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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?

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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.

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    $\begingroup$ The first statement is not entirely true. Since the pi-bond influences the bond distance it influences the overlap of the sigma-bond forming orbitals, hence also influencing the sigma system, even if the orbitals are orthogonal. An observational explanation is given with Bent's rule. $\endgroup$ – Martin - マーチン Jan 1 '15 at 16:35
  • $\begingroup$ @Martin Excellent point Martin, I'll edit my answer. $\endgroup$ – ron Jan 1 '15 at 16:38
  • $\begingroup$ Do pi electrons repel the CH sigma electrons? $\endgroup$ – Dissenter Jan 1 '15 at 18:11
  • $\begingroup$ @Dissenter Yes they do. $\endgroup$ – ron Jan 1 '15 at 18:12
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    $\begingroup$ @Dissenter The electron density in the "top" of the pi cloud will tend to repel the C-H bonding electrons and push the down; but the "bottom" of the pi cloud will push them up. By symmetry these effects must cancel and the result is no net up-down C-H deformation due to the pi cloud. Also due to symmetry the pi cloud cannot push the C-H bonds out or in. $\endgroup$ – ron Jan 1 '15 at 18:18
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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
Figure 1: Ethane’s structure including bond lengths and angles. Taken from Wikipedia, where a full list of authors is available.

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