Microwave spectroscopy (Instrumentation)

Microwave spectroscopy is a powerful analytical technique that analyses the rotational transitions of molecules. Operating in the microwave region of the electromagnetic spectrum, this method provides highly precise information about molecular structure, bond lengths, bond angles, and even the distribution of electrons within a molecule.

The diagram above outlines a conventional microwave spectrometer. A stable and tunable microwave source is essential for probing molecular energy transitions. This is fulfilled by a device known as a Klystron amplifier, which is a vacuum tube engineered to generate and amplify narrow-band microwave signals that can be precisely tuned to the desired frequency. It was invented in 1937 by American electrical engineers Russell and Sigurd Varian.

A Klystron amplifier includes an electron gun, two resonant cavities (buncher and catcher), a drift tube and a collector (see diagram above). The electron gun operates on the principle of thermionic emission, where electrons are emitted from a heated coil (cathode) and accelerated towards an anode. The buncher cavity is connected to a weak external AC supply (with frequency in the microwave range), which creates an oscillating flow of free electrons within the cavity walls. This, in turn, generates an alternating electric field across the perforated section (grid) of the cavity.

As electrons pass through the grid, they are either accelerated or decelerated depending on the polarity of the electric field. When the entrance grid is negative and the exit grid is positive, electrons are accelerated. When the polarity reverses, they are decelerated. This periodic acceleration and deceleration causes the electrons to bunch together as they travel through the drift tube, forming dense clusters that correspond to the frequency of the external AC source.

When these bunched electrons enter the catcher cavity, they induce a stronger alternating current (due to their higher density) at the same frequency as the AC input. Through an inductor, this energy is converted into strong monochromatic microwave radiation, which is then propagated along a waveguide.

 

Question

Is the frequency of the generated microwave radiation tunable? How does induction occur in the catcher cavity?

Answer

The resonant cavities are precisely engineered in both shape and size to resonate at a narrow band of microwave frequencies. Each Klystron amplifier module has a narrow operational bandwidth that can be tuned by mechanically adjusting the cavity dimensions using tuning screws. To cover a wider frequency range, multiple modules with different cavity dimensions must be employed.

Electromagnetic theory states that a flowing current generates a magnetic field, and that a changing magnetic field induces an alternating current in a conductor. The varying density of electrons flowing through the catcher cavity constitutes a varying current, which in turn creates a time-varying magnetic field. This results in the induction of an alternating current within the cavity walls.

 

The waveguide, constructed using copper of stainless steel, is a hollow evacuated tube designed to confine and direct the microwaves to the sample cell, which is often a section of the waveguide. Gases to be analysed are introduced via a gas inlet system, and the cell is often equipped with pumps and valves to control pressure and flush previous samples.

If the monochromatic microwave radiation matches the energy of a specific rotational transition of the molecules, a portion of it will be absorbed by the sample. The remaining (transmitted) microwave signal then reaches the detector, which typically consists of an inductive pickup (antenna), a diode rectifier, capacitors, and an amplifier. The inductor acts as an antenna that couples the microwave radiation into an alternating current (AC), which is then rectified by the diode to produce a pulsating direct current (DC) signal. A low-pass filter, often just a capacitor, smooths the rectified signal into a more stable DC voltage. This process is repeated as the microwave frequency is swept or incremented across a range. At each frequency step, the resulting DC voltage, which is proportional to the transmitted microwave intensity, is amplified and sent to a computer for data acquisition and analysis. Finally, the amplitudes of the DC voltages are plotted as a function of frequency, producing a microwave absorption spectrum that reveals the rotational transitions of the sample.

 

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Rotational selection rules

Rotational selection rules for molecules determine the probabilities of rotational state transitions observed in spectroscopy.

According to the time-dependent perturbation theory, the transition probability between the orthogonal rotational states and within a given vibrational state of a molecule, as observed by microwave spectroscopy, is proportional to , where is the operator for the molecule’s electric dipole moment. In other words, a molecule must possess a permanent electric dipole moment () to exhibit a rotational spectrum. Homonuclear diatomic molecules and spherical rotors with no net permanent dipole moment (such as H2 and CH4 in their ground states) are generally rotationally inactive.

