Rotational and Vibrational Spectra Questions and Answers

Q: The expression for the second overtone frequency in the vibrational absorption spectra of a diatomic molecule in terms of the harmonic frequency \(\omega_{e}\) and anharmonicity constant \(x_{e}\) is

The second overtone frequency is \(3\omega_{e}(1 - 4x_{e})\).The anharmonic oscillator levels are \(G(v) = \omega_{e}(v + \tfrac{1}{2}) - \omega_{e}x_{e}(v + \tfrac{1}{2})^{2}\). The second overtone is the transition \(v = 0 \to 3\):\[ \begin{aligned} G(3) - G(0) &= 3\omega_{e} - \omega_{e}x_{e}\left(\tfrac{49}{4} - \tfrac{1}{4}\right) \\ &= 3\omega_{e} - 12\omega_{e}x_{e} = 3\omega_{e}(1 - 4x_{e}) \end{aligned} \]The fundamental (\(0 \to 1\)) is \(\omega_{e}(1 - 2x_{e})\) and the first overtone (\(0 \to 2\)) is \(2\omega_{e}(1 - 3x_{e})\), which is the distractor for those who count the fundamental as the first overtone. \(3\omega_{e}(1 - 2x_{e})\) multiplies the fundamental by three, ignoring the growing anharmonic correction, and \(2\omega_{e}(1 - x_{e})\) fits no transition.

Q: It is given that the electronic ground state of a diatomic molecule \(X_{2}\) has even parity and the nuclear spin of \(X\) is 0. Which one of the following is the CORRECT statement with regard to the rotational Raman spectrum (\(J\) is the rotational quantum number) of this molecule?

Only lines of even \(J\) values are present.The nuclei have spin 0, so they are bosons and the total molecular wavefunction must be symmetric when the two nuclei are exchanged. With nuclear spin 0 the nuclear-spin part is symmetric, and the electronic ground state is even, so the rotational part must be symmetric too. The rotational wavefunction has exchange symmetry \((-1)^{J}\), so only even \(J\) levels exist.Rotational Raman lines obey \(\Delta J = 0, \pm 2\), which links even levels only to even levels, so every line comes from an even \(J\) and the odd-\(J\) lines are missing altogether, as in the Raman spectrum of CO2, whose oxygen nuclei also have spin 0.Lines of all \(J\) would need a heteronuclear molecule. The 3 : 1 intensity alternation belongs to spin-1/2 nuclei such as in H2. Only odd \(J\) would require the opposite electronic or nuclear symmetry.

Q: The rotational spectrum of a molecule is sensitive to isotopic substitution of atoms in the molecule. If the ratio of the rotational constant B′ of 13C16O to the constant B of 12C16O is 0·956; and if the first rotational line for 12C16O is observed at 3·84 cm−1, that for 13C16O will be observed at :

The line for \( ^{13}\text{C}^{16}\text{O} \) is at 3·67 cm−1.The first rotational line, \( J = 0 \to 1 \), lies at \( 2B \), so it scales directly with the rotational constant:\[ \bar{\nu}' = 3.84 \times 0.956 = 3.67\ \text{cm}^{-1} \]The heavier \( ^{13}\text{C} \) raises the moment of inertia and lowers B, so the line must move to a lower wavenumber. 4·79 and 3·87 cm−1 move it up, and 2·89 cm−1 is far too large a shift for a ratio of 0·956.

Q: A typical PR contour for vibration-rotational spectrum for CO molecule shows \(\Delta\bar{\nu}\) = 55 cm\(^{-1}\). The associated value of rotational constant B is (Given; Boltzmann const. k = 1.38 × 10\(^{-23}\) J/K, T = 300, h = 6.6 × 10\(^{-34}\) J-s) :

The rotational constant is about 1.8 cm\(^{-1}\).In a vibration-rotation band the most intense lines of the P and R branches come from \(J_{max} \approx \sqrt{kT/2hcB}\), and the separation between the two maxima is about \(4BJ_{max}\):\[ \begin{aligned} \Delta\bar{\nu} &= \sqrt{\frac{8BkT}{hc}} \\ B &= \frac{hc(\Delta\bar{\nu})^{2}}{8kT} \end{aligned} \]With c in cm/s:\[ \begin{aligned} \frac{hc}{kT} &= \frac{(6.6 \times 10^{-34})(3 \times 10^{10})}{1.38 \times 10^{-23} \times 300} \\ &\approx 4.78 \times 10^{-3}\ \text{cm} \end{aligned} \]\[ \begin{aligned} B &= \frac{4.78 \times 10^{-3} \times 55^{2}}{8} \\ &= \frac{14.5}{8} \approx 1.8\ \text{cm}^{-1} \end{aligned} \]This is close to the known value for CO (about 1.9 cm\(^{-1}\)). The other options do not follow from the given separation.

