Question

The structure $I{F}_5$ can be best described as:
(a)

(b)

(c)

(d) none of these

Answer 1

Hint: When two different halogens react with each other, forms subordinate to halogens are termed as interhalogen compounds. These are prepared by direct mixing of halogens and by a reaction of halogen with lower interhalogen compounds.

Complete step by step solution:
The general composition formula of interhalogen compounds are $XY,X{Y}_3,X{Y}_5,\mathrm{\&}X{Y}_7$, where X is less electronegative halogen which is a larger size of halogen, and Y is more electronegative with a smaller size of halogen.
Radius ratio $={\displaystyle \frac{radiusofbiggerhalogenparticle(X)}{radiusofsmallerhalogenmolecule(Y)}}$
As the radius proportion increases, the number of atoms per molecule also increases. Hence, iodine (VII) fluoride should have a maximum number of toms as the radius ratio between I and F would be maximum. Thus its molecular formula is $I{F}_7$.
Given $I{F}_5$, iodine (V) fluoride belongs to $X{Y}_5$ the category of interhalogen compounds. The structure of this molecule hybridization $s{p}^3{d}^2$ with one lone pair. The number of bond pairs is 5 with $s{p}^3{d}^2$ hybrid orbitals and one lone pair which does not participate in the hybridization. So, due to one lone pair, the structure distorted from trigonal bipyramidal to square pyramidal and with F-I-F bond angle ${90}^{o}C$.

Therefore, the correct answer is option C.

Note: All interhalogen molecules are covalent molecules with diamagnetic. These are volatile solids or liquids at 298K except that ClF is gas. All molecular structures of these compounds can be explained based on the VSEPR theory.