Advanced Chemistry - Mono Mole

August 1, 2025October 26, 2025

Centrifugal distortion

Centrifugal distortion refers to the deformation of a molecule’s structure that occurs due to the rotational motion of the molecule.

As a molecule rotates, the centrifugal force, which acts outward from the axis of rotation, causes the molecule’s bond lengths and bond angles to stretch and distort, particularly at higher rotational speeds. This effect is most pronounced in larger, heavier molecules, where the rotational energy is sufficient to cause a noticeable shift in the geometry. Centrifugal distortion is an important factor in molecular spectroscopy, as it can influence the rotational energy levels of a molecule, leading to shifts in the observed spectral lines, and must be accounted for when interpreting high-resolution rotational spectra.

What is centrifugal force, and how is it different from centripetal force?

Centripetal force and centrifugal force are both associated with circular motion, but they are described in different frames of reference. Centripetal force is described in an inertial frame of reference or in a frame where the object is experiencing actual physical forces that are measurable and observable. In the inertial frame, the force that keeps the object in circular motion is real and necessary to counteract the object’s tendency to move in a straight line (due to inertia).

Consider a particle moving around a circle (see diagram above) of radius $r$ with an instantaneous velocity $v_1$ at $t=t$ and $v_2$ at $t=t'$ , where $\vert v_1\vert=\vert v_2\vert=v$ . If $\Delta\theta$ is small, the arc length $\Delta s$ is approximately equal to $\Delta r$ and hence to $\Delta v$ . Since the ratio of corresponding sides of two similar triangles is always equal, $\frac{\Delta v}{v}=\frac{\Delta s}{r}$ . Dividing both sides of this equation by $\Delta t$ and rearranging (noting that $a=\frac{\Delta v}{\Delta t}$ and $v=\frac{\Delta s}{\Delta t}$ ) gives:

$a=\frac{v^2}{r}\;\;\;\;\;\;\;\;85$

Substituting $v=r\omega$ (see this article for details) into eq85 yields:

$a=r\omega^2\;\;\;\;\;\;\;\;86$

Therefore the centripetal force $F$ acting on the rotating particle of reduced mass $\mu$ is

$F=\mu r\omega^2\;\;\;\;\;\;\;\;87$

Centrifugal force, on the other hand, is experienced in a non-inertia frame of reference (rotating frame). It appears to act outward on the rotating particle. However, it does not correspond to any real physical force but is instead a fictitious force introduced to account for the observed effects of acceleration in the rotating frame. Since the magnitude of centrifugal force is the same as that of the centripetal force, the magnitude of the centrifugal force experienced by a diatomic molecule of reduced mass $\mu$ and bond length $r$ rotating at angular velocity $\omega$ is $\mu r\omega^2$ .

Question

Explain how the fictitious centrifugal force accounts for the observed effects of acceleration in the rotating frame?

Answer

Consider a ball tied to a string, rotating in a horizontal plane. In the inertia frame of reference, the only force acting on the ball is the centripetal force, directed towards the axis of rotation. In the rotating frame of reference, the centripetal force still acts on the ball, even though the ball appears stationary. This would imply that the ball should accelerate towards the centre, contradicting the observation that it remains at rest in the rotating frame. Therefore, a fictitious force, equal in magnitude to the centripetal force but acting in the opposite direction, must be introduced to restore consistency with Newton’s second law.

At equilibrium, the centrifugal force (or the centripetal force) equals the restoring force given by Hooke’s law:

$\mu r\omega^2=k(r-r_0)\;\;\;\;\;\;\;\;88$

where $k$ is the force constant and $r_0$ is the equilibrium bond length.

For small displacements, the motion expressed by eq88 is simple harmonic, with the following classical Hamiltonian for the system:

$H=\frac{J^2}{2\mu r^2}+\frac{1}{2}k(r-r_0)^2\;\;\;\;\;\;\;\;89$

where the kinetic energy term is given by the classical form of eq4a and the potential energy term is derived from this article.

