# Describe the formation of single, double, and triple bonds by orbital overlap?

Jun 1, 2016

Single bonds are made when two orbitals overlap head-on (such as two $s$ or two $2 {p}_{z}$ orbitals along the $z$ axis) to make a sigma ($\sigma$) bond. In other words, a single bond consists of one $\sigma$ bond.

Example:

• Let a $\text{H"-"H}$ bond to be made be along the $z$ axis.

${\text{H}}_{2}$ uses one $1 s$ orbital from each hydrogen to make a $\text{H"-"H}$ bond through a $1 s - 1 s$ head-on orbital overlap. A head-on overlap must occur to make the only component of a single bond.

A double bond incorporates an additional $\pi$ bond, which is sidelong orbital overlap (such as a $2 {p}_{x}$ with a $2 {p}_{x}$ orbital). So, a double bond consists of one $\sigma$ and one $\pi$ bond.

Example:

• Let a $\text{C"="C}$ bond to be made be along the $z$ axis.

The $2 {p}_{z}$ orbital of each carbon overlaps head-on to make one $\sigma$ bond.

Then, the $2 {p}_{x}$ of each carbon overlaps sidelong to generate a $\pi$ bond. This gives us both components of the double bond.

A triple bond just incorporates one more $\pi$ bond than a double bond, so it consists of one $\sigma$ and two $\pi$ bonds.

Example:

• Let a $\stackrel{\left(-\right)}{: \text{C")-=stackrel((+))("O} :}$ bond to be made be along the $z$ axis.

As before, the $2 {p}_{z}$ orbital of each atom overlaps head-on to generate a $\sigma$ bond. This is always necessary as a first step.

Then, a $2 {p}_{x}$ orbital from each atom overlaps as before to generate one $\pi$ bond.

The $2 {p}_{y}$ orbital is finally incorporated from each atom to generate the second $\pi$ bond. That gives us all three components of a triple bond.