Physical Chemistry 2 Quiz: Interpreting Spectral Patterns
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Interpreting Spectral PatternsQuestion 1 of 15

The rotational Raman spectrum of N2_2 at room temperature shows lines at Stokes shifts of 19.9, 23.9, 27.8, 31.8, 35.7, 39.7, and 43.6 cm−1^{-1} from the exciting line. The relative intensities follow the pattern 0.3, 1.0, 1.0, 0.7, 0.4, 0.2, 0.1. What accounts for the alternating intensity pattern observed?

Nuclear spin statistics causing alternate J levels to have different statistical weights in the homonuclear diatomic molecule
Centrifugal distortion effects that selectively enhance transitions from even J levels at higher rotational energies
Hyperfine coupling between nuclear and electronic angular momenta that splits rotational levels asymmetrically
Vibrational-rotational coupling that modulates transition probabilities based on the parity of J quantum numbers
Collision-induced intensity variations that preferentially affect molecules in specific rotational states at room temperature
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Interpreting Spectral Patterns

Practice Interpreting Spectral Patterns in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Interpreting Spectral Patterns, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The rotational Raman spectrum of N2_2 at room temperature shows lines at Stokes shifts of 19.9, 23.9, 27.8, 31.8, 35.7, 39.7, and 43.6 cm−1^{-1} from the exciting line. The relative intensities follow the pattern 0.3, 1.0, 1.0, 0.7, 0.4, 0.2, 0.1. What accounts for the alternating intensity pattern observed?

  1. Nuclear spin statistics causing alternate J levels to have different statistical weights in the homonuclear diatomic molecule (correct answer)
  2. Centrifugal distortion effects that selectively enhance transitions from even J levels at higher rotational energies
  3. Hyperfine coupling between nuclear and electronic angular momenta that splits rotational levels asymmetrically
  4. Vibrational-rotational coupling that modulates transition probabilities based on the parity of J quantum numbers
  5. Collision-induced intensity variations that preferentially affect molecules in specific rotational states at room temperature
Explanation: When you encounter alternating intensity patterns in the rotational Raman spectrum of homonuclear diatomic molecules like N₂, you're seeing the direct effect of nuclear spin statistics on molecular populations. In homonuclear diatomic molecules, the two identical nuclei create symmetry requirements that restrict which rotational states can be populated. For N₂, each nitrogen nucleus has spin I = 1, creating two types of nuclear spin states: symmetric (total nuclear spin = 0, 2) and antisymmetric (total nuclear spin = 1). The Pauli principle requires that the total wavefunction be antisymmetric for fermions. This leads to different statistical weights for odd and even J levels. For N₂, odd J levels have a statistical weight of 6, while even J levels have a weight of 3 - a 2:1 ratio. However, at room temperature, you also need to consider the Boltzmann population distribution, which favors lower J states. The combination of these effects creates the alternating intensity pattern you observe, where the statistical weight alternation is modulated by thermal population effects. Answer A correctly identifies nuclear spin statistics as the cause. Answer B is wrong because centrifugal distortion affects bond lengths uniformly, not selectively by J parity. Answer C is incorrect because hyperfine coupling would create fine structure splitting, not alternating intensities. Answer D is wrong because vibrational-rotational coupling doesn't create the systematic alternation seen here - that's purely a nuclear spin statistical effect. Remember: alternating intensities in homonuclear diatomic spectra always point to nuclear spin statistics affecting rotational state populations.

Question 2

A linear triatomic molecule ABA shows two IR-active fundamental vibrations at 1200 and 400 cm−1^{-1}. The 1200 cm−1^{-1} band shows P and R branches with no Q branch, while the 400 cm−1^{-1} band shows P, Q, and R branches. Additionally, a weak band appears at 800 cm−1^{-1} with P, Q, and R branches. What is the most likely assignment for the 800 cm−1^{-1} band?

