When two different halogens react with each other, interhalogen compounds are formed. They can be assigned general compositions as XX’ , XX’_3, XX’₅ and XX’₇ where X is halogen of larger size and X’ of smaller size and X is more electropositive than X’. As the ratio between radii of X and X’ increases, the number of atoms per molecule also increases. Thus, iodine (VII) fluoride should have maximum number of atoms as the ratio of radii between I and F should be maximum. That is why its formula is IF₇ (having maximum number of atoms).

What is the hybridization of interhalogen compounds of the type IF₇?

sp³d³

The correct answer is option 4. \(sp^3d^3\).

The electronic configuration of \(I\) is

\(_{53}I = [_{36}Kr]4d^{10}5s^25p^5\)

The electronic configuration of \(F\) is

\(_{9}F = [_2He]2s^22p^5\)

Valence electrons in iodine = 7

Valence electrons in fluorine = 7

Total number of valence electrons in \(IF_7\)

\(= 7 + 7 \times 7\)

\(= 7 + 49\)

\(= 56\)

Number of orbitals used in hybridization \(= \frac{56}{8} = 7\)

Since there are 7 hybrid orbitals,out of which there is 1 s-orbital, 3 p-orbital and 3 d-orbitals.

So, the hybridization of \(IF_7\) is \(sp^3d^3\)

Thus, the structure will be as: