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I would really appreciate any amount of detail/depth. I do not have a background in quantum, which is probably why I don't know the answer to this. However, any answer referencing quantum concepts is welcome and appreciated; I will follow up on any resources provided.

My question is basically as follows: Why does IR absorption require a change in dipole moment, and why is it that other types of spectroscopy (UV Vis, microwave) don't have this requirement? What causes this requirement specifically for IR?

Thank you!

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All the types of spectroscopy you mention have a similar condition for a spectrum to be observed. This is that the molecule has to be able to interact with the radiation in the first place and then the radiation has to be of the correct energy to change the energy of the molecule from one level to another.

A plain (classical) explanation is as follows. (We implicitly assume that the radiation has the correct energy). As radiation is electric dipole in nature (it consists of oscillating electric and magnetic fields) if the molecule also generates an oscillating electric field it can interact with the radiation; energy is taken from the radiation field and enters the molecule. Thus a molecule may gain a quantum of vibrational or rotational energy or an excited state may be produced.

In the case of a vibrating diatomic molecule only if it is hetero-nuclear e.g. HCl, will it have a dipole and thus an oscillating dipole and oscillating electric field as it vibrates. Thus hetero-nuclear diatomics show an IR spectrum, homo-nuclear diatomics do not. In a polyatomic molecule vibrations can distort the molecule, e.g. benzene is distorted so that it is no longer 6 fold symmetric, so that a dipole is produced and an IR spectrum may result. What form this takes depends on the molecule's symmetry and group theory point groups are used to determine this.

In a rotating diatomic molecule we usually use a rigid-rotor model. If the molecule has a dipole then this clearly changes as the molecule rotates, produced an oscillating electric field, and so a microwave (rotational) spectrum is produced. A homo-nuclear diatomic e.g. $\ce{O2}$ has no microwave spectrum.

In a UV/visible transition an excited state may be produced but only if the electron distribution in the excited state is different from that of the ground state, i.e. a dipole is produced that can interact with the radiation. A molecule such a benzene has a forbidden first electronically excited state due to its symmetry (which is however observed weakly due to other effects) and so absorbs very little compared to aniline (amino-benzene) or many dye molecules where symmetry does not limit any electronic dipole formation.

Thus although the details vary, the underlying ideas are the same. You should understand now that the statements about spectra in your question are not all correct. The rules used to determine what transitions are allowed or not allowed (called forbidden) are called 'selection rules'.

You can read about this in more detail, as a quantum version, in most undergraduate phys. chem. textbooks.

You may realise that the vibrational frequencies of homo-nuclear molecules are known even though they have no ir spectrum and this is because they can be measured by Raman spectroscopy. This depends on the change in shape/size of the 'electron cloud' (polarisability) a molecule possesses. Electro-magnetic radiation also possesses an oscillating magnetic field but this interacts far more weakly with molecules than its electric counterpart and can often but not always be ignored, the exception being NMR spectroscopy.

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    $\begingroup$ Good explanation. $\endgroup$ – ankit7540 Jan 13 '17 at 6:58
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Lets begin with the absorption.

The probability of ANY absorption taking place is given by the the Einstein's B coefficient, which is \begin{equation} B_{n^{\prime}n^{\prime\prime}}=\frac{8 \pi^3}{3h^2}[|\int \psi_{n^\prime}\mu_{X} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Y} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Z} \psi_{n^{\prime\prime}}d \tau|^2] \end{equation}

Above equation is in complete generality, and the wave function $\psi$ is the complete wave function. $\mu_{X}$ is the dipole moment component (in X, using Cartesian coordinates). Integration is over all space.

RULE: if the dipole moment components for a given transition are zero, the absorption will not take place. (By components I mean the integral. If the integral goes to zero then no absorption for that component, if all three integrals are zero then no absorption at all) This determines the selection rules for all absorption processes (and also induced emission, but not discussing here !).

