Dibromomethane bond angles

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.

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)

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:

• And yes, I'd personally consider these kinds of species to really be stretching the Bent's rule analogy. I think that's Weinhold's point. – Geoff Hutchison Sep 17 '14 at 3:20
• "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.)" - Right, after consulting the experimental data, I find that simply applying Bent's rule leads me astray. It seems that both me and my professor are incorrect here. Indeed, it is the H-C-Br bond angle that is smallest and Br-C-Br bond angle that is biggest. So it is indeed a combination of electronics and sterics at play here. – Dissenter Sep 17 '14 at 5:39
• Also @Geoff Hutchinson - how do you suggest predicting or rationalizing bond angles? My professor likes to stick with Coulombic effects exclusively, dismissing Bent's rule as baloney. Although it's clear to me now that I can't apply Bent's rule blindly without any other consideration, Bent's rule does seem to have a fair bit of utility and seems to be another useful tool in the arsenal. – Dissenter Sep 17 '14 at 5:45
• I really do not see, how $\ce{CH2Br2}$ can be of $D_{2d}$ symmetry. Or were you referring to a geometrical squashed tetrahedron having that symmetry? – Martin - マーチン Sep 17 '14 at 6:18
• @Martin I was referring to the general "squashed tetrahedron". I'll clarify that. The molecule, obviously, is not. – Geoff Hutchison Sep 17 '14 at 13:04

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.

• would you say that the Cl in HCl is nearly unhybridized with respect to the H-Cl bond? I can't imagine why chlorine would want to use higher energy electrons to bond with the relatively low energy H 1s orbital. – Dissenter Sep 17 '14 at 22:27
• The phosphorus atom in phosphine is nearly unhybridized, my guess would be that the chlorine atom in HCl would behave in a similar manner. What do you mean by "higher energy electrons"? There is no total electron energy difference between all of the bonding electrons in a hybridized or unhybridized situation. – ron Sep 17 '14 at 22:46
• 1) sp3 is higher than pure s, but... 4 sp3's are not higher than s + 3 p's. In HCl my guess is that Cl is using a p-orbital to bond with hydrogen. 2) yes, s-character is concentrated in orbitals holding lone pairs, when possible. 3) In PH3, the H-P-H angle is closed down because the molecule is unhybridized, maybe that is a result of Bent's rule, it is certainly consistent with the rule. – ron Sep 17 '14 at 23:02
• PH3 uses p orbitals to bond to H, that's why the H-P-H bond angle is ~90. P orbitals are used here for bonding because they are directional and provide much more effective overlap than an s orbital would; plus you save the s for use with a full lone pair rather than an unevenly shared bonding pair. Guess I'm saying that a lone pair is more electropositive than a bonding pair (in most cases?). – ron Sep 17 '14 at 23:10
• The opposite of electronegativity (I'm being serious). – ron Sep 17 '14 at 23:12