Q: Which of the following is not correct (for number of fundamental modes of vibration) ?

The incorrect statement is that a non-linear molecule has 3N-3 modes.A molecule of N atoms has 3N degrees of freedom. Three are translations. A non-linear molecule also has three rotations, leaving 3N - 6 vibrations; a linear molecule has only two rotations (rotation about its axis does not count), leaving 3N - 5.3N - 3 subtracts only the translations and forgets the rotations. A spherical top such as CH4 is non-linear, so 3N - 6 applies to it, as option D says.

Q: The rotational spectrum of 12C16O and 13C16O provide the values of rotational constants B and B′ respectively. If B is 1.92 cm-1, what is the value of B′ ?(h = 6.6 × 10-34 Js)

The rotational constant of \(^{13}C^{16}O\) is B′ = 1.83 cm\(^{-1}\).The bond length is the same for both isotopes, so \(B = \frac{h}{8\pi^{2}c\mu r^{2}}\) is inversely proportional to the reduced mass \(\mu\):\[ \begin{aligned} \mu &= \frac{12 \times 16}{28} = 6.857 \\ \mu' &= \frac{13 \times 16}{29} = 7.172 \\ B' &= B\frac{\mu}{\mu'} = 1.92 \times 0.956 \\ &= 1.836\ \text{cm}^{-1} \end{aligned} \]The heavier isotope has the larger moment of inertia and so the smaller B, which rules out 2.12 cm\(^{-1}\) and 4.24 cm\(^{-1}\). 1.06 cm\(^{-1}\) is far too large a change for one extra neutron. (The value of h is not needed for a ratio.)

Q: If the average separation between the adjacent lines in the rotational IR spectrum of CO molecule is 3.844 cm\(^{-1}\), then the value of rotational constant is :

The rotational constant is \(B = 1.922\) cm\(^{-1}\).For a rigid rotor the rotational levels are \(F(J) = BJ(J + 1)\), and the allowed transitions \(J \rightarrow J + 1\) give lines at\[ \bar{\nu} = 2B(J + 1) \]so adjacent lines are separated by 2B. Hence\[ B = \frac{3.844}{2} = 1.922\ \text{cm}^{-1} \]Taking the spacing itself as B gives 3.844, twice the spacing gives 7.688, and 5.766 is 3B; none is the rotational constant.

The expression for the second overtone frequency in the vibrational absorption spectra of a diatomic molecule in terms of the harmonic frequency \(\omega_{e}\) and anharmonicity constant \(x_{e}\) is

A.\(2\omega_{e}(1 - x_{e})\)
B.\(2\omega_{e}(1 - 3x_{e})\)
C.\(3\omega_{e}(1 - 2x_{e})\)
D.\(3\omega_{e}(1 - 4x_{e})\)

Exam Year(s)

GATE Physics - 2018,

Answer: D

The second overtone frequency is \(3\omega_{e}(1 - 4x_{e})\).

The anharmonic oscillator levels are \(G(v) = \omega_{e}(v + \tfrac{1}{2}) - \omega_{e}x_{e}(v + \tfrac{1}{2})^{2}\). The second overtone is the transition \(v = 0 \to 3\):

\[ \begin{aligned} G(3) - G(0) &= 3\omega_{e} - \omega_{e}x_{e}\left(\tfrac{49}{4} - \tfrac{1}{4}\right) \\ &= 3\omega_{e} - 12\omega_{e}x_{e} = 3\omega_{e}(1 - 4x_{e}) \end{aligned} \]

The fundamental (\(0 \to 1\)) is \(\omega_{e}(1 - 2x_{e})\) and the first overtone (\(0 \to 2\)) is \(2\omega_{e}(1 - 3x_{e})\), which is the distractor for those who count the fundamental as the first overtone. \(3\omega_{e}(1 - 2x_{e})\) multiplies the fundamental by three, ignoring the growing anharmonic correction, and \(2\omega_{e}(1 - x_{e})\) fits no transition.

