Advanced Chemistry - Mono Mole

May 16, 2026July 3, 2026

Electronic selection rules for molecules

Electronic selection rules for molecules are quantum-mechanical conditions (based on changes in quantum numbers and symmetry) that determine whether transition between energy levels in the molecules are allowed or forbidden.

According to the time-dependent perturbation theory, the transition probability between the initial and final states, $\Psi$ and $\Psi'$ , of a molecule is proportional to the square of the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , where $\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$ is the operator for the molecule’s electric dipole moment.

Consider a homonuclear diatomic molecule. Although it has no permanent dipole moment in a stationary state, it can acquire a transient, time-dependent dipole moment (just as an atom does) when interacting with oscillating incident radiation. Like multi-electron atoms, $\Psi$ and $\Psi'$ must satisfy the Pauli exclusion principle and can be represented by Slater determinants built from orthogonal spin-orbitals $\chi(\boldsymbol{\mathit{x}})=\psi(\boldsymbol{\mathit{r}})\sigma(\omega)$ , with $\psi(\boldsymbol{\mathit{r}})$ and $\sigma(\omega)$ being the spatial and spin wavefunctions respectively, and $\sigma=\alpha\;or\;\beta$ . Since the dipole moment operator does not act on the spin wavefunctions, which are identical for the initial and final states, the total spin quantum number does not change, i.e. $\Delta S=0$ .

The selections rules involving angular momentum are determined using the two vanishing integral rules in group theory:

Rule 1

If $\boldsymbol{\mathit{f_j}}$ and $\boldsymbol{\mathit{g_l}}$ transform according to two different non-equivalent irreducible representations $\boldsymbol{\mathit{\Gamma_f}}$ and $\boldsymbol{\mathit{\Gamma_g}}$ respectively, the integral of their product over all space is necessarily zero

Rule 2

If a function transforms according to an irreducible representation that is not the totally symmetric representation of a group, its integral over all space is necessarily zero

For linear molecules belonging to the $C_{\infty v}$ point group (see this link for details), the electric dipole moment operator can be resolved into parallel and perpendicular components and transforms according to either the basis function $z$ or the pair $(x,y)$ , the latter forming a linear combination of $x$ and $y$ . If the operator transforms as $z$ , then the product $\hat{\boldsymbol{\mathit{\mu}}}\Psi$ in $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ transforms according to a direct product representation. In this case, the product transforms as the same irreducible representation as $\Psi$ , since $z$ transforms as the totally symmetric irreducible representation $\Sigma^+$ . Therefore, Rule 1 requires that the initial and final states transform according to the same irreducible representation ( $\Sigma^+\leftrightarrow \Sigma^+$ and $\Sigma^-\leftrightarrow \Sigma^-$ ) for the integral to be non-zero. This also implies that the transition $\Sigma^+\leftrightarrow \Sigma^-$ is forbidden. Since states transforming as $\Sigma^{\pm}$ correspond to $\Lambda=0$ , it follows that

$\Delta\Lambda=0$

If the dipole moment operator transforms as $(x,y)$ , we have the following cases:

Case 1: $\Psi_{\Lambda=0}\leftrightarrow\Psi'_{\Lambda=1}$ with $\Delta\Lambda=\pm 1$

The direct product of $\Psi'\hat{\boldsymbol{\mathit{\mu}}}\Psi$ is $\Gamma_a=E_1\otimes E_1\otimes A_1$ or $\Gamma_b=E_1\otimes E_1\otimes A_2$ , both corresponding to the reducible representation

where we have used the identity $2cos^2x=1+cos2x$ .

Decomposing the reducible representation gives: $A_1\oplus A_2\oplus E_2$ . Since $\int\psi_{\Gamma_1}\oplus \psi_{\Gamma_2}\oplus\cdots d\tau=\biggr$\int\psi_{\Gamma_1}d\tau\biggr$\oplus\biggr$\int\psi_{\Gamma_2}d\tau\biggr$\oplus\cdots$ (see this link for proof), the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is non-zero according to Rule 2 because it includes the term $\int\psi_{A_1}d\tau$ .

Case 2: $\Psi_{\Lambda=0}\leftrightarrow\Psi'_{\Lambda=2}$ with $\Delta\Lambda=\pm 2$

The direct product of $\Psi'\hat{\boldsymbol{\mathit{\mu}}}\Psi$ is $\Gamma_c=E_2\otimes E_1\otimes A_1$ or $\Gamma_d=E_2\otimes E_1\otimes A_2$ , both corresponding to the reducible representation

where we have used the identity $cosAcosB=\frac{1}{2}[cos(A+B)+cos(A-B)]$ .

Decomposing the reducible representation gives: $E_1\oplus E_3$ . Since $\int\psi_{\Gamma_1}\oplus \psi_{\Gamma_2}\oplus\cdots d\tau=\biggr$\int\psi_{\Gamma_1}d\tau\biggr$\oplus\biggr$\int\psi_{\Gamma_2}d\tau\biggr$\oplus\cdots$ , the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is zero according to Rule 2. In fact, if $\Delta\Lambda>1$ or $\Delta\Lambda<-1$ , the character for the rotation operator is always $\chi(C_{\infty})=4cos\phi cos(\Lambda\phi)=2[cos(\Lambda+1)\phi+cos(\Lambda-1)\phi)$ . Hence, $\chi(C_{\infty})$ , for $\Delta\Lambda>1$ , will never include the $cos(0)$ term, which is necessary for the $A_1$ component to exist.

Therefore, the selection rules involving $\Lambda$ are:

$\Delta\Lambda=0,\pm 1\;\;\;(with\;\Sigma^+\leftrightarrow\Sigma^-\;forbidden)$

For linear molecules, $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ becomes $\Omega=\vert\Lambda+\Sigma\vert$ (see this link for explanation), where

$\Omega$ is the projection of the total electronic angular momentum onto the molecular axis.
$\Lambda$ is the projection of the total electronic orbital angular momentum ( $\boldsymbol{\mathit{L}}$ ) onto the molecular axis.
$\Sigma$ is the projection of the total electronic spin angular momentum ( $\boldsymbol{\mathit{S}}$ ) onto the molecular axis.

Since the values of $\Sigma$ range from $-S$ to $+S$ , and $\Delta S=0$ , we have $\Delta\Sigma=0$ . This implies that $\Delta\Omega=\Delta\Lambda$ or equivalently,

$\Delta\Omega=0,\pm 1\;\;\;(with\;\Sigma^+\leftrightarrow\Sigma^-\;forbidden)$

Repeating the same analysis for linear molecules belonging to the $D_{\infty h}$ point group (see this link for details) yields the same angular momentum-based selection rules as for molecules belonging to the $C_{\infty v}$ point group. However, because $D_{\infty h}$ possesses inversion symmetry, additional selection rules apply:

$g\leftrightarrow u\;\;\;(allowed)$

$g\leftrightarrow g,\;\;u\leftrightarrow u\;\;\;(forbidden)$

These arise from the Laporte selection rule, which states that the electric dipole transition matrix element is non-zero only if the overall integrand is symmetric (even) under spatial inversion over all space. Since the dipole moment operator transforms as $u$ , the direct product of the initial and final states symmetries must also be $u$ . Consequently, the initial and final states must have opposite parity for the integrand to be overall $g$ , allowing the transition. In summary,

For non-linear polyatomic molecules, $\Delta S=0$ remains applicable to all systems. However, the Laporte selection rule applies only to molecules possessing an inversion centre (e.g. those belonging to $D_{2h}$ or $O_h$ ). The determination of whether the transition moment integral $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is nonzero follows the same logic as described above: one first identifies the point group of the molecule, then determines the symmetries of the initial and final states, and finally applies the vanishing integral rules.

Question

Does the selection rule $\Delta n=0$ apply to molecules?

Answer

No. $n$ is not a good quantum number for molecular states. Instead, molecular states are described using term symbols and symmetry labels, so no selection rule involving $\Delta n$ applies.

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Electronic spectroscopy

Electronic spectroscopy is the study of how atoms or molecules absorb or emit light due to transitions of electrons between different energy levels.

When atoms or molecules absorb energy, typically in the ultraviolet or visible regions of the spectrum, electrons can be promoted from lower-energy (ground) states to higher-energy (excited) states. Conversely, when these excited electrons return to lower energy levels, energy is released, often in the form of light. These absorption and emission processes form the basis of electronic spectra.

The resulting spectra are highly characteristic of the substance being studied. Because each element or compound has a unique arrangement of electrons and energy levels, the wavelengths of light they absorb or emit act like a fingerprint. This makes electronic spectroscopy a powerful analytical tool in chemistry, physics, and materials science for identifying substances and studying their structure.

An important branch of electronic spectroscopy is atomic spectroscopy, which focuses specifically on isolated atoms, usually in the gas phase. Unlike molecules, atoms have simpler electronic structures, so their spectra consist of sharp, well-defined lines rather than broad bands. Techniques such as atomic absorption spectroscopy (AAS) and atomic emission spectroscopy (AES) measure the specific wavelengths of light absorbed or emitted by atoms as their electrons transition between discrete energy levels. Because each element produces a unique line spectrum, atomic spectroscopy is widely used for precise elemental analysis in fields such as environmental testing, metallurgy and forensic science.

There are several other types of electronic spectroscopy, with ultraviolet-visible (UV-Vis) spectroscopy being one of the most common. In UV-Vis spectroscopy, a sample absorbs light in the ultraviolet or visible region, causing electronic transitions, particularly in molecules with conjugated systems or transition metal complexes. The intensity and position of absorption bands can provide information about concentration (via Beer–Lambert law), molecular structure and the nature of chemical bonds.

Another important form is fluorescence spectroscopy, where a substance absorbs light at one wavelength and then emits light at a longer wavelength. This technique is especially useful in biological and environmental studies due to its high sensitivity.

Overall, electronic spectroscopy serves as a bridge between light and matter, revealing the electronic structure and properties of substances through their interaction with radiation.

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Atomic absorption and emission spectroscopy

Atomic Absorption Spectroscopy (AAS) and Atomic Emission Spectroscopy (AES) are analytical techniques used to determine the concentration of elements by measuring the interaction of free atoms in the gaseous state with electromagnetic radiation.

A typical atomic absorption spectrometer consists of a light source, such as a hollow cathode lamp (HCL), which emits light characteristic of the element being analysed (see diagram above). The light is split into two beams by mirrors, with one beam passing through an atomiser, where the sample atoms are introduced using a nebuliser and then vaporised by an air-acetylene flame. The other beam is directed through a vacuum reference cell.

Question

Explain how the HCL works.