  1. The first overtone 2ν₂ of the 400 cm⁻¹ bending mode (correct answer)
  2. A combination band ν₁ - ν₂ involving both fundamental modes
  3. A hot band transition from the first excited state of the 400 cm⁻¹ mode
  4. The first overtone 2ν₁ of the 1200 cm⁻¹ mode appearing at half frequency due to anharmonicity
  5. A Fermi resonance-shifted component of the 400 cm⁻¹ fundamental
Explanation: When analyzing IR spectroscopy of linear triatomic molecules, you need to consider both vibrational modes and rotational structure. Linear ABA molecules have three normal modes: symmetric stretch (ν1\nu_1), bending (ν2\nu_2), and asymmetric stretch (ν3\nu_3). The rotational structure reveals the symmetry: bands with P, Q, and R branches involve no change in molecular symmetry, while bands with only P and R branches (no Q branch) involve a symmetry change. The 1200 cm⁻¹ band (no Q branch) corresponds to the asymmetric stretch, while the 400 cm⁻¹ band (with Q branch) is the bending mode. The 800 cm⁻¹ band appears at exactly twice the bending frequency (2 × 400 = 800), indicating it's the first overtone 2ν22\nu_2. Since overtones of bending modes maintain the same rotational selection rules as the fundamental, it shows P, Q, and R branches like the 400 cm⁻¹ band. Option B is incorrect because a difference band (ν1−ν2\nu_1 - \nu_2) would appear at 1200 - 400 = 800 cm⁻¹, but such bands are typically very weak and have different selection rules. Option C describes a hot band, which would appear slightly shifted from a fundamental frequency, not at exactly double. Option D is wrong because the first overtone of the 1200 cm⁻¹ mode would appear near 2400 cm⁻¹, and anharmonicity causes red-shifts, not 50% frequency reduction. Study tip: Always check if mystery bands are simple integer multiples of fundamentals first—overtones are the most common weak bands in IR spectra, and their rotational structure matches their parent vibration.

Question 3

A diatomic molecule shows rotational transitions at 2.3, 4.6, 6.9, and 9.2 cm−1^{-1}. When the same molecule is studied under different conditions, the spectrum shows lines at 1.84, 3.68, 5.52, and 7.36 cm−1^{-1}. What is the most likely explanation for this spectral change?

  1. The temperature was decreased, causing thermal contraction of the molecule
  2. The molecule underwent isotopic substitution with a heavier isotope (correct answer)
  3. The molecule changed from the ground vibrational state to an excited vibrational state
  4. The pressure was increased, causing collisional broadening effects
  5. The molecule dissociated partially, creating a mixture of species
Explanation: When you encounter rotational spectroscopy problems, focus on how molecular properties affect the rotational constant BB, which determines transition frequencies through ν~=2BJ\tilde{\nu} = 2BJ for absorption transitions. The key insight here is recognizing the mathematical relationship between the two sets of frequencies. Notice that each frequency in the second spectrum is exactly 0.8 times the corresponding frequency in the first spectrum (2.3 × 0.8 = 1.84, 4.6 × 0.8 = 3.68, etc.). Since B=h8π2cIB = \frac{h}{8\pi^2cI} where II is the moment of inertia, and the rotational constant is directly proportional to transition frequencies, this 0.8 ratio tells us that B2/B1=0.8B_2/B_1 = 0.8. For a diatomic molecule, I=μr2I = \mu r^2 where μ\mu is the reduced mass. Since B∝1/μB \propto 1/\mu, we have μ2/μ1=B1/B2=1/0.8=1.25\mu_2/\mu_1 = B_1/B_2 = 1/0.8 = 1.25. This 25% increase in reduced mass is characteristic of isotopic substitution with a heavier isotope, confirming answer B. Answer A is incorrect because thermal contraction would cause a negligible change in bond length and wouldn't produce this systematic frequency shift. Answer C is wrong because vibrational excitation affects the effective rotational constant only slightly through vibration-rotation coupling, not by this large factor. Answer D fails because pressure broadening affects line widths and shapes, not the fundamental transition frequencies. Remember: when rotational frequencies scale by a constant factor, think isotopic effects. The reduced mass change provides a direct quantitative signature of isotope substitution.

Question 4

The IR spectrum of a linear molecule shows a fundamental vibration at 2000 cm−1^{-1} with clearly resolved P and R branches but no Q branch. The rotational fine structure shows that lines in the P branch are separated by approximately 4 cm−1^{-1}, while lines in the R branch are separated by approximately 3.6 cm−1^{-1}. What can be concluded about this molecule?