Discuss in Forum

It is given that the electronic ground state of a diatomic molecule \(X_{2}\) has even parity and the nuclear spin of \(X\) is 0. Which one of the following is the CORRECT statement with regard to the rotational Raman spectrum (\(J\) is the rotational quantum number) of this molecule?

A.Lines of all \(J\) values are present
B.Lines have alternating intensity in the ratio of 3: 1
C.Lines of only even \(J\) values are present
D.Lines of only odd \(J\) values are present

Exam Year(s)

GATE Physics - 2023,

Answer: C

Only lines of even \(J\) values are present.

The nuclei have spin 0, so they are bosons and the total molecular wavefunction must be symmetric when the two nuclei are exchanged. With nuclear spin 0 the nuclear-spin part is symmetric, and the electronic ground state is even, so the rotational part must be symmetric too. The rotational wavefunction has exchange symmetry \((-1)^{J}\), so only even \(J\) levels exist.

Rotational Raman lines obey \(\Delta J = 0, \pm 2\), which links even levels only to even levels, so every line comes from an even \(J\) and the odd-\(J\) lines are missing altogether, as in the Raman spectrum of CO₂, whose oxygen nuclei also have spin 0.

Lines of all \(J\) would need a heteronuclear molecule. The 3 : 1 intensity alternation belongs to spin-1/2 nuclei such as in H₂. Only odd \(J\) would require the opposite electronic or nuclear symmetry.

Discuss in Forum

The mean distance between the two atoms of HD molecule is \(r\), where H and D denote hydrogen and deuterium, respectively. The mass of the hydrogen atom is \(m_{H}\). The energy difference between two lowest lying rotational states of HD in multiples of \(\hbar^{2}/(m_{H}r^{2})\) is

A.\(\dfrac{3}{2}\)
B.\(\dfrac{2}{3}\)
C.6
D.\(\dfrac{4}{3}\)

Exam Year(s)

GATE Physics - 2024,

Answer: A

The energy difference is \(\dfrac{3}{2}\) in units of \(\hbar^{2}/(m_{H}r^{2})\).

Rotational levels are E_J = ℏ²J(J + 1)/(2μr²). With m_D = 2m_H, the reduced mass of HD is

\[ \begin{aligned} \mu &= \frac{m_{H} \cdot 2m_{H}}{3m_{H}} = \tfrac{2}{3}m_{H} \\ E_{1} - E_{0} &= \frac{2\hbar^{2}}{2\mu r^{2}} = \frac{3}{2}\,\frac{\hbar^{2}}{m_{H}r^{2}} \end{aligned} \]

2/3 is the reduced mass ratio itself, not the energy. 6 and 4/3 come from using J(J + 1) = 6 for the J = 2 level, or from treating the molecule with the full mass instead of the reduced mass.

Discuss in Forum

The rotational spectrum of a molecule is sensitive to isotopic substitution of atoms in the molecule. If the ratio of the rotational constant B′ of ¹³C¹⁶O to the constant B of ¹²C¹⁶O is 0·956; and if the first rotational line for ¹²C¹⁶O is observed at 3·84 cm⁻¹, that for ¹³C¹⁶O will be observed at :

A.4·79 \( \text{cm}^{-1} \)
B.2·89 \( \text{cm}^{-1} \)
C.3·87 \( \text{cm}^{-1} \)
D.3·67 \( \text{cm}^{-1} \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The line for \( ^{13}\text{C}^{16}\text{O} \) is at 3·67 cm⁻¹.

The first rotational line, \( J = 0 \to 1 \), lies at \( 2B \), so it scales directly with the rotational constant:

\[ \bar{\nu}' = 3.84 \times 0.956 = 3.67\ \text{cm}^{-1} \]

The heavier \( ^{13}\text{C} \) raises the moment of inertia and lowers B, so the line must move to a lower wavenumber. 4·79 and 3·87 cm⁻¹ move it up, and 2·89 cm⁻¹ is far too large a shift for a ratio of 0·956.

Discuss in Forum

A typical PR contour for vibration-rotational spectrum for CO molecule shows \(\Delta\bar{\nu}\) = 55 cm\(^{-1}\). The associated value of rotational constant B is (Given; Boltzmann const. k = 1.38 × 10\(^{-23}\) J/K, T = 300, h = 6.6 × 10\(^{-34}\) J-s) :

A.1.8 cm\(^{-1}\)
B.1.6 cm\(^{-1}\)
C.2.0 cm\(^{-1}\)
D.1.4 cm\(^{-1}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The rotational constant is about 1.8 cm\(^{-1}\).

