Fluorine exhibits only -1 oxidation state while iodine exhibits oxidation states of -1, + 1, + 3, +5 and + 7. This is due to :
flourine being a gas
available d-orbitals in iodine
non-availability of d-orbitals in iodine
none of the above.
The correct option is C available $$d$$-orbitals in iodine Due to absence of d-orbitals, $$F$$ can not expand its valency but $$I$$ can expand its valency (as shown in the figure) and therefore, exhibts variable oxidation states.