Absorption of light by molecules and the reasons for this

[RoboNerd]

Homework Statement

N2 molecules absorb ultraviolet light, but not visible light. I2 molecules absorb both visible and ultraviolet light. Which of the following statements explains the observations?

a) More energy is required to make N2 molecules vibrate than is required to make I2 molecs. vibrate
b) More energy is required to remove an electron from an I2 molecule than is required to remove an electron from a N2 molecule.
c) visible light does not produce transitions between electronic energy levels in the N2 molecule but does produce transitions in the I2 molecule.
d) the molecular mass of I2 is greater than the molecular mass of N2.

In my AP Chem class, we did not cover the absorption of light by molecules, but I found this problem above in the course description on the collegeboard site.

Could anyone explain why C is right and the other answers are wrong [and direct me at the theory needed to solve]?

Homework Equations

no equations

The Attempt at a Solution

I thought that since UV light has more energy than visible light and N2 can absorb only energy from UV light, then the energy from the visible light is inadequate for exciting it. I2 molecules absorb both visible and uv light successfully, so they require less energy to excite.

I know... this is a half-baked guess.

Anyways, I am looking forward to hearing your inputs.
Thanks in advance and for the time.

 

[Borek]

First of all: it is mostly about energies. As a rule of thumb:

Vibrational energies lie in the IR range.

Ionization requires quite high energies, UV at least.

Excitation of electrons requires energies from the visible spectrum and close UV.

 

[RoboNerd]

Thanks for that information and for the help.

However, I am still confused as to why each individual option is correct/incorrect.

Could you please go through each option A-D and explain why each one is right/wrong the way it is?

Thanks?

 

[Borek]

Start with "a" - can it explain the observation? Why yes, why not?

 

[RoboNerd]

a) More energy is required to make N2 molecules vibrate than is required to make I2 molecs. vibrate

OK. so we know that vibrating energies lie in IR range. N2 molecules are smaller in terms than I2, so they need more energy to get vibrating?

This is just a guess

 

[Borek]

If the vibrational energies are in the IR range, can they be responsible for observations made in the UV and visible light ranges?

 

[RoboNerd]

No. The N2 molecules would not vibrate given that they did not absorb IR light. Right?

 

[Borek]

Yes. That means a is out.

What about b?

 

[RoboNerd]

b is false as I2 has a lower ionization energy

 

[Borek]

But does it matter at all? Ionization is related to which wavelengths?

 

[RoboNerd]

Ionization energy is related to higher UV wavelength. But both molecules absorb UV, right? This means that that option does not make sense.

 

[Borek]

Even if the statement about ionization is right, does it describe what we observe in a visible light?

 

[RoboNerd]

Even if the statement about ionization is right, does it describe what we observe in a visible light?

What do you mean "what we observe in visible light?"

 

[Borek]

Processes that we observe when we do the experiments using visible light.

In other words: can we explain our observations done when recording the visual light spectra with the behavior of the molecule in the UV range?

 

[RoboNerd]

I honestly have no idea what would happen if we were recording the visual light spectra with the behavior of the molecule in the UV range... this is just a bit over my head at the moment.

 

[Borek]

I honestly have no idea what would happen if we were recording the visual light spectra with the behavior of the molecule in the UV range... this is just a bit over my head at the moment.

This is not different from the previous point - the one about vibrations (IR range) and visible/UV ranges. You are making it harder than it is.

 

[RoboNerd]

I probably am, but I do not know what would happen.

All I know is that in visible wavelength I will excite the electrons and in the infrared I will vibrate them.

What is your opinion?

 

[Borek]

I do not know what would happen.

What about "nothing"?

You had no problems realizing that IR - which is responsible for vibrational spectra - has nothing to do with the visible light spectra. How is it different from the fact that ionization - which requires UV - has nothing to do with the observations made in the visible range?

 

[RoboNerd]

You had no problems realizing that IR - which is responsible for vibrational spectra - has nothing to do with the visible light spectra. How is it different from the fact that ionization - which requires UV - has nothing to do with the observations made in the visible range?

I do not know what I am doing right now... but I think I might be on to something.

The I2 molecules absorb both visible and UV light... so I am guessing we are exciting the electron. If we were removing an electron, then we would just do UV light. N2 absorbs UV light but not visible light, so I am guessing that we only have ionization there.

Thus according to this approach, B is out as we are not removing electrons from both molecules, but rather from one.

Do you think this approach of mine is legit? Thanks!

 

[Borek]

The I2 molecules absorb both visible and UV light... so I am guessing we are exciting the electron. If we were removing an electron, then we would just do UV light. N2 absorbs UV light but not visible light, so I am guessing that we only have ionization there.

Yes.

Thus according to this approach, B is out as we are not removing electrons from both molecules, but rather from one.

Actually we don't remove electrons in a visible light at all, in neither case (which you have correctly stated above). And the information about energies required to remove electrons in the UV range doesn't tell us anything abut what is happening in the visible light, these things are completely independent. You can't conclude from the information given in b what will happen in the visible light, thus it is not a valid explanation.