Rearranging eq88 results in $r=\frac{1}{1-\mu\omega^2/k}r_0$ , and therefore

$\frac{1}{r^2}=\frac{1}{r_0^2}-\frac{2\mu\omega^2}{kr^2_0}+\frac{\mu^2\omega^4}{k^2r^2_0}\;\;\;\;\;\;\;\;90$

For small displacements, $r\approx r_0$ , which implies from $r=\frac{1}{1-\mu\omega^2/k}r_0$ that $1-\frac{\mu\omega^2}{k}\approx 1$ , and hence $\vert\frac{\mu\omega^2}{k}\vert\ll 1$ . So, we can ignore the last term in eq90 as a first approximation, giving:

$\frac{1}{r^2}=\frac{1}{r_0^2}-\frac{2\mu\omega^2}{kr^2_0}\;\;\;\;\;\;\;\;91$

Substituting $J=I\omega=\mu r^2\omega$ (see this article for details) into RHS of eq91 gives:

$\frac{1}{r^2}=\frac{1}{r_0^2}-\frac{2J^2}{k\mu r^4r^2_0}\;\;\;\;\;\;\;\;92$

Substituting eq92 and eq88 into eq89 yields:

$H=\frac{J^2}{2\mu r_0^2}-\frac{J^4}{k\mu^2r^4r_0^2}+\frac{J^4}{2k\mu^2 r^6}\;\;\;\;\;\;\;\;93$

Since $r\approx r_0$ for small displacements, eq93 becomes:

$H=\frac{J^2}{2\mu r_0^2}-\frac{J^4}{2k\mu^2r_0^6}$

Using the quantum mechanics postulate that “to every observable quantity in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics”, and replacing the classical angular momentum $J$ with the quantum mechanical operator gives

$\hat{H}=\frac{\hat{J}^2}{2\mu r_0^2}-\frac{\hat{J}^4}{2k\mu^2r_0^6}\;\;\;\;\;\;\;\;94$

The corresponding eigenvalues of eq94 (derived using orbital angular momentum ladder operators) are:

$E=\frac{J(J+1)\hbar^2}{2\mu r_0^2}-\frac{J^2(J+1)^2\hbar^4}{2k\mu^2r_0^6}$

$E=hcBJ(J+1)-hcDJ^2(J+1)^2\;\;\;\;\;\;\;\;95$

where $B=\frac{\hbar}{4\pi c\mu r_0^2}$ is the rotational constant and $D=\frac{\hbar^3}{4\pi ck\mu^2r_0^6}$ is the centrifugal distortion constant.

However, rotational energies are typically reported as wavenumbers in molecular spectroscopy. Substituting $E=h\nu$ and $c=\nu\lambda$ into eq95 yields $\tilde{\nu}=BJ(J+1)-DJ^2(J+1)^2$ , which is a function of $J$ . Therefore,

$F(J)=BJ(J+1)-DJ^2(J+1)^2\;\;\;\;\;\;\;\;96$

Question

Can centrifugal distortion be observed from an inertia frame?

Answer

Yes, centrifugal distortion can indeed be observed from an inertial frame, though it’s important to clarify that there are no “fictitious forces” like centrifugal force in an inertial frame. However, centrifugal distortion is not a fictitious effect. It is a real, physical phenomenon that arises due to the rotation of the molecule, and this can be observed in an inertial reference frame.

In a non-rotating molecule, the bonds between atoms are at their equilibrium lengths, representing the most stable configuration where the bond is neither stretched nor compressed. When the molecule begins to rotate, each atom wants to move in a straight line (due to its inertia) tangentially away from the center of rotation. To keep the atoms moving in a circular path, the chemical bond must provide an inward pulling force (the centripetal force).

Just like a spring, a chemical bond can only exert a restoring force when it is displaced from its equilibrium position. To generate a small centripetal force (restoring force) during low-speed rotation, the bond needs to stretch only a little. For faster rotation, the bond must stretch more to provide a larger centripetal force. This stretching of the bond beyond its equilibrium length is the physical phenomenon of centrifugal distortion, which can be observed and measured in an inertia frame.

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August 1, 2025February 10, 2026

Microwave spectra

Microwave spectra represent the electromagnetic radiation absorbed or emitted by molecules when they undergo transitions between discrete rotational energy levels. These spectra are a key component in the study of molecular structure and dynamics. They provide valuable insights about molecular properties, such as bond lengths and moments of inertia.

An ideal, pure absorption rotational spectrum of a rigid linear molecule, generated in the microwave frequency range of 3 to 600 GHz (1 to 200 cm^-1), typically exhibits a series of absorption lines that correspond to the rotational transitions between discrete rotational energy levels (see diagram below). In the absence of fine or hyperfine splitting, these transitions follow the selection rule $\Delta J=+1$ , meaning that a molecule can only transition from one rotational level to the next higher level.

