Decreasing order of electron density in benzene ring among the halobenzenes
Firstly, I would like to talk about the order of withdrawing effect among halogens. Since all the four halogens are electron-rich, they tend to donate electrons (+M effect). However, the high electronegativity of these elements hinders the donation by applying an even greater electron-withdrawing effect (-I effect) that overcomes the +M (resonance) effect.
In the case of halogens, the -I effect varies as -$\ce{F}$ &gt; -$\ce{Cl}$ &gt; -$\ce{Br}$ &gt; -$\ce{I}$ due to the decreasing electronegative character as we move from fluorine to iodine.
The +M effect also varies as -$\ce{F}$ &gt; -$\ce{Cl}$ &gt; -$\ce{Br}$ &gt; -$\ce{I}$ due to the decreasing overlap between the 2p orbitals of carbon and the 2p, 3p, 4p and 5p orbitals of the halogens respectively. Since the -I effect is stronger than the +M effect as mentioned earlier, the final order of electron-withdrawing capability is
-$\ce{F}$ &gt; -$\ce{Cl}$ &gt; -$\ce{Br}$ &gt; -$\ce{I}$
Equipped with this knowledge, we come back to the question of comparing the electron densities of mono halogen benzene derivatives. From the above-mentioned trend, we can infer that the order of electron density is
$\ce{Ph-F}$ &lt; $\ce{Ph-Cl}$ &lt; $\ce{Ph-Br}$ &lt; $\ce{Ph-I}$
This trend is validated by the order of reactivity of these compounds towards electrophilic substitution reactions.
Hope it helps!