JEE Main 2021 · 25 February · morning · Single correct · +4 / −1
The correct statement about B₂H₆ is :
- (a) Its fragment, BH₃, behaves as a Lewis base.
- (b) All B-H-B angles are of 120°.
- (c) The two B-H-B bonds are not of same length.
- (d) Terminal B-H bonds have less p-character when compared to bridging bonds.
Answer and solution
Answer (d) Terminal B-H bonds have less p-character when compared to bridging bonds.
Solution
Chemistry ma’am · 3 steps
In diborane, boron gives more s-character to the terminal bonds and more p-character to the bridges.
- Structure. Each B in B₂H₆ is roughly sp³ hybridised. Four terminal H atoms make ordinary 2-centre 2-electron bonds. Two bridging H atoms each hold the two borons together through a 3-centre 2-electron (banana) bond.
- Where s-character goes. The four hybrids on each boron are not identical. The terminal H-B-H angle opens to about 120°, so those orbitals carry more s-character. The bridging H-B-H angle closes to about 97°, so those orbitals carry more p-character, which is also why terminal B-H (119 pm) is shorter than bridging B-H (133 pm).
Option Fact Verdict (a) BH₃ has only 6 electrons on B and accepts a pair, so it is a Lewis acid false (b) B-H-B angle is about 83°, far from 120° false (c) Both bridges are equivalent by symmetry, so they have the same length false (d) Terminal bonds have more s, so less p-character, than the bridging bonds true - So. The correct statement is (d): terminal B-H bonds have less p-character than the bridging bonds.
The idea to keep. Wider bond angle means more s-character, and more s-character means a shorter, stronger bond. In B₂H₆ the terminal bonds win on all three counts.
Answer from the NTA final answer key. Solution by PrepWiser.