Molecular Spectroscopy Questions and Answers

Answer: A

The rotational constant is B = 1.2 cm^-1.

The ¹⁶O nucleus has spin zero, and the ground state of O₂ is ³Σ_g^-. Nuclear spin statistics then allow only odd rotational levels (J = 1, 3, 5 ...). The first Stokes line is therefore J = 1 to J = 3, not J = 0 to J = 2. With rotational energies BJ(J + 1):

\[ \begin{aligned} \Delta\tilde{\nu} &= B(3 \cdot 4 - 1 \cdot 2) = 10B \\ 10B &= 20491 - 20479 = 12\ \text{cm}^{-1} \\ B &= 1.2\ \text{cm}^{-1} \end{aligned} \]

2.0 cm^-1 is what the usual first-line shift of 6B (J = 0 to 2) would give, which is the trap here. 3.0 and 6.0 come from taking the shift as 4B or 2B.

Answer: B

The next Stokes line is at 20469 cm^-1.

The next allowed transition is J = 3 to J = 5, since only odd J levels are populated. Using B = 1.2 cm^-1 from the previous question:

\[ \begin{aligned} \Delta\tilde{\nu} &= B(5 \cdot 6 - 3 \cdot 4) = 18B = 21.6\ \text{cm}^{-1} \\ \tilde{\nu} &= 20491 - 21.6 = 20469.4\ \text{cm}^{-1} \end{aligned} \]

Successive lines of ¹⁶O₂ are 8B apart, about 9.6 cm^-1, because every other level is missing. 20467 assumes the next line is a further 12 cm^-1 away. 20471 and 20475 fit no allowed transition for B = 1.2 cm^-1: the J = 3 to 5 line must lie 18B below the exciting line.

Answer: C

The allowed transition is \(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{+}\).

Electric dipole transitions in a diatomic molecule obey three rules that matter here:

1. ΔS = 0: the spin multiplicity must not change;
2. g ↔ u: parity must change (Laporte rule);
3. for Σ to Σ transitions, + ↔ + and - ↔ -, never + ↔ -.

\(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{+}\) satisfies all three. The two transitions to \(^{3}\Sigma_{u}\) states change the multiplicity from singlet to triplet, so they are spin-forbidden. \(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{-}\) keeps the spin and changes parity, but goes from + to -, which is forbidden.

Answer: C

The most intense stretching band comes from O-H.

An infrared band's intensity is proportional to the square of the change in dipole moment during the vibration, (dμ/dr)². The larger the electronegativity difference across the bond, the larger that change. Oxygen is the most electronegative of the four atoms bonded to hydrogen here, so the O-H stretch, near 3200-3600 cm^-1, is strong and broad.

N-H is less polar and gives a weaker band. C-H is only slightly polar, and S-H, with sulfur's electronegativity close to that of carbon, gives one of the weakest bands, near 2550 cm^-1.

Answer: C

Peak III appears at the lowest energy.

Peak I is the strongest and does not change with temperature: it is the fundamental, v = 0 to 1, from the ground level that holds almost all molecules. Peak II is ten times weaker and also temperature-independent: it is the first overtone, v = 0 to 2, which lies at almost twice the energy of the fundamental.

Peak III grows as the temperature rises, so it starts from a level whose population increases with temperature: it is a hot band, v = 1 to 2. Because of anharmonicity the vibrational levels get closer together as v increases, so the 1 to 2 gap is smaller than the 0 to 1 gap and the hot band lies just below the fundamental.

So in energy, peak III lies just below peak I and peak II lies highest. That makes 'peak I at the highest energy' wrong, and neither I nor II is the lowest.

Answer: B

The correct match is P-3, Q-2, R-1, S-4.

• Benzene, C₆H₆, is a flat symmetric top: the two in-plane moments are equal and the moment about the six-fold axis is the largest, so it is an oblate symmetric rotor.
• CH₃Cl has a three-fold axis along C-Cl; the moment about that axis is the smallest and the other two are equal and larger, a prolate symmetric rotor.
• SiH₄ is tetrahedral, with all three moments equal: a spherical rotor.
• CH₃OH has no axis of three-fold or higher symmetry, so all three moments differ: an asymmetric rotor.

CO₂ is linear and is the extra item. Options that call SiH₄ oblate, benzene spherical, or CO₂ or CH₃OH prolate misclassify at least one molecule.

Answer: B

The correct statement is (I) is infrared inactive while (II) and (III) are infrared active.

A mode is IR active only if it changes the dipole moment. SO₃ is trigonal planar (D_3h).

• (I): all three S-O bonds stretch in phase (the symmetric stretch, A₁'). The molecule keeps its full symmetry and its dipole stays zero, so this mode is IR inactive (it is Raman active).
• (II): S moves out of the plane one way and the three O atoms the other (A₂''). This creates a dipole along the three-fold axis, so it is IR active.
• (III): one S-O bond shortens while the other two lengthen, an asymmetric stretch (E'). The in-plane dipole changes, so it is IR active.

A zero permanent dipole does not make every mode inactive: what matters is whether the vibration makes the dipole oscillate, so the 'none of the modes' option is wrong, and the options calling (III) inactive or (I) active are wrong too.

Answer: C

The structure is Y as the central atom in linear XY₂, that is X-Y-Y.

A vibration-rotation spectrum with only P and R branches and no Q branch is characteristic of a linear molecule (parallel bands). Being microwave active, the molecule must also have a permanent dipole moment.

A linear molecule with X in the middle, Y-X-Y, would be centrosymmetric and have no dipole, so it would be microwave inactive (like CO₂). The unsymmetrical linear arrangement X-Y-Y has a dipole, as N₂O (NNO) does. Bent molecules show a Q branch in their vibration-rotation bands, which rules out both bent options.

Answer: C

k, l and m are 1, 2 and 3.

• Symmetric stretch of CO₂: no dipole change, polarisability changes: Raman only.
• Stretch of H₂: homonuclear, no dipole: Raman only.
• Bend of CO₂: dipole changes but, by the mutual exclusion rule for a centrosymmetric molecule, Raman inactive: IR only.
• O-H symmetric stretch of H₂O, the HCl stretch and the N-H symmetric stretch of NH₃: molecules without a centre of symmetry, both IR and Raman active.

That gives k = 1 (IR only), l = 2 (Raman only) and m = 3 (both). The other options swap these counts.

Answer: B

The molecule is the trans-planar conformer of H₂O₂.

The question describes the rule of mutual exclusion: in a molecule with a centre of inversion, vibrations that are IR active (odd, u) are Raman inactive, and Raman-active (even, g) ones are IR inactive. Planar trans H-O-O-H belongs to \(C_{2h}\), which contains i, so the rule applies.

None of the others has a centre of inversion. H₂O is \(C_{2v}\), and all three of its modes are both IR and Raman active. Eclipsed ethane is \(D_{3h}\) (it is the staggered form, \(D_{3d}\), that is centrosymmetric). CH₄ is \(T_{d}\), and its t₂ modes are active in both spectra.