organic chemistry - What's the H-C-H bond angle in ethene?

The carbon is $\mathrm{sp^2}$ hybridised and is therefore planar and should also, theoretically be $120^\circ$. However, VSEPR theory suggests that the π bond would "need more space" due to greater electron repulsion. As a consequence the $\ce{H-C-H}$ bond angle would be smaller. However, since the π bond is out of the plane of the molecule does this actually happen?

Answer

The H-C-H bond angle in ethene is ca. 117 degrees and the H-C-C angle is ca. 121.5 degrees. There are two reasons that combine to explain this angular deformation in ethene.

First, from these bond angles and Coulson's Theorem (ref_1, ref_2) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.

From these hybridization indices (the index is the exponent "n" in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma electrons will tend to flow towards this lower energy C-C bond. Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds. Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions.

Second, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.