MCQ
$\ce{B_2H_6}$ on heating with $\ce{NH_3}$ gases gives:
- ✓$ \mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$
- B$ \mathrm{~B}_2 \mathrm{H}_6 \cdot \mathrm{NH}_3 $
- C$\mathrm{H}_6 \cdot 3 \mathrm{NH}_3 $
- D$\mathrm{~B}_2 \mathrm{~N}_4 \mathrm{H}_{10} $
✓
Answer
Correct option: A.
$ \mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$
$3\text{B}_2\text{H}_6+6\text{NH}_3\rightarrow2\text{B}_3\text{N}_3\text{H}_6+12\text{H}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Inorganic benzene)}$
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