Isotopic Effects on Spectra - Physical Chemistry 2

1. In the rotational Raman spectrum of $^{14}N_2$, the Stokes lines appear at frequency shifts of 19.9, 33.2, 46.4, and 59.7 cm$^{-1}$ from the Rayleigh line. When the same experiment is performed with $^{15}N_2$, what characteristic change would be most prominent in the spectrum?
   Answer: The spacing between adjacent Stokes lines decreases to approximately 12.6 cm$^{-1}$ due to the increased moment of inertia In Raman spectroscopy, the spacing between adjacent Stokes lines equals $4B$ where $B$ is the rotational constant. From the given data, the spacing is approximately 13.3 cm$^{-1}$, so $B_{^{14}N_2} ≈ 3.33$ cm$^{-1}$. Since $B \propto I^{-1}$ and $I \propto \mu r^2$ (where $r$ is the bond length, which doesn't change significantly), we have $\frac{B_{^{15}N_2}}{B_{^{14}N_2}} = \frac{\mu_{^{14}N_2}}{\mu_{^{15}N_2}} = \frac{14/2}{15/2} = \frac{14}{15} = 0.933$. Therefore, $B_{^{15}N_2} = 3.33 \times 0.933 = 3.11$ cm$^{-1}$, giving a spacing of $4 \times 3.11 = 12.4$ cm$^{-1}$ (closest to 12.6). Choice A gives an incorrect calculation. Choice C incorrectly assumes no frequency change. Choice D is wrong because isotopic substitution doesn't significantly affect the population distribution.
2. A student analyzes the IR spectrum of HDO vapor and observes three fundamental vibrational modes at 3707, 2727, and 1595 cm$^{-1}$. When comparing this to H$_2$O (with fundamentals at 3657, 3756, and 1595 cm$^{-1}$), which statement best explains the observed isotopic shifts?
   Answer: The symmetric and antisymmetric O-H stretches in H$_2$O become distinguishable in HDO because the deuterium breaks the molecular symmetry In H$_2$O, the C$_{2v}$ symmetry leads to symmetric (3657 cm$^{-1}$) and antisymmetric (3756 cm$^{-1}$) O-H stretching modes. HDO has C$_s$ symmetry, so the two O-H bonds are no longer equivalent—one is trans to deuterium, the other cis. This breaks the degeneracy, and we observe separate O-H (3707 cm$^{-1}$) and O-D (2727 cm$^{-1}$) stretches. The O-D stretch is lower due to the $\sqrt{2}$ mass ratio effect. Choice B incorrectly suggests the O-H frequency is unchanged when it actually shifts due to vibrational coupling changes. Choice C incorrectly applies molecular mass effects to vibrational frequencies (reduced masses matter, not total mass). Choice D is wrong because the bending mode does shift slightly due to mass distribution changes affecting the normal mode composition.
3. A researcher studies the electronic absorption spectrum of gaseous $^{79}Br_2$ and observes that the (v'=0, v''=0) transition of the A$^3\Pi_{1u} \leftarrow$ X$^1\Sigma_g^+$ band system appears at 18,345 cm$^{-1}$. The vibrational progression in the excited state shows spacings of 285 cm$^{-1}$. When the same experiment is repeated with $^{81}Br_2$, which prediction is most accurate?
   Answer: The (0,0) band will remain at 18,345 cm$^{-1}$ and the excited state vibrational spacing will become 280 cm$^{-1}$ Electronic transition energies (band origins) are primarily determined by electronic energy differences and are largely independent of isotopic substitution—the electronic states themselves don't change significantly. Therefore, the (0,0) band remains near 18,345 cm$^{-1}$. However, vibrational spacings do change with isotopic substitution. The vibrational frequency scales as $\omega \propto \mu^{-1/2}$ where $\mu$ is the reduced mass. For Br$_2$: $\mu = m/2$, so $\frac{\omega_{^{81}Br_2}}{\omega_{^{79}Br_2}} = \sqrt{\frac{79}{81}} = 0.988$. Thus, the new spacing is $285 \times 0.988 = 282$ cm$^{-1}$ (closest to 280). Choice A incorrectly predicts an electronic energy shift. Choice C incorrectly assumes no vibrational frequency change. Choice D incorrectly assumes complete independence from nuclear mass effects.
4. A researcher observes that the fundamental vibrational frequency of $^{12}C^{16}O$ is 2143 cm$^{-1}$, while that of $^{13}C^{16}O$ is 2096 cm$^{-1}$. If the same researcher then examines $^{12}C^{18}O$, which prediction about its vibrational frequency is most consistent with the harmonic oscillator model?
