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:


$\ce{NH2}$
$\ce{OCH3}$
$\ce{Ph}$
$\ce{F}$
$\ce{Cl}$
$\ce{NO2}$


and the order was given as: 

$1 &gt; 2 &gt; 3 &gt; 5 &gt; 4 &gt; 6$

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 &#x3c0; bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ &#x3c0; bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.

The question is comparing the electron-donating capabilities of fluorine and chlorine, and that italicised word there alone should ring a bell. Fluorine is much more electronegative than chlorine which means that a carbon bound to fluorine has a much smaller electron density which destabilises cations much more than could be offset by the overall weak &#x3c0; donating capacity of fluorine.

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.