# List of H20 breakdown substances

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero). Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realise that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.

I'm assembling a molecular laboratory. I need to acquire an IR Spectrometer. But the cheapest ones available (in fact mostly available) are the NIR spectrometer with range of wavelength of 650 to 2500 nm. The fundamental IR active modes are all above 2500nm. The ones below are just the overtones. I want to ask about overtones and how any external energy introduced into the molecules can show up in the spectrums from 650 to 2500nm.

2,700 nm corresponds to about 3700 cm-1

15,000 nm corresponds to about 650 cm-1

In details. NIR spectrometers have primarily a range of 900nm to 2400nm (or 11,111 cm-1 to 4166 cm-1). In the following table or list of the molecules functional groups, the range is between 2700nm to 15,000 nm (3700 cm-1 and 650 cm-1). Outside of the range of NIR spectrometers.

https://www.sigmaaldrich.com/technical-

the rest the list is in the url above.

So if I get an IR spectrometers that can only detect the overtones of the fundamental IR dipole modes. And external energy is introduced into the molecules. How much will it show up in the 650 to 2500 nm wavenength range? How much can external energy affect the overtones? Or not visible at all? I won't get NIR Spectrometers to study functional groups but just to see if the overtones can be affected by any external energy. What is the wavelengh of higher rotational and vibrational states?

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero).

To clarify something. You were describing polarizability in the second sentence, right? If you were, then you didn't get my second to last message. I wasnt asking about polarizability anymore which we have discussed in many messages already.

What I was inquiring in second to last message was about the equipartition of energy where the energy can be distributed in rotational, vibrational or electronic modes of vibrations.

The fundamental modes of molecular permanent dipole moments are in the range of 3 to 15 micron. Most IR spectrometers I saw are in the NIR range which is 0.78 to 2.5 micron. And these can only detect the overtones of the fundamental modes.

Chemistry: Notes about IR spectroscopy (openchemistryhelp.blogspot.com)

So in my question to you. I wasn't asking about polarizability anymore but how rotational, vibrational, electron degree of freedom in the equipartition of energy is related to the permanent dipole moments. You said not significant. So what wavelength are these rotational, vibrational electronic degree of freedom?

Also I was googling about "thermal imaging and dipole moments". I couldn't find any hits. Thermal imaging seems to involve blackbody radiation. But what molecular modes contribute to blackbody radiation? If you can see the wavelength range above. The thermal modes overlap with the permanent dipole moments modes.

I already owned a thermal imaging system that can probe 7.5 to 13micron wavelength. So I need to understand everything about the infrared wavelength to understand it all. I indeed tried to google and read a lot. So I'm not seeking a tutorial here but just clarifications. That's all. Thanks.

Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realise that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.

mjc123
Homework Helper
To clarify something. You were describing polarizability in the second sentence, right?
No. I was describing a dipole moment that changes during vibration due to the changing geometry of the molecule. It is this (and I repeat, NOT a "permanent dipole moment") that is necessary for a vibration to be IR active.
Polarisability refers to the production of an induced dipole moment by the application of an external electric field. Raman activity requires a change in polarisability during the vibration.

No. I was describing a dipole moment that changes during vibration due to the changing geometry of the molecule. It is this (and I repeat, NOT a "permanent dipole moment") that is necessary for a vibration to be IR active.
Polarisability refers to the production of an induced dipole moment by the application of an external electric field. Raman activity requires a change in polarisability during the vibration.

Thanks for the clarifications. So the IR wavelength being seen by thermal cameras/imagers is the same dipole moment that changes due to the changing geometry of the molecule? Note the wavelength of these thermal images are similar to the IR spectroscopy fundamental functional groups wavelength, both about 3 to 15 microns.

But you said "Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).".

I still don't know how to relate the blackbody radiation and dipole moments of molecules. Not a single site out of nearly half a hundred sites I read mentioned the connections.

Is it that they are not exactly related because the temperature seen by the thermal imager is the peak of the following Blackbody and not the molecules dipole moment wavelengths? What other vibrational states other than dipole moments and rotations that contribute to thermal vibrations?