   Answer: The frequency will be approximately 2090 cm$^{-1}$ because the reduced mass change is similar to the $^{13}C^{16}O$ case The vibrational frequency depends on $\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$ where $\mu$ is the reduced mass. For $^{12}C^{16}O$: $\mu_1 = \frac{12 \times 16}{12 + 16} = 6.86$ amu. For $^{13}C^{16}O$: $\mu_2 = \frac{13 \times 16}{13 + 16} = 7.17$ amu. For $^{12}C^{18}O$: $\mu_3 = \frac{12 \times 18}{12 + 18} = 7.20$ amu. Since $\nu \propto \mu^{-1/2}$, the frequency ratios give: $\frac{\nu_3}{\nu_1} = \sqrt{\frac{6.86}{7.20}} = 0.976$, so $\nu_3 \approx 2143 \times 0.976 = 2092$ cm$^{-1}$. Choice A incorrectly assumes oxygen dominates the mass effect. Choice C underestimates the oxygen isotope effect. Choice D incorrectly assumes the force constant changes with isotopic substitution.
5. Resonance Raman spectroscopy of $^{16}O_2$ using 488 nm excitation shows an intense progression in the $\nu_1$ vibrational mode with spacings of 1580 cm$^{-1}$ in the excited electronic state. The progression extends to v' = 12 before intensity falls below detection limits. When the same experiment is performed on $^{16}O^{18}O$, which change in the vibrational progression is most likely?
   Answer: The progression spacing decreases to 1540 cm$^{-1}$ and extends to v' = 13 due to the reduced vibrational frequency When analyzing isotope effects in vibrational spectroscopy, you need to consider how mass changes affect both vibrational frequencies and quantum mechanical selection rules. The key insight is that isotopic substitution changes the reduced mass, which directly impacts vibrational behavior. For $^{16}O^{18}O$, the reduced mass increases compared to $^{16}O_2$, causing the vibrational frequency to decrease according to $\nu \propto \sqrt{k/\mu}$. The new frequency becomes approximately 1540 cm$^{-1}$ (a decrease of about 2.5%, consistent with the mass change). Simultaneously, the lower vibrational frequency means that for a given amount of vibrational energy deposited during electronic excitation, more vibrational quanta can be accommodated, extending the observable progression to higher v' values. Option A incorrectly assumes resonance enhancement eliminates isotopic effects—while resonance does enhance intensity, it doesn't override fundamental vibrational physics. Option B wrongly suggests decreased maximum v' and attributes the change to anharmonicity rather than the primary isotopic mass effect. The anharmonicity doesn't significantly change between isotopologues. Option C correctly identifies that spacing remains unchanged (which is wrong) but mentions Franck-Condon factors—while these do change slightly with isotopic substitution, the primary effect is the frequency shift. Remember that isotopic substitution in vibrational spectroscopy always affects frequencies through reduced mass changes, and lower frequencies generally allow access to higher vibrational quantum numbers for a given energy range. This principle applies broadly across vibrational spectroscopic techniques.
6. In high-resolution NMR spectroscopy, the $^1H$ chemical shift of the methyl group in $^{12}CH_3$-$^{12}COOH$ appears at 2.08 ppm. When this compound is replaced with $^{13}CH_3$-$^{12}COOH$, which effect on the $^1H$ NMR spectrum is most likely to be observed?
   Answer: The $^1H$ chemical shift changes to approximately 2.06 ppm due to the secondary isotope effect on electronic shielding Secondary isotope effects occur when isotopic substitution at a site adjacent to the observed nucleus causes small changes in chemical shift, typically 0.01-0.1 ppm for $^{13}C$ substitution. This arises from changes in bond lengths, vibrational frequencies, and electronic distribution. The $^{13}C$ substitution slightly alters the C-H bond properties and electronic shielding of the protons, typically causing an upfield shift (smaller ppm value). Choice B is incorrect because while $^{13}C$-$^1H$ coupling does occur ($^1J_{CH} ≈ 125$ Hz), it would be present in natural abundance $^{13}C$ anyway and doesn't explain the isotope effect. Choice C incorrectly predicts a downfield shift. Choice D is wrong because secondary isotope effects are well-documented and measurable with high-resolution NMR.
7. Mass spectrometry of $^{32}S^{16}O^{35}Cl_2$ shows a molecular ion peak at m/z = 119. In the fragmentation pattern, a prominent peak appears at m/z = 84 corresponding to loss of Cl. If the same analysis is performed on $^{32}S^{18}O^{35}Cl_2$, which fragmentation pathway becomes thermodynamically favored compared to the $^{16}O$ analog?
   Answer: Loss of $^{18}O$ becomes more favorable because the heavier oxygen isotope has a lower zero-point vibrational energy When analyzing mass spectrometry fragmentation patterns involving isotopic substitution, you need to consider how isotope effects influence bond strengths and vibrational energies. The key principle here is that heavier isotopes have lower zero-point vibrational energies, which affects the energy required for bond breaking. In the $^{18}O$ isotopologue, the S-$^{18}O$ bond has a lower zero-point energy compared to the S-$^{16}O$ bond because the heavier oxygen atom vibrates at a lower frequency. This lower vibrational energy means less energy is required to break the S-$^{18}O$ bond, making loss of $^{18}O$ thermodynamically more favorable than loss of $^{16}O$ in the original molecule. Answer C correctly identifies this isotope effect. Answer A is incorrect because isotopic substitution doesn't significantly strengthen the S-O bond relative to S-Cl bonds—the primary effect is on vibrational energies, not relative bond strengths between different atom types. Answer B misunderstands the mechanism; simultaneous loss of both chlorines isn't favored simply because $^{18}O$ increases stability—in fact, the opposite occurs since the $^{18}O$ bond becomes easier to break. Answer D incorrectly assumes isotope effects are negligible; while bond dissociation energies change only slightly, the zero-point energy differences can significantly affect fragmentation preferences. Remember that isotope effects in mass spectrometry primarily arise from differences in zero-point vibrational energies, not from major changes in bond strengths. Heavier isotopes always have lower vibrational frequencies and thus lower zero-point energies, making their bonds slightly easier to break.
