# Confusion about comparing the charge densities of benzene and fluorobenzene

I understand that the functional group $$\ce{-F}$$ is an electron-withdrawing group. It is expected that the difficulties of an electrophile to attack the carbon atoms are $$\ce{PhH}<\ce{2,4of PhF}<\ce{3 of PhF}\tag{1}$$Therefore, I want to demonstrate it quantitatively by computing the charge densities of carbon atoms.
Using command # opt freq hf/3-21g pop=nbo geom=connectivity and the molecule models in Gaussian 09, I obtained the charge densities of $$\ce{PhH}$$ and $$\ce{PhF}$$. To my surprise, the result is:
$$\begin{array}c\text{molecule}&\text{ortho}&\text{meta}&\text{para}\\\ce{PhH}&-0.239&-0.239&-0.239\\\ce{PhF}&-0.306&-0.222&-0.261\end{array}$$ By observing this result, the fact that the electrophilic substitution of $$\ce{PhF}$$ is likely to occur in o- and p- is verified, or the right half of $$(1)$$ is verified.

But why would the charge density of position 2,4 of $$\ce{PhF}$$ is more negative than that of benzene, given the fact that $$\ce{-F}$$ decreases rate of the electrophilic substitution reactions?

My speculation
i) $$(1)$$ may be false.
ii) The NBO charge densities between two different molecules cannot be compared.
iii) The charge density cannot indicate the difficulty of the attacking of electrophile.

Which one, or something else, can explain this strange observation? Can another method to order electron-withdrawing ability of some functional groups computationally be suggested?