Application of the electronegativity effect on bond angles

Question

There's a rule in determining molecular geometry that says that if the electronegativity of the central atom increases, the bond angle of the molecule increases.

Is this is only applicable in the cases where the molecular geometry of the said compounds are similar? For example, considering $\ce{ClF3, BF3,}$ and $\ce{ PF3}$, they're primarily differentiated on the basis of them having different geometries.

Answer 1

(I will simply be converting my comments above into an answer here)

Yes, generally, you can apply this in cases with similar geometry. You can see why from the following rationale:

• Imagine that the bond pairs look like a ring-shaped cloud, and one is worn on your index finger and another on your middle finger, both towards the fingertip. For example, to represent the two $\ce{O-H}$ bonds of $\ce{H2O}$.
• More electronegative central atoms will pull the ring towards the knuckles and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger

Here's an illustrative image for $\ce{H2O}$, with blue rings representing bond pair density :

Hence, you can see why the bond angle of $\ce{H2O}$ would be bigger than, say, $\ce{OF2}$ and $\ce{H2S}$.

In case of the former, the fluorines won't led the ring move towards the depression easily, hence the index and middle fingers can come closer to each other in contrast to $\ce{H2O}$

For the latter, $\ce{S}$ is less electronegative than $\ce{O}$, so again, the rings will show less movement towards the knuckles.

This will also help you to understand that you can't do this in cases of radically different molecular geometries, as this visualization will not be very useful there