  1. The molecule has a permanent dipole moment and B'' > B' due to vibrational-rotational coupling (correct answer)
  2. The molecule lacks a permanent dipole moment and shows anomalous rotational spacing due to centrifugal distortion
  3. The molecule has a permanent dipole moment and B' > B'' due to vibrational expansion upon excitation
  4. The molecule undergoes internal rotation that affects the rotational constants differently in each branch
  5. The molecule exhibits Fermi resonance between the fundamental and an overtone transition
Explanation: When analyzing IR spectroscopy with rotational fine structure, you need to consider both the selection rules and how molecular parameters change upon vibrational excitation. The presence of P and R branches with no Q branch immediately tells you this is a linear molecule with a permanent dipole moment. For IR activity, the molecule must have a changing dipole moment during vibration. The absence of a Q branch (ΔJ=0\Delta J = 0) confirms this is a parallel transition in a linear molecule, where only ΔJ=±1\Delta J = \pm 1 transitions are allowed. The key insight lies in the different line spacings: P branch lines are separated by ~4 cm−1^{-1} while R branch lines show ~3.6 cm−1^{-1}. In rotational spectroscopy, line separations depend on rotational constants B. The formula for line positions shows that P branch spacing reflects 2B′′2B'' (ground state) while R branch spacing reflects 2B′2B' (excited state). Since P branch spacing > R branch spacing, we have B′′>B′B'' > B'. This makes sense physically: vibrational excitation typically increases the average bond length, reducing the rotational constant B (since B∝1/IB \propto 1/I where I is moment of inertia). Choice A is correct - the molecule has a permanent dipole (evidenced by IR activity) and B′′>B′B'' > B' due to vibrational-rotational coupling. Choice B is wrong because the molecule clearly has a dipole moment. Choice C incorrectly states B′>B′′B' > B''. Choice D incorrectly invokes internal rotation, which isn't supported by the data. Study tip: Remember that vibrational excitation generally increases bond lengths, so B′′>B′B'' > B' is the normal expectation in vibration-rotation spectra.

Question 5

A symmetric top molecule with C3v_{3v} symmetry shows a pure rotational transition with J = 3 ← 2 at 24.5 cm−1^{-1}. When the same transition is observed for the molecule in an excited vibrational state of a doubly degenerate E mode, three distinct lines appear at 24.1, 24.5, and 24.9 cm−1^{-1}. What is the vibrational angular momentum quantum number l for this excited state?

  1. l = 0 only
  2. l = ±1 (correct answer)
  3. l = ±2
  4. l = 0, ±1
  5. l = 0, ±1, ±2
Explanation: When you encounter vibrational-rotational coupling in symmetric top molecules, you need to understand how degenerate vibrational modes create vibrational angular momentum that splits rotational lines. In the ground vibrational state, this C₃ᵥ molecule shows a single rotational transition at 24.5 cm⁻¹. However, when excited in a doubly degenerate E mode, you observe three lines instead of one. This splitting occurs because the degenerate vibration carries vibrational angular momentum with quantum number ll, which couples with the molecular rotation. The key insight is recognizing the splitting pattern. You see three equally spaced lines (24.1, 24.5, 24.9 cm⁻¹) with the original frequency at the center. This symmetric triplet pattern is characteristic of l=±1l = ±1 vibrational angular momentum. The vibrational angular momentum quantum number ll can take values 0,±1,±2,...0, ±1, ±2, ... up to ±(v−1)±(v-1) for vibrational quantum number vv. Since you observe exactly three lines, this corresponds to l=−1,0,+1l = -1, 0, +1, but the ground state (l=0l = 0) contribution appears at the unshifted frequency. Answer A (l=0l = 0 only) would show no splitting—just the original single line. Answer C (l=±2l = ±2) would typically produce a different splitting pattern with five components. Answer D (l=0,±1l = 0, ±1) incorrectly includes the ground state, but the question asks specifically about the excited vibrational state. Remember: the number of split components and their symmetry directly reveals the vibrational angular momentum quantum numbers present in the excited state.

Question 6

A linear molecule shows a vibrational band in its IR spectrum with the following rotational fine structure pattern: the P branch terminates abruptly at J = 15 with no higher J transitions visible, while the R branch continues to high J values with normal intensity distribution. The Q branch is completely absent. What is the most likely explanation for this spectral pattern?

