Answer:
Iodide > Bromide > Chloride > Fluoride
Step-by-step explanation:
For the case of the Sn1 reactions, let's remember that in this kind of reactions, we form a carbocation intermediate. This formation is done when the electrons of the carbon - halogen bond is transfered to the halogen we are talking. When this happens, and the halogen leaves the molecule, the halogen could be either strong or weak. Is the base is weaker, then this means that the carbocation is formed faster.
According to this, if we have the HI, HCl, HBr and HF, it's clearly obvious that the conjugate base of these acids, are I-, Br-, Cl, F-. And because the F- is the most electronegative element of the periodic table, this do not contribute a lot to the location of electrons in the base, and it's a very small atom compared to the Cl, Br and I. Therefore, The F atom would be the weaker and the Iodine would be the strongest.
In a Sn2 reaction, we do not form carbocation but the strength of the reaction comes with the strength in the carbon - halogen bond. Is the bond is weaker, this promoves a faster Sn2 reaction. As the Carbon - Fluoride is the most stronger bond because of Fluoride electronegativity, this makes the bond stronger to break and that's why it still the slowest leaving group.
Therefore the order of leaving groups for either Sn1 and Sn2 is:
I > Br > Cl >>F