I was studying VSEPR and hybridization theory and one of the popular questions in many textbooks was comparison of bond angle of $\ce{NH3}$, $\ce{NF3}$ and $\ce{NCl3}$.

$\ce{NF3}$ had smaller bond angle as electrons were nearer to the more electronegative fluorine atom, thereby reducing bond pair - bond pair repulsion (of N-F bonds) and thus allowing contraction of bond angle.

Further, in $\ce{NCl3}$ bond angle was larger than both $\ce{NH3}$ and $\ce{NF3}$ due to steric repulsion between large Cl atoms. So the order was $\ce{NCl3}$>$\ce{NH3}$>$\ce{NF3}$.

Now, a question that I later saw in one of my textbooks was the analysis of bond angles of $\ce{COCl2}$ and $\ce{$\ce{COBr2}$}$, which have trigonal planar geometry. According to my textbook, $\ce{COCl2}$ has smaller $\ce{O-C-Cl}$ angle, less than $\mathrm{120^{\circ}}$, (and thereby greater $\ce{Cl-C-Cl}$ angle). The angles for $\ce{COBr2}$ were not given.

My Approach

Now if we apply the reasoning given for $\ce{NCl3}$ and $\ce{NH3}$, $\ce{NF3}$ in this question comparing $\ce{COCl2}$ and $\ce{COBr2}$, we would find that the $\ce{C-X}$ bond pair electrons would be more polarized towards X, if X is Cl, and thus analogously to $\ce{NF3}$, it should contract leading to a smaller bond angle than in $\ce{COBr2}$.

Also, due to steric repulsion, $\ce{C-X}$ bond angle in $\ce{COBr2}$ should be greater than $\ce{COCl2}$.

Solution given in the book

The author has stated that there are four electrons in the $\ce{C-O}$ bond

Thus, bond pair repulsion between electrons of $\ce{C-O}$ bond and $\ce{C-X}$ bond would be much greater than any repulsion between the electrons of the $\ce{C-X}$ bonds.

As electrons in the $\ce{C-X}$ bond would be closer to carbon atom (and thus $\ce{C-O}$ bond pair electrons) in $\ce{COBr2}$, therefore a greater amount of repulsion would be present between them. This repulsion would lead to contraction of Br-C-Br bond angle leading to an angle less than that of $\ce{COCl2}$.

However, no data was provided in support or against it. Also no reasoning was given why they would be greater or less than $\mathrm{120^{\circ}}$, only the above reasoning for comparing the bond angle was provided.

I have not been able to find any bond angle data. My teachers are also not sure. Also, the textbook I referred to is infamous for having lots of errors.

Which reasoning is correct, or is it something else entirely?

What is the actual data?

  • $\begingroup$ I believe your textbook is incorrect with regard to COCl2 bond angles. It is the Cl-C-Cl bond angle that is less than 120; it is approx. 111 degrees (making the O-C-Cl angle 124), at least in the solid phase. The crystal structure was reported in 1952 (Acta Cryst 5:833). The Schupf computational lab at Colby College gives the same bond angles for COBr2 as for COCl2, but I don't know if that is experimentally verified or not. $\endgroup$ – Andrew Jul 31 '20 at 18:00
  • $\begingroup$ This data is available on the Computational Chemistry Comparison and Benchmark DataBase.These would be the values of $\ce{COCl2}$, $\ce{COBr2}$ and $\ce{COF2}$ $\endgroup$ – Safdar Jul 31 '20 at 19:31

The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles of $\ce{COCl2}$ and $\ce{COBr2}$.

To circumvent these issues, a better method(less prone to failures) would be using Bent's rule. Now, there has been a lot of discussion on this website over what this rule actually is (see here) and why it is physically useful (see this). The Wikipedia article on the same has a wonderful intuitive discussion under the "Justification" tab for why this rule exists. However, I will not be delving into the origins of this rule, and just state outright that the basic premise of this rule is that not all hybrid orbitals are equivalent under a given hybridization scheme. The percentage s-character and p-character (that is, the amount of s-orbital or p-orbital present in a given hybrid orbital as a percentage of the whole) varies as per the electronegativities of the substituents.

A statement of this rule was provided by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

The idea is that the amount of p character will increase (alternatively, the amount of s character will decrease) in those bonds which have more electronegative substituents (that is, the atom other than carbon in the bond), while simultaneously the amount of p character will decrease(alternatively, the amount of s character will increase) in bonds towards less electronegative substituents in such a manner that the net sum of percentage s and p characters over all the bonds in the molecule comes out to be $100$%. This is essentially a redistribution of the orbital energies across bonds to minimize the energy of the system. Further details on this rule can be found in the links mentioned above.

