Identify the correct statement for $B_{2} H_{6}$ from those given below:
(A) In $B_{2} H_{6}$ , all $B - H$ bonds are equivalent.
(B) In $B_{2} H_{6}$ , there are four 3-centre-2-
electron bonds.
(C) $B_{2} H_{6}$ is a Lewis acid.
(D) $B_{2} H_{6}$ can be synthesized from both
$B F_{3}$ and $N a B H_{4}$
(E) $B_{2} H_{6}$ is a planar molecule.
Choose the most appropriate answer from the options given below:

(A) and (E) only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(B), (C) and (E) only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B (C) and (D) only
Structure of $B_{2} H_{6}$

It has two 3-centre-2-electron 2-centre-2-electron bonds.
Hence, all $B - H$ bonds are not equivalent.
It is an electron deficient compound as the octet of boron is incomplete.
Hence, it can behave as a Lewis acid.
It can be synthesized from both $B F_{3}$ , and $N a B H_{4}$
$2 B F_{3} + 6 N a H 450 K - -- \to B_{2} H_{6} + 6 N a F$
$2 N a B H_{4} + I_{2} ⟶ B_{2} H_{6} + 2 N a l + H_{2}$

It is a non-planar molecule.
Hence, only Statements (C) and (D) are correct.

Suggest Corrections

Similar questions

B F_{3} + L i H \to

B_{2} H_{6} + H_{2} O \to

N a H + B_{2} H_{6} \to

H_{3} B O_{3} △ \to

A l + N a O H \to

B_{2} H_{6} + N H_{3} \to

B_{2} H_{6}

B_{2} H_{6}

(B_{2} H_{6})

Join BYJU'S Learning Program