Rotational and Vibrational Spectra Questions and Answers

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.

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.

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.

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.

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.

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.

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.)

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.

Answer: C

The values are \(\nu = 89.6 \times 10^{12}\) Hz and \(\bar{\nu} = 2987\ cm^{-1}\).

The vibration is that of the reduced mass on a spring of constant \(k\):

\[ \begin{aligned} \mu &= \frac{m_{H}m_{Cl}}{m_{H} + m_{Cl}} \\ &= 1.628 \times 10^{-27}\ \text{kg} \\ \nu &= \frac{1}{2\pi}\sqrt{\frac{516}{1.628 \times 10^{-27}}} \\ &= 8.96 \times 10^{13}\ \text{Hz} \\ \bar{\nu} &= \frac{\nu}{c} = \frac{8.96 \times 10^{13}}{3 \times 10^{10}\ \text{cm/s}} \\ &\approx 2987\ cm^{-1} \end{aligned} \]

Because chlorine is so much heavier, \(\mu\) is close to the mass of the hydrogen atom.

The option pairing 89.6 THz with 1890 cm⁻¹ has an inconsistent wavenumber, and the options with 179.2 THz or 44.8 THz double or halve the frequency with wavenumbers scaled the same way, so they do not follow from \(k\) and \(\mu\).

Answer: C

The bond length is 1.131 Å.

For a rigid rotor the \(J = 0 \to 1\) line lies at \(\bar{\nu} = 2B\), so \(B = 1.921\ cm^{-1} = 192.1\ m^{-1}\). With \(B = h/(8\pi^{2}cI)\) and \(I = \mu r^{2}\):

\[ \begin{aligned} I &= \frac{h}{8\pi^{2}cB} = 1.456 \times 10^{-46}\ \text{kg m}^{2} \\ \mu &= \frac{19.921 \times 26.561}{46.482} \times 10^{-27} \\ &= 1.138 \times 10^{-26}\ \text{kg} \\ r &= \sqrt{I/\mu} = 1.131 \times 10^{-10}\ \text{m} \end{aligned} \]

The key step is halving the line position: using \(\bar{\nu}\) itself as \(B\) gives \(r\) smaller by \(\sqrt{2}\), about 0.80 Å, which is not an option, while the other listed lengths (0.956, 1.483 and 1.762 Å) do not follow from the given masses and line position.