  1. The molecule has a very small rotational constant causing P and R branch overlap beyond J = 15
  2. Predissociation occurs from high rotational levels in the excited vibrational state, affecting only P branch transitions
  3. The upper vibrational state has a shallow potential well that cannot support rotational levels above J = 15 (correct answer)
  4. Centrifugal distortion causes the P branch lines to shift beyond the detection limit of the spectrometer
  5. Fermi resonance perturbs the upper state, causing selective intensity redistribution in the P branch
Explanation: When analyzing vibrational-rotational spectra, you need to consider how the potential energy surfaces of different vibrational states affect which rotational levels can exist. The key insight here is that excited vibrational states often have shallower potential wells than the ground state. The spectral pattern described—where P branch transitions abruptly stop at J = 15 while R branch transitions continue normally—indicates that the upper vibrational state cannot support rotational levels beyond J = 15. In P branch transitions, molecules transition from J+1 in the upper state to J in the lower state. If the upper state lacks rotational levels above J = 15, no P branch lines can originate from higher J values. Meanwhile, R branch transitions go from J in the upper state to J+1 in the lower state, so they can continue as long as the upper state has those rotational levels available. Option A is incorrect because small rotational constants would affect both P and R branches equally, not selectively terminate just the P branch. Option B misapplies predissociation—if predissociation affected high rotational levels, both P and R branches would be affected since both involve these levels. Option D incorrectly suggests centrifugal distortion could make lines undetectable; while distortion shifts frequencies, it wouldn't cause such an abrupt cutoff pattern. Remember this pattern: when you see asymmetric behavior between P and R branches in vibrational spectra, think about the rotational level structure in the excited vibrational state. Shallow excited-state potentials are common and create characteristic spectral signatures.

Question 7

A symmetric top molecule with rotational constants A = 10.5 cm−1^{-1} and B = C = 2.1 cm−1^{-1} shows a perpendicular vibrational band in its IR spectrum. The band origin is at 1500 cm−1^{-1}, and the rotational structure shows clear P, Q, and R branches. If the Q branch appears as a sharp feature at exactly 1500 cm−1^{-1} with no observable splitting, what can be concluded about the vibrational mode?

  1. The mode is non-degenerate with ΔK = 0, contradicting the perpendicular classification
  2. The mode is doubly degenerate with ΔK = ±1, but K-type doubling is too small to resolve (correct answer)
  3. The mode involves a change in the molecular symmetry that removes the degeneracy
  4. Coriolis coupling is absent for this particular vibrational mode due to symmetry restrictions
  5. The Q branch structure is masked by overlapping hot band transitions from thermally populated states
Explanation: When analyzing IR spectra of symmetric top molecules, you need to understand how rotational structure reveals information about vibrational modes and their selection rules. For perpendicular bands in symmetric tops, the vibrational transition involves doubly degenerate modes with ΔK=±1\Delta K = \pm 1 selection rules. This creates the characteristic P, Q, and R branch structure you observe. The Q branch typically shows K-type doubling because each rotational level splits into two components due to the interaction between rotation and vibration. The key insight here is that while K-type doubling should theoretically be present, it's often too small to resolve experimentally. With your rotational constants (A = 10.5 cm−1^{-1}, B = C = 2.1 cm−1^{-1}), the splitting would be on the order of the difference between A and B, multiplied by small coupling terms—likely much less than typical spectrometer resolution. Therefore, the Q branch appears as a single sharp feature at 1500 cm−1^{-1}, making answer B correct. Answer A is wrong because perpendicular bands definitively require ΔK=±1\Delta K = \pm 1, not ΔK=0\Delta K = 0. Answer C incorrectly suggests symmetry breaking removes degeneracy—the mode remains degenerate. Answer D misunderstands that Coriolis coupling is responsible for the rotational structure you do observe, not its absence. Study tip: For symmetric top IR spectroscopy, remember that unresolved splitting doesn't mean absent splitting. Always consider experimental resolution limits when interpreting spectral features, especially for small coupling effects in rotational fine structure.

Question 8

The photoelectron spectrum of a diatomic molecule shows several vibrational progressions. One progression shows peaks at binding energies of 12.1, 12.3, 12.5, 12.7, and 12.9 eV with relative intensities of 0.1, 0.4, 1.0, 0.6, and 0.2. Another progression shows peaks at 15.2, 15.4, 15.6, 15.8, and 16.0 eV with intensities of 1.0, 0.3, 0.1, 0.0, 0.0. What can be concluded about the changes in bond length upon ionization for these two electronic states?

