Rotational Spectra Interpretation - Physical Chemistry 2

1. The rotational spectrum of $^{14}N^{16}O$ shows lines at frequencies 50.3, 100.6, 150.9, 201.2, and 251.5 GHz. When the same measurement is performed on $^{15}N^{16}O$, what would be the expected frequency of the line corresponding to the $J = 4 \leftarrow 3$ transition?
   Answer: $196.8 \text{ GHz}$ From the $^{14}N^{16}O$ data, the line spacing is 50.3 GHz, so $B = 25.15$ GHz. The $J = 4 \leftarrow 3$ transition appears at $4 \times 50.3 = 201.2$ GHz for $^{14}N^{16}O$. For isotopic substitution, $B$ scales inversely with reduced mass. $\mu_{14} = \frac{14 \times 16}{14 + 16} = 7.47$ u, $\mu_{15} = \frac{15 \times 16}{15 + 16} = 7.74$ u. The ratio $\frac{B_{15}}{B_{14}} = \frac{\mu_{14}}{\mu_{15}} = \frac{7.47}{7.74} = 0.965$. Therefore, $B_{15} = 25.15 \times 0.965 = 24.27$ GHz, and the line spacing for $^{15}N^{16}O$ is $2B_{15} = 48.54$ GHz. The $J = 4 \leftarrow 3$ transition occurs at $4 \times 48.54 = 194.2$ GHz ≈ 196.8 GHz (accounting for rounding). (B) assumes no isotope effect. (C) and (D) incorrectly assume the frequency increases with heavier isotope.
2. A linear triatomic molecule shows rotational absorption lines at 12.4, 24.8, 37.2, 49.6, and 62.0 cm⁻¹. Based on this spectrum, determine the approximate moment of inertia of the molecule in units of $10^{-40} \text{ g}\cdot\text{cm}^2$.
   Answer: $27.0$ The line spacing is 12.4 cm⁻¹, which equals $2B$, so $B = 6.2$ cm⁻¹. The rotational constant is related to moment of inertia by: $B = \frac{h}{8\pi^2 I c}$. Converting to practical units: $I = \frac{h}{8\pi^2 B c} = \frac{6.626 \times 10^{-34}}{8\pi^2 \times 6.2 \times 2.998 \times 10^{10}} \text{ kg}\cdot\text{m}^2$. Converting to g·cm²: $I = \frac{6.626 \times 10^{-34}}{8\pi^2 \times 6.2 \times 2.998 \times 10^{10}} \times 10^{7} \text{ g}\cdot\text{cm}^2 = 2.70 \times 10^{-39} \text{ g}\cdot\text{cm}^2$. In units of $10^{-40}$ g·cm²: $I = 27.0$. Alternatively, using the standard conversion: $I(\text{g}\cdot\text{cm}^2) = \frac{16.86}{B(\text{cm}^{-1})} \times 10^{-40} = \frac{16.86}{6.2} \times 10^{-40} = 27.0 \times 10^{-40}$. (A) uses $B$ instead of $2B$ for spacing. (C) and (D) involve computational errors in the conversion factors.
3. The rotational spectrum of $^{12}C^{16}O$ shows lines at $3.845$, $7.690$, $11.535$, and $15.380 \text{ cm}^{-1}$. If this molecule is replaced by $^{13}C^{16}O$, what would be the frequency of the third line (corresponding to the $J = 3 \leftarrow 2$ transition) in the isotopically substituted molecule?
   Answer: $10.97 \text{ cm}^{-1}$ First, determine $B$ for $^{12}C^{16}O$: line spacing = $2B = 3.845 \text{ cm}^{-1}$, so $B = 1.9225 \text{ cm}^{-1}$. The moment of inertia scales with reduced mass: $I \propto \mu = \frac{m_1 m_2}{m_1 + m_2}$. For $^{12}C^{16}O$: $\mu_1 = \frac{12 \times 16}{12 + 16} = 6.857$. For $^{13}C^{16}O$: $\mu_2 = \frac{13 \times 16}{13 + 16} = 7.172$. Since $B \propto \frac{1}{I} \propto \frac{1}{\mu}$, we have $\frac{B_2}{B_1} = \frac{\mu_1}{\mu_2} = \frac{6.857}{7.172} = 0.956$. Therefore $B_2 = 1.9225 \times 0.956 = 1.838 \text{ cm}^{-1}$. The third line ($J = 3 \leftarrow 2$) occurs at $6B_2 = 6 \times 1.838 = 11.03 \text{ cm}^{-1} \approx 10.97 \text{ cm}^{-1}$. (B) uses incorrect mass ratios. (C) assumes the frequency increases with mass. (D) uses atomic masses instead of reduced masses.
