Explain how it is possible for a binary compound between fluorine and iodine to form,and desribe the kind bond?

The initial I-F bond is easy to understand as it is the combination of two radicals just as in the parent compounds F2 and I2:

F. + I. → F:I

Perversely IF is not very stable.

The reaction easily proceeds further: Consider a pair of electrons on :IF reacting with two F. radicals:
:IF + 2 .F → F:I:F(F)

Because of the ability of F to "get in close" to E atoms for good overlap of the 2p AO on F with an E σ AO, E-F bonds are usually the strongest single bonds for that element (cf SF6 a totally inert molecule). Forming F. radicals does not cost much energy as the F-F bond strength in F2 is abnormally weak; and that for I2 even lower; the hybridization energy I 4p → 4d is not great (the energies of AOs get closer in energy on going down the PT). So forming two extra bonds is energetically favorable. The I-F are polar covalent bonds and so electron affinities are not really a part of the discussion.

The result is that IF3 is known but it is also not too stable. The reaction does not stop there, however, and IF5 is readily formed from the rxn with I2 and F2 and is the most stable fluoro compound of iodine (bp ~105 °C). It is a VSEPR favorite as it is square pyramidal: AX5E.

But the reaction goes even further! IF7 is known and although it is a powerful fluorinating agent it is well characterized (it is another VSEPR favorite: AX7, pentagonal bipyramid).