Answer

A HCL emits narrow wavelengths of light because its operation is based on the specific electronic energy levels of a single element. Inside the lamp, the cathode is made of the element to be analysed (for example, copper or sodium), and the tube is filled with an inert gas such as neon or argon. When a high voltage is applied, the inert gas becomes ionised, producing positively charged gas ions. These ions accelerate towards the cathode and strike its surface, knocking out atoms of the cathode material in a process called sputtering. Some of these sputtered atoms are excited by collisions within the lamp. When their electrons return from higher energy levels to lower ones, they emit light. Because the energy differences between electronic levels in atoms are fixed and discrete, the emitted photons have very specific energies (wavelengths). As a result, the lamp produces a spectrum consisting of sharp emission lines rather than a broad range of wavelengths. Since the cathode is made of only one element, the emitted lines correspond almost exclusively to that element, giving the hollow cathode lamp its narrow, characteristic output. For example, if the cathode is made of sodium, the emitted light involves the 3p → 3s energy levels, corresponding to:

²P_3/2 → ²S_1/2 ( ≈ 589.0 nm: D2 line)

²P_1/2 → ²S_1/2 ( ≈ 589.6 nm: D1 line)

N.B: Click this link to understand how the above atomic term symbols are derived.

As the light passes through the cloud of gaseous atoms in the atomiser, it is absorbed at specific wavelengths corresponding to electronic energy level transitions. The transmitted light then passes into a monochromator, which isolates the specific wavelength absorbed by the element of interest, removing any unwanted radiation.

The two beams, emerging from the monochromator and the reference cell respectively, are subsequently directed to separate detectors, typically photomultiplier tubes. Here, the intensities of the transmitted and reference light are measured. The signals are then compared by a signal processor, which compensates for fluctuations in the light source and instrumental noise. Finally, the processed signal is converted into an electrical output and displayed, typically as absorbance, allowing the concentration of the element in the sample to be determined.

To analyse the concentrations of multiple elements in a sample (see above diagram), the light source may adopt one of the following designs:

- Lamp turret: Several HCLs, each with a cathode made of a different element, are automatically rotated to emit different wavelengths of light. Measurements are then taken sequentially.
- Multi-element lamps: A single HCL, in which the cathode is made from an alloy or a mixture of metal powders (e.g. a combination of cobalt, chromium, copper, iron, manganeses and nickel), is used. Sensitivity is usually lower than that of single-element lamps.
- Continuum source: More modern instruments use a broadband light source, such as a xenon lamp. This provides a broad spectrum of light, allowing the detector to measure multiple elements at different wavelengths without changing lamps.

Atomic Emission Spectroscopy (AES) differs from AAS in that it does not use an external light source (see diagram above); instead, the sample itself is energetically excited using a high-energy source such as a flame, electric arc, spark or inductively coupled plasma (ICP). As the excited atoms spontaneously return to lower energy levels, they emit light at characteristic wavelengths, and this emitted radiation is measured directly. In AES, there is no need for a reference beam or a hollow cathode lamp, and the signal is based on emission intensity rather than absorption.

Modern AES instruments use monochromators with high resolving power, which can separate wavelengths that are extremely close together, enabling multi-element analysis. The computer records a composite spectrum but presents the results as individual element concentrations by filtering the data using its wavelength database. The emission spectra of individual elements can also be derived from the composite spectrum by the software (see diagram below).

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Electronic selection rules for atoms

Electronic selection rules for atoms are quantum-mechanical conditions (based on changes in quantum numbers and symmetry) that determine whether transitions between energy levels in the atoms are allowed or forbidden.

According to the time-dependent perturbation theory, the transition probability $\vert\mu_{fi}\vert^2$ between the states $\psi(r,\theta,\phi)_{n,l,m_l}$ and $\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}$ of an atom is proportional to $\vert\langle\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\psi(r,\theta,\phi)_{n,l,m_l}\rangle\vert^2$ , where $\hat{\boldsymbol{\mathit{\mu}}}$ is the operator for the atom’s electric dipole moment. Although an atom has no permanent dipole moment in a stationary state, it can acquire a transient, time-dependent dipole moment when interacting with an oscillating electromagnetic field. During this interaction, the atomic state is described as a time-dependent superposition of the initial and final states: $\Psi(t)=c'(t)\psi_{n^{'},l^{'},m_l^{'}}+c(t)\psi_{n,l,m_l}$ . When the incident frequency is far from resonance, the coefficient $c^{'}(t)$ remains very small. Near resonance, however, the mixing between the states becomes significant ( $c'\approx c$ ). This coherent superposition (for example between $\psi_{n^{'},l^{'},m_l^{'}}=\psi_p$ and $\psi_{n,l,m_l}=\psi_s$ ) leads to interference between the wavefunctions of the two states, creating an asymmetric, time-dependent electron distribution. As a result, the centre of negative charge oscillates relative to the nucleus, generating a transient dipole moment that oscillates at (or near) the driving frequency of the incident radiation.

Hydrogenic atoms

Consider a hydrogenic atom, for which $\psi(r,\theta,\phi)_{n,l,m_l}=R_{n,l}(r)Y_l^{m_l}(\theta,\phi)$ and $\mu_{fi}\propto I=\langle\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\psi(r,\theta,\phi)_{n,l,m_l}\rangle$ . For a plane-polarised electromagnetic wave with its electric field oscillating along the $z$ -direction, only the $z$ -component of the dipole operator contributes. Thus, $\hat{\boldsymbol{\mathit{\mu}}}=-ez=-ercos\theta$ and

$I=-e\int_0^{\infty}R_{n^{'},l^{'}}(r)rR_{n,l}(r)r^2dr\int_0^{\pi}\int_0^{2\pi}Y_{l^{'}}^{m^{'}_l*}(\theta,\phi)cos\theta Y_l^{m_l}(\theta,\phi)sin\theta d\theta d\phi$

The angular integral involving the spherical harmonics is non-zero only when (see this link for derivation):

$\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

This leaves the radial integral:

$I'=\int_0^{\infty}R_{n^{'},l^{'}}(r)rR_{n,l}(r)r^2dr$

where $l'=l\pm 1$ .

Question

Show that $rR_{n,l}(r)$ is square-integrable.

Answer

For $rR_{n,l}(r)$ to be square-integrable,

$\int_0^{\infty}\vert rR_{n,l}(r)\vert^2r^2dr=\int_0^{\infty}\vert R_{n,l}(r)\vert^2r^4dr<\infty$

When $r\rightarrow 0$ , we have $R_{n,l}(r)=e^{-\frac{r}{a_0n}}\frac{r^l}{(na_0)^{l+1}}L^{2l+1}_{n-l-1}\biggr$\frac{2r}{na_0}\biggr$\propto r^l$ . So, $\vert R_{n,l}(r)\vert^2r^4\propto r^{2l+4}$ . Consider the integral $\int^b_0r^adr$ . If $a\neq -1$ , then $\int^b_0r^adr=\frac{b^{a+1}}{a+1}$ . If $a=-1$ , then $\int^b_0\frac{1}{r}dr=lnb$ . As $b\rightarrow 0$ ,

Thus, $\int_0^{\infty}\vert R_{n,l}(r)\vert^2r^4dr\propto\int_0^{\infty}r^{2l+4}dr$ converges as $r\rightarrow 0$ because $2l+4>-1$ . Furthermore, $R_{n,l}(r)=e^{-\frac{r}{a_0n}}\frac{r^l}{(na_0)^{l+1}}L^{2l+1}_{n-l-1}\biggr$\frac{2r}{na_0}\biggr$=e^{-\frac{r}{a_0n}}f(r)$ , so that $\vert R_{n,l}(r)\vert^2r^4=g(r)^2e^{-\frac{2r}{a_0n}}$ , where $f(r)$ and $g(r)$ are polynomial functions of $r$ . As $r\rightarrow \infty$ , the exponential term decays to zero faster than $g(r)^2$ , ensuring that $\vert R_{n,l}(r)\vert^2r^4$ converges. Therefore, $\int_0^{\infty}\vert rR_{n,l}(r)\vert^2r^2dr<\infty$ .

Since $rR_{n,l}(r)$ is square-integrable with respect to $r^2dr$ , it belongs to the radial Hilbert space. Additionally, $\{R_{n',l'}(r)\}$ forms a complete set for the radial Hilbert space for a fixed $l'$ . We can therefore expand $rR_{n,l}(r)$ as a linear combination of $\{R_{n',l'}(r)\}$ for any fixed $l'$ :

$rR_{n,l}(r)=\sum_{n'}c_{n'}R_{n',l'}(r)$

with

$c_{n'}=\langle R_{n',l'}(r)\vert r\vert R_{n,l}(r)\rangle$

If there were a restriction such as $n'=n\pm 1$ , the expansion $\sum_{n'}c_{n'}R_{n',l'}(r)$ would contain only a finite number of terms. However, multiplying $R_{n,l}(r)$ by $r$ generally produces a function with a different shape that requires infinitely many basis functions to represent. This corresponds to an infinite number of nonzero coefficients $c_{n'}$ , and hence infinitely many possible values of $n'$ . In other words, $c_{n'}=\langle R_{n',l'}(r)\vert r\vert R_{n,l}(r)\rangle=I'$ is nonzero for infinitely many values of $n'$ , resulting in no selection rule for $\Delta n$ . Therefore, the combined selection rules for electronic transitions in hydrogenic atoms are:

$\Delta n\;(no\;restriction)\;\;\;\;\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

Multi-electron atoms

For multi-electron atoms, the initial and final states, $\Psi$ and $\Psi'$ , must satisfy the Pauli exclusion principle. This can be accomplished by representing them using Slater determinants built from orthogonal spin-orbitals $\chi(\boldsymbol{\mathit{x}})$ :

$\Psi_a=\frac{1}{\sqrt{N!}}\begin{vmatrix} \chi_1(1)&\chi_2(1)&\cdots &\chi_N(1)\\ \chi_1(2)&\chi_2(2)& \cdots &\chi_N(2)\\ \vdots&\vdots &\ddots &\vdots \\ \chi_1(N)&\chi_2(N)&\cdots &\chi_N(N) \end{vmatrix}$

where

$\chi(\boldsymbol{\mathit{x}})=\psi(\boldsymbol{\mathit{r}})\sigma(\omega)$ , with $\psi(\boldsymbol{\mathit{r}})$ and $\sigma(\omega)$ being the spatial and spin wavefunctions respectively, and $\sigma=\alpha\;or\;\beta$ .
$\Psi_a$ denotes the antisymmetric form of $\Psi$ and $\Psi'$ , i.e. $\Psi_a$ or $\Psi'_a$ .
$\Psi_s$ denotes the symmetric form of $\Psi$ and $\Psi'$ , i.e. $\Psi_s$ or $\Psi'_s$ .

