# Electron probability in oscillating hydrogen molecule

1. Mar 30, 2005

### sid_galt

In a hydrogen molecule, the electron probability is the highest between the two nuclei of the hydrogen molecule. Let's say that the nuclei of the hydrogen molecule start oscillating.
When they get closer to each other, will the electron probability decrease or increase between the two nuclei?

2. Mar 30, 2005

### Gokul43201

Staff Emeritus
1. This is not a chemistry question.

2. Why is this important ?

3. This is a non-trivial calculation. You want to apply a time dependent perturbation to a many-particle hamiltonian (with a non-central potential) and determine the evolution of the angular part of the wave function.

4. Not doing the above, I'd go with 'decrease'. (consider the limiting case)

3. Mar 30, 2005

### Staff: Mentor

It is not. Plot of electron density looks like a two-humped camel.

Last edited by a moderator: Aug 13, 2013
4. Mar 30, 2005

### Gokul43201

Staff Emeritus
The "two humped camel" is the plot of probability density vs. position along the internuclear line. So it tells you nothing about the angular dependence of the density. For this you must look at a 2d plot with the density designated by the packing of dots (cloud pictures) or an angular plot of the density.

5. Mar 30, 2005

### Staff: Mentor

I know, just 'along the internuclear line' was beyond scope of my English at the moment of writing

But it doesn't change the fact that statement "the electron probability is the highest between the two nuclei of the hydrogen molecule" is false. Unless "between the two nuclei" doesn't mean "on the internuclear line"

Last edited by a moderator: Aug 13, 2013
6. Mar 30, 2005

### Gokul43201

Staff Emeritus
Then at what angular position is it the highest (roughly) ?

7. Mar 31, 2005

### Staff: Mentor

It seems I was partially wrong, or rather not precise enough when thinking. Angular position has nothing to do with the hydrogen particle. 'Camel plot' goes through the maximum values, but - stricly speaking - maximum is not between nuclei. There are two maximums, both near the nuclei. So it is not difficult to find a place in the particle with the electron density higher then in the cencter of the particle. That's what I was referring to.

Unless I am wrong again...

Last edited by a moderator: Aug 13, 2013
8. Mar 31, 2005

### Gokul43201

Staff Emeritus
Yes, there are two maxima, but they both lie along the internuclear line (ie : between the nucleii, but not in the middle of the separation) . Perhaps I was being unclear with my previous remarks.

In any case this is only a side issue and and the real question asks what happens to the wave function as the internuclear spacing changes.

9. Apr 2, 2005

### SpaceTiger

Staff Emeritus
A very loose bond would have a near-zero probability at the center of the internuclear line. However, as they were brought closer together, you would be more likely to find an electron in between them because they'd be more strongly bonded.

10. Apr 2, 2005

### Gokul43201

Staff Emeritus
The equilibrium internuclear spacing (bond length) corresponds to a local minimum in PE and hence a local maximum in bond strength. Bringing the atoms closer than equilibrium spacing will not increase the bond strength.

11. Apr 2, 2005

### SpaceTiger

Staff Emeritus
You're right, how sloppy of me! Bond strength does refer to the energy required to separate it, not the proximity. The rest of what I said holds, though.

12. Apr 4, 2005

### Gokul43201

Staff Emeritus
SpaceTiger, what would you imagine happened in the limiting case where the nuclei were infinitesimally close to each other. Would you still expect a larger electron density between them, rather than on the outsides ?

13. Apr 4, 2005

### SpaceTiger

Staff Emeritus
At that point, it's not really a molecule anymore, it's an atom. So the wave function would be of a different character...

14. Apr 5, 2005

### Gokul43201

Staff Emeritus
What you are suggesting is a discontinuity in the density with separation.

As separation decreases, density increases monotonically even at very small separations. Then when the separation goes to zero, the density suddenly drops to zero.

What I'm suggesting is that there is instead a continuous variation of density with separation; the density being zero for infinite as well as zero separation, with a maximum somewhere in between.

Which looks "nicer" to you ?

15. Apr 5, 2005

### SpaceTiger

Staff Emeritus
When they're that close to one another, you have a change in the shape of the potential because the nuclei bond to one another by the strong force. Up until that point, the potential is double-peaked and will give different wave functions. It's not a discontinuity because they don't have to be infinitesmally close, but on the scales of an atom, it's practically one.

Even if your arguments were correct, I still wouldn't be convinced that the probability density would change near that critical point (that is, near the stable separation). I'll concede, however, that my intuition may not be entirely correct on this issue and would defer the final word to someone who has actually solved the SE for this.

Last edited: Apr 5, 2005
16. Apr 5, 2005

### Gokul43201

Staff Emeritus
Neither am I. I do not claim to know the location of the maximum, but am making a top-of-my-hat guess based purely on intuition.