Since , the components of the dipole moment in polar coordinates are:

Suppose the perturbation on the molecule is caused by a plane-polarised electromagnetic wave with an electric component oscillating in the -direction. We have , with

Substituting the explicit expression of the spherical harmonics wavefunction into eq63 gives:

where , and and are functions of .

When , the integral with respect to is . When , it evaluates to . Since eq63 must be non-zero for a transition to be probable in the -direction, or

Substituting back into eq64 and noting that yields:

where .

Substituting the polar coordinate form into eq370 (a recurrence relation of the associated Legendre polynomials) results in:

Substituting eq67 into eq66 gives:

For , either integral in eq68 must be non-zero. Each integral is non-zero when the corresponding pair of spherical harmonics is not orthogonal. This occurs if and for the first integral and and  for the second integral. In other words, if

or using rotational spectroscopy notations:

For an electric component oscillating in the -direction, eq63 becomes

Substituting the explicit expression of the spherical harmonics wavefunction and into eq70 gives:

When , the integral with respect to equals zero. When , it evaluates to . Since eq71 must be non-zero for a transition to be probable in the -direction, it must satisfy:

Substituting back into eq71, and noting that , yields:

Substituting the polar coordinate form and  into eq371 (another recurrence relation of the associated Legendre polynomials) results in:

Substituting eq74 into eq73 gives:

For , either integral in eq75 must be non-zero. Each integral is non-zero when the corresponding pair of spherical harmonics is not orthogonal. This occurs if

Integrals Condition 1 Cases Condition 2 Results
1st if
if
if
if
2nd if
if
if
if

Combining the results, when and , or in rotational spectroscopy notations:

Repeating the derivation for , we arrive at the same selection rules expressed by eq76. Therefore, the rotational selection rules for polar molecules (linear rotors and spherical rotors) subjected to isotropic radiation are:

 

Question

Is eq77 applicable both to transitions between rotational levels within a single vibrational state (e.g. ) and to transitions between rotational levels of two different vibrational states (e.g. and )?

Answer

Yes, the selection rule applies to both pure rotational transitions within a vibrational state and to rotational transitions accompanying vibrational transitions (rovibrational transitions).

 

For symmetric rotors, the electric dipole moment lies along the principal molecular axis (-axis). However, there is a third quantum number to consider.  If  also lies along the principal axis, the wavefunction can be approximated as the product of three functions , where depends solely on the quantum number and  is one of three Euler angles (see diagram above). The matrix element can then be expressed as a product of three integrals over the Euler angles, with one of the integrals being . Since is independent of ,

For this integral to be non-zero, we must have . In other words, the rotational transition selection rules for symmetric rotors are:

 

Question

Is the effect of nuclear statistics on rotational states different from rotational selection rules?

Answer

The effect of nuclear statistics on rotational states is a separate, yet related, concept from the general rotational selection rules. While both influence which rotational states are observed in a spectrum, they operate based on different fundamental principles. The key distinction is that rotational selection rules dictate the possible transitions, while nuclear statistics determine the relative populations of the initial states. A transition must be both “allowed” by the selection rules and originate from a “populated” state to be observed. Therefore, nuclear statistics modify the intensity of the allowed transitions without changing the fundamental rule of which transitions are allowed.

 

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Molecular rotational energies

Molecular rotational energies are quantised states arising from rotation about principal axes, determined by the moments of inertia. These levels are fundamental in molecular spectroscopy.

 

Diatomic rigid rotor

For a diatomic rigid rotor, the rotational Hamiltonian is given by eq4a, . Here, is the square of the total angular momentum operator, with the eigenvalues (derived using orbital angular momentum ladder operators), where is the rotational quantum number and . Therefore, the rotational energy levels of a diatomic molecule are given by:

Eq44 can also be expressed in terms of the rotational constant :

where is the Planck constant and is the speed of light.

However, rotational energies are typically reported as wavenumbers in molecular spectroscopy. Substituting and into eq45 yields , which is a function of . Therefore,

This indicates that the separation between adjacent rotational levels increases linearly with (see above diagram), as given by:

Furthermore, each energy level (i.e. eq44 for a particular value of ) is -fold degenerate due to the allowed eigenvalues of the component of angular momentum, which are , where . In the absence of an external magnetic field, these sublevels are degenerate.