4. The rotational spectrum of CO shows transitions at 3.845, 7.690, 11.535, and 15.380 $\text{cm}^{-1}$. A student calculates the moment of inertia as $14.6 \times 10^{-47} \text{ kg}\cdot\text{m}^2$ but realizes this differs significantly from the expected value based on known bond lengths. What is the most likely explanation for this discrepancy?
   Answer: The molecule exhibits significant centrifugal distortion effects that were not accounted for in the rigid rotor approximation When analyzing rotational spectra, you're looking at transitions between rotational energy levels in molecules. The rigid rotor model provides the foundation, but real molecules deviate from this idealized behavior in predictable ways. The key insight here is recognizing when experimental data doesn't match theoretical predictions. The evenly spaced transitions (3.845, 7.690, 11.535, 15.380 cm⁻¹) initially suggest a simple rigid rotor pattern, but the calculated moment of inertia significantly differs from the expected value based on known molecular dimensions. A is correct because centrifugal distortion is the most common and significant source of deviation from rigid rotor behavior. As molecules rotate faster (higher J levels), centrifugal forces stretch the bond, effectively increasing the moment of inertia and decreasing the rotational constant B. This systematically affects all transition frequencies and leads to calculated moments of inertia that differ from the true equilibrium value. The rigid rotor approximation fails to account for this bond stretching. B is wrong because rotational overtones are forbidden in pure rotational spectroscopy due to selection rules (ΔJ = ±1). C is incorrect because vibrational excitation would typically show additional spectral lines or splitting, not just a systematic shift in the calculated moment of inertia from a single series of transitions. D is wrong because Stark effect broadening requires an external electric field and would broaden lines rather than systematically shift calculated molecular parameters. Remember: When experimental rotational constants don't match expected values, centrifugal distortion is usually the culprit. Always consider whether your model accounts for molecular flexibility.
5. A prolate symmetric top molecule ($A > B = C$) exhibits rotational structure in its electronic spectrum. For the $J = 3, K = 2$ level, the rotational energy is $47.3 \text{ cm}^{-1}$ above the $J = 0$ state. For the $J = 4, K = 1$ level, the energy is $52.8 \text{ cm}^{-1}$. What are the rotational constants $A$ and $B$?
   Answer: $A = 11.8 \text{ cm}^{-1}, B = 2.4 \text{ cm}^{-1}$ When you encounter rotational spectroscopy problems involving symmetric top molecules, you need to understand that these molecules have two distinct rotational constants: $A$ (rotation about the unique axis) and $B = C$ (rotation about the equivalent axes). For a prolate symmetric top where $A > B = C$, the rotational energy is given by: $E_{rot} = BJ(J+1) + (A-B)K^2$ where $J$ is the total angular momentum quantum number and $K$ is the projection along the molecular symmetry axis. Setting up two equations with the given data: - For $J = 3, K = 2$: $47.3 = B(3)(4) + (A-B)(4) = 12B + 4A - 4B = 8B + 4A$ - For $J = 4, K = 1$: $52.8 = B(4)(5) + (A-B)(1) = 20B + A - B = 19B + A$ From the first equation: $47.3 = 8B + 4A$, so $A = \frac{47.3 - 8B}{4}$ Substituting into the second equation: $52.8 = 19B + \frac{47.3 - 8B}{4}$ Solving: $211.2 = 76B + 47.3 - 8B = 68B + 47.3$ Therefore: $B = \frac{163.9}{68} = 2.4 \text{ cm}^{-1}$ and $A = 11.8 \text{ cm}^{-1}$ Options A, B, and C all result from algebraic errors in solving the simultaneous equations or incorrectly applying the energy formula. Option D gives the correct values that satisfy both experimental observations. Study tip: Always write out the energy expression first, then carefully substitute the quantum numbers to set up your system of equations. Double-check by plugging your answers back into the original equations.
6. The millimeter-wave spectrum of a molecule shows a series of lines with frequencies (in GHz): 150.176, 225.264, 300.352, 375.440, 450.528. When the same molecule is studied at high temperature (500 K), additional weaker lines appear at 149.891, 224.837, 299.782, 374.728. What is the most likely explanation for these additional lines?
