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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.

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    $\begingroup$ Yes, in generality, you can apply this in those cases with similar geometry. You can see why due to the rationale it stems from: imagine hypothetically that the bond pair looks like a ring-shaped cloud, and that ring is on your index finger(for example, let us take one O-H bond of H2O). Now, imagine that the O is present on the aforementioned finger towards the knuckles, while the H is sitting towards the fingernail side. Now, more will the electronegativity of the central atom, more will the ring be pulled towards the knuckles..... $\endgroup$ – Yusuf Hasan Apr 6 at 9:53
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    $\begingroup$ ….and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger(note that the middle finger here rougly represent the second O-H bond, and more precisely, the oxygen of the H2O would be placed on the depression between your index and middle finger). Hence, you can see why the bond angle of H2O would be bigger than, say, OF2 and H2S. 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 $\endgroup$ – Yusuf Hasan Apr 6 at 10:01
  • $\begingroup$ chemistry.stackexchange.com/questions/15620/what-is-bents-rule $\endgroup$ – Mithoron Apr 6 at 19:05
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(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 :

Figure 1. Bent's Rule

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

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