# Selection rules in rotational spectroscopy- water molecule

The University of Liverpool maintains a ChemTube 3D page, where it says

http://www.chemtube3d.com/spectrorotcd0-CE-TEST-ROTATE-ALL.html

*> Highly symmetric polyatomic molecules, such as carbon dioxide, also

have no net dipole moment - the dipoles along the C-O bonds are always equal and opposite and cancel each other out. It is important to recognise also that if a molecule has a permanent dipole, but this dipole lies along the main rotation axis, then the molecule will not have a rotational spectrum - such as for a water molecule.

1) A chemistry student brought this question to me, and I am also surprised by their claim. Why can't water have a rotational spectrum? It defies the very purpose of microwave oven and the gas phase rotational spectrum of water is all over the spectroscopic literature. I would like to confirm the with the opinion of spectroscopists that this is incorrect or not.

2) Another question from the student electromagnetic waves have an sinusoidal electric and a magnetic field of microwaves. How does the magnetic field interact with the permanent electric dipole moment of a molecule say HCl? I have checked many physical chemistry texts and they are quiet about the interaction from varying magnetic component and the dipole moment of the molecule. Electric field interaction is discussed in detail.

Thank you.

• It is interesting as to what you have written in that the animation on the web page shows possible rotations of the carbon dioxide molecule. – Farcher Mar 9 at 6:33
• Yes, the page is itself contradictory. CO2 can have pure rotations but since it does have a permanent dipole, its rotations can be probed by Raman spectroscopy. – M. Farooq Mar 9 at 6:38
• – Farcher Mar 9 at 7:40
• Thanks, it says the strength is different but the energies are same. Wouldn't a electrical dipole interact with magnetic field? – M. Farooq Mar 9 at 14:00

For question 2. from EMM we knew that $$B$$ filed was very weak and can be ignored during the calculation for those cases. (If you accept that C-O bound does not change length, then you should accept $$B$$ field just doesn't do anything as well.)