Question

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

Answer 1

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.