# How does polarity affect bond angles?

Aug 21, 2014

Theoretically, the bond angle X-A-X should be bigger in a compound ${E}_{2} A {X}_{2}$ or $A {X}_{2} {Y}_{2}$ than, respectively, Z-A-Z angle in ${E}_{2} A {Z}_{2}$ or $A {Z}_{2} {Y}_{2}$ if the A-X bonds were more polar than A-Z bonds.

In this notation (from Gillespie-Nyholm's VSEPR's theory ) A = central atom; E = lone pair; X, Y, Z = atoms bonded to A.

This "effect" should be related to the electrostatic repulsion between polarized X atoms (negatively-shielded or positively-unshielded atoms) that would be less than the electrostatic repulsion between Z & Z.

My answer-explanation is also theoretical because it would be necessary to find three A, X, Z atoms in which the A-X and A-Z bond should have the same lenght, the X and Z atoms would have the same covalent radii, the only difference between X and Z being their electronegativity.

To check this answer we should find some specific couples of $A {X}_{2}$ - $A {Z}_{2}$ bent molecules or tetrahedral $A {X}_{2} {Y}_{2}$ - $A {Z}_{2} {Y}_{2}$ molecules where the above said conditions would hold, and to know the true X-ray (X-diffractometrically measured) X-A-X & Z-A-Z angles.

If the question you have submitted is up to you, you deserve an A+ for it. If you can find similar molecules and data to check the theoretical answer, you deserve a double A++. If the question comes from the teacher, he or she should know the proper example to let you checking out :-)

Tentatively, I would try $C {H}_{2} {F}_{2}$ vs $C {H}_{2} B {r}_{2}$ as a first attempt. Bromine is much less electronegative than fluorine and the small negative charge is diffuse in a bigger surface over the bromine atoms. Hence, if the F-C-F angle would be bigger than Br-C-Br in the two dihalomethanes, the hypothesis would have been proven, in at least one case. If F-C-F < Br-C-Br, you could still search for a better couple of test molecules, because the prevaling steric hindrance of Br atoms could explain the test failure.