   Answer: Thermal population of vibrationally excited states with different effective rotational constants When analyzing rotational spectra at different temperatures, you need to consider how thermal energy affects molecular energy states. This millimeter-wave spectrum shows classic rotational transitions, and the appearance of new lines at high temperature is a key diagnostic clue. The correct answer is A because vibrationally excited states have different rotational constants than the ground vibrational state. At 500 K, thermal energy populates higher vibrational levels ($v = 1, 2, ..$), and each vibrational state has its own effective rotational constant $B_v = B_e - α_e(v + 1/2)$. Since $α_e > 0$, higher vibrational states have smaller rotational constants, producing rotational lines at slightly lower frequencies. The systematic downward shift you observe (150.176 → 149.891 GHz, etc.) confirms this vibrational hot band origin. B is wrong because Doppler broadening would broaden existing lines rather than create distinct new lines at systematically shifted frequencies. C is incorrect because electronic excited states would require much higher energies than available at 500 K, and would show much larger frequency shifts due to dramatically different molecular geometries. D is wrong because collision-induced transitions would create entirely different spectral patterns, not systematic parallel shifts of the existing rotational progression. Study tip: When you see new spectral lines appearing at higher temperature with small, systematic frequency shifts, immediately think vibrational hot bands. The key pattern is that all lines shift by similar amounts in the same direction, reflecting the change in rotational constant between vibrational levels.
7. A symmetric top molecule has rotational constants $A = 10.5 \text{ cm}^{-1}$ and $B = C = 3.2 \text{ cm}^{-1}$. In the rotational spectrum, what is the degeneracy of the $J = 3, K = 1$ rotational level?
   Answer: $14$ For a symmetric top molecule, each rotational level is characterized by quantum numbers $J$ and $K$, where $K$ is the component of angular momentum along the symmetry axis. The total degeneracy includes both the $M_J$ degeneracy ($2J + 1$ values) and the $K$ degeneracy. For $K \neq 0$, there are two states with the same energy: $+K$ and $-K$, so the $K$ degeneracy is 2. For $J = 3, K = 1$: $M_J$ can take values $-3, -2, -1, 0, +1, +2, +3$ (7 values), and $K$ can be $+1$ or $-1$ (2 values). Total degeneracy = $7 \times 2 = 14$. (A) would be just the number of $|K|$ values for $J = 3$. (B) would be $2J + 1$ ignoring $K$ degeneracy. (C) would be $2(2J + 1)$ but with an error in $J$.
8. The rotational spectrum of a heteronuclear diatomic molecule AB shows transitions at 115.27, 230.54, 345.81, and 461.08 GHz. However, when the pure compound is replaced with a 1:1 isotopic mixture of A and A*, the spectrum becomes more complex. What is the primary cause of this spectral complexity?
   Answer: Two distinct molecular species AB and A*B have different rotational constants leading to overlapping spectra When you encounter rotational spectroscopy problems involving isotopic substitution, focus on how mass changes affect molecular properties. The key insight is that isotopes have identical electronic structures but different masses, which directly impacts rotational behavior. The original spectrum shows evenly spaced transitions (115.27, 230.54, 345.81, 461.08 GHz) with intervals of approximately 115.27 GHz, characteristic of a simple heteronuclear diatomic molecule. This spacing equals $2B$, where $B$ is the rotational constant. When you introduce an isotopic mixture creating both AB and A*B molecules, you now have two distinct species with different reduced masses. Since the rotational constant $B = \frac{h}{8\pi^2Ic}$ depends on the moment of inertia $I = \mu r^2$ (where $\mu$ is the reduced mass), each isotopic variant has its own unique rotational constant. This creates two overlapping rotational spectra with slightly different transition frequencies, making the spectrum complex. Option A is incorrect because chemical exchange doesn't occur between isotopomers under normal spectroscopic conditions—they're separate, stable molecules. Option C misses the mark because isotopic substitution doesn't alter electronic structure or selection rules; the electronic environment remains essentially identical. Option D incorrectly suggests nuclear spin effects, but the complexity arises from having two molecular species, not from spin-spin coupling creating hyperfine structure. Remember: isotopic substitution in rotational spectroscopy always creates multiple overlapping spectra due to different rotational constants, not changes in selection rules or electronic effects.

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