lone pair and polarity

[lamborghiniMD]

I've read that a compound with 2 lone pairs is most polar than any other compounds of comparable weight. But if a compound had 3 or 4 lone pairs, would that compound be even more polar, theoretically?

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[LoLCareerGoals]

4 lone pairs would all cancel each other out if set up tetrahedrally, so it'd be completely non-polar (not sure I can imagine such a "compound"). Are we talking like sulfur with 2 bonds and 4 non-bonding pairs?
Bring up a specific example instead of making up vague hypotheticals.

 

[shaboobly]

polarity depends on the shape of a molecule. the molecule with 2 lone pairs, like water, is extremely polar because of the 2 OH bonds and the resulting dipole. the dipole results because of the vector sum of the individual polar bonds.

when a molecule has a lone pair, it doesnt have a bond and wont be able to have a dipole in that direction because there is no bond. it's not because of the lone pairs that make a molecule polar, it's the types of bonds.

 

[lamborghiniMD]

I was just asking for a generalization, no need to be rude.

 

[gettheleadout]

polarity depends on the shape of a molecule. the molecule with 2 lone pairs, like water, is extremely polar because of the 2 OH bonds and the resulting dipole. the dipole results because of the vector sum of the individual polar bonds.

when a molecule has a lone pair, it doesnt have a bond and wont be able to have a dipole in that direction because there is no bond. it's not because of the lone pairs that make a molecule polar, it's the types of bonds.

This answer is incomplete. In general, it is the electron group geometry, not the molecular geometry alone, that determines polarity in accorsance with electronegtivity differences. Regarding water, yes, the polarity is partially a product of the O-H polar bond and the resulting dipole, but the two lone pairs on the O not only cause a bent molecular geometry an prevent the O-H bonds from canceling each other out as vectors, but provide a great amount of electron density in the oxygen's area to contribute to the overall dipole. Imagine H2O with one less lone pair (fictional molecule but just imagine). In this case the O-H bonds still do not cancel, but the region of electron density around the O would be less negative simply because half as many electrons occupy this space.


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[shaboobly]

This answer is incomplete. In general, it is the electron group geometry, not the molecular geometry alone, that determines polarity in accorsance with electronegtivity differences. Regarding water, yes, the polarity is partially a product of the O-H polar bond and the resulting dipole, but the two lone pairs on the O not only cause a bent molecular geometry an prevent the O-H bonds from canceling each other out as vectors, but provide a great amount of electron density in the oxygen's area to contribute to the overall dipole. Imagine H2O with one less lone pair (fictional molecule but just imagine). In this case the O-H bonds still do not cancel, but the region of electron density around the O would be less negative simply because half as many electrons occupy this space.


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correct, the shape of the molecule depends on the electron geometry because electrons repel each other and cause the bent configuration as in water. when i said the shape of the molecule it implies all the necessary points, including electron geometry.

 

[dmf2682]

correct, the shape of the molecule depends on the electron geometry because electrons repel each other and cause the bent configuration as in water. when i said the shape of the molecule it implies all the necessary points, including electron geometry.

If I can referee it sounds like you're talking vector direction while gtlo is talking magnitude. You're both right, you've to do a vector sum which ends up being the net dipole . Gtlo, thanks for that explanation, it was succinct.