Q.State the hybridisation, bond angle and geometry at the carbon atoms of ethene, and explain why rotation about the $\text{C=C}$ bond is restricted.

Each carbon of ethene is $sp^2$ hybridised (one $s$ + two $p$ orbitals mixed), giving three coplanar sigma-bonding orbitals at $\approx 120°$ to each other -- a trigonal planar arrangement -- plus one unhybridised $p$ orbital left perpendicular to that plane on each carbon. The $\text{C=C}$ double bond is one sigma bond, from head-on overlap of the two carbons' remaining $sp^2$ orbitals, PLUS one pi bond, from sideways overlap of their two parallel, perpendicular $p$ orbitals above and below the molecular plane. This sideways overlap is only effective while the two $p$ orbitals remain parallel; rotating one carbon relative to the other about the $\text{C--C}$ axis would twist the $p$ orbitals out of alignment, reducing and eventually destroying their overlap and so breaking the pi bond -- an energetically very costly process (breaking a real chemical bond) compared with the essentially free rotation possible about a …

Identify $sp^2$ hybridisation from three sigma-bonded groups per carbon; separate the double bond conceptually into its sigma component (free to exist at any relative rotation) and pi component (which spe …

Thinking restricted rotation is about sterics/crowding rather than the electronic requirement that the p orbitals stay parallel for pi overlap; stating the …