Question

Compound X on reduction with LiAH₄ gives a hydride Y containing 21.72% hydrogen along with other products. The compound Y reacts with air explosively resulting in boron trioxide. Identify X and Y. Give balanced equations involved in the formation of Y and its reaction with air. Draw the structure of Y.

Answer 1

Step 1: To determine the molecular formula and structure of compound Y. 

1. Since the hydride Y reacts with air forming boron trioxide, therefore, Y must be an hydride of boron.

2. %H = 21.72% (Given)

Element	%	Atomic Weight	Atomic Ratio \(\frac{1/2}{Mol.Wt}\)	Simplest Ratio = \(\frac{At\, Rate}{Small\, Value}\)
B	78.28	11	\(\frac{78.28}{11}\) = 7.12	\(\frac{7.12}{7.12}\) = 1
11	21.72	1	\(\frac{21.72}{1}\) = 21.72	\(\frac{21.72}{7.12}\) = 3

∴ Emperical formula of Y = BH₃

Since boron forms two types of hydrides, i.e., B_nH_n+4 (nidoboranes) and B_nH_n+6 (arachnoboranes), therefore, Y must be a nidoborane with n = 2. Thus, M.F of Y = B₂H₆. If Y is B₂H₆ (diborane), then its structure must be as follows:

Bridges B…H = 134 pm Terminal B – H = 119 pm

Step 2: To determine the structure of the compound X. Since compound Y i.e., B₂H₆ is formed by reduction of compound X with LiAH₄,

therefore, X must either BCl₃ or BF₃.

The equation representing the reaction of Y with O₂ may be written as follows:

B₂H₆ + 3O₂ → 2B₂O₃ + 3H₂O

Diborane,Y

Thus X = BF₃ and Y = B₂H₆