# Ethylene hybridization [duplicate]

In the carbon atom, there are electrons in 2s, 2Px, 2Py, and 2Pz. Out of these orbitals, only 2Px, 2Py, and 2s are hybridized. The 2Pz orbital is left. When making a bond there is a pi bond between two carbon atoms which is formed by the 2 Pz orbitals of the each carbon atom. In $\ce{CH4}$, 2Pz orbital also hybridized.

The orbital 2Px and 2s always make sigma bonds. Each carbon atom connects with two hydrogen atoms in $\ce{C2H4}$. Therefore I like to imagine that 2py and 2pz orbitals can make sigma bonds with 2hydrogen atoms. Then only left orbitals are 2px and 2s. They can make double bonds between two carbon atoms. 2px and 2s always form as sigma bonds and then the double bond between carbon atoms, one is sigma bond and another also a sigma bond.

I know that according to the VSEPR theory there should be one sigma bond between two atoms. But atoms are hybridized for their comfortability. But according to my above imagination making 4 sigma bonds is more comfortable than making 3 sigma bonds and one pi bond for the $\ce{C2H4}$ molecule. Therefore, why is it not trying to make such a bond?

• Such bonding produces an equivalent description of ethene. See this earlier answer. – ron May 7 '17 at 13:53

Suppose the carbon was $sp^3$ hybridised. The carbon would have 4 hybridized $sp^3$ orbitals which would point towards the corners of a tetrahedron; they are $109.5^\circ$ degrees apart from each other.
You have to form two bonds between the same two carbon atoms. Suppose you use one $sp^3$ hybridised orbital to form one $\sigma$-bond. How are you going to form another bond? The other bonds of the carbon are $109.5^\circ$ apart. This is not feasible.
The other option for the carbon is to assume $sp^2$ hybridization and have three hybridized orbitals stay $120^\circ$ apart. The one remaining $p$-orbital can be in any direction ($x$-axis or $y$-axis) except along the bonding axis ($z$-axis). In this way, you can form a double bond using one $sp^2$ hybridized orbital and one $p$-orbital.