Question

Why are single bonds weaker than double?

Answer 1

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.

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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.