# How can you tell the difference between sp3, sp2, and sp hybridization?

Jan 20, 2016

At its core, the meaning of $s {p}^{n}$ is that one $s$ orbital mixes with $n$ number of $p$ orbitals close in energy to form degenerate (same-energy) hybridized atomic orbitals that can allow access to more electrons than available from "pure" ($s$, $p$, etc) atomic orbitals for bonding.

• $s {p}^{3}$ bonding involves using four $s {p}^{3}$-hybridized atomic orbitals, so it must have four electron groups. EX: ${\text{CH}}_{4}$

• $s {p}^{2}$ bonding involves using three $s {p}^{2}$-hybridized atomic orbitals, so it must have three electron groups. EX: ${\text{BH}}_{3}$, ${\text{H"_2"C"="CH}}_{2}$

• $s p$ bonding involves using two $s p$-hybridized atomic orbitals, so it must have two electron groups. EX: $\text{H"-"C"-="C"-"H}$, $: \text{C"-="O} :$

I've explained $s {p}^{3}$ and $s {p}^{2}$ hybridization below, and from that, I think you can imply what $s p$ hybridization is.

$\setminus m a t h b f \left(s {p}^{2}\right)$-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, and the one with two electrons for incorporating one $\sigma$ and one $\pi$ bond 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.

I think from here, you can imply what $s p$ hybridization means. (Hint: It can also be called $s {p}^{1}$ hybridization.)