Q.Describe the hybridisation, geometry and bond angles at the double-bond carbons of ethene, and explain why the molecule is planar.

Each carbon in ethene, $\text{CH}_2{=}\text{CH}_2$, mixes one $s$ and two $p$ orbitals into three equivalent $sp^2$ hybrid orbitals, arranged in a single plane at $120°$ to one another (trigonal planar geometry), leaving one unhybridised $p$ orbital perpendicular to that plane. Two of each carbon's three $sp^2$ orbitals form sigma bonds to its two hydrogens, and the third forms a sigma bond, head-on, to the other carbon's $sp^2$ orbital. Because all these sigma bonds lie in the same plane on both carbons, the whole $\text{H}_2\text{C=CH}_2$ skeleton -- both carbons and all four hydrogens -- is forced into one plane, with $\text{H--C--H}$ and $\text{H--C--C}$ bond angles close to the ideal $120°$. Separately, the two carbons' unhybridised $p$ orbitals, both perpendicular to this plane, overlap sideways to form the pi bond; this sideways overlap can only be maintained if the two $p$ orbitals stay parallel to each o …

Identify the hybridisation from the number of sigma-bonded groups plus lone pairs (3, here = $sp^2$); note that $sp^2$ orbitals are inherently coplanar at $120°$; connect the pi bond's parallel …

Saying the bond angle is exactly $120°$ (it is close to, but not exactly, $120°$ due to the slightly different steric/electronic demands of the double vs single bonds); confusing the sigma bond (head-on, in-pla …