Lewis structures of H₂(C)xCH₂ molecules

I need to calculate the angles between CCC, CCH, HCH in the following molecules:

• ethene ($\ce{H2CCH2}$),
• propa-1,2-diene ($\ce{H2CCCH2}$), and
• buta-1,2,3-triene ($\ce{H2CCCCH2}$).

If I'm not mistaken, the Lewis structures of all of these molecules have the carbons in the middle, connected by double bonds to each other, and the hydrogen atoms connected to the carbons in the edge. For example:

H           H
\         /
C = C = C
/         \
H           H

The carbons at the edge are $\mathrm{sp^2}$ and the middle carbon is always $\mathrm{sp}$ .

This means that the angle CCC is always 180°. But how can I determine the angles CCH, HCH?

In addition, how can I determine if the hydrogen atoms in the three molecules lie on the same plane?

• Be aware, that in $\ce{H2C=CH2}$ all hydrogens are in plane, and the same holds for $\ce{H2C=(C=C)_{$n$}=CH2}$ because of symmetry reasons. For any $\ce{H2C=(C)_{$n$}=CH2}$ the alignment of the hydrogens is perpendicular. You can explain this with the pi orbitals. Feb 23 '15 at 4:16