[PicturesOfLilly]

I get the basics; about how a molecule will try and minimise repulsion. And I get why the repulsion between two lone pairs is greater than two bond pairs.

I don't see how a lone pair of electrons can contribute to the shape of a molecule, the bonds pairs yes, but not the lone pairs. In other words, I'd expect something like NH3 to be planar (like boron trichloride is). In NH3, the lone pair is associated purely with the N atom, so shouldn't the extra net force be in all directions, giving the same shape as boron trichloride?

Or is it that the lone pair electrons move slower when at a certain side of the N atom or what? And if this is the case, I don't see why they would?

Thank you

[Borek]

shouldn't the extra net force be in all directions, giving the same shape as boron trichloride?

It is in all directions, which is exactly why it pushes all H atoms away - and the only way they can increase the distance from the lone pair is by leaving the planar surface and going under.

[PicturesOfLilly]

shouldn't the extra net force be in all directions, giving the same shape as boron trichloride?

It is in all directions, which is exactly why it pushes all H atoms away - and the only way they can increase the distance from the lone pair is by leaving the planar surface and going under.

Thanks Borek,

If the net force of the lone pair is in all directions, I can't see how this would change the bond angles. I could understand how it might increase the distances between the atoms.

I don't understand how the lone pair can be treated the same as a bond pair! As I already said, it's almost as if the lone pair electrons are spending more time at a certain side of the central atom!

Still a bit confused.

[Borek]

Imagine a flat molecule with three hydrogens lying on the surface. Now, stick the lone pair to the central atom, perpendicular to the surface. It is lobe shaped, it sticks in one direction only, in this aspect it is not much different from the bonded hydrogens. It tries to push away all atoms, and the only thing hydrogens can do to increase the distance from the lone pair, is to go under the surface.

[PicturesOfLilly]

Imagine a flat molecule with three hydrogens lying on the surface. Now, stick the lone pair to the central atom, perpendicular to the surface. It is lobe shaped, it sticks in one direction only, in this aspect it is not much different from the bonded hydrogens. It tries to push away all atoms, and the only thing hydrogens can do to increase the distance from the lone pair, is to go under the surface.

Thanks Borek,

That makes more sense, assuming it's a p orbital.

To ask a different question, will these electrons always be found in this lobe? Because if this is the case, that means that these electrons don't actually go around the nucleus? With an s orbital, that is spherically shaped, it is easier to imagine the electrons going around the nucleus.

[Borek]

To ask a different question, will these electrons always be found in this lobe? Because if this is the case, that means that these electrons don't actually go around the nucleus? With an s orbital, that is spherically shaped, it is easier to imagine the electrons going around the nucleus.

That's where your problem starts - you try to imagine the orbital as if the electron was kind of a small, charged sphere, flying inside of the orbital. It is not - it is a wave, so it is not orbiting the nucleus, rather it is present around the nucleus in all places at the same time (but with a different density).

[PicturesOfLilly]

To ask a different question, will these electrons always be found in this lobe? Because if this is the case, that means that these electrons don't actually go around the nucleus? With an s orbital, that is spherically shaped, it is easier to imagine the electrons going around the nucleus.

That's where your problem starts - you try to imagine the orbital as if the electron was kind of a small, charged sphere, flying inside of the orbital. It is not - it is a wave, so it is not orbiting the nucleus, rather it is present around the nucleus in all places at the same time (but with a different density).

Thanks Borek,

Are you saying that when referring to s orbitals? Do you mean "present around the nucleus" in all places within the shape (lobe, sphere, etc)

Because if that description you just gave applies to p orbitals too, then how in the name of God did they come up with such weird shapes.

Do you know of any on line demonstration that actually explains in such a way that someone can get their head around it. Because I've always thought it was badly thought and that most students don't understand what they're regurgitating in exams. I've read a bit about Heisenberg and Schrodinger, but there's still something I'm not getting.

[Burner]

They apply to all kind of orbitals(s,p,d,...)

[Borek]

Are you saying that when referring to s orbitals?

No, to every orbital.

Do you mean "present around the nucleus" in all places within the shape (lobe, sphere, etc)

More or less - yes. But please remember that these shapes are not bounded (that is: they have no well defined border). Electron is - to some extent - everywhere, what differs is the "fraction" of the complete wave in a given volume element. These volumes with the highest charge density ("probability of finding the electron") are typically drawn as orbitals.

Because if that description you just gave applies to p orbitals too, then how in the name of God did they come up with such weird shapes.

The only answer I know to this kind of question is "when you calculate it using QM, you got that". That is - there are no "reasons" why they look like that other than that's just the way it is.

Do you know of any on line demonstration that actually explains in such a way that someone can get their head around it. Because I've always thought it was badly thought and that most students don't understand what they're regurgitating in exams. I've read a bit about Heisenberg and Schrodinger, but there's still something I'm not getting.

I am afraid there is no "one size fits all" site that will explain it to everyone. We all have our ways of learning, what works for me doesn't have to work for you.