Once you are aware of this rule, applying it to your case becomes a fairly simple task. I will try to compare the $\ce{Cl-C-Cl}$ and $\ce{Br-C-Br}$ angles here, the orders for the $\ce{O-C-Cl}$ and $\ce{O-C-Br}$ angles can be easily inferred from there. From a classical hybridization perspective, the carbon in both these molecules uses $\ce{sp^2}$ hybrid orbitals for bonding, and all the three hybrid orbitals thus generated should possess equivalent atomic orbital's character ($33.33$% s character and $66.66$% p character approximately). But as per Bent's rule, since $\ce{Cl}$ and $\ce{Br}$ are both more electronegative than carbon, they will increase p character towards their direction (that is, in the $\ce{C-Cl}$ and $\ce{C-Br}$ bonds respectively). Since $\ce{Cl}$ is even more electronegative than $\ce{Br}$, the p character increase in the $\ce{C-Cl}$ bonds will be more than that what will happen in the $\ce{C-Br}$ bonds.

Hence, we can say that more p character will be directed in the two $\ce{C-Cl}$ bonds as compared to the two $\ce{C-Br}$ bonds. This directly bears a relation to the bond angle: more is the p character in the constituent bonds, lesser will be the bond angle(you can see this from a fairly easy observation: $\mathrm{sp}$ hybrid orbitals are generally at an angle of $180°$ with $50$% p character, the $\mathrm{sp^2}$ hybrid orbitals are generally at an angle of $120°$ with $66.66$% p character and the $\mathrm{sp^3}$ hybrid orbitals are generally at an angle of $109.5°$ with $75$% p character).

Hence, Bent's rule predicts that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ should be lesser than the $\ce{Br-C-Br}$ of $\ce{COBr2}$. Experimental data says that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ is $111.83 ± 0.11°$, while the $\ce{Br-C-Br}$ of $\ce{COBr2}$ is $112.3 ± 0.4°$. Hence, the prediction made by Bent's rule is confirmed.

Notice that the difference between the given bond angles is quite small (around $1°$). This is also somewhat expected, as the electronegativity difference between chlorine and bromine is also quite small (approximately $0.2$).

The advantage of employing this method is that you don't have to consider conflicting factors like those you stated in your approach. The rule explains these angle variations based on Coulson's theorem, from which the following relation pops out:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$ for $\theta \in (90^\circ,180^\circ)$

($\theta$ stands for the bond angle while ${P}$ and ${S}$ stand for p character and s character respectively)

As you can clearly see, when the p character goes up, the value of $\cos \theta$ also goes up if you closely observe the simplified form $\cos \theta = 1 - \frac{1}{P}$. This means that the value of $\theta$ will decrease for the given range of $\theta$ (as $\cos \theta$ is a decreasing function for $\theta \in (90^\circ,180^\circ)$)

Hence, you can generally predict things quite accurately with Bent's rule without getting confused with contrasting factors, as they are usually taken care of while determining the atomic orbital characters.

  • $\begingroup$ Ok great answer. One thing I want to clarify is whether the double bond with oxygen has any role or not? Does it affect bent's rule? Also if we take formic acid, would you expect any significant difference in p character of both oxygen atoms? $\endgroup$ – Manit Agarwal-El psy congroo Aug 1 '20 at 2:54
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    $\begingroup$ @YusufHasan Equivalence depends on the timescale that you are talking about, see my answer at chemistry.stackexchange.com/a/134072/16683. Unless specifically in a context where the proton transfer is considered "fast", I would not say that the oxygens are equivalent. That's certainly not the case here, because when you talk about s- and p-character, you're essentially referring to an optimised ground-state geometry (any sort of vibrational motion that changes the bond lengths or angles will affect the s- and p-character)... $\endgroup$ – orthocresol Aug 1 '20 at 4:34
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    $\begingroup$ ... indeed, these optimised geometries are exactly the kind of "instantaneous" or "static" picture that I mention in that answer. Even though proton transfer is fast compared to some other chemical processes, it still takes a finite time for it to happen, and so cannot be incorporated into the "static" picture. Reusing the photography analogy: a person can walk between two places many times, but it's not ever possible to take one photo where they're at both places simultaneously. $\endgroup$ – orthocresol Aug 1 '20 at 4:35
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    $\begingroup$ The thing is, you don't need to go to these lengths to talk about formic acid, because the entirety of your comment applies to the formate ion. (I notice you describe it as a "resonance hybrid", which is true for formate, but certainly not formic acid.) And your answer itself is quite fine; I was only commenting about your comment. $\endgroup$ – orthocresol Aug 1 '20 at 4:42
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    $\begingroup$ @orthocresol Ok,I guess I understood after reading your comments and answer, that while resonance in the formate ion is something that creates equivalence even in the instantaneous or "static" picture for the carboxylate anion, proton transfers are subjected to a finite transfer time,and so they don't reflect equivalence in a static frame for the carboxylic acid.So, different p characters should be directed towards the two oxygens in formic acid,while my original comment above would apply to the "formate anion" more suitably than "formic acid" itself. I hope I got this correctly,thank you :) $\endgroup$ – Yusuf Hasan Aug 1 '20 at 4:59

