Terminal B—H bonds have less p-character when compared to bridging bonds.
Its fragment, BH$_3$, behaves as a Lewis base.
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The Correct Option isC
Solution and Explanation
Step 1: Diborane ($B_2H_6$) has two types of bonds: 2-center-2-electron (terminal) and 3-center-2-electron (bridging).
Step 2: Bridging bonds ($B-H-B$) are longer and weaker, requiring more $p$-character from Boron.
Step 3: Terminal $B-H$ bonds are shorter and stronger, possessing more $s$-character (and thus less $p$-character).
Step 4: (A) is wrong; angles are approx 97$^\circ$ and 120$^\circ$. (B) is wrong; they are symmetric. (D) is wrong; $BH_3$ is electron-deficient (Lewis acid).