# Vibrational Frequency .

1. Nov 10, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
It's just that I am curious from the vibrational frequency table in the textbook--could someone tell/explain to me why H2 molecule's vib. frequency is about 4400cm-1, whereas H2+ is 2300cm-1? It just seems bizarre to me that such similar molecule should have different vibrational frequency.

2. Relevant equations
maybe v = ω/(2πc)? I am not sure, but I think the equation is related to my question above.

3. The attempt at a solution
What I am sure of now is that H2's vib. frequency is about x2 that of H2+....so I wonder if this problem arises from H2+ having less electron, thus more delocalization between two H nuclei?

2. Nov 10, 2014

### Staff: Mentor

There are two electrons, to some extent each is responsible for half energy of the bond. When you remove one, there is only "half bond" left.

3. Nov 10, 2014

### terp.asessed

Thank you for the reply! Aside, so, since H2+ has only one electron responsible for the bond, b/c it has to take responsiblity of ALL the bond energy, one electron taking the role of two at the same time, its vibrational frequency is twice the H2? So, it has nothing to do with angular frequency?

4. Nov 10, 2014

### Staff: Mentor

Vibrational vs rotational, are you sure you are not confusing them?

5. Nov 10, 2014

### OldEngr63

What kind of frequency has the units cm^(-1)?

6. Nov 11, 2014

### Staff: Mentor

It is not a frequency. cm-1 is a unit of describing the wavelength in spectroscopy (in a way similar to Hz describing frequency), and wavelength and frequency are related by $c=\nu\lambda$. A little bit obscure when you are not in the field, but should be clear for every chemist :)

7. Nov 11, 2014

### Dr.D

Thank goodness I'm not a chemist!!

It must be extremely confusing to be required to call everything by the wrong name. Why not call it wave length and be done with it?

8. Nov 11, 2014

### Staff: Mentor

I don't think it is strictly limited to chemistry, more like to spectroscopy (being a chemist I am a little bit skewed). Wavenumber is more convenient than the wavelength as it is directly proportional to energy (which is particularly convenient in the case of the initial question - 2300 cm-1 is almost exactly half of the 4400 cm-1. Bingo!