Question #d5075

Sep 15, 2016

Here's my explanation.

Explanation:

The geometry of ${\text{B"_2"H}}_{6}$ is

Each $\text{B}$ atom has a tetrahedral arrangement of bonds, so it is reasonable to assume an $s {p}^{3}$ hybridization.

(From wps.prenhall.com)

But it is a distorted tetrahedron. The bridging $\text{H-B-H}$ bond angles are 97°, but the outer $\text{H-B-H}$ bond angles are 120 °.

The bridging bonds are banana bonds.

To understand the bond angles, recall the discussion of banana bonds in the structure of cyclopropane.

The four bonds between the two $\text{B}$ atoms and the two bridging $\text{H}$ atoms "want" to get close to a square, so the internal bond angle will settle on some value between 90° and 109.5°.

The inner $s {p}^{3}$ orbitals get a little more $p$ character, and the internal $\text{H-B-H}$ bond angles decrease by about 12°.

Thus, the outer $s {p}^{3}$ orbitals get a little more $s$ character, and the outer $\text{H-B-H}$ bond angles increase by about 11° to 120°.

That the angle is the same as the $s {p}^{2}$ bond angle is just a coincidence.

Molecular Orbital theory gives a better explanation of the structure of ${\text{B"_2"H}}_{6}$, as in the video below.