Question

What is the difference between an absorption spectrum and an emission spectrum?

Answer

An external radiation is required to produce an absorption spectrum. The detector measures the amount of energy absorbed by the sample at each frequency, corresponding to transitions with $\Delta J=+1$ .

In an emission spectrum, the sample itself acts as the source. It must first be excited to higher-energy rotational states, for example by heating or applying an electrical discharge. The detector then measures the amount of energy emitted by the sample, corresponding to transitions with $\Delta J=-1$ .

Both types of spectra can be obtained using the same spectrometer, with different settings. Suppose the spectrometer is configured to record the absorption spectrum of a sample. If two rotational states are populated under the experimental conditions, the sample can both absorb and emit radiation at the same frequency, since both processes involve transitions with $\Delta J=\pm 1$ between the same pair of quantised energy levels. However, the conditions for measuring an absorption spectrum are typically set to ensure that net absorption occurs.

The rotational energy levels are given by eq44 or eq45 and the frequencies of the transitions are directly related to the rotational constant $B$ , and as such, the spacing between the spectral lines provides information about the moment of inertia and the molecular structure. Using the definition $\tilde{\nu}=BJ(J+1)$ , we have

$\Delta\tilde{\nu}=B[(J+1)(J+2)-J(J+1)]=2B(J+1)\;\;\;\;\;\;\;\;80$

Therefore, the first peak, obtained by substituting $J=0$ into eq80, lies at the point $2B$ along the horizontal axis. Similarly, the second and third peaks are at $4B$ and $6B$ respectively. In other words, the peaks in an ideal rotational spectrum (without centrifugal distortion) of a rigid linear rotor are equally spaced at $2B$ .

Question

Calculate the moment of inertia and bond length of HCl if the line spacing in the rotational spectrum is 21.2 cm^-1?

Answer

Since the spacing is $2B$ , we have $B=10.6$ cm^-1. Using the formula $B=\frac{\hbar}{4\pi cI}$ gives $I=2.64\times 10^{-47}$ kg m². Furthermore, $I=\mu r^2=\biggr$\frac{35.5}{1+35.5}\biggr$\frac{1\times 10^{-3}}{N_A}r^2$ . Therefore, the bond length is $r=1.28\times 10^{-10}$ m.

The intensity of each transition $J\rightarrow J+1$ depends on the population of the initial level $J$ , which is governed by the Boltzmann distribution, with higher- $J$ states being less populated at lower temperatures:

$\frac{n_J}{N}=\frac{e^{-\frac{\varepsilon_J}{kT}}}{\sum_Je^{-\frac{\varepsilon_J}{kT}}}$

where

$k$ is the Boltzmann constant.
$n_J$ is the number of particles in the energy state $\varepsilon_J$ .
$N$ is the total number of particles in the system.

Since $\frac{n_j}{N}$ represents the fraction of particles in the state $\varepsilon_J$ , the intensity of a spectral line is proportional to $e^{-\frac{\varepsilon_J}{kT}}$ , where the value $\varepsilon_J$ corresponds to a specific energy level. However, there are $(2J+1)$ degenerate states associated with each value of $J$ in a rigid linear molecule. This degeneracy increases the statistical weight of higher $J$ levels, redistributing the population towards more degenerate states, which become thermally accessible at typical laboratory temperatures. In other words, at a given temperature, more particles will occupy a higher $J$ state with greater degeneracy at equilibrium than they would if the state were non-degenerate. Therefore, when a sample is exposed to an external microwave field, the intensity $I$ of each allowed transition in the pure rotational spectrum is proportional to the population of the initial state, which is the product of $e^{-\frac{\varepsilon_J}{kT}}$ and $(2J+1)$ . This gives:

$I=N'(2J+1)e^{-\frac{hcBJ(J+1)}{kT}}\;\;\;\;\;\;\;\;81$

where $N'$ is a proportionality constant and $\varepsilon_J$ is given by eq45.