The electric dipole operator is a sum of one-electron operators:

$\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$

Determining the selection rules requires evaluating $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , in which the initial and final states must differ. Let us assume that $\Psi$ and $\Psi'$ differ by one spin orbital, i.e. $\Psi_s=\psi_1(x_1)\sigma_1(\omega_1)\cdots\psi_j(x_j)\sigma_j(\omega_j)\cdots\psi_N(x_N)\sigma_N(\omega_N)$ and $\Psi'_s=\psi_1(x_1)\sigma_1(\omega_1)\cdots\psi_k(x_j)\sigma_k(\omega_j)\cdots\psi_N(x_N)\sigma_N(\omega_N)$ . Applying the Slater-Condon rule for one-electron operators, where $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle=\langle\Psi'_s\vert\hat{\boldsymbol{\mathit{\mu}}}\hat{A}^2\vert\Psi_s\rangle$ , in which $\hat{A}=\frac{1}{\sqrt{N!}}\sum_{r=1}^{N!}(-1)^rP_r$ is the antisymmetriser, gives:

$\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle=-e\sum_{j=1}^N\langle\Psi'_s\vert\boldsymbol{\mathit{r}}_j\sum_{t=1}^{N!}(-1)^t\hat{P}_t\Psi_s\rangle$

Thus,

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\int\cdots\int[\psi_1(x_1)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&[\psi_1(x_1)\psi_2(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)-\\&\psi_2(x_1)\psi_1(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)+\cdots]dx_1\cdots dx_N\times\\&\int\cdots\int\sigma_1^*(\omega_1)\sigma_1(\omega_1)\cdots\sigma^*_k(\omega_j)\sigma_j(\omega_j)\cdots\sigma^*_N(\omega_N)\sigma_N(\omega_N)d\omega_1\cdots d\omega_N\end{align}$

Due to spin orthogonality, $\int\sigma_i^*(\omega)\sigma_j(\omega)d\omega=\delta_{ij}$ . So, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle\neq 0$ only if $\sigma_k(\omega_j)=\sigma_j(\omega_j)$ . This implies $\Delta m_s=0$ and therefore $\Delta M_S=0$ because the electron spins in a multi-electron atom are coupled such that $M_S=\sum_im_{s,i}$ , where $M_S=-S,-S+1,\cdots,S-1,S$ . Since the spin wavefunctions of the initial and final states are identical, this further requires that the total spin quantum number does not change, i.e. $\Delta S=0$ . It follows that:

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\biggr\{\int\cdots\int[\psi_1(x_1)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&\psi_1(x_1)\psi_2(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)-\\&\int\cdots\int[\psi_1(x_1)\psi_2(x_2)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&[\psi_2(x_1)\psi_1(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)+\cdots]\biggr\}dx_1\cdots dx_N\end{align}$

Due to the orthogonality of the spatial wavefunctions, only the first term on the RHS survives, giving:

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\int\psi_1^*(x_1)\psi_1(x_1)dx_1\cdots\int\psi_k^*(x_j)\boldsymbol{\mathit{r}}_j\psi_j(x_j)dx_j\\&\cdots\int\psi_N^*(x_N)\psi_N(x_N)dx_N\end{align}$

All terms vanish except for $j=k$ , yielding

$\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =-e\int\psi_k^*(x_k)\boldsymbol{\mathit{r}}_k\psi_k(x_k)dx_k$

It is evident from the integrals in the derivation above that if $\Psi$ and $\Psi'$ differ by more than one spin orbital, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =0$ . This is why double or multiple electon excitations are forbidden in electric dipole transitions. Therefore, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ reduces to a single one-electron integral, whose selection rules are those of the hydrogenic atom:

$\Delta n\;(no\;restriction)\;\;\;\;\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

$\Delta n$ remains unrestricted for multi-electron atoms. The dipole operator acts on a single electron and enforces the hydrogenic selection rule $\Delta l=\pm 1$ . In a multi-electron atom, the total orbital angular momentum is obtained by vector coupling $\boldsymbol{\mathit{L}}\sum_i\boldsymbol{\mathit{l}}_i$ . A change in one electron’s orbital angular momentum by one unit therefore changes the coupled total orbital angular momentum by at most one unit, leading to the selection rule $\Delta L=0,\pm 1$ (with $L=0\leftrightarrow L'=0$ forbidden). Since $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ and $\Delta S=0$ , any change in the total angular momentum between the initial and final states ( $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{J}}'=\boldsymbol{\mathit{L}}'+\boldsymbol{\mathit{S}}'$ ) must arise entirely from the change in $\boldsymbol{\mathit{L}}$ , which is governed by $\Delta L=0,\pm 1$ . Therefore, $\Delta J=0,\pm 1$ , with $J=0\leftrightarrow J'=0$ forbidden. Furthermore, since $J_z=L_z+S_z$ , it follows that $M_J=M_L+M_S$ and hence $\Delta M_J=\Delta M_L+\Delta M_S$ . Because $\Delta M_S=0$ and $\Delta M_L=\sum_i\Delta m_{l,i}$ in the Slater-Condon reduction, we obtain $\Delta M_J=0,\pm 1$ .

The final selection rule for multi-electron atoms is based on the Laporte selection rule, which states that the electric dipole transition matrix element $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ is non-zero only if the integrand is even under spatial inversion over all space. This follows from the fact that if the integrand is odd under the transformation $(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)\rightarrow(-\boldsymbol{\mathit{r}}_1,\cdots,-\boldsymbol{\mathit{r}}_N)$ , then each configuration has a corresponding inverted configuration contributing equal magnitude with opposite sign, leading to cancellation and a vanishing integral.

Since $\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$ is odd under inversion ( $\boldsymbol{\mathit{r}}_j\rightarrow -\boldsymbol{\mathit{r}}_j$ ), the matrix element $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ is non-zero only if the initial and final states have opposite parity. For a multi-electron atom, the parity of a Slater determinant is given by $(-1)^{\sum_i^Nl_i}$ . So, the wavefunction is even if $\sum_i^Nl_i$ is even and odd if $\sum_i^Nl_i$ is odd.

The combined selection rules for multi-electron atoms are:

$\Delta n\;(no\;restriction)$
$\Delta S=0$
$\Delta L=0,\pm 1\;(with\;L=0\leftrightarrow L'=0\;forbidden)$
$\Delta J=0,\pm 1\;(with\;J=0\leftrightarrow J'=0\;forbidden)$
$\Delta M_J=0,\pm 1$
$Parity\;of\;states\;must\;change\;(g\leftrightarrow u)$

Question

Why is $L=0\leftrightarrow L'=0$ forbidden? Evaluate whether the transition $^3P_1\rightarrow ^3P_2$ is allowed (see this link for how atomic term symbols are derived).

Answer

Electric dipole transitions in centrosymmetric atoms involve the generation of a transient dipole moment, which is directional, leading to the redistribution of charge density. Such a transition cannot occur if both the initial and final states are spherically symmetric. Since an $L=0$ state is spherically symmetric and has no directional dependence, transitions with $L=0\leftrightarrow L'=0$ are forbidden. The same reasoning explains why $J=0\leftrightarrow J'=0$ transitions are forbidden.

For the transition $^3P_1\rightarrow ^3P_2$ ,

1. $\Delta n=0$ : allowed, since there is no restriction on $\Delta n$ .
2. $\Delta S=1-1=0$ : satisfies the selection rule.
3. $\Delta L=1-1=0$ : allowed, since $L\neq 0$ .
4. $\Delta J=2-1=+1$ : satisfies the selection rule.
5. For $J=1$ and $J'=2$ , the magnetic quantum numbers are $M_J=-1,0,+1$ and $M_J=-2,-1,0,+1,+2$ respectively. Transitions are allowed only for $\Delta M_J=0,\pm 1$ .
6. The parity of both the initial and final states is $(-1)^{\sum_i^Nl_i}=(-1)^{1+1}=+1$ , which violates the parity rule.

Therefore, the transition $^3P_1\rightarrow ^3P_2$ is forbidden.

The selection rules for hydrogenic and multi-electron atoms are best illustrated by Grotrian diagrams (see above Grotrian diagram for hydrogen).

In summary, electronic transitions in hydrogenic and multi-electron atoms produce discrete spectral lines (see diagram above), even when fine structures is taken into account. However, in molecules, each electronic transition is accompanied by many possible simultaneous vibrational and rotational transitions. These thousands of closely-spaced lines overlap, creating the appearance of a continuous band.

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May 16, 2026May 16, 2026

Vibronic and ro-vibronic selection rules for molecules

Vibronic and ro-vibronic selection rules specify which combined electronic, vibrational, and rotational transitions in molecules are allowed based on quantum number and symmetry constraints.

An electronic transition in a molecule is often accompanied by vibration-rotation transitions. To determine the selection rules for these transitions, we refer to the Born-Oppenheimer approximation, which states that the total wavefunction $\Psi$ of the molecule can be expressed as:

$\Psi=\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\psi_v(Q)\psi_r(\tau_r)$

where

$\psi_e$ is the electronic wavefunction that is a function of nuclear coordinates $\boldsymbol{\mathit{R}}$ and electronic coordinates $\boldsymbol{\mathit{r}}$ .
$\psi_v$ is the vibrational wavefunction that is a function of normal coordinates $Q$ .
$\psi_r$ is the rotational wavefunction that is a function of rotational coordinates $\tau_r$ , typically the Euler angles.

According to the time-dependent perturbation theory, the transition probability between the initial and final states, $\Psi$ and $\Psi'$ is proportional to the square of $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , where the electric dipole moment operator $\hat{\boldsymbol{\mathit{\mu}}}$ is the sum of all charge–position contributions in the molecule:

$\hat{\boldsymbol{\mathit{\mu}}}=\hat{\boldsymbol{\mathit{\mu}}}_N+\hat{\boldsymbol{\mathit{\mu}}}_{el}$

where

$\hat{\boldsymbol{\mathit{\mu}}}_N=\sum_AZ_Ae\boldsymbol{\mathit{R}}_A$
$\hat{\boldsymbol{\mathit{\mu}}}_{el}=-\sum_ie\boldsymbol{\mathit{r}}_i$
$Z_Ae$ and $\boldsymbol{\mathit{R}}_A$ are the charge and position of nucleus $A$ .
$e$ and $\boldsymbol{\mathit{r}}_i$ are the charge and position of electron $i$ .

The total transition matrix $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is:

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\vert\psi_e\psi_v\psi_r\rangle+\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{el}\vert\psi_e\psi_v\psi_r\rangle$

Although $\psi_e$ depends parametrically on the nuclear coordinates, the nuclear dipole operator $\hat{\boldsymbol{\mathit{\mu}}}_N$ does not act on the electronic coordinates. Therefore, we can write:

$\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\vert\psi_e\psi_v\psi_r\rangle=\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{N}\langle\psi_{e^{'}}\vert\psi_e\rangle\vert\psi_v\psi_r\rangle$

Since electronic spectroscopy involves transitions between different electronic states, and the eigenfunctions in the set $\{\psi_e\}$ are orthonormal, $\langle\psi_{e^{'}}\vert\psi_e\rangle=0$ and thus this term vanishes. In contrast, for microwave or infrared spectroscopy, where no change in electronic state occurs, $\langle\psi_{e^{'}}\vert\psi_e\rangle=1$ , and so:

$\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\langle\psi_{e^{'}}\vert\psi_e\rangle\vert\psi_v\psi_r\rangle=\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{N}\vert\psi_v\psi_r\rangle$

which corresponds to vibration-rotation transitions.

