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Compound (X) on reduction with LiAIH(4)...

Compound (X) on reduction with `LiAIH_(4)` gives a hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Compounds X and Y are repectively.

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`(X) +LiAlH_(4) overset("Reduction")rarr (Y) + "Other products Hydride", 2.72 % H`
`(Y) +Air overset("Explosive")underset("reaction") rarr B_(2) O_(3)`
Determination of the molecular formula and structure of compound `(Y)` : Since the hydride `(Y)` reacts with air forming boron trioxide, therefore `(Y)` must be hydride of boron.
Given `% H = 21.72 %`
:. `% B = (100 - 21.72)% = 78.28 %`
`B :H = (78.28)/(11) : (21.72)/(1) = 1:3`
Hence, `B` and `H` are present in the ratio `1 : 3` Simplest boron hydride corresponding to this ratio is `B_(2) H_(6)`. Hence, `(Y)` is diborane `B_(2) H_(6)`.

Determination of `(X)` : Since `B_(2)H_(6)` is formed by the reduction of `(X)` with `LiAlH_(4)`, therefore `(X)` is either `BF_(3)` or `BCl_(3)`.
`underset((X)) (4BCl_(3)) + 3LiAlH_(4) rarr underset((Y))(2B_(2)H_(6)) + 3LiAlF_(4)`
The equation representing the reduction of `(Y)` with oxygen is
`B_(2)H_(6) rarr underset("Bororn trioxide")(B_(2) O_(3) )+3H_(2)O`
Hence, `(X)` is `BCl_(3)` and `(Y)` is `B_(2) H_(6)`.
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