Question

Compare x and y bond angles for the above given molecules.

A. x > y
B. y > x
C. x=y
D. None of these

Answer 1

Observe the bonding atoms of bond angles x and y. The fluorine atom is more electronegative than the oxygen atom. Determine the type of repulsion present in the given molecules. Lower bond pair-bond pair repulsion lower is the bond angle.

Complete step by step answer:
The structures of molecules given to us are:

In the first structure bond angle ${\text{O = I = O}}$ is represented by ‘x’. In the second structure bond angle ${\text{F - I = O}}$ is represented by ‘y’. So in these two structures, there is a difference of one bond. So we can determine the difference in bond angle by comparing the electronegativity of the F and O atom. As the F atom is more electronegative than the O atom it will attract the bond pair of electrons towards itself. So in the second structure bond pair-bond pair repulsion ${\text{F - I}}$ and ${\text{I = O}}$ is less compare to bond pair-bond pair repulsion of two ${\text{I = O}}$ bonds in the first structure.

The bond angle of a molecule having greater bond pair-bond pair repulsion is greater.
So, angle x>y.

**Thus, the correct option is (A) x>y

Additional information: **
The reported bond angle${\text{O = I = O}}$ is ${\text{100}}^\circ$ and reported bond angle of ${\text{F - I = O}}$ is $98^\circ$.
The bond angle depends on the following 3 types of repulsions.
1. Lone pair-lone pair repulsion
2. Bond pair-bond pair repulsion
3. Lone pair-bond pair repulsion
The strongest among all these three repulsions is Lone pair-lone pair repulsion.

Note:
Bond angle of two bonds is based on bonding atoms, lone pair on bonding atoms, type of bonds and electronegativity of bonding atoms. As electronegativity increases from left to right across the period so F atom is more electronegative than O atom. The greater the repulsion of bonding atoms greater is the bond angle.