The reasoning provided in the book conforms with the computational data and hence can be presumed correct. As the $\ce{C-X}$ bond lengths of increases, the position of electrons moves away from the $\ce{CO}$ (carbonyl electrons), and hence they are repelled less. Another proven method for the claim can be using the bent's rule provided by Yusuf Hasan. To support the above claims:

Here I present the $\ce{X-C-X}$ bond angles as calculated by Geometrical optimization by method Hartree-Fock on the basis set: Basic 3-21G using GPU accelerated engine provided as TeraChem. (The computations were iterated for lowest possible error by computational analysis)

The computation finished in {1.92sec,5.20sec,12.63sec,38.80sec} for all halogens {F,Cl,Br,I}

The data below each image corresponds to the Natural Bond Orbital Analysis as suggested by Orthocresol. The currently essential p-character data for Carbon-Halogen and Halogen-Carbon bonds have been depicted as below. The computations were made by Engine: Gaussian, using method Hartree-Fock on the basis set: Basic 3-21G. (Here BD stands for a 2-centered bond with the labeled atoms[i])

  1. $\ce{COF2}$ enter image description here enter image description here
  2. $\ce{COCl2}$ enter image description here enter image description here
  3. $\ce{COBr2}$ enter image description here enter image description here
  4. $\ce{COI2}$ enter image description here enter image description here

Here we can observe that the p-character for $\ce{X-C}$ increases quite significantly (as per considerations of Electronegativity) as we go from Fluorine to Iodine and for $\ce{C-X}$ bond, the p-character decreases distinctly enough as we go from Fluorine to Iodine.

This computational evidence is in support with the arguments stated in the book, hence they are supposedly accurate.

Note:The data above has been calculated and the experimental error (considering the experiments as true value falls below 1.5% for Phosgene$^{[1]}$ and 0.33% for carbonyl fluoride$^{[2]}$ which seems a reasonable and acceptable precision.

1.Nakata, M.; Kohata, K.; Fukuyama, T.; Kuchitsu, K. (1980). "Molecular Structure of Phosgene as Studied by Gas Electron Diffraction and Microwave Spectroscopy. The rz Structure and Isotope Effect". Journal of Molecular Spectroscopy. 83: 105–117. doi:10.1016/0022-2852(80)90314-8. 2.https://pubchem.ncbi.nlm.nih.gov/compound/Carbonyl-fluoride

  • $\begingroup$ @Safdar shows 111.300 ...I have another from a univ. colby.edu/chemistry/webmo/COCl2.html $\endgroup$ – user96208 Jul 31 '20 at 19:11
  • $\begingroup$ @Safdar Also, the links you have provided except for the wikipedia one are goverened by computations (that's the thing we werent trusting)... For a counter I refined my computational observations and provided a pubchem cite. $\endgroup$ – user96208 Jul 31 '20 at 19:14
  • $\begingroup$ @Safdar My computations conforms to the methods of computation adopted by Tera chem. As a known fact, the bond angle is always subject to change due to uncontrollable environmental factors, that does mean that the computational data is correct for idealized environment. COI2 doesnt exist, that does not mean an optimization cannot be predicted. $\endgroup$ – user96208 Jul 31 '20 at 19:20
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    $\begingroup$ It's one thing to optimise the geometries, find the bond angles, and say that it matches the trend. It's another thing to claim that the rationalisation given (about electron repulsion etc.) is valid. Your results only show that the bond angles increase going from F -> Cl -> Br -> I. They don't show why this is so. Showing why something happens can be a rather involved process that often necessitates further analysis of the computational results, e.g. orbital analysis (NBO), etc. $\endgroup$ – orthocresol Aug 1 '20 at 4:15
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    $\begingroup$ Also why would optimize with 3-21g? If you have TeraChem and GPU it seems a waste $\endgroup$ – Cody Aldaz Aug 1 '20 at 4:58

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