Therefore, the total transition matrix reduces to:

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\int_Q\int_{\tau_r}\psi_{v^{'}}^{*}(Q)\psi_{r^{'}}^{*}(\tau_r)\biggr\[\int_r\psi^{*}_{e^{'}}(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\hat{\boldsymbol{\mathit{\mu}}}_{el}\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})d\boldsymbol{\mathit{r}}\biggr\]\psi_v(Q)\psi_r(\tau_r)dQd\tau_r$

Because the integral in square brackets is performed over all electronic coordinates for each set of nuclear coordinates $\boldsymbol{\mathit{R}}$ , we can define it as a function of $\boldsymbol{\mathit{R}}$ : $\hat{\mu}_{el}(\boldsymbol{\mathit{R}})=\int_r\psi^{*}_{e^{'}}(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\hat{\boldsymbol{\mathit{\mu}}}_{el}\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})d\boldsymbol{\mathit{r}}$ . To further simplify the total transition matrix, we apply the Condon approximation, which assumes that electronic transitions occur so rapidly (on the order of 10^-15 s) that the nuclei can be considered stationary during the transition. Evaluating at the equilibrium nuclear geometry $\boldsymbol{\mathit{R}}_{eq}$ (i.e. an average internuclear separation), $\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})$ can be treated as a constant. Thus,

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\psi_r(\tau_r)d\tau_r$

Question

Is $\hat{\boldsymbol{\mathit{\mu}}}_{el}=-\sum_ie\boldsymbol{\mathit{r}}_i$ defined in the molecular frame or the laboratory frame?

Answer

It is defined in the molecular frame. In the Born-Oppenheimer approximation, the electronic coordinates $\boldsymbol{\mathit{r}}_i$ are defined relative to the nuclear centre of mass.

Although $\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})$ is defined in the molecular frame $(a,b,c)$ , the rotational wavefunctions are defined with respect to the laboratory frame $(x,y,z)$ . For the integral to be physically meaningful, the dipole operator and the wavefunctions must be expressed in the same coordinate system (usually the laboratory frame) as that is where the interaction with the external electromagnetic field occurs. To simplify the integral, we shall consider the projection of $\hat{\boldsymbol{\mathit{\mu}}}_{el}$ onto the lab $z$ -axis (see this article for derivation):

$\hat{\mu}_{z}=\mu_a\Phi_a+\mu_b\Phi_b+\mu_c\Phi_c$

where $\Phi_a=sin\theta cos\chi$ , $\Phi_b=-sin\theta sin\chi$ , $\Phi_c=cos\theta$ .

Since $\mu_a$ , $\mu_b$ , $\mu_c$ are constants (Condon approximation),

$\langle\Psi'\vert\hat{\mu}_z\vert\Psi\rangle=\sum_{g=a,b,c}\int_{\tau_{el}}\psi^*_{e^{'}}\hat{\mu}_g\psi_ed\tau_{el}\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$

with $d\tau_r=sin\theta d\theta d\phi d\chi$ (see this article for derivation).

$\int_{\tau_{el}}\psi^*_{e^{'}}\hat{\mu}_g\psi_ed\tau_{el}$ defines the electronic selection rules. As explained in the previous article, these rules are governed by

1. Symmetry: One first identifies the point group of the molecule, then determines the symmetries of the initial and final states, and finally applies the vanishing integral rules.
2. Parity: The Laporte selection rule applies only to molecules possessing an inversion centre (e.g. those belonging to $D_{2h}$ or $O_h$ ).
3. Spin: $\Delta S=0$ , assuming spin-orbit coupling is weak.

For example (see previous article for derivation),

Unlike pure molecular vibrational motion in infrared spectroscopy, where the initial and final states are part of a complete set of eigenfunctions of the same Hamiltonian (resulting in the pure vibrational selection rule $\Delta v=\pm 1$ ), the initial state in $\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ$ (known as the Frank-Condon overlap integral) belongs to a complete set of eigenfunctions of a Hamiltonian that is different from that of the final state. This is because the two states correspond to different electronic states, each with a different potential-energy function. Therefore, the final vibrational state may not be orthogonal to the initial vibrational state. However, both complete sets of eigenfunctions span the same Hilbert space, which implies that we can expand $\psi_{v^{'}}^{*}(Q)$ as a linear combination of $\psi_v(Q)$ :

$\psi_{v^{'}}^{*}(Q)=\sum_vc_v\psi_v(Q)$

with

$c_v=\langle\psi_v(Q)\vert\psi_{v^{'}}^{*}(Q)\rangle$

If there were a restriction such as $v=v'\pm 1$ , the expansion $\sum_vc_v\psi_v(Q)$ would contain only a finite number of terms. However, the excited state generally requires infinitely many initial state basis functions to represent. This corresponds to an infinite number of nonzero coefficients $c_v$ and hence infinitely many possible values of $v$ , resulting in no selection rule for $\Delta v$ . In other words,

$\Delta v=0,\pm 1,\pm 2,\cdots$

For linear molecules, $\mu_a=0$ and $\mu_b=0$ , while $\mu_c\neq 0$ , even for homonuclear diatomic molecules, because of the generation of a transient dipole moment in electronic spectroscopy. Therefore, $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$ in $\langle\Psi'\vert\hat{\mu}_z\vert\Psi\rangle$ simplifies to $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)cos\theta\psi_r(\tau_r)d\tau_r$ , where the spherical harmonics are the wavefunctions, and the volume element reduces to $d\tau_r=sin\theta d\theta d\phi$ because the spherical harmonics do not depend on $\chi$ . As shown in another article, this integral yields the selection rules: $\Delta J=0,\pm 1$ . The inclusion of $\Delta J=0$ reflects the fact that conservation of total angular momentum is satisfied during a photon-mediated electronic transition in linear molecules when $\Delta \Omega=\pm 1$ , even though $\Delta J=0$ is forbidden for pure rotational transitions in linear molecules. More specifically, the ro-vibronic transition selection rules for linear molecules are:

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=\pm 1\;\;\;\;for\;\Delta\Omega=0\;transitions$

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1\;\;\;\;for\;\Delta\Omega=\pm 1\;transitions$

This is why homonuclear diatomic molecules are vibrationally and rotationally inactive in standard infrared and microwave spectroscopy, but vibrationally and rotationally active in electronic spectroscopy.

For symmetric rotors, the rotational wavefunctions are represented by the Wigner D-functions $D^J_{M_JK}(\phi,\theta,\chi)=e^{-i\chi K}\langle J,M_J\vert e^{-i\phi\hat{J}_z}e^{-i\theta\hat{J}_y}\vert J,K\rangle$ . As shown in another article, $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$ results in $\Delta J=0,\pm 1,\Delta K=0$ for parallel transitions, and $\Delta J=0,\pm 1,\Delta K=\pm 1$ for perpendicular transitions. Therefore, for allowed electronic transitions,

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1,\;\;\;\Delta K=0\;\;\;\;for\;parallel\;transitions$

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1,\;\;\;\Delta K=\pm 1\;\;\;\;for\;perpendicular\;transitions$

Although a spherical top is rotationally inactive in pure rotational spectroscopy, it is rotationally active in ro-vibronic transitions due to the generation of a transcient dipole moment. Since a spherical top is a special case of a symmetric top for which the energy does not depend on $K$ , the selection rules for allowed electronic transitions are simply:

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1$

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May 16, 2026July 3, 2026

Vibronic and ro-vibronic spectra

A vibronic spectrum records simultaneous changes in the electronic and vibrational energies of molecules, whereas a ro-vibronic spectrum additionally includes resolved rotational energy changes.

Unlike an atomic absorption spectrometer, which typically uses element-specific line sources such as hollow-cathode lamps, the spectrometers used to measure vibronic and ro-vibronic absorption spectra generally employ broadband continuous radiation sources (e.g. deuterium or tungsten lamps) together with monochromators to isolate the desired wavelength range (see diagram below).

Vibronic and ro-vibronic spectra can also be observed as emission spectra. In such cases, the sample molecules can be excited using:

1. Electric discharges: Passing a high-voltage current through a low-pressure gaseous sample.
2. Inductively coupled plasma arcs: Heating the sample to extremely high temperatures.
3. Laser excitation: Using a laser tuned to a specific transition. The molecules absorb the laser light and then emit it as fluorescence, allowing for very “clean” ro-vibronic spectra.
4. Electron beams: Bombarding the gas with high-energy electrons.

Electronic spectroscopic analyses of very large or heavy molecules, such a protein, usually produce vibronic spectra. In such cases, the rotational constants are so small that the rotational energy levels are closer together than the natural linewidth imposed by the Heisenberg uncertainty principle. Studies of molecules in the liquid or solid phase also result in broad vibronic bands because collisions and intermolecular interactions shorten the lifetimes of the excited states and broaden the spectral lines, preventing the individual rotational transitions from being resolved. Conversely, small molecules in the gaseous phase, like H₂ or CO, have larger rotational constants and more widely spaced rotational energy levels, making ro-vibronic spectrum observable.

Consider the homonuclear diatomic molecule N₂. Its first few electronic states are: $X^1\Sigma_g^+$ (ground state), $A^3\Sigma_u^+$ (first excited state), $B^3\Pi_g$ (next higher excited state), $C^3\Pi_u$ (another higher excited state), and so on (click this link for derivation of term symbols). The transition $X^1\Sigma_g^+\leftrightarrow A^3\Sigma_u^+$ is forbidden due to the selection rule $\Delta S=0$ . Therefore, the first two observed emission spectroscopy transitions are $B^3\Pi_g\rightarrow A^3\Sigma_u^+$ and $C^3\Pi_u\rightarrow B^3\Pi_g$ , which, along with their associated vibrational and rotational fine structures, are known as the first positive system and the second positive system respectively (see diagram above). For illustration purposes, three vibrational transitions ( $v=0\leftrightarrow v=0$ , $v=0\leftrightarrow v=1$ and $v=0\leftrightarrow v=2$ ), along with their associated rotational fine structures, are shown for each system. These emissions lines typically group into bands, with the wavelength corresponding to each vibrational transition located approximately at the band head (the point of highest intensity within a band). Notably, the parity rule $g\leftrightarrow u$ is satisfied for both systems, but the Q-branches are absent in the second positive system because $\Delta J=0$ is forbidden for $\Delta\Omega=0$ transitions (see this link for details). In practice, these bands may overlap, forming a “forest of peaks” rather than distinct, well-separated bands.

The intensities of the bands of any ro-vibronic spectrum vary according to the Frank-Condon principle, which states that because atomic nuclei are much more massive than electrons, an electronic transition occurs so rapidly that the nuclear configuration of the molecule remains practically unchanged during the process. In general, a molecule’s nuclear configuration is described by vibrational wavefunctions, each expressed as a function of the internuclear separation $R$ corresponding to a given electronic state, where the equilibrium separation $R_e$ satisfies the relationship: $R_e(ground)\neq R_e(1st\;excited)\neq R_e(2nd\;excited)\neq\cdots$ . For example, the $\Delta v=+1$ band for the $\vert\Omega=0,v=0\rangle\leftrightarrow\vert\Omega'=1,v'=1\rangle$ transition in the diagram below is expected to have a higher intensity than the $\Delta v=0$ band.

