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I've learnt about VSEPR theory and have a general idea of how lone electron pairs affect the molecular geometry by repulsions. But my teacher told me that when the angle between bond pairs or lone pairs is 120 degrees, there is no repulsion. $\ce {SO2}$ has $sp^2$ hybridization and thus has trigonal planar electron geometry. Thus, angle between the lone pair and each of the $\ce {S-O}$ bonds is 120 degrees. This means there should be no repulsions and the bond angle between the two $\ce {S-O}$ bonds should remain 120, instead of becoming less than that. So then why is it, in fact, less than 120?

Edit: The possible duplicate to my question is answered by saying that (a) VSEPR isn't perfect and (b) in this case it actually does do a pretty good job of explaining the bond angle.

...the results give a bond angle ∠(O−S−O)≈119∘. This is indeed very close to 120∘ so you can say that the VSEPR method gives a good approximation of the SO2 bond angle.

The essence of my question revolves around the fact that lone and/or bond pairs separated by an angle of 120 degrees do not create repulsions, so considering that, why is the $\ce{O-S-O}$ bond angle compressed down to 119?

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    $\begingroup$ It is written not as $\ce SO_2$ ($\ce SO_2$), but as $\ce{SO2}$ ($\ce{SO2}$) // Note that CH SE prefers plain text in titles. $\endgroup$
    – Poutnik
    Commented Sep 15, 2022 at 14:58
  • $\begingroup$ @Mithoron I've edited my question to explain why that is not a duplicate. $\endgroup$
    – AVS
    Commented Sep 16, 2022 at 9:53

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Under the VSEPR model, rationalization of bond angle compression to less than 120$^\circ$ can be made by examining how diffuse a lone-pair is relative to a bond. In a bond, the electrons are confined to the band of space between the two atoms—the presence of the positive nuclei on each end attracts electron density into a tube, which makes up the bond(s). A lone-pair does not have this same restriction; instead, a lone-pair is only bound to a single atom (by definition) so further from the nucleus, the electrons can diffuse across a larger space. There is also less positive charge as you move away from the nucleus, and so the negative charge of electron density over there is more potent. This has the net effect of a lone-pair being more repulsive than a chemical bond, which causes the $\ce{O—S—O}$ bond angle to compress.

In reality, most bond angles are a slurry of effects from various lone-pairs and bonds influencing what angles are most stable. Addressing the nature and impact of these factors on chemical reactivity is called sterics.

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  • $\begingroup$ I understand why there is repulsion in general, but shouldn't the 120 angle between lone pair and $\ce{S-O}$ bond mean there shouldn't be repulsion at all, and thus no compression of bond angle? (See edited question) $\endgroup$
    – AVS
    Commented Sep 15, 2022 at 16:19
  • $\begingroup$ A lone pair is more cumbersome than a covalent bond. It repels more strongly the covalent bonds. $\endgroup$
    – Maurice
    Commented Sep 15, 2022 at 16:34
  • $\begingroup$ @AVS probably best not to think about it as "no repulsion at all" and think of 120 degrees as "where 3 equally repelling things will be relative to each other because that is the farthest they can get from each other" $\endgroup$ Commented Sep 15, 2022 at 22:57
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    $\begingroup$ Oh, so then since the 3 "things" in this case are not equally repelling (two bond pairs and one lone pair) the repulsion still takes place. Have I understood this right? $\endgroup$
    – AVS
    Commented Sep 16, 2022 at 9:55
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    $\begingroup$ @AVS mostly—i would still say when the three "things" are equal, there is repulsion. it's just the least repulsion possible or "minimal repulsion" which you could consider as a definition for no repulsion. if you add all the force vectors, there is no net repulsion, but the force vectors are still there. $\endgroup$ Commented Sep 17, 2022 at 0:58

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