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The reactivity of halide ion in alkyl halide is:
A. F > Cl > Br >I
B. F < Cl < Br C. I > CI > Br >F
D. Br >I > Cl>F

Answer
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Hint: Substitution nucleophilic bimolecular $S{{N}_{2}}$ reaction-
Nucleophilic substitution bimolecular or $S{{N}_{2}}$ is a single step bimolecular reaction in which the incoming nucleophile attacks the C-atom of substrate in a direction opposite to the outgoing nucleophile.
The reaction passes through a transition state in which both the incoming and outgoing nucleophiles are bonded to the same C-atom.

Complete step by step answer:
In the transition state, the C-atom is $s{{p}^{2}}$ hybridised with a p-orbital whose one lobe overlaps with an orbital of incoming nucleophile and the other lobe overlaps with an orbital of outgoing nucleophile. The three non-reacting atoms or groups attached to the C-atom are nearly coplanar at an angle of ${{120}^{0}}$. The reaction is completed when the outgoing nucleophile leaves with the bond pair of electrons and simultaneously the incoming nucleophile binds to the C-atom. As the reaction progresses, the configuration of the C-atom under attack is inverted. An $S{{N}_{2}}$ reaction is always accompanied by inversion of configuration. The inversion in configuration implies change in configuration from R to S or S to R (provided the incoming nucleophile and outgoing nucleophile have same priority) and not from (+) to (-) or (-) to (+).
Steric hindrance plays a very vital role in an $S{{N}_{2}}$ reaction. As steric hindrance increases the rate of $S{{N}_{2}}$ reaction decreases. Thus for the same halogen the reactivity order of alkyl halides towards the reaction is as under
Methyl halide > ${{1}^{0}}$-Alkyl halide > ${{2}^{0}}$-Alkyl halide > ${{3}^{0}}$-Alkyl halide
For the same alkyl group, the reactivity of alkyl halides increases with the decrease in C-X bond dissociation energy. Therefore, R-I>R-Cl > R-Br> R-F.

So the correct option is C.
Note:
Group 17 elements are called halogens and they consist of seven valence electrons in their valence shell. $S{{N}_{2}}$ results in inversion of configuration, whereas, in $S{{N}_{1}}$, retention of the structure occurs.