Compound X on reduction with LiAlH_(4) gives a hydride Y containing 21.72% hydrogen and other products. The compound Y reacts with air expolosively resulting in boron trioxide. What are X and Y respectively ?

Compound X on reduction with LiAlH_(4) gives a hydride Y containing 21

1.	Compound X on reduction with LiAlH_(4) gives a hydride Y containing 21.72% hydrogen and other products. The compound Y reacts with air expolosively resulting in boron trioxide. What are X and Y respectively ?
Answer» `BCl_(3), B_(2)H_(6)` `PCl_(3), B_(2)H_(6)` `B_(2)H_(6), BCl_(3)` `LiAlH_(4),PCl_(3)` Solution :The reaction SCHEME is From the reaction scheme and the properties, it is clear that the COMPOUND Y is boron hydride. The boron hydride that contains 21.72% hydrogen is `B_(2)H_(6)`. So, Y is `B_(2)H_(6)` (dibrorane). `B_(2)H_(6)` is obtained by the reduction of `BCl_(3)` by `LiAlH_(4)`. So, X is `BCl_(3)`. Reactions : `underset((X))(4BCl_(3))+3LiAlH_(4)rarr 3LiCl+underset((Y))(2B_(2)H_(6))+3AlCl_(3)` `underset((Y))(B_(2)H_(6))+underset("from AIR")(3O_(2))overset("EXPLOSIVE")rarr B_(2)O_(3)+3H_(2)O`