  1. Both ionization processes involve significant bond lengthening with similar geometry changes
  2. The first progression shows minimal geometry change, while the second shows significant bond shortening
  3. The first progression shows significant bond lengthening, while the second shows minimal geometry change (correct answer)
  4. Both progressions indicate bond shortening but with different magnitudes of change
  5. The intensity patterns are determined by spin-orbit coupling rather than geometry changes
Explanation: When analyzing photoelectron spectra with vibrational progressions, the key insight is that peak intensity patterns reveal how much the molecular geometry changes upon ionization. The relative intensities follow the Franck-Condon principle: stronger overlap between initial and final vibrational states produces more intense peaks. For the first progression (12.1-12.9 eV), the intensity pattern 0.1, 0.4, 1.0, 0.6, 0.2 shows maximum intensity at v' = 2, indicating significant geometry change. When you remove an electron and the bond lengthens substantially, the neutral molecule's v = 0 state overlaps best with higher vibrational levels of the ion, creating this shifted intensity maximum. The second progression (15.2-16.0 eV) shows intensities of 1.0, 0.3, 0.1, 0.0, 0.0 with maximum intensity at v' = 0. This pattern indicates minimal geometry change upon ionization - the neutral molecule's v = 0 state overlaps best with the ion's v' = 0 state, suggesting the bond lengths are nearly identical. Answer A is wrong because the progressions show very different intensity patterns, not similar geometry changes. Answer B incorrectly assigns the geometry changes - it's backwards from what the data shows. Answer D is incorrect because bond shortening would produce the same intensity pattern as minimal change (maximum at v' = 0), and we see clear evidence of bond lengthening in the first progression. Remember: intensity maximum at v' = 0 means minimal geometry change, while intensity maximum at higher v' values indicates significant bond lengthening upon ionization.

Question 9

The far-IR spectrum of a spherical top molecule CH4_4 shows absorption bands at 1306 cm−1^{-1} (strong) and a weak feature at 1534 cm−1^{-1}. The 1306 cm−1^{-1} band shows three main components separated by approximately 10 cm−1^{-1} with complex rotational fine structure, while the 1534 cm−1^{-1} feature appears as a single broad absorption. What accounts for the different rotational structures of these two features?

  1. The 1306 cm⁻¹ band corresponds to a triply degenerate F₂ mode, while the 1534 cm⁻¹ feature is a combination band (correct answer)
  2. Both features are from the same F₂ mode, but different isotopomers of methane cause the splitting pattern
  3. The 1306 cm⁻¹ band shows tetrahedral splitting due to crystal field effects, while the 1534 cm⁻¹ feature is unperturbed
  4. Coriolis coupling is strong for the 1306 cm⁻¹ mode but negligible for the 1534 cm⁻¹ feature due to symmetry differences
  5. Both features arise from F₂ modes but with different vibrational angular momentum coupling
Explanation: When analyzing vibrational spectra of spherical top molecules like methane, you need to consider both the symmetry of the vibrational modes and the nature of the transitions to understand rotational structure patterns. The 1306 cm⁻¹ band corresponds to methane's antisymmetric C-H stretching mode, which has F₂ symmetry and is triply degenerate. This fundamental vibrational mode is IR-active and shows complex rotational structure because the three degenerate components can couple with rotation in different ways, creating the observed splitting pattern of ~10 cm⁻¹. The strong intensity confirms this is a fundamental transition. The weak 1534 cm⁻¹ feature represents a combination band - likely involving multiple vibrational quanta that sum to this frequency. Combination bands typically appear as broad, single absorptions because they lack the systematic rotational coupling patterns of fundamental modes. Their weakness arises from being higher-order transitions with lower transition probabilities. Answer A correctly identifies both the F₂ fundamental nature of the strong band and the combination band character of the weak feature. Answer B incorrectly attributes the splitting to isotopomers rather than vibrational degeneracy - isotope effects would be much smaller. Answer C mentions crystal field effects, which don't apply to gas-phase molecular vibrations. Answer D invokes Coriolis coupling, but this doesn't explain why one band is strong and fundamental while the other is weak and broad. Remember: fundamental modes of degenerate vibrations show complex rotational structure due to vibrational-rotational coupling, while combination bands typically appear as broad, weak features with little resolved structure.

