{\text {Other products }} $
Molar mass of $B _2 H _6=2 \times 11+6=28$
$ \begin{aligned} & % \text { of } H \text { in } B _2 H _6=\frac{6}{28} \times 100=21.5 \approx 21.72 \\ & \underset{Y}{B _2 H _6}+3 O _2 \longrightarrow B _2 O _3+3 H _2 O+\text { Heat } \\ \end{aligned} $
Structure of $\boldsymbol{Y}\left(B _2 H _6\right)$
(a) There are $4$ terminal $B-H$ bonds.
(b) There are two $3$-centre-$2$-electron $B-H-B$ bridged bonds.
(c) Terminal $H-B-H$ planes are perpendicular to bridged $B-H-B$ bonds.
BETA