Lord Jestocost
Gold Member
I still don't know how to relate the blackbody radiation and dipole moments of molecules.

There is no relation. Have a look at https://www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/blackbody.html:

"The situation is similar in a glowing solid or liquid. Instead of the single atoms or ions, in this case one has to consider the whole bulk system, as the particles are in permanent interaction with their neighbours. Thermal motion of electrons and ions is again irregular enough so that the above arguments are applicable.......
...... It is remarkable how little the intensity and its distribution over the different wavelengths depend on the glowing substance. Radiation emerging from a small opening in a cavity depends only on the temperature and not at all on the material of the cavity's walls. The opening of a cavity appears black at low temperatures and represents an almost ideal black body; the radiation emitted is called black-body radiation or cavity radiation. Given the temperature, the black-body radiation is completely specified. Light of the sun is similar to blackbody radiation."

There is no relation. Have a look at https://www.itp.uni-hannover.de/fileadmin/arbeitsgruppen/zawischa/static_html/blackbody.html:

"The situation is similar in a glowing solid or liquid. Instead of the single atoms or ions, in this case one has to consider the whole bulk system, as the particles are in permanent interaction with their neighbours. Thermal motion of electrons and ions is again irregular enough so that the above arguments are applicable.......
...... It is remarkable how little the intensity and its distribution over the different wavelengths depend on the glowing substance. Radiation emerging from a small opening in a cavity depends only on the temperature and not at all on the material of the cavity's walls. The opening of a cavity appears black at low temperatures and represents an almost ideal black body; the radiation emitted is called black-body radiation or cavity radiation. Given the temperature, the black-body radiation is completely specified. Light of the sun is similar to blackbody radiation."

Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to 15 micron?

an example of IR spectrum

and reflect (pun unintended) this

DrClaude
Mentor
Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to 15 micron?
The blackbody spectrum is derived from a gas of photons at thermal equilibrium. It turns out that a perfect blackbody, an idealized material that absorbs all EM radiation that falls onto it, has the same emission spectrum as the gas of photons at the same temperature.

Many materials are good approximations to blackbodies. When you look at their spectrum, their emission intensity as a function of wavelength follows more or less what is expected of a blackbody, so that emission can be used to measure their temperature, and this is the basis for thermal imaging.

It turns out that, at temperatures we humans consider normal, the photon energy of the blackbody spectrum peaks in the IR part of the spectrum. This is why thermal imaging is done using IR cameras. It also turns out that from a molecular and solid state point of view, this range of photon energy corresponds to vibrations of bonded atoms. In molecules, the vibrational modes are discrete, which leads to the kind of spectra you just showed, while for the solid state the vibrational modes (phonons) are much closer in energy and are basically continuous.

Note that since homonuclear diatomic molecules are not active in the IR, as written above, the main components of air, N2 and O2, are transparent in the IR, which is why thermal imaging cameras can work.

So there is ultimately a link between polar bonds and blackbody radiation, but there are many layers of understanding that go between the two.

Lord Jestocost and jake jot
Ok thanks. But why do the wavelengths of dipole moments and thermal imaging coincide at 3 to `15 micron?

an example of IR spectrum

View attachment 274241
Now reflect on this spectrum and imagine a much thicker layer of the same substance.
The top line around 80 % represents reflection of apparatus. The rest is the spectrum of sample.
It is a thin layer. Look at, for example, the double peak at 1440 and 1400 cm-1.
About 25 % transmission at 1440/cm, 35 % at 1400/cm and 45 % in between, at 1420/cm
Normalized for the 80 % reflection, it is roughly 31 % at 1440/cm, 44% at 1400/cm and 56 % at 1420/cm
Double the thickness and the numbers are 10 %, 19 % and 31 %. Quadruple the thickness and the numbers are just 1%, 4 % and 10 %.

In thick layers, the IR lines disappear because the weaker absorption between the peaks is already nearly 100 %. There will be little visible difference between light where 99 % is absorbed in the first 1 mm, between peaks, and light where 99,9 % is absorbed in the first 0,1 mm, at peak. Likewise, the emitted light will be almost as strong as black body radiation both at spectral peaks and non-peaks.

jake jot