[cabaal]

I understand the main concept behind VSEPR - that electrons want to get as far away from one another as possible. But I don't understand why lone pairs radically alter geometries?



Why are lone pairs more repulsive than the electron pairs in the above example? I tried reading explanations on Wikipedia and my textbook, but I can't seem to grasp either one. Could someone try explaining it to me?

Also, what's the proper way of finding hybridization? I've seen the method where the sum of lone pairs and bonds (lone pair = 1 unit, double bond = 1 unit, triple bond = 1 unit, etc) equal the sum of the subscripts in s¹p^x. I've also seen the method of just memorizing the hybridization that goes with the accompanying VSEPR shape. Which of these is the "more correct" method?

Thank you.

[renge ishyo]

Why are lone pairs more repulsive than the electron pairs in the above example? I tried reading explanations on Wikipedia and my textbook, but I can't seem to grasp either one. Could someone try explaining it to me?

Lone pairs are more repulsive than the electrons involved in bonds because the lone pairs are only pulled towards the centeral atom whereas the electrons involved in bonds are pulled out *between* the central atom and the outer atom that the central atom is bonded to. Because the lone pairs are not "tugged away" from the central atom by an outside atom, they can exert a greater repulsive force on the central atom than they would be able to do if they were  instead involved in bonds that pulled the electrons away from the central atom.

For example, with CH₄ there are no lone pairs on the central atom so the bond angles are 109 degrees roughly. However, on NH₃ a single lone pair exists on nitrogen. This pair is not pulled away by an outer atom, but is instead localized on the nitrogen atom where it can push away at the other electrons involved in bonds. The result is that the bond angles change a bit to about 107 degrees. Now, if you look at water (H₂0), the central atom oxygen has two lone pairs on it. These push against the other bonds on the central atom as well as each other so that the bond angle is further compressed to about 104 degrees.

Also, what's the proper way of finding hybridization? I've seen the method where the sum of lone pairs and bonds (lone pair = 1 unit, double bond = 1 unit, triple bond = 1 unit, etc) equal the sum of the subscripts in s1px. I've also seen the method of just memorizing the hybridization that goes with the accompanying VSEPR shape. Which of these is the "more correct" method?

I do not know which is more correct, but I prefer drawing out the lewis structure and counting the connections + lone pairs to the central atom. If there are two connections, it is sp; 3 connections, sp²; 4 connections equals sp³, etc. I only get into trouble with this simplified approach when the possibility of d-orbitals comes into play, but fortunately that only matters for row three elements and below (the aforementioned method is good enough for basic organic chemistry).

[cabaal]

That makes perfect sense. Thanks again.

[stewie griffin]

As an organic chemist I use the same method that renge ishyo prefers, and it's worked well for me. You can still use this idea for the d orbitals in some cases... particularly when it comes to phosphorus and sulfur. For example, in SF₆ the sulfur has six "units" around it (it has no lone pairs left in this example). So that's 1 s + 3 p's + 2 d's so we can say that sulfur is sp³d² hybridized. However, once we get to the transition metals (which of course is all about d orbitals), almost everybody prefers to switch to molecular orbital theory to explain the bonding and geometry rather than hybridization theory.