# What does sp2 and sp3 hybridization mean?

Dec 13, 2015

$s {p}^{2}$ and $s {p}^{3}$ indicate the number of $s$ and $p$ orbitals mixed to create new, degenerate hybrid orbitals.

$\setminus m a t h b f \left(s {p}^{2}\right)$-hybridized bonding

For instance, ${\text{H"_2"C"="CH}}_{2}$ involves two $\sigma$ bonds (one for each single bond), and then one $\sigma$ and one $\pi$ bond (used in one double bond), so three electron groups are needed, but 4 electrons need to be donated by carbon.

Since carbon has 4 valence electrons, but its $p$ orbitals (which are highest in energy) only contain 2, it needs to mix two of the three $2 p$ orbitals with the $2 s$ orbital to make use of 2 more valence electrons.

This is favorable because it involves the lowering of the energies for two of the $2 p$ orbitals, increasing stability.

This results in the usage of three $s {p}^{2}$ hybrid orbitals to bond: the ones with one electron for $\sigma$ bonding to hydrogen or the other carbon, and the $2 {p}_{z}$ for $\pi$ bonding with the other carbon.

1 $2 s$ orbital had been incorporated, and 2 $2 p$ orbitals had been incorporated, so it is called $s {p}^{2}$, having 33% $s$ character and 66% $p$ character.

$\setminus m a t h b f \left(s {p}^{3}\right)$-hybridized bonding

A similar reasoning follows for $s {p}^{3}$ bonding. Let's take ${\text{CH}}_{4}$ as an example. It needs four electron groups, and it needs to make four IDENTICAL $\sigma$ bonds (one for each single bond).

4 valence electrons are needed from carbon, but only 1 electron needs to be contributed per $\sigma$ bond. So, we need four separate degenerate hybrid orbitals to make each $\sigma$ bond.

Therefore, all three $2 p$ orbitals must mix with the $2 s$ orbital and stabilize in energy overall to get four degenerate hybrid orbitals.

This results in the usage of four $s {p}^{3}$ hybrid orbitals to bond: the ones with one electron allow $\sigma$ bonding to hydrogen.

1 $2 s$ orbital had been incorporated, and 3 $2 p$ orbitals had been incorporated, so it is called $s {p}^{3}$, having 25% $s$ character and 75% $p$ character.