Question

Which of the following statements is/are correct regarding $B_2{H}_6$?

• A. banana bonds are longer but stronger than normal B-H bonds
• B. $B_2{H}_6$ is also know as 3c-2e compound
• C. the hybrid state in $B_2{H}_6$ is $s{p}^3$ while that of $s{p}^2$ in $B{H}_3$
• D. it cannot be prepared by reacting $B{F}_3$ woth $LiB{H}_4$ in the presence of dry ether

Answer 1

Answer: (A), (B), (C)

The correct options are
A banana bonds are longer but stronger than normal B-H bonds
B $B_2{H}_6$ is also know as 3c-2e compound
C the hybrid state in $B_2{H}_6$ is $s{p}^3$ while that of $s{p}^2$ in $B{H}_3$

$B_2{H}_6$ has banana bonds which are $3c-2e$ (3 center 2 electron) bonds. These bonds are longer but stronger than normal $B-H$ bonds. The hybrid state in $B_2{H}_6$ is $s{p}^3$ while that of $s{p}^2$ in $B{H}_3$. This can be shown from figure. When $B{F}_3$ reacts with $LiB{H}_4$ in the presence of dry ether diborane is the product as shown in reaction.
$8B{F}_3+6LiB{H}_4\to B_2{H}_6+6LiB{F}_4$