This is because, classically, the amplitude of the wavefunction of an oscillator in an excited vibrational state is greatest near the turning points, where the nuclei move most slowly. In contrast, the ground-state wavefunction has its largest amplitude near $R_e$ . Since the vibrational transition intensity is proportional to the square of the overlap integral $\int_Q\psi^*_{v^{'}}(Q)\psi_v(Q)dQ$ , it is determined, according to the Frank-Condon principle, by the extent of overlap between the initial-state wavefunction and the final-state wavefunction lying vertically above it. Since $\int_Q\psi^*_{v^{'}}(Q)\psi_v(Q)dQ$ is constant for every rotational line within a given vibrational band, it acts as a global scaling factor. Consequently, the intensities of rotational fine structures within a higher-intensity vibrational band are generally higher than those within a lower-intensity vibrational band for a given electronic transition.

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April 7, 2026April 8, 2026

Why lithium-ion batteries sometimes catch fire

Lithium-ion batteries power everything from smartphones and laptops to electric vehicles and energy storage systems. They are efficient, lightweight and rechargeable. However, under certain conditions, they can overheat, expand and even catch fire. Understanding why this happens comes down to how these batteries are built and what can go wrong inside them.

A lithium-ion battery consists of three key components: anode, cathode and the electrolyte (see diagram above). Separating the anode and cathode is a thin porous membrane called the separator, which plays the critical role of keeping the two electrodes apart while still allowing ions to pass through. As the battery charges and discharges, lithium ions shuttle back and forth between the electrodes, storing and releasing energy.

Under normal conditions, lithium-ion batteries are safe. Problems arise when internal or external factors disrupt their delicate balance, with the most serious outcome being thermal runaway — a chain reaction in which rising temperature accelerates further heat generation. The main trigger for a thermal runaway is an internal short circuit, which occurs when the anode and cathode come into direct contact. When this happens, charges no longer flow through the higher-resistance electrolyte, but instead travel via a very low-resistance path between the electrodes. This surge in internal current flow results in rapid heat generation that can lead to combustion.

So what causes the anode and cathode to come into direct contact? Physical damage, such as dropping, crushing or puncturing a battery can deform its internal structures. Manufacturing defects may leave contaminants, like metal particles or dust, inside the battery that can pierce the separator. Another cause is lithium plating, which can lead to dendrite growth.

Lithium plating occurs when lithium ions deposit as metallic lithium on the surface of the anode, Li⁺(aq) + e^– → Li(s), instead of moving into the anode’s internal structure (a process known as intercalation). Fast charging drives lithium ions rapidly towards the anode and can overload its surface. If the intercalation rate is slower than the incoming flux, the excess ions tend to accumulate and are reduced to metallic lithium on the anode surface. Low operating temperatures (< 273.15 K), which increase the viscosity of the electrolyte and decrease ionic mobility, can also promote lithium plating (the slower diffusion of ions into the graphite layers causes them to accumulate and plate onto the surface).

Initially, this plated lithium layer is often relatively uniform, so the anode and cathode remain physically separated by the separator, although the battery capacity may be affected. However, if the plated lithium grows unevenly over time, it can form needle-like structures called dendrites. These body-centred cubic structures can pierce the separator and reach the cathode, creating a direct, low-resistance path between the electrodes and leading to an internal short circuit.

Finally, gas buildup in a lithium-ion battery, caused by electrolyte decomposition over time, can also result in thermal runaway. This decomposition involves the battery’s electrolyte, which typically consists of the lithium salt LiPF₆ dissolved in an organic solvent such as ethylene carbonate or dimethyl carbonate. Under stress (e.g. overcharging or high temperature), LiPF₆ can decompose to PF₅ (LiPF₆ → LiF + PF₅), which then reacts with trace amounts of water to produce hydrogen fluoride gas (PF₅ + 4H₂O → H₃PO₄ + 5HF). The organic solvents can also oxidise or decompose, producing gases like CO, CO₂ or hydrocarbons, leading to battery swelling commonly observed as bulging laptop casings. The increased internal pressure may displace the electrodes, causing an internal short circuit, or even rupture the battery. This significantly increases the risk of explosion, especially if the gases are flammable.

To mitigate such risks, high-quality lithium-ion battery designs use a combination of materials, architecture and protective systems to prevent lithium plating, dendrite formation, gas buildup, and short circuits. For instance, high-quality graphite, silicon-graphite composites or lithium titanate anodes ensure even lithium intercalation, reducing the risk of plating. Additives are also included in the electrolyte to regulate lithium deposition and suppress dendrite growth. Furthermore, multi-layer microporous polymer separators can resist puncture by dendrites.

Additionally, devices powered by lithium-ion batteries incorporate electronic circuitry that monitor battery voltage, current and temperature (battery management systems) to prevent overcharging and over-discharging. Cooling systems, such as personal computer fans, also help reduce overheating and electrolyte decomposition.

Question

Do electric cars (EVs) also use lithium-ion batteries? If so, why do they rarely catch fire?

Answer

Yes, most EVs use lithium-ion batteries. While the chemistry is similar to that of batteries in phones or laptops, EV battery systems are much more advanced, with high-quality electrodes and robust casings that protect against physical damage. Sophisticated battery management and thermal management systems are also employed to prevent overheating (see diagram below).

In summary, lithium-ion batteries catch fire not because they are inherently unsafe, but because their high energy density makes them sensitive to internal failures and external stress. When key safeguards (separator, electrolyte stability or controlled ion movement) are compromised, a cascade of reactions can lead to thermal runaway. However, with continued advances in materials science, battery design and management systems, the risks are being steadily reduced. When properly designed, manufactured and used, lithium-ion batteries remain a safe and indispensable technology in modern life.

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April 5, 2026April 9, 2026

Inductive effect

The inductive effect is the shift of electron density in $\sigma$ -bonds caused by electron-withdrawing or electron-donating groups, resulting in polarised bonds.

This effect occurs because different atoms or groups influence electron density according to their electronegativity relative to the atoms to which they are bonded. Atoms or groups that are more electronegative tend to withdraw electron density, while those that are less electronegative can donate electron density through the $\sigma$ -bond framework. In organic chemistry, the inductive effect is often discussed for substituents attached to carbon chains, where electron density is shifted along the bonds. As a result, atoms near an electron-withdrawing group may become slightly electron-deficient and develop a partial positive charge, whereas atoms near an electron-donating group may become slightly electron-rich and develop a partial negative charge. Whether a substituent is labelled electron-withdrawing (–I) or electron-donating (+I) depends entirely on how its electronegativity compares to a hydrogen atom bonded to that same position.

Electron-withdrawing groups (-I)

To quantum-mechanically analyse the inductive effect caused by electron-withdrawing substituents (classified as –I substituents), we consider an atom in which the energy of an electron can be expressed as two main contributions:

$H=T+V$

where $H$ is the Hamiltonian, $T$ is the kinetic energy of the electron and $V$ is the electrostatic potential energy arising from attraction to the nucleus.

The potential energy is approximately

$V=-\frac{Z_{eff}e^2}{4\pi\varepsilon_0r}$

where $Z_{eff}$ is the effective nuclear charge and $r$ is the electron-nucleus distance.

Consider the heteronuclear diatomic molecule HCl, in which the Cl atom is more electronegative than the H atom. Since $Z_{eff}(Cl)>Z_{eff}(H)$ , the potential energy of an electron on Cl is more negative than that on H, lowering the valence orbital energy of Cl relative to that of H. In other words, the more electronegative atom or group usually has a lower-energy valence orbital.

When the two atoms combine to form a $\sigma$ -bond, the bonding molecular orbital (MO) wavefunction $\psi$ is

$\psi=c_1\phi_1+c_2\phi_2$

where $\phi_1$ and $\phi_2$ are the atomic orbital (AO) wavefunctions of H and Cl respectively, and $c_1$ and $c_2$ are real-valued coefficients that are key to explaining the inductive effect.

If $\psi$ is normalised and the AOs are orthogonal, the electron density of the $\sigma$ -bond is given by

$\int\psi^*\psi d\tau=c_1^2+c^2_2=1$

This implies that $c_1$ and $c_2$ determine the contribution of each AO to the electron density of the bond. The larger coefficient corresponds to a greater contribution and therefore greater electron density in the region of that atom. Because Cl is more electronegative, we expect $\vert c_2\vert>\vert c_1\vert$ . To justify this, we multiply the eigenvalue equation $\hat{H}\psi=E\psi$ on the left by $\psi^*$ and integrate to give

$E=\frac{\langle\psi\vert\hat{H}\vert\psi\rangle}{\langle\psi\vert\psi\rangle}=\frac{c_1^2\alpha_1+c^2_2\alpha_2+2c_1c_2\beta_{12}}{c_1^2+c_2^2}\;\;\;\;\;\;\;\;380$

where $\alpha_1=\langle\phi_1\vert\hat{H}\vert\phi_1\rangle$ , $\alpha_2=\langle\phi_2\vert\hat{H}\vert\phi_2\rangle$ , $\beta_{12}=\langle\phi_1\vert\hat{H}\vert\phi_2\rangle$ and we have

1. used the Hermitian property of the Hamiltonian $\langle\phi_i\vert\hat{H}\vert\phi_j\rangle=\langle\phi_j\vert\hat{H}\vert\phi_i\rangle^*$
2. assumed that the set $\{\phi_i\}$ is orthonormal and expectation values are real, so that $\langle\phi_j\vert\hat{H}\vert\phi_i\rangle^*=\langle\phi_j\vert\hat{H}\vert\phi_i\rangle$ .

Applying the variational principle to eq380 by setting the partial derivative of $E$ with respect to each coefficient to zero, we obtain a set of simultaneous equations known as secular equations:

$(\alpha_1-E)c_1+\beta_{12}c_2=0\;\;\;\;\;\;\;\;382$

$(\alpha_2-E)c_2+\beta_{12}c_1=0\;\;\;\;\;\;\;\;383$

Expressing eq382 and eq383 in matrix form yields:

$\begin{pmatrix}\alpha_1-E&\beta_{12}\\\beta_{12}&\alpha_2-E\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}=0\;\;\;\;\;\;\;\;384$

Eq384 is a linear homogeneous equation with non-trivial solutions only if

$\begin{vmatrix}\alpha_1-E&\beta_{12}\\\beta_{12}&\alpha_2-E\end{vmatrix}=0$

Expanding the determinant gives the characteristic equation:

$E^2-(\alpha_1+\alpha_2)E+\alpha_1\alpha_2-\beta^2_{12}=0$

with two solutions corresponding to the bonding MO ( $E_-$ ) and antibonding MO ( $E_+$ ):

$\begin{align}E_{\pm}&=\frac{\alpha_1+\alpha_2\pm\sqrt{(\alpha_1+\alpha_2)^2-4\alpha_1\alpha_2+4\beta^2_{12}}}{2}\\&=\frac{\alpha_1+\alpha_2\pm\sqrt{(\alpha_1-\alpha_2)^2+4\beta^2_{12}}}{2}\;\;\;\;\;\;\;\;385\end{align}$

The energy separation between these two MOs is

$\Delta E=E_+-E_-=2\sqrt{\biggr$\frac{\alpha_1-\alpha_2}{2}\biggr$^2+\beta^2_{12}}$

If $\beta_{12}=0$ , then $\Delta E=\alpha_1-\alpha_2$ , which implies that the two AOs do not interact and no splitting occurs. It follows that $\vert\beta_{12}\vert$ increases the separation between the two MOs, with a larger $\vert\beta_{12}\vert$ resulting in a greater separation, and hence, a greater stabilisation of the bonding MO.