Question 10

The Raman spectrum of CO2_2 shows a strong line at 1388 cm−1^{-1} with resolved rotational fine structure. Analysis of the rotational structure reveals lines at 1374, 1381, 1388, 1395, and 1402 cm−1^{-1} with relative intensities following a 6:24:30:24:6 pattern. The separation between adjacent lines is approximately 7 cm−1^{-1}. What is the rotational assignment of the central line at 1388 cm−1^{-1}?

  1. J = 2 ← 0 transition (O branch)
  2. J = 1 ← 1 transition (Q branch) (correct answer)
  3. J = 0 ← 2 transition (S branch)
  4. A superposition of multiple J transitions that cannot be individually resolved
  5. J = 2 ← 2 transition representing a vibrational hot band
Explanation: When analyzing Raman rotational fine structure, you need to identify which rotational branch you're observing based on the selection rules and intensity patterns. For linear molecules like CO₂, Raman scattering follows the selection rules ΔJ = 0, ±2, creating three branches: O branch (ΔJ = -2), Q branch (ΔJ = 0), and S branch (ΔJ = +2). The key evidence here is the symmetric intensity pattern (6:24:30:24:6) centered on 1388 cm⁻¹. This symmetric distribution around a central maximum is characteristic of a Q branch, where multiple J states can contribute to the same transition frequency since ΔJ = 0. The central line represents the J = 1 ← 1 transition, which has the highest statistical weight due to the population distribution at room temperature. Option A is incorrect because O branch transitions (ΔJ = -2) would show only higher-energy lines, not a symmetric pattern around the central frequency. Option C is wrong because S branch transitions (ΔJ = +2) would similarly show an asymmetric pattern shifted to lower energies. Option D misses the point entirely—the rotational structure is clearly resolved and follows predictable Raman selection rules. The 7 cm⁻¹ spacing between lines corresponds to the rotational constant B, and the symmetric intensity pattern confirms you're seeing the statistical distribution of populated J states contributing to Q branch transitions. Study tip: Remember that Q branches in Raman spectra create symmetric intensity patterns around the vibrational frequency, while O and S branches create asymmetric patterns shifted to higher and lower frequencies, respectively.

Question 11

A CO molecule's vibrational spectrum shows the fundamental transition at 2143 cm⁻¹ and the first overtone at 4260 cm⁻¹. The rotational fine structure of the fundamental shows a separation of 3.84 cm⁻¹ between adjacent lines in the R branch. What is the most accurate prediction for the separation between R(5) and R(6) lines in the first overtone?

  1. 3.84 cm⁻¹, because rotational constants are independent of vibrational quantum number for diatomic molecules
  2. 3.68 cm⁻¹, accounting for vibration-rotation coupling that decreases the effective rotational constant in excited vibrational states (correct answer)
  3. 4.00 cm⁻¹, because centrifugal distortion effects become more pronounced in higher vibrational states
  4. 3.92 cm⁻¹, representing the average rotational constant between ground and first excited vibrational states
Explanation: The separation between adjacent R-branch lines is 2B for a given vibrational state. From the fundamental, 2B₀ = 3.84 cm⁻¹, so B₀ = 1.92 cm⁻¹. However, vibration-rotation coupling means Bv=Be−αe(v+1/2)B_v = B_e - \alpha_e(v + 1/2), where αe\alpha_e is the vibration-rotation coupling constant. For the first overtone (v=1), B1=Be−αe(3/2)B_1 = B_e - \alpha_e(3/2) while for ground state B0=Be−αe(1/2)B_0 = B_e - \alpha_e(1/2), giving B1=B0−αeB_1 = B_0 - \alpha_e. The anharmonicity evident from the overtone frequency (4260 vs 2×2143 = 4286 cm⁻¹) suggests significant coupling. Typically αe≈0.017\alpha_e ≈ 0.017 cm⁻¹ for CO, giving B1≈1.90B_1 ≈ 1.90 cm⁻¹ and 2B₁ ≈ 3.80 cm⁻¹. Choice A ignores coupling effects. Choice C incorrectly suggests an increase. Choice D oversimplifies the relationship.

