Question

The product obtained in the reaction of diborane with excess of ammonia at low temperature is:

• A. $3B_2{H}_6\cdot 2N{H}_3$
• B. $(BN{)}_2$
• C. $B_3{N}_3{H}_6$
• D. $(BN{)}_4$

Answer 1

The correct option is A $3B_2{H}_6\cdot 2N{H}_3$

The product obtained in the reaction of diborane with excess of ammonia at low temperature is $3B_2{H}_6\cdot 2N{H}_3$$B_2{H}_6+N{H}_3\left(\text{excess}\right)\stackrel{\text{low temperature}}{\to }3B_2{H}_6.2N{H}_{3}or\left[H_{2}B\right(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}$
Hence correct option is a.