Rearranging eq383 and taking absolute values on both sides yield:

$\frac{\vert c_2\vert}{\vert c_1\vert}=\frac{\vert\beta_{12}\vert}{\vert\alpha_2-E\vert}\;\;\;\;\;\;\;\;386$

Substituting $E_-=E$ in eq385 into $\vert\alpha_2-E\vert$ gives:

$\vert\alpha_2-E\vert=\biggr\vert\sqrt{\biggr$\frac{\alpha_1-\alpha_2}{2}\biggr$^2+\beta^2_{12}}-\frac{\alpha_1-\alpha_2}{2}\biggr\vert$

Multiplying and dividing the RHS by its conjugate $=\biggr\vert\sqrt{\biggr$\frac{\alpha_1-\alpha_2}{2}\biggr$^2+\beta^2_{12}}+\frac{\alpha_1-\alpha_2}{2}\biggr\vert$ yields

$\vert\alpha_2-E\vert=\frac{\vert\beta_{12}^2\vert}{\biggr\vert\sqrt{$\frac{\alpha_1-\alpha_2}{2}$^2+\beta^2_{12}}+\frac{\alpha_1-\alpha_2}{2}\biggr\vert}$

Since $\vert\sqrt{\beta_{12}^2}\vert=\vert\beta_{12}\vert$ , then $\biggr\vert\sqrt{\biggr$\frac{\alpha_1-\alpha_2}{2}\biggr$^2+\beta^2_{12}}\biggr\vert>\vert\beta_{12}\vert$ . $\alpha_1$ is the expectation value of the hydrogen AO, i.e. $\alpha_1\approx E(1s_H)$ , which is less negative than $\alpha_2\approx E(3p_{Cl})$ . So, $\alpha_1-\alpha_2>0$ and $\biggr\vert\sqrt{\biggr$\frac{\alpha_1-\alpha_2}{2}\biggr$^2+\beta^2_{12}}+\frac{\alpha_1-\alpha_2}{2}\biggr\vert>\vert\beta_{12}\vert$ . Therefore, $\vert\alpha_2-E\vert=\frac{\vert\beta_{12}\vert\vert\beta_{12}\vert}{m\vert\beta_{12}\vert}$ , where $m>1$ , or equivalently,

$\vert\beta_{12}\vert>\vert\alpha_2-E\vert$

Substituting this into eq386 results in

$\vert c_2\vert>\vert c_1\vert$

This proves that the coefficient of the more electronegative atom corresponds to a greater contribution to the $\boldsymbol{\mathit{\sigma}}$ -bond, and therefore, greater electron density in the region of that atom.

In terms of polyatomic aliphatic molecules, the inductive effect is transmitted along covalent bonds in the chain, but its strength decreases rapidly with distance. To illustrate this, consider the C-C-Cl $\sigma$ -framework in the molecule CH₃CH₂Cl, with the following bonding MO wavefunction:

$\psi=c_1\phi_1+c_2\phi_2+c_3\phi_3$

where 1 denotes the terminal carbon, 2 denotes the carbon bonded to Cl and 3 denotes the chlorine atom.

The difference in electronegativity between chlorine and the adjacent carbon is significantly larger than that between the two carbon atoms. We would therefore expect $\vert c_3\vert>\vert c_2\vert$ , with the electron density skewed towards chlorine, leaving the middle carbon slightly electron-deficient and effectively more electronegative relative to the terminal carbon atom. Applying similar reasoning to the C-C bond, we obtain $\vert c_2\vert>\vert c_1\vert$ , and consequently,

$\vert c_3\vert>\vert c_2\vert>\vert c_1\vert$

This result implies that the inductive effect transmitted along covalent bonds in a chain decreases in strength with increasing distance from the substituent. It also means that the C-Cl moiety may be regarded as an electron withdrawing group. More generally, a group containing several atoms may behave collectively as an electron-withdrawing group if its internal bonding renders an atom electron-deficient relative to the surrounding network and capable of attracting electron density from neighbouring bonds.

For example, in the formyl group (–CHO), the carbonyl oxygen strongly stabilises the $\pi$ system of the C=O bond, lowering the energy of orbitals centred on the carbonyl carbon. As a result, the carbonyl carbon becomes relatively electron-deficient and can draw electron density from the neighbouring $\sigma$ framework. Such redistribution of electron density can significantly influence the chemical behaviour of molecules. For instance, inductive effects can alter the acidity of different amino acids by stabilising or destabilising charged intermediates, and can also affect reaction mechanisms by increasing the electrophilicity of certain atoms, making them more susceptible to nucleophilic attack.

Electron-donating groups (+I)

A group that has the opposite effect on a chain of atoms compared to an electron-withdrawing group is known as an electron-donating group.

The same molecular orbital reasoning can be used to explain electron-donating groups. The key idea is that the direction of electron density shift depends on the relative energies of the interacting atomic orbitals, which are reflected in the Coulomb integrals $\alpha_i$ . For an electron-donating group, the situation is reversed. The atom attached to the carbon chain has higher-energy valence orbitals, corresponding to a larger Coulomb integral than that of the neighbouring carbon atom. When the bonding molecular orbital is formed, the coefficient on this substituent becomes relatively smaller than that on the adjacent carbon atom. Consequently, electron density is pushed into the carbon skeleton. Thus, the substituent effectively donates electron density through the $\sigma$ framework, producing a positive inductive effect (+I).

Alkyl substituents provide a simple example. Because carbon is slightly more electronegative than hydrogen, the electron density in C-H bonds of an alkyl group such as CH₃– is shifted towards the carbon atom. This leaves the carbon atom relatively electron-rich and its valence orbitals at slightly higher energy than those of a methylene carbon (CH₂). When these units combine to form the ethyl group CH₃CH₂-, the methyl fragment behaves as an electron-donating substituent. Comparing H₃C-CH₂– and H-CH₂-, we expect the methylene carbon in the ethyl group to be more electron-rich than that in the methyl group, and hence the ethyl group to be a stronger electron-donating group than the methyl group. It follows that, in terms of increasing +I effect:

$t-Bu>i-Pr>Et>Me>H$

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Mesomeric effect

The mesomeric effect is the electron-donating or electron-withdrawing effect of a substituent transmitted through a conjugated π-system by resonance.

Resonance, or the resonance effect, typically occurs when atoms are connected by alternating single and multiple bonds, where electrons can be shared across several atoms instead of being localised. For instance, the π-electrons in benzene are dislocalised over the entire ring rather than confined to three double bonds depicted in a Lewis structure.

In other words, the mesomeric effect is an electron-donating (+M) or electron-withdrawing (-M) effect via resonance. It requires a functional group attached to a conjugated system and focuses on how the group changes the electron distribution.

Question

Is the mesomeric effect different from the inductive effect?

Answer

Yes, the inductive effect is the displacement of electron density through $\sigma$ -bonds along a chain of atoms due to the presence of an electronegative or electropositive atom or group. In contrast, the mesomeric effect involves the delocalisation of electrons through π-bonds. Although the mesomeric effect is governed mainly by resonance and orbital interactions, the electronegativities of atoms in substituents may be considered when analysing the mesomeric effect of a molecule (e.g. via the Hückel method). Nevertheless, the two effects are generally different. For example, lone pairs on atoms such as O or N often give rise to a +M effect. However, these same atoms may exert a –I effect at the same time. Therefore, the mesomeric and inductive effects can act in the same direction or in opposite directions in a molecule with a substituent.

-M effect

The Hückel method is the standard semi-empirical approach to quantify the mesomeric effect in a conjugated system. Consider the acrolein molecule CH₂=CH-CHO, which contains two π-bonds, one of which belongs to an electron-withdrawing formyl group (-CHO). In general, we can express the one-electron molecular orbital (MO) wavefunction of acrolein as:

$\psi=c_1\phi_1+c_2\phi_2+c_3\phi_3+c_4\phi_4$

where $c_n$ and $\phi_n$ represent the coefficients and the p atomic orbital wavefunctions respectively; $n=1$ corresponds to the oxygen atom, $n=2$ corresponds to the carbonyl carbon, $n=3$ corresponds to the middle ethylene carbon and $n=4$ corresponds to the terminal ethylene carbon.

One quick way to analyse the MO energies of acrolein is to refer to the secular determinant of butadiene:

$\begin{vmatrix}\alpha-E&\beta &0&0\\\beta&\alpha-E &\beta&0\\0&\beta &\alpha-E&\beta\\0&0 &\beta&\alpha-E\end{vmatrix}=0$

which represents a four- atom framework with two alternating π-bonds.

To account for the electronegativity of oxygen, we modify the Hückel parameters as follows:

$\begin{vmatrix}\alpha_O-E&\beta_{CO} &0&0\\\beta_{CO}&\alpha_{CO}-E &\beta&0\\0&\beta &\alpha-E&\beta\\0&0 &\beta&\alpha-E\end{vmatrix}=0\;\;\;\;\;\;\;\;390$

where

$\alpha_O=\alpha+h_O\beta$ is the Coulomb integral for oxygen, with $h_O$ being a factor that lowers the orbital energy.
$\alpha_{CO}=\alpha+h_{CO}\beta$ is the Coulomb integral for the carbonyl carbon.
$\beta_{CO}=\langle\phi_C\vert\hat{H}_{eff}\vert\phi_O\rangle=k_{CO}\beta$ is the resonance integral for overlap between C and O.
$\alpha=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_i}\rangle$ is the Coulomb integral for carbon atoms.
$\beta=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_j}\rangle$ is the resonance integral between carbon atoms.
$E$ is the energy of an MO.

The next step involves estimating trial values for $h_O$ , $h_{CO}$ and $k_{CO}$ . Since oxygen is more electronegative than carbon, we set $h_O=2.0$ . This results in C=O being a stronger bond compared to C-C. Consequently, we set $k_{CO}=1.4$ . The corresponding inductive effect by oxygen on the the carbonyl carbon causes it to be slightly electronegative relative to the adjacent ethylene carbon. We can therefore set $h_{CO}=0.2$ . It is important to note that although the electronegativity of oxygen helps stabilise the system, it is not the primary cause of the mesomeric effect, which is fundamentally due to conjugation and orbital overlap.

Using the matrix identity of $\vert cA\vert=c^n\vert A\vert$ , and multiplying both sides of eq390 by $\frac{1}{\beta^4}$ , give

$\begin{vmatrix}x+2.0&1.4 &0&0\\\1.4&x+0.2 &1&0\\0&1 &x&1\\0&0 &1&x\end{vmatrix}=0$

where $x=\frac{\alpha-E}{\beta}$ .

