Question:

The correct statement about B$_2$H$_6$ is :

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In $B_2H_6$, the four terminal $H$ atoms and the two $B$ atoms lie in one plane, while the two bridging $H$ atoms lie above and below this plane.
Updated On: Jan 9, 2026
  • All B—H—B angles are of 120$^\circ$.
  • The two B—H—B bonds are not of same length.
  • 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 is C

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).
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