Explain the hybridization of carbon and the bond structure of ethene $\left( C_2H_4 \right)$ $OR$ Explain: Ethene is a planar molecule.

(N/A) Ground state carbon: $[He] \ 2s^{2} \ 2p^{2}$
Excited state carbon: $[He] \ 2s^{1} \ 2p_{x}^{1} \ 2p_{y}^{1} \ 2p_{z}^{1}$
In $sp^{2}$ hybridization,one half-filled $2s$ orbital and two $2p$ orbitals of the excited carbon atom combine to form three $sp^{2}$ hybrid orbitals. These three $sp^{2}$ orbitals are planar and oriented at an angle of $120^{\circ}$. In this planar arrangement,one unhybridized $2p$ orbital remains perpendicular to the plane.
The three $sp^{2}$ orbitals are arranged at $120^{\circ}$ angles in one plane at the corners of a triangle.
One $sp^{2}$ orbital from each carbon atom overlaps to form a $C-C$ sigma bond. The remaining two $sp^{2}$ orbitals on each carbon overlap with the $1s$ orbital of hydrogen atoms to form four $C-H$ sigma bonds. Consequently,the two carbon atoms and the four hydrogen atoms all lie in the same plane,making ethene a planar molecule.