Expanding the determinant gives the characteristic equation:

$x^4+2.2x^3-3.56x^2-4.2x+1.56=0$

with solutions $x=-2.86,-1.16,0.31,1.52$ , which when substituted back into $x=\frac{\alpha-E}{\beta}$ yields:

MO	$\boldsymbol{\mathit{x}}$	$\boldsymbol{\mathit{E}}$	Type
$\pi_4^*$	1.52	$\alpha-1.52\beta$	Antibonding
$\pi_3^*$	0.31	$\alpha-0.31\beta$	LUMO
$\pi_2$	-1.16	$\alpha+1.16\beta$	HOMO
$\pi_1$	-2.86	$\alpha+2.86\beta$	Bonding

Question

Show that the isolated C=O Hückel bond energy is $2\alpha+5.44\beta$ .

Answer

The secular determinant for the C=O bond is

$\begin{vmatrix}x&1.4 \\1.4&x+2.0\end{vmatrix}=0$

Solving for $x$ gives $x\approx -2.72,0.72$ . Therefore, the bonding MO energy is $2(\alpha+2.72\beta)=2\alpha+5.44\beta$ .

The total ground state π-electron energy, $2\pi_1+2\pi_2=4\alpha+8.04\beta$ , is lower than the sum of the energies of isolated C=C ( $2\alpha+2\beta$ ) and C=O ( $2\alpha+5.44\beta$ ) bonds by $0.60\beta$ , indicating the delocalisation energy provided by the mesomeric effect. To further understand the electron distribution of the molecule, we determine the coefficients of the $\pi_i$ wavefunctions as follows:

$\begin{pmatrix}x+2.0&1.4 &0&0\\\1.4&x+0.2 &1&0\\0&1 &x&1\\0&0 &1&x\end{pmatrix}\begin{pmatrix}c_1\\c_2\\c_3\\c_4\end{pmatrix}=0$

which implies:

$(x+2.0)c_1+1.4c_2=0$

$1.4c_1+(x+2.0)c_2+c_3=0$

$c_2+xc_3+c_4=0$

$c_3+xc_4=0$

The simultaneous equations for $\pi_1$ can be solved by substituting $x=-2.86$ into them and expressing all coefficients in terms of $c_4$ :

$c_3=2.86c_4$

$c_2=-c_4+2.86c_3=-c_4+2.86^2c_4\approx 7.18c_4$

$c_1=\frac{1.4c_2}{0.86}\approx 11.69c_4$

Normalisation gives:

$(11.69c_4)^2+(7.18c_4)^2+(2.86c_4)^2+c_4^2=1$

$c_4\approx 0.07$

Therefore, $c_1\approx 0.82,c_2\approx 0.50,c_3\approx 0.20,c_4\approx 0.07$ . Repeating the calculations for $\pi_2$ and $\pi_3$ yields:

Coefficient	$\boldsymbol{\mathit{\pi_1}}$ (bonding)	$\boldsymbol{\mathit{\pi_2}}$ (HOMO)	$\boldsymbol{\mathit{\pi_3}}$ (LUMO)	Ground state electron density
c₄	0.07	0.60	0.67	0.73
c₃	0.20	0.70	-0.21	1.06
c₂	0.50	0.21	-0.60	0.59
c₁	0.82	-0.35	0.37	1.59

where the ground state electron density is given by $c_i=2[(c_{i,\pi_1})^2+(c_{i,\pi_2})^2]$ for two electrons in each MO.

The value of 1.59 at oxygen indicates significant accumulation of π-electron density. This arises from the –M effect of the carbonyl group, where electron density is delocalised towards the oxygen atom. In contrast, the terminal carbon and the carbonyl carbon are electron deficient due to this delocalisation (see resonance structures below). While oxygen’s electronegativity contributes to the overall polarisation of the molecule, the calculated electron densities primarily reflect the mesomeric effect rather than the inductive effect. In this case, both effects act in the same direction.

The calculations also suggest regioselectivity of acrolein in reactions with nucleophiles. For example, “hard” (ionic) nucleophiles, such as LiAlH₄ or Grignard reagents, are driven mainly by electrostatic attraction and attack the most electron-deficient carbonyl carbon (with ground state electron density of 0.59), resulting in 1,2-addition reactions (see diagram below). On the other hand, “soft” (covalent) nucleophiles, like enolates, thiols or organocuprates, are governed by orbital interactions. They attack the atom where the LUMO has the largest “lobe” ( $(0.67)^2>(-0.21)^2,(0.60)^2,(0.37)^2)$ ), which is the terminal ethylene carbon of acrolein, leading to 1,4-addition (Michael addition) reactions.

+M effect

The +M effect is typically attributed to a substituent with lone pair electrons. Consider vinylamine CH₂=CH-NH₂, with the general one-electron MO wavefunction:

$\psi=c_1\phi_1+c_2\phi_2+c_3\phi_3$

where $\phi$ represents the p atomic orbital wavefunctions; 1 denotes the nitrogen atom, 2 denotes the middle ethylene carbon and 3 denotes the terminal ethylene carbon.

The secular determinant is given by:

$\begin{vmatrix}\alpha_N-E&\beta_{CN} &0\\\beta_{CN}&\alpha-E &\beta\\0&\beta &\alpha-E\end{vmatrix}=0\;\;\;\;\;\;\;\;391$

where

$\alpha_N=\alpha+h_N\beta$ is the Coulomb integral for nitrogen, with $h_N$ being a factor that lowers the orbital energy.
$\beta_{CN}=\langle\phi_C\vert\hat{H}_{eff}\vert\phi_N\rangle=k_{CN}\beta$ is the resonance integral for overlap between C and N.
$\alpha=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_i}\rangle$ is the Coulomb integral for carbon atoms.
$\beta=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_j}\rangle$ is the resonance integral between carbon atoms.
$E$ is the energy of an MO.

Let’s set $h_N=1.5$ due to the electronegativity of N, and $k_{CN}=0.8$ because the C-N bond is slightly weaker than the C=C bond. This translates to:

$\begin{vmatrix}x+1.5&0.8 &0\\0.8&x &1\\0&1 &x\end{vmatrix}=0$

with the corresponding characteristic equation:

$x^3+1.5x^2-1.64x-1.5=0$

Substituting the solutions $x=-1.95,-0.68,1.13$ back into $x=\frac{\alpha-E}{\beta}$ yields:

MO	$\boldsymbol{\mathit{x}}$	$\boldsymbol{\mathit{E}}$	Type
$\pi_3^*$	1.13	$\alpha-1.13\beta$	LUMO
$\pi_2$	-0.68	$\alpha+0.68\beta$	HOMO
$\pi_1$	-1.95	$\alpha+1.95\beta$	Bonding

Unlike acrolein, in which one electron from each of the four atoms populates the MOs, vinylamine has two nitrogen electrons (lone pair) and one electron from each carbon atom contributing to its bonding and highest occupied MOs. The total ground state π-electron energy, $2\pi_1+2\pi_2=4\alpha+5.26\beta$ , is lower than the sum of the energies of isolated C=C bond ( $2\alpha+2\beta$ ) and the nitrogen lone-pair ( $2\alpha+3.0\beta$ ), indicating a delocalisation energy of $0.26\beta$ provided by the +M effect.

Question

Show that the isolated nitrogen lone pair energy is $2\alpha+3.0\beta$ .

Answer

Since nitrogen’s lone pair is not yet delocalised into the π-system, its energy corresponds to twice the Coulomb integral $\alpha_N=\alpha+1.5\beta$ , which is $2\alpha+3.0\beta$ .

The coefficients of the $\pi_i$ wavefunctions are given by:

$\begin{pmatrix}x+1.5&0.8 &0\\0.8&x &1\\0&1 &x\end{pmatrix}\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}=0$

Substituting the three values of $x$ into the matrix equation and solving it gives:

Coefficient	$\boldsymbol{\mathit{\pi_1}}$ (bonding)	$\boldsymbol{\mathit{\pi_2}}$ (HOMO)	$\boldsymbol{\mathit{\pi_3}}$ (LUMO)	Ground state electron density
c₃	0.24	0.72	-0.65	1.15
c₂	0.48	0.49	0.73	0.94
c₁	0.85	-0.48	-0.22	1.91

Since the ground state electron density of a neutral, non-polarised one-electron atom is 1.0, the ground state electron density of the nitrogen lone pair of 1.91 is lower than its unpolarised value of 2.0 (density = number of electrons per unit volume). In contrast, the electron density of the terminal carbon of 1.15 is higher than its non-polarised value of 1.0. This shift of electron density from N to the terminal C is a consequence of the +M effect of the amine group, even though N exerts a –I effect on its neighbours. In this case, mesomeric and inductive effects act in opposite directions. Furthermore, the relatively electron-rich terminal carbon and nitrogen atoms (ground state electron density > 1.0) make the molecule susceptible to electrophilic attack.

In terms of regioselectivity, a “hard” electrophile like H⁺ (small, concentrated positive charge) is attracted to N, which has the highest ground state electron density of 1.91. Protonation at N forms an ammonium ion. H⁺ can also attack the terminal carbon (electron density = 1.15) to yield an iminium ion. The relative proportion of ammonium and iminium ions formed depends on temperature and reaction time.

The ammonium ion forms quickly at relatively low temperatures through fast collisions and is considered the kinetic product. However, it is not energetically favourable because formation of the $\sigma$ -bond between H⁺ and N disrupts the conjugated system, reducing the molecule’s delocalisation energy. Given enough time for equilibration at room temperature, the resonance-stabilised iminium ion, in which the positive charge is delocalised between carbon and nitrogen, becomes the thermodynamic product and predominates. If water is present, this iminium ion quickly hydrolyses to acetaldehyde and an ammonium salt, which explains why vinylamines (and enamines) are often used as intermediates to functionalise carbonyls.

On the other hand, a “soft” electrophile such as CH₃I, in which the the partially positive carbon is diffuse due to iodine’s polarisability, seeks the largest HOMO lobe for orbital overlap. This leads to formation of a new carbon-carbon bond at the terminal carbon of vinylamine.

Overall, the reactivity of vinylamine illustrates how electron density, substituent effects and the nature of the electrophile work together to determine both regioselectivity and product distribution. Understanding these effects allows chemists to predict and control reaction outcomes, making vinylamines valuable intermediates in organic synthesis.

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Ortho/para and meta directing groups

Ortho/para-directing and meta-directing groups are substituents that guide incoming electrophiles to the ortho and para positions or to the meta position, respectively, on an aromatic ring.

This directing effect arises from how the substituent interacts with the ring’s electron density. Ortho/para-directing groups are typically electron-donating (e.g. –OH, –NH₂, –CH₃); they increase electron density at the ortho and para positions through resonance or inductive effects, stabilising the intermediate carbocation during electrophilic aromatic substitution. In contrast, meta-directing groups are usually electron-withdrawing (e.g., –NO₂, –COOH, –CN); they pull electron density away from the ring, making the ortho and para positions less stable for substitution and thus favouring attack at the meta position.

Ortho/para-directing groups

A classic example of an ortho/para-directing group is the hydroxyl group in phenol. Although it is inductively withdrawing (-I) due to the electronegativity of the oxygen atom, the lone pair of electrons on oxygen exerts a +M effect on the benzene ring, overwhelming the inductive effect and activating the ortho and para positions for electrophilic substitution.

