Which Molecules Are Suitable for Purely Rotational Spectroscopy?

AI Thread Summary
Purely rotational spectroscopy can be used to study molecules with a permanent dipole moment. Among the listed molecules, NH3 and CS are suitable candidates due to their dipole moments. In contrast, N2 lacks a dipole moment, making it unsuitable for this type of spectroscopy. The discussion highlights that non-polar molecules like N2 can be studied using Raman spectroscopy instead. Understanding the molecular geometry and dipole characteristics is crucial for determining suitability for rotational spectroscopy.
adiabatman
Messages
3
Reaction score
0
Which of the following molecules can be studied by purely rotational spectroscopy?

a) NH3 b) N2 c) CH4 d) SF6 e) CS f) CS2

The attempt at a solution[/b]

To my understanding, using only rotational spectroscopy could be used to study b) N2 and e) CS but I am not sure about my answer. Would the lack of a dipole for N2 make it unfit? Any help would be greatly appreciated.
 
Physics news on Phys.org
adiabatman said:
Which of the following molecules can be studied by purely rotational spectroscopy?

a) NH3 b) N2 c) CH4 d) SF6 e) CS f) CS2

The attempt at a solution[/b]

To my understanding, using only rotational spectroscopy could be used to study b) N2 and e) CS but I am not sure about my answer. Would the lack of a dipole for N2 make it unfit? Any help would be greatly appreciated.

http://en.wikipedia.org/wiki/Rotational_spectroscopy
The rotational spectra of non-polar molecules cannot be observed by those methods, but can be observed and measured by Raman spectroscopy.
Which of the molecules above posses dipole momentum?

ehild
 
  • Like
Likes 1 person
Thank you! Now I feel confident that e) CS is the answer
 
What about NH3?

The NH3 molecule has a large dipole moment, and this is consistent with its geometry, a triangular
pyramid.
http://en.wikipedia.org/wiki/Ammonia

Structure

The ammonia molecule has a trigonal pyramidal shape as predicted by the valence shell electron pair repulsion theory (VSEPR theory) with an experimentally determined bond angle of 106.7°.[13] The central nitrogen atom has five outer electrons with an additional electron from each hydrogen atom. This gives a total of eight electrons, or four electron pairs that are arranged tetrahedrally. Three of these electron pairs are used as bond pairs, which leaves one lone pair of electrons. The lone pair of electrons repel more strongly than bond pairs, therefore the bond angle is not 109.5°, as expected for a regular tetrahedral arrangement, but is measured at 106.7°.[13] The nitrogen atom in the molecule has a lone electron pair, which makes ammonia a base, a proton acceptor. This shape gives the molecule a dipole moment and makes it polar.

ehild
 
  • Like
Likes 1 person
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top