Question

What are X and Y in the reaction?
$3B_2{H}_6+6X\to 3\left[B{H}_2\right(X{)}_2{]}^{+}[B{H}_4{]}^{-}\stackrel{Heat}{\to }Y+12H_2$

• A. $X=N{H}_3,Y=B_3{N}_3{H}_6$
• B. $X=CO,Y=B{H}_{3}CO$
• C. $X=NaH,Y=NaF$
• D. $X=N{F}_3,Y=B_3{N}_3$

Answer 1

Answer: (A)

$X=N{H}_3,Y=B_3{N}_3{H}_6$

Answer:- (A) $X=N{H}_3,Y=B_3{N}_3{H}_6$

Diborane on reaction with any Lewis base forms a weak complex. Thus, X is a Lewis base $(i.e.N{H}_{,}P{H}_3,etc.)$

$\therefore 3B_2{H}_6+6N{H}_3⟶3{\left[B{H}_2\right(N{H}_3{)}_2]}^{+}[B{H}_4{]}^{-}$

This weak complex on further heating breaks down into borazine $\left(B_3{N}_3{H}_6\right)$ and hydrogen$\left(H_2\right)$.

$3{\left[B{H}_2\right(N{H}_3{)}_2]}^{+}[B{H}_4{]}^{-}\underset{180-190°C}{\overset{\left(Heat\right)}{\to }}\underset{\left(Y\right)}{2B_3{N}_3{H}_6}+12H_2$

Hence, $X=N{H}_3\&Y=B_3{N}_3{H}_6$.