Question

Compound (X) on reduction with $LiAl{{H}_{4}}$ gives hydride (Y) containing 21.72% hydrogen along with all other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Compounds X and Y are respectively:
a.) $BC{{l}_{3}},{{B}_{2}}{{H}_{6}}$
b.) ${{B}_{2}}{{H}_{6}},BC{{l}_{3}}$
c.) $B{{F}_{3}},A{{l}_{2}}{{O}_{6}}$
d.) ${{B}_{2}}{{H}_{6}},B{{F}_{3}}$

Answer 1

Hint: X gets reduced by $LiAl{{H}_{4}}$and gives Y which when reacted with air gives boron trioxide. Since boron trioxide is formed and Y has hydride in it so compound Y must have 6 hydrogen in it to get replaced by 3 O and X gets reduced by $LiAl{{H}_{4}}$ therefore, it should be boron halide.

Complete answer:
Compound Y reacts with oxygen present in air and the formation of boron trioxide as the main product takes place. Oxygen present in the air reacts with Y.
So, the overall reaction is
Diborane reacts with oxygen to give boron trioxide or oxide of borane with the release of water. It is an exothermic process.
${{B}_{2}}{{H}_{6}}+3{{O}_{2}}\to B{{O}_{3}}+3{{H}_{2}}O+heat$
Therefore, the compound Y must contain boron and hydrogen. Therefore, the Y compound is hydride of boron.
Compound Y is obtained by the reduction of X with $LiAl{{H}_{4}}$so the compound X must be boron trihalide that can be either or
$4B{{X}_{3}}+3LiAl{{H}_{4}}\to 2{{B}_{2}}{{H}_{6}}+3LiX+3Al{{X}_{3}}~(X=CI~or~F)$
Molecular weight of $({{B}_{2}}{{H}_{6}})$Y= 21.76+6=27.76
Percentage of hydrogen present in Y= $\dfrac{6}{27.76}\times 100=21.76$
Therefore, compound Y is ${{B}_{2}}{{H}_{6}}$and compound X is any boron trihalide. Therefore, from all the above option A fits the criteria. Hence, $BC{{l}_{3}},{{B}_{2}}{{H}_{6}}$ are the compounds respectively.
The correct answer is A.

Note: $LiAl{{H}_{4}}$ is quite a strong reducing agent. It is not so selective in nature. It readily reduces the polar double bonds. It is the source of ${{H}^{-}}$ ion. It provides ${{H}^{-}}$ion to borane and readily form diborane. Therefore, it readily reduces boron trihalide to form $({{B}_{2}}{{H}_{6}})$which upon oxidation form oxide of borane. Therefore $({{B}_{2}}{{H}_{6}})$ on oxidation from boron trioxide.