The factor $(2J+1)$ in eq81 increases linearly with $J$ , while the exponential term decreases exponentially. As a result, the graph of $I$ versus $J$ forms a skewed bell curve (see diagram above), with a maximum found by treating $J$ as a continuous variable and taking the derivative of $I$ with respect to $J$ :

$\frac{dI}{dJ}=N'e^{-\frac{hcBJ(J+1)}{kT}}\biggr\[2-(2J+1)^2\frac{hcB}{kT}\biggr\]$

Setting $\frac{dI}{dJ}=0$ , and noting that $e^{-\frac{hcBJ(J+1)}{kT}}\neq 0$ for any finite $J$ , gives

$J=\sqrt{\frac{kT}{2hcB}}-\frac{1}{2}\;\;\;\;\;\;\;\;82$

Since $J$ must be an integer, the maximum intensity occurs at the value of $J$ closest to the result of eq82.

Question

Is the intensity of each transition dependent on the electric dipole moment?

Answer

Although the transition probability between adjacent rotational states is proportional to $\vert\langle\psi(\theta,\phi)_{l',m'_l}\vert\boldsymbol{\mathit{\hat{\mu}}}\vert\psi(\theta,\phi)_{l,m_l}\rangle\vert^2$ , which depends on the magnitude of the electric dipole moment, the sample molecules in the waveguide of a typical microwave spectrometer are randomly oriented. As a result, the effect transition dipole moment does not vary significantly from one $J\rightarrow J+1$ transition to the next for a given molecule. Therefore, while the absolute intensity depends on the dipole moment, it is often treated as a constant factor when comparing transitions within the same molecule.

For a symmetric rotor, $\tilde{\nu}=\tilde{B}J(J+1)+K^2(\tilde{A}-\tilde{B})$ and $\Delta\tilde{\nu}$ is also given by eq80. However, unlike a linear rotor, each allowed transition receives contributions from multiple $K$ states. These overlapping contributions increase the intensity of the lines but do not change their positions for an ideal spectrum. Consequently, the overall intensity envelope of the spectrum still resembles a skewed bell curve, with a maximum at the value of $J$ corresponding to the highest population.

In practice, the peaks in a real rotational spectrum are not discrete vertical lines but have finite width and shape, appearing as broadened curves. This broadening arises, under typical laboratory conditions, from several effects:

- Doppler broadening, due to the thermal motion of molecules, causes a spread in observed frequencies as molecules move towards or away from the detector.
- Instrumental broadening, resulting from the finite resolution of the spectrometer itself.
- Lifetime broadening, which is based on the energy-time uncertainty relation.

Beyond broadening, other phenomena affect the detailed structure and spacing of the spectral lines:

- Centrifugal distortion causes deviations from the ideal rigid rotor model. As rotational speed increases at higher $J$ levels, the molecular bond stretches slightly, increasing the moment of inertia and decreasing the rotational constant $B$ . This leads to uneven spacing between lines, especially at higher $J$ , and is accounted for using a correction term.

- Isotopic substitution alters the moment of inertia due to changes in atomic mass, thereby changing the rotational constant $B$ . Different isotopologues of the same molecule produce distinct sets of rotational lines, each with slightly different spacings. If multiple isotopes are present in the sample, the resulting spectrum may show clusters or duplications of lines, corresponding to each isotopologue.

Together, these effects lead to a rotational spectrum that is richer and more complex than the idealised, evenly spaced series of lines predicted by the rigid rotor model.

Question

Do large molecules, such as polymers, enzymes and DNA, undergo rotational transitions when radiated by microwaves?

Answer

Yes, large molecules can undergo rotational transitions when exposed to microwave radiation. However, such molecules are flexible and do not behave like rigid rotors, which invalidates some of the assumptions used in microwave rotational spectroscopy. If we assume that $\Delta \tilde\nu=2B(J+1)$ still applies, the rotational energy levels will be very closely spaced because the moment of inertia $I=\mu r^2$ is large, and $B=\frac{\hbar}{4\pi cI}$ becomes very small. As a result, individual rotational transitions for large molecules often overlap, leading to broad, unresolved bands or a complete loss of identifiable peaks.

Furthermore, instead of undergoing uniform rotation, different parts of these complex molecules can move independently in response to the microwave field, including side-chain rotations and wobbling. These movements cause the molecules to collide with neighbouring molecules, converting the absorbed microwave energy into kinetic energy and heat. This general heating effect may denature the large molecules, leading to further changes in their structure and, consequently, variations in the observed spectra.

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