# What would happen if there was no hybridization in "CH"_4?

The answer would take a ${90}^{\circ}$ shape, even more contracted than water.
This structure would not happen because $C {H}_{4}$, methane, requires 4 bonds.
The pure $2 {s}^{2}$ electrons would be nonbonding, so they have no ability to form more bonds. Then, the pure $2 p$ orbitals would only form $: C {H}_{2}$ (with an empty $2 {p}_{z}$), not $C {H}_{4}$.