Compound `X’ `Y’ [which contains 21.72% hydrogen and other products]
Compound `Y’ + O2 (air) → B2O3
Therefore compound `Y’ is hydride of boron and it is obtained by the reduction of `X’ with LiAlH4. Hence, X is BCl3.
Now,
Mol. wt. of compound Y (B2H6) = 21.76 + 6 = 27.76
Structure of Y (B2H6)
In this case, two electrons of a B-H bond are involved in formation of three center bond (Banana bond) these two bonds are represented by dotted lines in figure, in which one lies above and second lies below to the main plane.