Question

A hydride of 2^nd period alkali metal (A) on reaction with a compound of Boron (B) to give a reducing agent (C). identify A, B and C.

Answer 1

A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).

Lithium hydride (A) reacts with diborane (B) to give lithium borohydride (C) which is acts as a reducing agent.

\[\ce{\underset{\text{[Diborane (B)]}}{B2H6} + \underset{\text{[Lithium hydride (A)]}}{2LiH} ->[Ether] \underset{\text{[Lithium borohydride (C)]}}{2LiBH4}}\]

(A) Lithium hydride – LiH
(B) Diborane – B₂H₆
(C) Lithium borohydride – LiBH₄