In the following reactions: $$ \begin{aligned} & \mathrm{B}_2 \mathrm{H}_6+\mathrm{NH}_3 \text { (excess) } \stackrel{\Delta}{\longrightarrow} X+\mathrm{H}_2 \\ & \mathrm{NaH}+\mathrm{BF}_3 \stackrel{450 \mathrm{~K}}{\longrightarrow} Y+\mathrm{NaF} \\ & \mathrm{B}_2 \mathrm{H}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Z}+\mathrm{H}_2 \end{aligned} $$ $X, Y$ and $Z$ are respectively.

• A

  $\mathrm{B}_2 \mathrm{H}_6, \mathrm{LiBH}_2, \mathrm{H}_3 \mathrm{BO}_3$
• Correct answer
  B

  $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6, \mathrm{~B}_2 \mathrm{H}_6, \mathrm{H}_3 \mathrm{BO}_3$
• C

  $(\mathrm{BN})_n, \mathrm{LiBH}_4, \mathrm{HBO}_2$
• D

  $\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_2 \mathrm{H}_6, \mathrm{HBO}_2$

Solution

Therefore, $$ \begin{aligned} & X=\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6 \\ & Y=\mathrm{B}_2 \mathrm{H}_6 \\ & Z=\mathrm{H}_3 \mathrm{BO}_3 \end{aligned} $$ Hence, option (b) is correct.

Hence, the correct answer is option: 'B'.