What is the hybridization on #CH_4#, #NH_3#, #O_2#, #N_2#, and #H_2O#?

Orbital hybridization is required when one central atom and its surrounding atoms do not have matching orbital symmetries. This can only start occurring for molecules of #AX_2# or higher-order structures.

In other words, there has to be a need to mix orbitals to create degenerate orbitals that are compatible with the surrounding atoms. This need is not present for binary molecules like #"N"_2# or #"HCl"#.

OXYGEN AND NITROGEN

Neither #O_2# nor #N_2# require hybridization to make their bonds.

By convention, the internuclear axis is defined as the #z# axis. For homonuclear diatomic molecules, that means the orbitals can combine directly in space. Each #2p_x# can combine sidelong with the other #2p_x#, each #2p_y# sidelong with the other #2p_y#, and each #2p_z# head-on with the other #2p_z#.

So, what we have is:

• Each oxygen atom bonds to the other using a #2p_z# atomic orbital and either a #2p_x# or #2p_y# atomic orbital, generating one #sigma# and one #pi# bond, giving a double bond.
• Each nitrogen atom bonds to the other using a #2p_z# atomic orbital as well as both its #2p_x# and #2p_y# atomic orbitals, generating one #sigma# and two #pi# bonds, giving a triple bond.

Neither of them need to hybridize with their #2s# atomic orbitals to create compatible orbitals.

METHANE, AMMONIA, AND WATER

Notice how methane, ammonia, and water all have four electron groups on them. That correlates with a tetrahedral electron geometry.

Going off of methane, orbital hybridization is needed because carbon's #2p# orbitals are three different, specific "symmetries", and hydrogen's #1s# atomic orbital is actually not compatible with the #2p_x# or #2p_y# orbitals of carbon (though it is compatible with the #2p_z#...).

Carbon needs to create four identical orbitals that are of the same symmetry as the orbitals contributed by all four hydrogens as a group, so it hybridizes the #2s# with three #2p# orbitals to generate four degenerate #sp^3# hybrid orbitals that can all overlap head-on with hydrogen's #1s# orbital.

Similarly, ammonia and water have that same issue, and so they hybridize their #\mathbf(2s)# and #\mathbf(2p)# orbitals in similar manners to generate four degenerate #sp^3# hybrid orbitals as well that can all overlap head-on with hydrogen's #1s# orbital. They just don't use all four to bond.

As a result, methane, ammonia, and water all have #sp^3# hybridizations.