9 corrected a typo "non-degenerate" > "degenerate"
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However, what happenswill happen if $\Gamma_n$ is non-degeneratedegenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

However, what happens if $\Gamma_n$ is non-degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

However, what will happen if $\Gamma_n$ is degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

8 Corrected a typo in an equation, changed hyphens to en dashes where proper
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As already mentioned, the Jahn-TellerJahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule). $\require{begingroup} \begingroup \newcommand{\En}[1]{E_n^{(#1)}} \newcommand{\ket}[1]{| #1 \rangle} \newcommand{\n}[1]{n^{(#1)}} \newcommand{\md}[0]{\mathrm{d}} \newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

Distortions that arise due to the $\En{1}$ term are called first-order Jahn-TellerJahn–Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn-TellerJahn–Teller distortions.

2. The first-order Jahn-TellerJahn–Teller effect

$$E_n = \En{0} + q\En{1} + \cdots \tag{8}$$$$E_n = \En{0} + \color{red}{q\En{1}} + \cdots \tag{8}$$

In order for there to be a first-order Jahn-TellerJahn–Teller distortion, we therefore require that

In all point groups, for any non-degenerate irrep $\Gamma_n$, we find that $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

and the molecule is stable against a first-order Jahn-TellerJahn–Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn-TellerJahn–Teller distortions.

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn-TellerJahn–Teller distortion.

3. The second-order Jahn-TellerJahn–Teller effect

Pearson has written an article on second-order Jahn-TellerJahn–Teller effects.5

$$\En = \En{0} + q\En{1} + \frac{q^2}{2}\En{2} \cdots \tag{13}$$$$E_n = \En{0} + q\En{1} + \color{red}{\frac{q^2}{2}\En{2}} \cdots \tag{13}$$

Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn-TellerJahn–Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.

The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (For a QM exercise - try to proveThe proof of this! is left to the reader. Hint: use the fact that $\md H/\md q$ is hermitian.) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.

These symmetry requirements are much less restrictive than previously, and second-order Jahn-TellerJahn–Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn-TellerJahn–Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.

A second factor that favours the distortion is the extremely small 5d-6s5d–6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d-4s3d–4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.

The second-order Jahn-TellerJahn–Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn-TellerJahn–Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

(5) Pearson, R. G. The second-order Jahn-TellerJahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

As already mentioned, the Jahn-Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule). $\require{begingroup} \begingroup \newcommand{\En}[1]{E_n^{(#1)}} \newcommand{\ket}[1]{| #1 \rangle} \newcommand{\n}[1]{n^{(#1)}} \newcommand{\md}[0]{\mathrm{d}} \newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

Distortions that arise due to the $\En{1}$ term are called first-order Jahn-Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn-Teller distortions.

2. The first-order Jahn-Teller effect

$$E_n = \En{0} + q\En{1} + \cdots \tag{8}$$

In order for there to be a first-order Jahn-Teller distortion, we therefore require that

In all point groups, for any non-degenerate irrep $\Gamma_n$, $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

and the molecule is stable against a first-order Jahn-Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn-Teller distortions.

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn-Teller distortion.

3. The second-order Jahn-Teller effect

Pearson has written an article on second-order Jahn-Teller effects.5

$$\En = \En{0} + q\En{1} + \frac{q^2}{2}\En{2} \cdots \tag{13}$$

Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn-Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.

The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (For a QM exercise - try to prove this!) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.

These symmetry requirements are much less restrictive than previously, and second-order Jahn-Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn-Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.

A second factor that favours the distortion is the extremely small 5d-6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d-4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.

The second-order Jahn-Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn-Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

(5) Pearson, R. G. The second-order Jahn-Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

As already mentioned, the Jahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule). $\require{begingroup} \begingroup \newcommand{\En}[1]{E_n^{(#1)}} \newcommand{\ket}[1]{| #1 \rangle} \newcommand{\n}[1]{n^{(#1)}} \newcommand{\md}[0]{\mathrm{d}} \newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

Distortions that arise due to the $\En{1}$ term are called first-order Jahn–Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn–Teller distortions.

2. The first-order Jahn–Teller effect

$$E_n = \En{0} + \color{red}{q\En{1}} + \cdots \tag{8}$$

In order for there to be a first-order Jahn–Teller distortion, we therefore require that

In all point groups, for any non-degenerate irrep $\Gamma_n$, we find that $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

and the molecule is stable against a first-order Jahn–Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn–Teller distortions.

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn–Teller distortion.

3. The second-order Jahn–Teller effect

Pearson has written an article on second-order Jahn–Teller effects.5

$$E_n = \En{0} + q\En{1} + \color{red}{\frac{q^2}{2}\En{2}} \cdots \tag{13}$$

Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn–Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.

The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (The proof of this is left to the reader. Hint: use the fact that $\md H/\md q$ is hermitian.) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.

These symmetry requirements are much less restrictive than previously, and second-order Jahn–Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn–Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.

A second factor that favours the distortion is the extremely small 5d–6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d–4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.

The second-order Jahn–Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn–Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

(5) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

7 added 157 characters in body
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It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry. 

Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II). Nevertheless, it is encouraging to see that the axial elongation indeed transforms as $\mathrm{e_g}$ - which lends credence to the group theoretical analysis above.

which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since itthe numerator is a square modulus, the numeratorit is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.

Most of the time, the firstrelevant excited state $\ket{m}$ arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,

It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry. Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II).

which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since it is a square modulus, the numerator is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.

Most of the time, the first excited state arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,

It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry. 

Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II). Nevertheless, it is encouraging to see that the axial elongation indeed transforms as $\mathrm{e_g}$ - which lends credence to the group theoretical analysis above.

which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since the numerator is a square modulus, it is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.

Most of the time, the relevant excited state $\ket{m}$ arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,

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