 

Other linear rotors

A rigid linear triatomic molecule (e.g. HCN) behaves similarly as a diatomic rigid rotor when rotating about an axis perpendicular to its bond axis. The moment of inertia about the molecular axis is essentially zero due to the absence of perpendicular mass displacement. Therefore, the rotational energy levels are, per eq44:

where is given by eq15.

Similar to diatomic molecules, each is -fold degenerate in the absence of an external magnetic field.

 

Question

Is C2H2 also a linear rotor?

Answer

Yes. Its rotational energy levels are also given by eq44 and eq46.

 

 

Symmetric rotors

Symmetric rotors have two equal moments of inertia about mutually orthogonal axes. Each of these is perpendicular to a third rotational axis associated with a distinct, non-zero moment of inertia . If , the molecule is known as a prolate symmetric rotor (shaped like a cigar) and rotates more easily around the principal axis. Examples include NH3 and CHCl3. If , the molecule is called an oblate symmetric rotor (flattened like a disc), and it rotates more easily around an axis perpendicular to the disc. Examples include C6H6 and BF3.

The rotational Hamiltonian of a rigid symmetric rotor in the molecule-fixed frame is given by eq4d:

Conventionally, and . Furthermore, from eq75, . Therefore,

With reference to eq132, and replacing the notations , and with , and , where , we have . It follows that the rotational energy levels of a rigid symmetric rotor are given by:

 

Question

Is a quantum number? Is it the same as ?

Answer

In eq48, refers to the component of the angular momentum operator along the symmetry axis (or -axis) in the molecule-fixed frame, not the -axis in the laboratory frame. Although its eigenvalues have the same form as those of in the laboratory frame, it is a different quantum number from  (or ). An analogy can be made with the projections of the spin vector of a spinning top: its projection along its own axis differs in general from the projection along an arbitrary laboratory axis, unless the axes are aligned. While both projections can take similar ranges of values (assuming these values are quantised), they are not equal in general. Likewise,  and are related to projections in different frames and are different quantum numbers.

 

Substituting and into eq49 yields:

Substituting and into eq50 gives , which is a function of two rotational quantum numbers and . Therefore,

When , there is only one distinct moment of inertia, with eq51 reducing to eq46, where . In this case, each is -fold degenerate in the absence of an external magnetic field. If , is determined by the magnitudes of and . Since is the same for and for each in eq49, the energy levels for and are degenerate. Additionally, there are possible  substates for each value of . Therefore, the total degeneracy when is in the absence of an external magnetic field.

 

Spherical rotors

A spherical rotor has three equal moments of inertia about three mutually orthogonal axes. It belongs to a point group that is a parent group of that of a symmetric rotor. Therefore, the rotational energy levels of a spherical rotor can be described by eq49 with , which is equivalent the condition in eq50. This implies that remains a function of both and , and that for a given energy level, is not restricted to just two values (as in the case of a symmetric rotor), but can take on different values. It follows that each energy level is -fold degenerate in the absence of an external magnetic field.

 

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Moments of inertia of spherical rotors

A spherical rotor is a molecule in which the moments of inertia about all three principal axes are equal. This high degree of symmetry, typically found in molecules with tetrahedral (e.g. methane, CH₄), octahedral (e.g. sulfur hexafluoride, SF₆), or icosahedral geometries, leads to simplified rotational behaviour. Unlike asymmetric or symmetric rotors, spherical rotors exhibit degenerate energy levels due to their identical rotational constants along each axis. As a result, they serve as important models in quantum mechanics and spectroscopy, particularly for interpreting rotational spectra and understanding molecular symmetry.

The left diagram above shows an octahedral molecule of the type BA6 ( symmetry), where B is the central atom. The moment of inertia along the -axis, which is equal to those along the -axis and -axis, is or

 

Question

Why do the three atoms that lie along the -axis not contribute to ?

Answer

In general, . Since the three atoms along the rotational axis, the moment of inertia about this axis is effectively zero due to the absence of perpendicular mass displacement ().

 

To determine the moment of inertia of the tetrahedral molecule around the -axis, let atom B be at the centre of a cube (see diagram above), which is designated as the origin . The four A atoms (, , and ) are located at alternating vertices, with coordinates: , , and , where  is the half the length of an edge of the cube.

