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?




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](https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666), [ref\_2](https://chemistry.stackexchange.com/questions/19977/determine-the-bond-angle-in-a-compound/19987#19987)) 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.


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In terms of the strongly simplifying VSEPR model, you could arrive at the same conclusion by noting that a carbon atom is larger than a hydrogen atom, so the $\ce{CH2}$ group will request more space than each hydrogen. You can compare this to ethane $\ce{H3C-CH3}$, which has $\angle(\ce{H-C-C}) = 111.17^\circ$, slightly deviating from the ideal tetrahedral angle of $109.5^\circ$ but no π system to explain the deviation. If I did the maths correctly, that means that $\angle(\ce{H-C-H}) = 107.72^\circ$. The deviation in ethane is actually larger than that in ethene, suggesting that steric factors predominate.


[![Image of ethane with bond lengths and angles](https://i.stack.imgur.com/EVFj1.png)](https://i.stack.imgur.com/EVFj1.png)  

**Figure 1:** Ethane’s structure including bond lengths and angles. Taken from [Wikipedia](https://commons.wikimedia.org/wiki/File:Ethane-staggered-CRC-MW-dimensions-2D.png?uselang=en), where a full list of authors is available.


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