# Why are single bonds weaker than double?

Jun 11, 2016

Simply from how they're constructed. Since a pure double bond consists of 1 $\sigma$ bond and 1 $\pi$ bond, it is one $\setminus m a t h b f \left(\pi\right)$ bond's worth stronger than a single bond.

All pure single bonds consist of one $\sigma$ bond, i.e. due to one head-on orbital overlap.

Below is an example a constructive $2 {p}_{z} - 2 {p}_{z}$ head-on overlap that forms a ${\sigma}_{2 {p}_{z}}$ molecular orbital, where electron density lies in the white bulged region---between the atoms.

A nice example is the $\setminus m a t h b f \left(\sigma\right)$ bond in $\setminus m a t h b f \left(\text{Cl"-"Cl}\right)$.

All pure double bonds consist of an additional $\pi$ bond, i.e. due to a sidelong orbital overlap.

Below is an example of the $2 {p}_{x} - 2 {p}_{x}$ constructive/bonding overlap, where electron density lies in the white bulged region (above the atoms).

An explicit $\pi$ bond example is the $\setminus m a t h b f \left(\pi\right)$ bond in $\setminus m a t h b f \left(\text{O"="O}\right)$, a product of either a $2 {p}_{x} - 2 {p}_{x}$ sidelong overlap, or a $2 {p}_{y} - 2 {p}_{y}$ sidelong overlap (but not both), that forms a ${\pi}_{2 {p}_{x \text{/} y}}$ orbital, depending on which pair overlaps.

This $\pi$ bond is made in addition to the $\sigma$ bond that was already made upon forming the first $\text{O"-"O}$ bond.

Therefore, since a pure double bond consists of 1 $\sigma$ bond and 1 $\pi$ bond, it is one $\setminus m a t h b f \left(\pi\right)$ bond's worth stronger than a single bond.