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?
Answer & explanation
Correct answer: option 4
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:
