Question

From the $B_2{H}_6$ all the following can be prepared except :

• A. $H_{3}B{O}_3$
• B.
  $\left[B{H}_2\right(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}$
• C. $B_2(C{H}_3{)}_6$
• D. $NaB{H}_4$

Answer 1

The correct option is C $B_2(C{H}_3{)}_6$

(C) $C{H}_3$ group being larger can not form a bridge between two small sized boron atoms.

(A) $B_2{H}_6+6H_{2}O\to 2H_{3}B{O}_3+6H_2\uparrow$

(B)
$B_2{H}_6+2N{H}_3\to \left[B{H}_2\right(N{H}_3{)}_2{]}^{+}[B{H}_4{]}^{-}$
This is formed by unsymmetrical cleavage.
(D) $2NaH+B_2{H}_6\to NaB{H}_4$