Dibromomethane bond angles

Question

With regard to the above professor's claim, I disagree. Bent's rule, which states that atomic p-character is concentrated in orbitals directed toward more electronegative substituents, implies that the $\ce{Br-C-Br}$ bond angle is going to be smaller than 109.5 degrees since the "$\ce{sp^3}$" hybrid orbitals involved in C-Br bonding have more p-character than 75%.

Therefore, the $\ce{H-C-H}$ bond angle should be bigger than 109.5 degrees.

Questions

1) Does Bent's rule work here (i.e. am I correct?)

2) Does Bent's rule always work?

Revision

1. HCH bond angle is ~110 degrees.
2. HCBr bond angle is ~108 degrees.
3. Br-C-Br bond angle is 112 degrees.

Rationales

1. vdW repulsions play a role in Br-C-Br bond angle expansion, although the amount of p-character in the C-Br bonds might be >75%.
2. HCH angle is expanded because of Bent's rule; s-character in the H-C bond exceeds 25%.
3. The expansion of the Br-C-Br bond angle and the expansion of the H-C-H bond angle squeezes together the remaining H-C-Br bond angle.

Answer 1

Here's the B3LYP/6-31G* geometry. On something like this, I'd suspect this is fairly reliable, although I'll try a few other methods in a second. It gives the Br-C-Br angle as a healthy 113.4 degrees, and H-C-H as 112.5, both in agreement with experiment.

I don't see how the Br-C-Br angle is ever going to be smaller than 109.5. It's the Br-C-H angles that shrink (107.7 degrees), since breaking the symmetry forces the molecule into a "squashed" tetrahedron (point group $D_{2d}$ for the squashed tetrahedron, $C_{2v}$ for $CH_2Br_2$ in particular)

So your questions:

1) Does Bent's rule work here (i.e. am I correct?)

Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)

2) Does Bent's rule always work?

I typically consider VSEPR and Bent's rule to be very useful heuristics. Indeed, it's used in one of my favorite force field methods, UFF.

I'm holding back on "always" because chemistry is incredibly varied. A quick Google search of "Bent's rule fail" turns up this interesting nugget from Weinhold's book:

Answer 2

Hybridization occurs in response to a bonding interaction. Bent's rule adds that the hybridization will occur in a direction that will lower the energy of the electrons in the system. I can't imagine a case were a molecule will hybridize in a fashion to raise the energy of (all) the electrons in the system, so Bent's rule should always apply.

The water molecule has an H-O-H angle around 104.5 degrees; in the direction Bent's rule would suggest. Bromine is much less electronegative than oxygen (Br = 2.96, O = 3.44, C = 2.55), so for dibromomethane we might expect a Br-C-Br bond angle larger than 104.5 and less than 109.5. Then we would need to also factor in the steric repulsion between the two bulky bromines which would open the bond angle even more.

It turns out that experimental values for the H-C-H and Br-C-Br bond angles are all fairly close to the tetrahedral angle with the Br-C-Br angle around 112 degrees. (ref. 1, ref. 2). It's hard to separate out the effects due to sterics and hybridization, but I could believe that the experimental values reflect a slightly sub-tetrahedral angle (due to Bent's rule) being opened up by sterics to the experimentally observed value.