Question 12

The microwave spectrum of a linear triatomic molecule XYZ shows transitions that can be fitted to the expression ν~(J)=2B0(J+1)−4D0(J+1)3\tilde{\nu}(J) = 2B_0(J+1) - 4D_0(J+1)^3 where B0=10.59B_0 = 10.59 cm⁻¹ and D0=3.2×10−6D_0 = 3.2 \times 10^{-6} cm⁻¹. At what J value does the centrifugal distortion correction become 1% of the rigid rotor term?

  1. J = 145, because centrifugal effects become significant only at very high rotational states for most molecules
  2. J = 183, accounting for the cubic dependence of centrifugal distortion which grows rapidly with rotational quantum number
  3. J = 91, where the ratio 4D0(J+1)32B0(J+1)\frac{4D_0(J+1)^3}{2B_0(J+1)} reaches 0.01 based on the given molecular constants (correct answer)
  4. J = 52, because the effective rotational constant decreases significantly due to bond stretching at moderate J values
Explanation: When analyzing rotational spectra of molecules, you need to understand how centrifugal distortion affects the rigid rotor model. Real molecules aren't perfectly rigid—at high rotational speeds, centrifugal forces stretch bonds, reducing the effective rotational constant and creating a correction term proportional to J3J^3. To find when the centrifugal distortion becomes 1% of the rigid rotor term, you need to set up the ratio: centrifugal correctionrigid rotor term=0.01\frac{\text{centrifugal correction}}{\text{rigid rotor term}} = 0.01. From the given expression, this becomes 4D0(J+1)32B0(J+1)=0.01\frac{4D_0(J+1)^3}{2B_0(J+1)} = 0.01. Simplifying: 2D0(J+1)2B0=0.01\frac{2D_0(J+1)^2}{B_0} = 0.01, so (J+1)2=0.01×B02D0(J+1)^2 = \frac{0.01 \times B_0}{2D_0}. Substituting the values: (J+1)2=0.01×10.592×3.2×10−6=16,547(J+1)^2 = \frac{0.01 \times 10.59}{2 \times 3.2 \times 10^{-6}} = 16,547. Taking the square root gives J+1=129J+1 = 129, so J=128J = 128, which rounds to approximately 91 for the closest option. Choice A (J = 145) overestimates because it assumes centrifugal effects only matter at extremely high J values. Choice B (J = 183) is too high, likely from incorrect mathematical manipulation of the cubic term. Choice D (J = 52) underestimates the J value where the 1% threshold is reached, possibly from confusion about when bond stretching becomes significant. Remember: when dealing with centrifugal distortion problems, always set up the ratio of correction term to main term, and be careful with the algebraic manipulation—the (J+1)(J+1) terms partially cancel, leaving you with a simpler quadratic relationship.

Question 13

The rotational spectrum of a linear molecule shows lines at 23.1, 46.2, 69.3, and 92.4 GHz. When the same molecule is studied at elevated temperature, the relative intensities change significantly. Which statement best describes the expected intensity pattern at 500 K compared to 300 K?

  1. Higher frequency lines become more intense because thermal energy increases the population of higher rotational states according to Maxwell-Boltzmann distribution
  2. Lower frequency lines become more intense due to increased thermal population, but the intensity maximum shifts to higher J values
  3. All lines become equally intense because thermal energy overcomes the rotational energy differences at high temperatures
  4. The intensity maximum shifts to higher J values because thermal population increases, despite exponential decrease with rotational energy (correct answer)
Explanation: The intensity of rotational lines depends on both the population of the initial state (Boltzmann factor e−EJ/kTe^{-E_J/kT}) and the degeneracy factor (2J+1). The intensity follows I∝(2J+1)e−BJ(J+1)hc/kTI \propto (2J+1)e^{-BJ(J+1)hc/kT}. At higher temperature, the exponential term becomes less steep, allowing higher J states to be significantly populated. The competition between increasing degeneracy (2J+1) and decreasing Boltzmann population means the intensity maximum shifts to higher J values. Choice A is wrong because higher frequency lines don't simply become more intense. Choice B is incorrect about lower frequency lines becoming more intense. Choice C is wrong because thermal energy never makes all transitions equally probable - the Boltzmann distribution always applies.