Back to absorption, the wave-function for the molecule can be expressed as \begin{equation} \psi= \psi_{E} \psi_{V} \psi_{R} \psi_{T} \end{equation} where $\psi_{E}, \psi_{V}, \psi_{R}$ and $\psi_{T}$ are the electronic, vibrational, rotational and translational wave-functions. For vibrational analysis the electronic wave function does not matter and we can safely exclude it. (This is the wave function needed for the calculation of the integral. For general understanding consider the wave function as a mathematical function whose integral gives you information about the molecule.)

The translational selection rule is that the change in the quantum number should be $\Delta n =0$ that is the quantum number does not change (while translation in space). [At the stage we have separated the electronic and translational wave functions, and we deal with only vibrations, hence $\psi_{v}$.]

Coming to rotational and vibrational part,

If $X_{\alpha}$, $Y_{\alpha}$ and $Z_{\alpha}$ is the coordinate which moves with the molecule, then a new coordinate can be introduced which moves and also rotates with the molecule. This is say $x_{\alpha}$, $y_{\alpha}$ and $z_{\alpha}$. The relation between $X_{\alpha}$ and $x_{\alpha}$ are governed by angles expressed as trigonometric relations, also known as direction cosines. Advantage of this transformation is that we can separate vibration from rotation and discuss vibration.

The electric moment of the full system of molecule has three components in Cartesian coordinates, \begin{equation} \mu_{x}=\sum_{\alpha}e_{\alpha}x_{\alpha} \end{equation} \begin{equation} \mu_{y}=\sum_{\alpha}e_{\alpha}y_{\alpha} \end{equation} \begin{equation} \mu_{z}=\sum_{\alpha}e_{\alpha}z_{\alpha} \end{equation} Now, during vibration the charged entities are moving. To be more clear I would say the nuclei are moving. The dipole moment is dynamic and changing with the vibration. The changing dipole moment can be expressed a Taylor expansion so as to include the change from equilibrium position. This expansion is carried out using the normal coordinates( normal coordinates are coordinates of the atoms which change with the vibration, when all atoms move with same frequency but different amplitudes).\begin{equation} \mu_{x}=\mu_{x}^{0}+\mu_{x}^{1}Q_{1}+\mu_{x}^{2}Q_{2}+... \end{equation} where $\mu_{x}^{n}$ is the nth derivative. Generally, only the first derivative is taken and hence, \begin{equation} \mu_{x}=\mu_{x}^{0}+ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} \end{equation}

Above result is now used with the vibrational wave function.

\begin{equation} \int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau= \int \psi_{n^\prime} [ \mu_{x}^{0}+ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d\tau \end{equation} \begin{equation} \int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau=\mu_{x}^{0} \int \psi_{n^\prime} \psi_{n^{\prime\prime}} d \tau + \int \psi_{n^\prime} [ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d \tau \end{equation} The first term would go to zero since the vibrational wave functions are orthogonal. In the second term, the condition of non-zero derivative of dipole moment is required to have this term being non-zero. Hence the change if dipole moment with vibration is utmost necessary. Other comments: With similar approach one can derive, for UV and visible absorption that the dipole is required; and for microwave absorption we need a permanent dipole moment.

References: Physical Chemistry by Atkins, Paula.;

Modern Spectroscopy by Hollas ;

(More advanced reference. Molecular vibrations by Wilson, Decius, Cross.)

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  • $\begingroup$ I'll improve it further. Thanks for the 'wow'. $\endgroup$ – ankit7540 Jan 18 '17 at 15:45
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As per my knowledge there are two reasons one is: we use infrared radiation as a source such that it helps to induce vibration in the molecule that results in the difference in dipole moment . Second : due to the reduced mass(mu) in hookes law equation given below,

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  • $\begingroup$ This doesn't really answer the OP's question... $\endgroup$ – Zhe Jan 22 '17 at 3:13

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