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My teacher gave a practice exam concerning VSEPR and bond angles and in the following diagram, he wrote the bond angle as greater than 109.5. However, I thought that the ideal angle should be 109.5 exactly. It makes sense that since chlorine is more electronegative, the electron cloud density will be pulled closer to it and so the repulsion will be greater, however I can't seem to find any sources to support that claim. All I can seem to find is that double and triple bonds have the same effect as lone pair electrons and take up more space.


Is the angle 109.5 or is it greater than 109.5?


marked as duplicate by Mithoron, Community Mar 3 '18 at 21:11

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Running a search in the CCCBDB is a good tool for these questions. According to very high fidelity computational methods (CCSDT with aug-cc-pvtz basis set), the $\ce{Cl-C-Cl}$ angle in chloroform is $\gt 109.5 ^{\circ}$

The cause for this deviation from ideal tetrahedral structure is due to steric / electronic overlaps. A normal $\ce{Cl}-\ce{Cl}$ bond is about 1.99 Angstroms (as in the molecule $\ce{Cl2}$). When bonded to the sp$^3$ carbon in this molecule, the chlorine atoms are forced closer than that: about 1.68 Angstroms. Naturally, the repulsive interaction of the electrons in the Cl atoms amounts to a slightly larger angle in the tetrahedral equilibrium structure. Meanwhile the dispersion (attractive) interactions are sufficient to keep the molecule intact.

The below calculated distances are derived from ab-initio ("from the beginning") Quantum-Mechanical theory. But, these calculations are representative of a single gas-phase molecule (i.e. a very lonely molecule in empty space). Interactions with other chloroform molecules may likely change the average bond angles due to intermolecular interactions (e.g. a $\ce{Cl}$ from one molecule can pull the $\ce{H}$ from another). This could slightly change the expected or average bond angles.

The point of VESPR theory is provide a geometric basis for expected molecular geometry. The "ideal" is decribed by VESPR. Reality is usually a little more complicated, especially here, when the four substituents on the sp$^3$ carbon are not identical. For reference, identical calculations on methane, $\ce{CH4}$ reveal a $\ce{H-C-H}$ bond angle of 109.471$^\circ$ .

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  • $\begingroup$ I ran a search for "Chloroform" on that page and it displays a big table with lots of "geom" links all over it. How do you know which link to click on? A few of them display 500 server errors. $\endgroup$ – Gaurang Tandon Mar 4 '18 at 2:05
  • $\begingroup$ CCSDT with aug-cc-pvtz is usually the best $\endgroup$ – khaverim Mar 4 '18 at 2:12

For $\ce{CHCl3}$, according to VSEPR theory, the geometry will be tetrahedral which leads to bond angle of 109.5 degrees. However, we have three chlorine atoms. Taking into account the size of the chlorine atoms vs. the size of hydrogen atom, since the chlorine atoms are larger, the $\ce{Cl-C-Cl}$ bond angle will be slightly more than 109.5 degrees, and the $\ce{H-C-Cl}$ bond angle will then be slightly smaller.


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