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