Why are single bonds weaker than double?

1 Answer
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 \mathbf(pi) 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 2p_z-2p_z head-on overlap that forms a sigma_(2p_z) molecular orbital, where electron density lies in the white bulged region---between the atoms.

A nice example is the \mathbf(sigma) bond in \mathbf("Cl"-"Cl").

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

Below is an example of the 2p_x-2p_x constructive/bonding overlap, where electron density lies in the white bulged region (above the atoms).

An explicit pi bond example is the \mathbf(pi) bond in \mathbf("O"="O"), a product of either a 2p_x-2p_x sidelong overlap, or a 2p_y-2p_y sidelong overlap (but not both), that forms a pi_(2p_(x"/"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 "O"-"O" bond.

Therefore, since a pure double bond consists of 1 sigma bond and 1 pi bond, it is one \mathbf(pi) bond's worth stronger than a single bond.