AP EAMCET · Chemistry · p Block Elements (Group 13 & 14)
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 \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{LiBH}_2, \mathrm{H}_3 \mathrm{BO}_3\)
- 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 \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_2 \mathrm{H}_6, \mathrm{HBO}_2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6, \mathrm{~B}_2 \mathrm{H}_6, \mathrm{H}_3 \mathrm{BO}_3\)
Step-by-step Solution
Detailed explanation
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.
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