To quantum-mechanically analyse the mesomeric effect of the hydroxyl group, we consider the general molecular orbital (MO) wavefunction of phenol:

$\psi=c_0\phi_0+c_1\phi_1+c_2\phi_2+c_3\phi_3+c_4\phi_4+c_5\phi_5+c_6\phi_6$

where $c_n$ and $\phi_n$ represent the coefficients and the p atomic orbital wavefunctions respectively; $n=0$ corresponds to the oxygen atom, $n=1$ corresponds to the substituted carbon, and $n=2$ through $n=6$ correspond to the remaining carbons in an anticlockwise direction.

Applying the Hückel method gives the following secular determinant:

$\begin{vmatrix}\alpha_O-E&\beta_{CO} &0&0&0&0&0\\\beta_{CO}&\alpha-E &\beta&0&0&0&\beta\\0&\beta &\alpha-E&\beta&0&0&0\\0&0 &\beta&\alpha-E&\beta&0&0\\0&0&0&\beta&\alpha-E&\beta&0\\0&0&0&0&\beta&\alpha-E&\beta\\0&\beta&0&0&0&\beta&\alpha-E\end{vmatrix}=0$

where

$\alpha_O=\alpha+h_O\beta$ is the Coulomb integral for oxygen, with $h_O$ being a factor that lowers the orbital energy.
$\beta_{CO}=\langle\phi_C\vert\hat{H}_{eff}\vert\phi_O\rangle=k_{CO}\beta$ is the resonance integral for overlap between C and O.
$\alpha=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_i}\rangle$ is the Coulomb integral for carbon atoms.
$\beta=\langle\phi_{C_i}\vert\hat{H}_{eff}\vert\phi_{C_j}\rangle$ is the resonance integral between carbon atoms.
$E$ is the energy of an MO.

Incidentally, the minor $M_{11}$ of the above secular determinant is the secular determinant of benzene.

The next step involves estimating trial values for $h_O$ and $k_{CO}$ . Since oxygen is more electronegative than carbon, we set $h_O=1.0$ , lowering the MO relative to carbon. Furthermore, we set $k_{CO}=0.8$ , which is lower than the reference of $k=1$ for every resonance integral between carbon atoms, because the oxygen’s 2p orbitals are more contracted than carbon’s, leading to a less effective overlap between the p orbitals of oxygen and carbon versus that between two carbon atoms.

Using the matrix identity of $\vert cA\vert=c^n\vert A\vert$ , and multiplying both sides of the secular determinant equation by $\frac{1}{\beta^7}$ , give

$\begin{vmatrix}x+1&0.8 &0&0&0&0&0\\0.8&x &1&0&0&0&1\\0&1 &x&1&0&0&0\\0&0 &1&x&1&0&0\\0&0&0&1&x&1&0\\0&0&0&0&1&x&1\\0&1&0&0&0&1&x\end{vmatrix}=0$

where $x=\frac{\alpha-E}{\beta}$ .

Expanding the determinant yields the characteristic equation:

$x^7+x^6-6.64x^5-4.4x^4+10.28x^3+4.2x^2-3.36x+0.8=0$

with solutions

$x=-2.427,-1.419,-0.744,-0.213,0.617,1.192,1.994$

Substituting the solutions back into $x=\frac{\alpha-E}{\beta}$ yields:

MO	$\boldsymbol{\mathit{x}}$	$\boldsymbol{\mathit{E}}$	Type
$\pi_7^*$	1.994	$\alpha-1.994\beta$	Antibonding
$\pi_6^*$	1.192	$\alpha-1.192\beta$	Antibonding
$\pi_5^*$	0.617	$\alpha-0.617\beta$	LUMO
$\pi_4$	-0.213	$\alpha+0.213\beta$	HOMO
$\pi_3$	-0.744	$\alpha+0.744\beta$	Bonding
$\pi_2$	-1.419	$\alpha+1.419\beta$	Bonding
$\pi_1$	-2.427	$\alpha+2.427\beta$	Bonding

The reactivity of phenol is reflected in its HOMO energy level, $\alpha+0.213\beta$ , which is higher than that of benzene ( $\alpha+\beta$ ). Because the HOMO is higher in energy, the electrons are less tightly held and more nucleophilic than those in benzene towards electrophiles. To further understand the directing effects of the hydroxyl group, we determine the coefficients of the HOMO wavefunction as follows:

$\begin{pmatrix}x+1&0.8 &0&0&0&0&0\\0.8&x &1&0&0&0&1\\0&1 &x&1&0&0&0\\0&0 &1&x&1&0&0\\0&0&0&1&x&1&0\\0&0&0&0&1&x&1\\0&1&0&0&0&1&x\end{pmatrix}\begin{pmatrix}c_0\\c_1\\c_2\\c_3\\c_4\\c_5\\c_6\end{pmatrix}=0$

which implies:

$(x+1)c_0+0.8c_1=0$

$0.8c_0+xc_1+c_2+c_6=0$

$c_1+xc_2+c_3=0$

$c_2+xc_3+c_4=0$

$c_3+xc_4+c_5=0$

$c_4+xc_5+c_6=0$

$c_1+c_5+xc_6=0$

$c^2_0+c^2_1+c^2_2+c^2_3+c^2_4+c^2_5+c^2_6=1$

Using an optimising tool, such as Microsoft Excel Solver, the normalised HOMO coefficients are approximately:

Coefficient	(HOMO)	HOMO electron density
c₀	0.623	0.777
c₁ (ipso)	-0.323	0.208
c₂ (ortho)	-0.344	0.236
c₃ (meta)	0.121	0.030
c₄ (para)	0.415	0.344
c₅ (meta)	0.121	0.030
c₆ (ortho)	-0.344	0.236

where the HOMO electron density (see diagram below) is given by $c_i=2[(c_{i,\pi_4})^2]$ for two electrons.

The calculated values show that the ortho and para carbons have higher electron densities than the meta carbons, with the para carbon having the highest. This is due to the +M and –I effects of the hydroxyl group. When phenol reacts with an electrophile such as bromine at low temperatures in non-polar solvents (e.g. CS₂ or CCl₄), mono-brominated products are formed, giving a para-bromophenol to ortho-bromophenol ratio of about 80:20. At higher temperatures, phenol reacts with aqueous bromine to yield the fully substituted 2,4,6-tribromophenol. Therefore, the hydroxyl group is known as an ortho/para-directing group. Substituents with similar effects include the amino group (-NH₂), substituted amino groups (-NR₂), alkoxy groups (-OR), phosphino and thio groups (-PR₂ and -SR) and the phenyl group (-C₆H₅).

Meta-directing groups

An example of a meta-directing group is the ammonium group -NH₃⁺ in the anilinium ion C₆H₅-NH₃⁺. This substituent is purely inductively withdrawing (-I) due to the positively charged nitrogen atom. To analyse the inductive effect of the ammonium group quantum mechanically, we consider the general molecular orbital (MO) wavefunction of the anilinium system:

$\psi=c_1\phi_1+c_2\phi_2+c_3\phi_3+c_4\phi_4+c_5\phi_5+c_6\phi_6$

where $c_n$ and $\phi_n$ represent the coefficients and the p atomic orbital wavefunctions respectively; $n=1$ corresponds to the substituted carbon, and $n=2$ through $n=6$ correspond to the remaining carbons in an anticlockwise direction.

As the nitrogen atom lacks a lone pair of electrons to interact with the benzene π-system, the substituted carbon atom is treated as being perturbed by the electron-withdrawing group, leading to the following secular determinant:

$\begin{vmatrix}x+1&1 &0&0&0&1\\1&x &1&0&0&0\\0&1 &x&1&0&0\\0 &0&1&x&1&0\\0&0&0&1&x&1\\1&0&0&0&1&x\end{vmatrix}=0$

Expanding the determinant yields the characteristic equation:

$x^6+x^5-6x^4-4x^3+9x^2+3x-4=0$

with solutions

$x=-2.278,-1.317,-1.000,0.705,1.000,1.891$

Substituting the solutions back into $x=\frac{\alpha-E}{\beta}$ yields:

MO	$\boldsymbol{\mathit{x}}$	$\boldsymbol{\mathit{E}}$	Type
$\pi_6^*$	1.891	$\alpha-1.891\beta$	Antibonding
$\pi_5^*$	1.000	$\alpha-\beta$	Antibonding
$\pi_4^*$	0.705	$\alpha-0.705\beta$	LUMO
$\pi_3$	-1.000	$\alpha+\beta$	HOMO
$\pi_2$	-1.317	$\alpha+1.317\beta$	Bonding
$\pi_1$	-2.278	$\alpha+2.278\beta$	Bonding

To further understand the directing effects of the ammonium group, we determine the coefficients of the HOMO wavefunction as follows:

$\begin{pmatrix}x+1&1 &0&0&0&1\\1&x &1&0&0&0\\0&1 &x&1&0&0\\0 &0&1&x&1&0\\0&0&0&1&x&1\\1&0&0&0&1&x\end{pmatrix}\begin{pmatrix}c_1\\c_2\\c_3\\c_4\\c_5\\c_6\end{pmatrix}=0$

Solving the above six simultaneous equations, along with the normalisation equation, using algebra for the HOMO coefficients and Excel Solver for the remaining coefficients, give:

Coefficient	$\boldsymbol{\mathit{\pi_1}}$	$\boldsymbol{\mathit{\pi_2}}$	$\boldsymbol{\mathit{\pi_3}}$ (HOMO)	Electron density
c₁ (ipso)	0.646	-0.517	0.000	1.370
c₂ (ortho)	0.413	-0.082	0.500	0.855
c₃ (meta)	0.295	0.409	0.500	1.008
c₄ (para)	0.259	0.621	0.000	0.904
c₅ (meta)	0.295	0.409	-0.500	1.008
c₆ (ortho)	0.413	-0.082	-0.500	0.855

where the electron density is given by $c_i=2[(c_{i,\pi_1})^2+(c_{i,\pi_2})^2+(c_{i,\pi_3})^2]$ for two electrons in each MO.

The calculated values show that the meta carbons have higher electron densities than the ortho and para carbons. When the anilinium ion reacts with an electrophile such as the nitronium ion NO₂⁺ at low temperatures, m-anilinium is the major product. Therefore, the ammonium group is known as a meta-directing group. Substituents with similar effects include the nitro group (-NO₂), sulfonyl groups (-SO₂R), the cyano group (-CN), formyl and acyl groups (-CHO and -COR) and the carboxyl group (-CO₂H).

Question

Why doesn’t the electrophile attack the ipso (latin for “itself” or “that very one”) carbon and displace the ammonium group?

Answer

Although the Hückel method indicates that the ipso carbon has the highest electron density, it considers only the π-electrons and does not account for constraints imposed by the σ-framework. Electrophilic attack at the ipso carbon would require either displacement of the ammonium group or formation of a stable σ-complex. Both pathways are energetically very unfavourable because -NH₃⁺ is a poor leaving group and the ipso carbon has already reached its maximum valency.

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