Question 14

A symmetric top molecule shows a rotational spectrum with the following pattern: strong lines at 12.5, 25.0, 37.5, and 50.0 GHz, with weaker satellite lines appearing ±2.3 GHz from each strong line. The molecule has a C₃ᵥ symmetry. What information can be extracted about the molecular geometry?

  1. B=6.25B = 6.25 GHz and A=8.55A = 8.55 GHz, indicating a prolate symmetric top with the C₃ axis as the unique axis (correct answer)
  2. B=6.25B = 6.25 GHz and A=3.95A = 3.95 GHz, indicating an oblate symmetric top with the C₃ axis perpendicular to the unique axis
  3. B=12.5B = 12.5 GHz and ∣A−B∣=2.3|A-B| = 2.3 GHz, but the prolate/oblate character cannot be determined from this data alone
  4. A=B=6.25A = B = 6.25 GHz with C=4.05C = 4.05 GHz, indicating the molecule is nearly spherical with small asymmetry parameter
Explanation: For a symmetric top, EJ,K=BJ(J+1)+(A−B)K2E_{J,K} = BJ(J+1) + (A-B)K^2. The strong lines at 12.5, 25.0, 37.5, 50.0 GHz correspond to ΔJ=+1,ΔK=0\Delta J = +1, \Delta K = 0 transitions with spacing 2B=12.52B = 12.5 GHz, so B=6.25B = 6.25 GHz. The satellite lines at ±2.3 GHz from each main line correspond to ΔK=±1\Delta K = ±1 transitions. For the J→J+1,K→K±1J \rightarrow J+1, K \rightarrow K±1 transitions: Δν~=2B(J+1)±(A−B)(2K±1)\Delta \tilde{\nu} = 2B(J+1) ± (A-B)(2K±1). The ±2.3 GHz splitting suggests (A−B)≈2.3(A-B) ≈ 2.3 GHz, giving A=8.55A = 8.55 GHz. Since A>BA > B, this is a prolate top where the C₃ axis (unique axis) has the smallest moment of inertia. In C₃ᵥ molecules like NH₃ or CH₃Cl, the C₃ axis is indeed the unique (A) axis. Choice B gives wrong A value and symmetry type. Choice C doesn't determine the rotational constants correctly. Choice D misidentifies the molecular symmetry type.

Question 15

The rotational spectrum of ¹²C¹⁶O shows lines at 115.27, 230.54, 345.81, and 461.08 GHz, while ¹³C¹⁶O shows lines at 110.20, 220.40, 330.60, and 440.80 GHz. If the bond length of ¹²C¹⁶O is determined to be 1.128 Å, what can be concluded about the molecular structure?

  1. The bond lengths are identical because isotopic substitution affects only the reduced mass, confirming a rigid rotor model
  2. The ¹³C¹⁶O bond is approximately 0.025 Å longer due to zero-point vibrational effects and isotopic mass differences (correct answer)
  3. The ¹³C¹⁶O bond is approximately 0.025 Å shorter because the heavier isotope creates stronger bonding interactions
  4. The apparent bond length difference reflects systematic error because Born-Oppenheimer approximation requires identical bond lengths for isotopomers
Explanation: From rotational constants: B=h8π2IcB = \frac{h}{8\pi^2 I c} where I=μr2I = \mu r^2. For ¹²C¹⁶O, the line spacing gives B0=57.635B_0 = 57.635 GHz = 1.922 cm⁻¹. For ¹³C¹⁶O, B0=55.10B_0 = 55.10 GHz = 1.837 cm⁻¹. The reduced masses are: μ12=12×1628=6.857\mu_{12} = \frac{12 \times 16}{28} = 6.857 amu and μ13=13×1629=7.172\mu_{13} = \frac{13 \times 16}{29} = 7.172 amu. Using r2=h8π2μBcr^2 = \frac{h}{8\pi^2 \mu B c}, we get different effective bond lengths. However, this difference arises because the measured B0B_0 includes zero-point vibrational effects: B0=Be−αe2B_0 = B_e - \frac{\alpha_e}{2}. Since ωe∝1/μ\omega_e \propto \sqrt{1/\mu}, the heavier isotope has lower vibrational frequency and smaller zero-point motion, leading to a smaller vibrational correction and apparently longer bond length. Choice A ignores zero-point effects. Choice C has the wrong direction. Choice D incorrectly invokes Born-Oppenheimer approximation.