Question #5a0a8

Dec 23, 2014

The electron configuration for $C$'s ground state is $1 {s}^{2} 2 {s}^{2} 2 {p}_{x}^{1} 2 {p}_{y}^{1} 2 {p}_{z}^{0}$.

In order for C to be able to form 4 bonds, one electron from the $2 s$ orbital is promoted to the empty $2 {p}_{z}$ orbital - this is referred to as the promoted state; this leads to the formation of hybrid orbitals - one s orbital mixes with the three p orbitals to form four $s {p}^{3}$ hybrid orbitals.

This is what happens to the $C$ atom in methane. Once the hybrid orbitals are formed, you can no longer refer to any of the three p orbitals. Here's a diagram of the formation of the $s {p}^{3}$ hybrid orbitals:

The four $s {p}^{3}$ orbitals now form sigma bonds with four $H$ atoms to produce the tetrahedral-shaped methane molecule.

So, to answer your question, the $2 {p}_{z}$ orbital is not drawn because the $C$ atom only has $s {p}^{3}$ orbitals in the $C {H}_{4}$ molecule.

Good explanation, nice figures can be found here: http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals