In this reaction sequence A, B and C compounds respectively are?
(A) \[{B_2}{H_6},{B_2}{O_3},B\]
(B) \[{B_2}{H_6},{H_3}B{O_3},{B_2}{O_3}\]
(C) \[{B_2}{H_6},{H_3}B{O_3},B\]
(D) \[HB{F_4},{H_3}B{O_3},{B_2}{O_3}\]

Hint: The above reaction is a redox reaction. In this reaction Boron trichloride acts as the oxidising agent and Lithium aluminium hydride acts as the reducing agent. By this reaction we can easily find the respective compound A, B and C.

Complete Solution :
In this reaction:
\[BC{l_3} + LiAl{H_4} \to A + LiCl + AlC{l_3}\]

- Boron trichloride reacts with Lithium aluminium hydride to give Diborane. Therefore, in the above reaction, A is Diborane\[({B_2}{H_6})\]. This is the laboratory method for the preparation of \[{B_2}{H_6}\]. Hence the reaction can be written as :
\[4BC{l_3} + 3LiAl{H_4} \to 2{B_2}{H_6} + 3LiCl + 3AlC{l_3}\]

This is the redox reaction. In this reaction, \[BC{l_3}\] acts as the oxidising agent and \[LiAl{H_4}\] acts as the reducing agent.
In the second reaction,
\[A + {H_2}O \to B + {H_2}\]

- We can say that Diborane \[{B_2}{H_6}\], will react with water molecules to produce Boric acid. Therefore, B is Boric acid \[({H_3}B{O_3})\].
The above reaction can be written as :
\[{B_2}{H_6} + 6{H_2}O \to 2{H_3}B{O_3} + 6{H_2}\]

In the third reaction,
\[B \to C\]
Orthoboric acid or boric acid \[({H_3}B{O_3})\] will undergo heating at various temperatures. At first boric acid is heated at \[170^\circ C\], water molecules will be lost and it gives meta-boric acid. Then when meta-boric acid is heated at an even higher temperature around \[300^\circ C\], it will again undergo loss of water molecules to give pyroboric acid. This formed pyroboric acid loses a water molecule to give Boron trioxide. Therefore, C is boron trioxide.

- At \[170^\circ C\],
\[{H_3}B{O_3}HB{O_2} + {H_2}O\]

- At \[300^\circ C\],
\[4HB{O_2} \to {H_2}{B_4}{O_7} + {H_2}O\]

\[{H_2}{B_4}{O_7}2{B_2}{O_3} + {H_2}O\]

Therefore, the overall reaction is:
\[2{H_3}B{O_3}{B_2}{O_3} + 3{H_2}O\]

In the following reactions sequence, A, B and C compounds respectively are \[{B_2}{H_6},{H_3}B{O_3},{B_2}{O_3}\].
So, the correct answer is “Option B”. \[{B_2}{H_6},{H_3}B{O_3},{B_2}{O_3}\]

Additional Information:
Properties of \[LiAl{H_4}\]
- Lithium aluminium hydride is a colourless solid.
- It is a very good reducing agent.
- It converts aldehydes, ketones, acids, esters, acyl chlorides into alcohols.
- It can convert quaternary ammonium salts into tertiary amine, tertiary to secondary and so on.
- It can convert nitrile, amide, nitro, imine, and oxime into an amine group.
So, the correct answer is “Option B”.

Note: Diborane, \[{B_2}{H_6}\] is colourless and highly toxic substance. It can catch fire very spontaneously on exposure to air. It will burn in presence of oxygen and readily release enormous amounts of smoke. Higher boranes will immediately react on exposure to air, so we have to be careful while handling diboranes.