In a vibration-rotation band the most intense lines of the P and R branches come from \(J_{max} \approx \sqrt{kT/2hcB}\), and the separation between the two maxima is about \(4BJ_{max}\):

\[ \begin{aligned} \Delta\bar{\nu} &= \sqrt{\frac{8BkT}{hc}} \\ B &= \frac{hc(\Delta\bar{\nu})^{2}}{8kT} \end{aligned} \]

With c in cm/s:

\[ \begin{aligned} \frac{hc}{kT} &= \frac{(6.6 \times 10^{-34})(3 \times 10^{10})}{1.38 \times 10^{-23} \times 300} \\ &\approx 4.78 \times 10^{-3}\ \text{cm} \end{aligned} \]

\[ \begin{aligned} B &= \frac{4.78 \times 10^{-3} \times 55^{2}}{8} \\ &= \frac{14.5}{8} \approx 1.8\ \text{cm}^{-1} \end{aligned} \]

This is close to the known value for CO (about 1.9 cm\(^{-1}\)). The other options do not follow from the given separation.

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Which of the following is not correct (for number of fundamental modes of vibration) ?

A.Non-linear molecule has 3N-6 modes
B.Non-linear molecule has 3N-3 modes
C.Linear molecule has 3N-5 modes
D.Spherical top molecule has 3N-6 modes

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The incorrect statement is that a non-linear molecule has 3N-3 modes.

A molecule of N atoms has 3N degrees of freedom. Three are translations. A non-linear molecule also has three rotations, leaving 3N - 6 vibrations; a linear molecule has only two rotations (rotation about its axis does not count), leaving 3N - 5.

3N - 3 subtracts only the translations and forgets the rotations. A spherical top such as CH4 is non-linear, so 3N - 6 applies to it, as option D says.

Discuss in Forum

The rotational spectrum of ¹²C¹⁶O and ¹³C¹⁶O provide the values of rotational constants B and B′ respectively. If B is 1.92 cm^-1, what is the value of B′ ?

(h = 6.6 × 10^-34 Js)

A.B′ = 2.12 \(\text{cm}^{-1}\)
B.B′ = 1.83 \(\text{cm}^{-1}\)
C.B′ = 1.06 \(\text{cm}^{-1}\)
D.B′ = 4.24 \(\text{cm}^{-1}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The rotational constant of \(^{13}C^{16}O\) is B′ = 1.83 cm\(^{-1}\).

The bond length is the same for both isotopes, so \(B = \frac{h}{8\pi^{2}c\mu r^{2}}\) is inversely proportional to the reduced mass \(\mu\):

\[ \begin{aligned} \mu &= \frac{12 \times 16}{28} = 6.857 \\ \mu' &= \frac{13 \times 16}{29} = 7.172 \\ B' &= B\frac{\mu}{\mu'} = 1.92 \times 0.956 \\ &= 1.836\ \text{cm}^{-1} \end{aligned} \]

The heavier isotope has the larger moment of inertia and so the smaller B, which rules out 2.12 cm\(^{-1}\) and 4.24 cm\(^{-1}\). 1.06 cm\(^{-1}\) is far too large a change for one extra neutron. (The value of h is not needed for a ratio.)

Discuss in Forum

If the average separation between the adjacent lines in the rotational IR spectrum of CO molecule is 3.844 cm\(^{-1}\), then the value of rotational constant is :

A.7.688 cm\(^{-1}\)
B.1.922 cm\(^{-1}\)
C.3.844 cm\(^{-1}\)
D.5.766 cm\(^{-1}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The rotational constant is \(B = 1.922\) cm\(^{-1}\).

For a rigid rotor the rotational levels are \(F(J) = BJ(J + 1)\), and the allowed transitions \(J \rightarrow J + 1\) give lines at

\[ \bar{\nu} = 2B(J + 1) \]

so adjacent lines are separated by 2B. Hence

\[ B = \frac{3.844}{2} = 1.922\ \text{cm}^{-1} \]

Taking the spacing itself as B gives 3.844, twice the spacing gives 7.688, and 5.766 is 3B; none is the rotational constant.

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