Since the B-A bond length is , we have , which when substituted back into the coordinates yields: , , and . Therefore, the moment of inertia of the tetrahedral molecule around the -axis is:

 

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Moments of inertia of a trigonal planar oblate symmetric rotor with \(D_{3h}\) symmetry

The moments of inertia of a trigonal planar oblate rotor with symmetry (e.g. BF3) are characterised by a unique moment of inertia around the principal axis and two equal moments of inertia perpendicular to the principal axis, where . They are derived using simple geometric considerations.

The diagram above shows a trigonal planar oblate rotor with its centre of mass located at atom B (of mass ), which is positioned at the origin . The three A atoms (1,2 and 3), each with mass , are equally spaced at 120° apart on an imaginary circle of radius .

The moment of inertia along the -axis, which is perpendicular to the plane of the diagram, is . Since the three B-A bonds have equal lengths of , we have

 

Question

Why do  not contribute to ?

Answer

In general, . Since atom B lie along the rotational axis, the moment of inertia about this axis is effectively zero due to the absence of perpendicular mass displacement ().

 

To derive , we can make use of the coordinates of the atoms that form the base of a trigonal pyramidal molecule mentioned in an earlier article, where and :

Atom Coordinates
Trigonal pyramidal Trigonal planar
A1
A2
A3

Therefore,  or

Similarly, .

 

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Moments of inertia of octahedral prolate symmetric rotor with \(D_{4h}\) symmetry

The moments of inertia of an octahedral prolate rotor with symmetry (e.g. Pt(NH3) 2Cl4) are characterised by a unique moment of inertia around the principal axis and two equal moments of inertia perpendicular to the principal axis, where . They are derived using simple geometric considerations.

The diagram above shows an octahedral molecule of the type BA4C2, where B is the central atom. The moment of inertia along the -axis is . Since the four B-A bonds have equal lengths of , we have

 

Question

Why do and  not contribute to ?

Answer

In general, . Since atoms B and C lie along the rotational axis, the moment of inertia about this axis is effectively zero due to the absence of perpendicular mass displacement ().

 

Similarly, the moment of inertia of the molecule about the -axis, which is equal to that about the -axis, is . Since the two B-C bonds have equal lengths of , we have

You’ll find that the expression for the moment of inertia about the -axis is the same as eq36.

 

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Moments of inertia of tetrahedral prolate symmetric rotor with \(C_{3v}\) symmetry

The moments of inertia of a tetrahedral prolate rotor with symmetry (e.g. CHCl3) are characterised by a unique moment of inertia around the principal axis and two equal moments of inertia perpendicular to the principal axis, where . They are derived using simple geometric considerations.

Since the rotation about the axis (-axis) of the molecule has the same symmetry as a trigonal pyramidal molecule, the moment of inertia for symmetric rotors like CHCl3 along the -axis is given by eq22:

To derive the expression for , place the origin at the centre of mass of the molecule. The -coordinates of atom C, atom B and each of the A atoms are , and respectively. Then, the centre of mass of the molecule satisfies or

with

Using the and  coordinates of the three A atoms defined in the previous article, the positions of the atoms are:

Atom Coordinates
C
B
A (point F)
A (point I)
A (point E)

Since , the moment of inertia of the molecule about the -axis is . Substituting the data from the above table and eq23 into this equation gives:

You’ll find that substituting the data from the above table into the expression for the moment of inertia of the molecule about the -axis , and then substituting eq23 into the resultant equation, yields the same expression as eq33. Therefore, eq22 and eq33 represent two distinct moments of inertia of a tetrahedral prolate rotor with symmetry.

 

Question

Does eq33 apply to CH3Cl, where the centre of mass of the molecule is between C and Cl?

Answer

Yes it does. Eq33 applies to any tetrahedral prolate rotor with symmetry. You can convince yourself by deriving the expression using the geometry of CH3Cl.

 

 

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Moments of inertia of trigonal pyramidal prolate symmetric rotor with \(C_{3v}\) symmetry

The moments of inertia of a trigonal pyramidal prolate rotor with symmetry (e.g. NH3) are characterised by a unique moment of inertia around the rotational axis (-axis) and two equal moments of inertia perpendicular to the axis, where . They are derived using simple geometric considerations.

The structure on the right of the above diagram illustrates the geometry of the central atom B (with mass ) and two of the three A atoms, each with mass . Let be the angle between an A-B-A bond (between FG and GI), be the angle between a B-A bond and the -axis (GH) and be the length of each B-A bond. According to the VSEPR theory, .

Applying the cosine rule on gives . Since , we have

Using the cosine rule again on and noting that yields . Substituting eq20 into this equation results in

Hence, the moment of inertia for symmetric rotors like NH3 along the -axis is:

To derive , we begin by noting that the centre of mass of the molecule (green sphere) lies along the -axis (one of three principal axes of rotation), between atom B and the plane formed by the three A atoms (see diagram below). The -axis and -axis are orthogonal to the -axis and to each other. They intersect at the centre of mass but do not intersect any of the B-A bonds.

The relationship between the angles and is established by letting atom B be the origin and the three A atoms at angles 0°, 120°, 240° around the -axis. Furthermore, let be the unit vector pointing from atom B to atom A at F, and be the unit vector pointing from atom B to atom A at I (see diagram below).

The dot product of the two vectors is given by or , which is equivalent to

Importantly, eq23 is independent of the choice of origin or reference frame even though it was derived with atom B as the origin.

Now, let’s place the origin at the centre of mass of the molecule. The -coordinates of atom B and each of the A atoms are and respectively. Then, the centre of mass of the molecule satisfies or

with  or

Since the position of the centre of mass relative to that of atom B only involves a shift in the -direction, the and coordinates of the three A atoms are those defined by the unit vectors above multiplied by the factor . Therefore, the positions of the atoms are:

Atom

Coordinates

B

A (point F)

A (point I)

A (point E)

 

Question

Derive the and  coordinates of atom A at point E.

Answer

With reference to the diagram below, we have .

 

Since , the moment of inertia of the molecule about the -axis is:

Substituting the data from the above table into eq26 and simplifying gives:

Substituting eq23 into eq27 and simplifying yields:

You’ll find that substituting the data in the above table into the moment of inertia of the molecule about the -axis results in the same expression as eq28. Therefore, eq22 and eq28 are the two distinct moments of inertia of a trigonal pyramidal prolate rotor with symmetry.

 

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Moments of inertia of symmetric rotors

The moments of inertia of symmetric rotors play a fundamental role in understanding their rotational dynamics and spectroscopic behaviour.

In such molecules, two of the three moments of inertia are equal, while the third is different. The unique moment of inertia, denoted by , describes rotation around an axis known as the principal axis. The other two, which are equal, are denoted by . If , the molecule is known as a prolate symmetric rotor (shaped like a cigar) and rotates more easily around the principal axis. Examples include NH3 and CHCl3. If , the molecule is called an oblate symmetric rotor (flattened like a disc), and it rotates more easily around an axis perpendicular to the disc. Examples include C6H6 and BF3.

 

Question

Is a trans-MA2B4 complex a prolate or an oblate symmetric rotor?

Answer

It is a prolate rotor if the axial atoms (A) are farther from the central metal M than the equatorial atoms (B), and an oblate rotor if the axial atoms are closer to M than the equatorial atoms.

 

In the next few articles, we will derive the moments of inertia of a few common prolate and oblate symmetric rotors.

 

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Moments of inertia of triatomic linear molecules

The moments of inertia of triatomic linear molecules play a fundamental role in understanding their rotational dynamics and spectroscopic behaviour.

In such molecules — composed of three atoms aligned along a straight line — the distribution of mass relative to the molecular axis determines how the molecule responds to rotational motion. Since all atoms lie along the same axis, the moment of inertia about this axis is effectively zero due to the absence of perpendicular mass displacement. However, the moment of inertia about axes perpendicular to the molecular axis becomes significant and is central to characterising the molecule’s rotational energy levels.

Consider a linear molecule with three atoms A, B and C with masses , and respectively, and bond lengths and (see diagram above). The molecule rotates the centre of mass axis , which passes through the centre of mass of the molecule. If , and are the distances between the centre of mass and the respective atoms, then the parallel axis theorem states that:

where is the moment of inertia about the axis , is the moment of inertia about the axis ,  and .

Substituting in eq11 gives

Let’s assume that the molecular axis is the -axis with the origin at B. By definition, the center of mass with respect to the -coordinate satisfies , which rearranges to:

Substituting eq13 in eq12 yields

Eq14 is the moment of inertia of a triatomic linear molecule where . If , then and eq14 becomes

 

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