# Comparison of Resonance Energies of some Carbocations

We were given the following carbocations and asked to compare their resonance energies.

$\ce{L-CH2+}$, where $\ce{L}$ is:

1. $\ce{NH2}$
2. $\ce{OCH3}$
3. $\ce{Ph}$
4. $\ce{F}$
5. $\ce{Cl}$
6. $\ce{NO2}$

and the order was given as:

$1 > 2 > 3 > 5 > 4 > 6$

I am clear with the other orders but I am unsure about $\ce{F}$ and $\ce{Cl}$. Why does $\ce{Cl-CH2+}$ have a higher Resonance Energy than its $\ce{F-CH2+}$ counterpart even though in resonance, $\ce{Cl}$ forms $\ce{{3p}-{2p}}\ \pi$ bonds compared to $\ce{{2p}-{2p}}$ ones in $\ce{F-CH2+}$, which are less stable due to less efficient overlapping? $\ce{F-CH2+}$ does have a higher electronegativity, but is it enough to overcome the higher stability of the $\ce{{2p}-{2p}}$ bond over the $\ce{{3p}-{2p}}$ one?

It is not like π bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ π bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.
Remember that for electrophilic aromatic substitutions, halide substituents are classified as deactivating — mainly due to their $-I$ effect — and only weakly ortho/para-directing — due to the smaller $+M$ effect. This should serve to show which effect predominates and it only gets more pronounced if instead of a large aromatic system you only consider the atom directly adjacent.