September 30, 2020, 09:22:34 PM
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### Topic: What is theselection rule for a vibrational mode of a molecule to be IR active?  (Read 222 times)

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#### AussieKenDoll

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##### What is theselection rule for a vibrational mode of a molecule to be IR active?
« on: August 20, 2020, 01:12:51 AM »
is it ΔV=±1 or what?
What is ΔV?

#### Corribus

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##### Re: What is theselection rule for a vibrational mode of a molecule to be IR active?
« Reply #1 on: August 20, 2020, 08:57:37 AM »
This is something you can just look up in a textbook or on google.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### AussieKenDoll

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##### Re: What is theselection rule for a vibrational mode of a molecule to be IR active?
« Reply #2 on: August 20, 2020, 09:07:19 AM »
Yea I know that The mode in question must involve a changing dipole moment during the vibration. This can involve a bending process as well as a stretching process.
But can I show this statement mathematically as ΔV=±1  instead of that explanation?

#### Corribus

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##### Re: What is theselection rule for a vibrational mode of a molecule to be IR active?
« Reply #3 on: August 20, 2020, 10:49:02 AM »
These are two completely different conditions - the requirement of a change in dipole is not the same as the requirement for a change in the quantum number of ± 1. Both requirements can be derived by evaluating the transition moment integral, but that's a fairly advanced mathematical exercise. The dipole change condition is easily shown by symmetry considerations.

You can see a little more here, although this link doesn't provide the actual derivation.

Note that the Δν = ±1 selection rule is only strictly valid in the harmonic approximation. In real systems that have anharmonic contributions, overtones with Δν = ±2, etc., become weakly allowed.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman