The order you provided describes the decreasing bond strength not reactivity. Bond strength of R-X decreases from Fluorine to Iodine as the orbitals become more diffuse and effectiveness of overlap decreases. Hence, due to the ineffective overlap, the covalent bond strength is weaker
I believe the reactivity of alkyl halides should be R-I > R-Br > R-Cl > R-F. This is because of the decreasing bond strength of the R-X bond which thus slows down the rate of Sn1 and Sn2 reactions.
Why does the strength of R-X bond affect the reactivity of halogenoalkane?
This is because, the slow step (i.e. rate determining step) of the Sn1 and Sn2 nucleophilic substitution mechanism involves the breaking of the R-X bond. Either to form a carbocation (for Sn1) or the tetrahedral compound (for Sn2). Thus, the stronger the bond, more energy would be required to break it, making it less reactive.
However, there is a contrasting effect: the polarity of the R-X bond decreases down the group. The R-I bond is less polar than the R-F bond. Hence it may seem as though the nucleophile would be more strongly attracted to the electron deficient carbon. But based on experimental results, it has shown that the bond strength is in fact the main factor that determines the reactivity of the halogenoalkane towards Sn2 reactions.
The magnitude of the partial positive charge of the carbon in the C-X bond does not affect the reactivity. The main factor that affects reactivity is bond strength
Note that we do not need to consider this point about bond polarity of the C-X bond when we are looking at Sn1 mechanism, because the rate determining step of Sn1 reactions does not involve the attack by a nucleophile.