Question

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

Answer 1

Answer :

A is lithium hydride $(LiH)$

B is boron trifluoride $(BF_3)$

C is lithium borohydride $(LiBH_4)$

Explanation :

As per question, a hydride of 2nd period alkali metal (lithium) that is lithium hydride on reaction with compound of Boron that is boron trifluoride to give a lithium borohydride as a reducing agent and lithium fluoride.

Lithium borohydride is used in organic synthesis as a reducing agent for esters.

The balanced chemical reaction will be,

$4LiH+BF_3\rightarrow LiBH_4+3LiF$

From this we conclude that, A is lithium hydride $(LiH)$, B is boron trifluoride $(BF_3)$, C is lithium borohydride $(LiBH_4)$

Answer 2

Answer:

Explanation:

LiH + B2H6  ------ether-------> LiBH4

 (A)       (B)                                 (C)

lithium    diborane                 lithium borohydride

hydride