8. In the photoelectron spectrum of $^{32}S^{16}O_2$, the first ionization peak shows vibrational fine structure with a spacing of 1350 cm$^{-1}$. If the same measurement is performed on $^{34}S^{16}O_2$, which factor most accurately predicts the change in vibrational spacing?
   Answer: The spacing will decrease by a factor of $\sqrt{\frac{32.8}{33.8}} = 0.985$ because the effective mass for the symmetric stretch involves all three atoms In SO$_2$, the symmetric stretching mode (which typically dominates photoelectron vibrational progressions) involves motion of all three atoms. The effective mass for this motion is approximately the reduced mass of the entire system, which can be estimated as $\mu_{eff} ≈ \frac{m_S + 2m_O}{3} = \frac{32 + 32}{3} = 21.3$ for $^{32}S^{16}O_2$ and $\frac{34 + 32}{3} = 22.0$ for $^{34}S^{16}O_2$. However, a more accurate treatment considers the center-of-mass motion, giving an effective mass ratio closer to $\frac{32.8}{33.8}$. The frequency ratio is then $\sqrt{\frac{32.8}{33.8}} = 0.985$, predicting a spacing of 1330 cm$^{-1}$. Choice A incorrectly uses only the sulfur mass ratio. Choice B incorrectly uses the total molecular mass. Choice D incorrectly assumes no vibrational effect from isotopic substitution.
9. ESR spectroscopy of the $^{12}CH_3$ radical shows a 1:3:3:1 quartet pattern due to hyperfine coupling with three equivalent protons ($a_H = 23.0$ G). When the same radical is generated using $^{13}CH_3$, which hyperfine pattern is expected?
   Answer: A 1:3:3:1 quartet ($a_H = 23.0$ G) split into doublets by $^{13}C$ coupling ($a_C ≈ 38$ G) When analyzing ESR hyperfine coupling patterns, you need to consider how each magnetic nucleus independently splits the electron spin signal. The key principle is that different nuclei with different coupling constants create separate, multiplicative splitting patterns. In $^{12}CH_3$, only the three equivalent protons ($I = 1/2$ each) cause hyperfine splitting, giving the observed 1:3:3:1 quartet with $a_H = 23.0$ G. When you switch to $^{13}CH_3$, you add $^{13}C$ ($I = 1/2$) as another magnetic nucleus, but the three protons remain and still couple with the same strength. The $^{13}C$ nucleus acts independently from the protons, splitting each line of the existing 1:3:3:1 quartet into a doublet. This creates a "quartet of doublets" pattern where the proton coupling ($a_H = 23.0$ G) determines the major splitting and the carbon coupling ($a_C ≈ 38$ G) creates the fine structure. Answer A incorrectly assumes the couplings combine additively to create four equivalent nuclei—they don't. Answer B ignores the magnetic properties of $^{13}C$; while $^{12}C$ is NMR-silent, $^{13}C$ definitely couples to the unpaired electron. Answer C incorrectly suggests the carbon coupling would dominate and obscure the proton splitting—both coupling patterns remain observable with different magnitudes. Remember: In ESR, each type of magnetic nucleus creates its own independent splitting pattern. When multiple nuclei are present, you see multiplicative splitting—one pattern split by another—not additive effects.
10. In the far-infrared spectrum of $^{16}O^{12}C^{32}S$, the first few rotational transitions (J+1 ← J) of the ground vibrational state appear at 24.3, 48.6, 72.9, and 97.2 cm$^{-1}$. A student calculates the bond lengths using the rigid rotor model and obtains r(C=O) = 1.17 Å and r(C=S) = 1.56 Å. If the same analysis is applied to $^{18}O^{12}C^{32}S$, which result is most consistent with the expected isotopic effect?
    Answer: The rotational transitions will appear at 23.1, 46.2, 69.3, and 92.4 cm$^{-1}$, and the calculated bond lengths will be r(C=O) = 1.17 Å and r(C=S) = 1.56 Å The rotational constant $B = \frac{h}{8\pi^2 I c}$ depends on the moment of inertia $I$. For OCS, $I = \sum m_i r_i^2$ where $r_i$ is the distance from each atom to the center of mass. Replacing $^{16}O$ with $^{18}O$ increases $I$, decreasing $B$. The transition spacing $\Delta\tilde{\nu} = 2B$ decreases proportionally. However, the actual bond lengths (C=O and C=S distances) remain unchanged—only the center of mass position shifts. The apparent change in calculated bond lengths in choice B would result from incorrectly assuming the center of mass position doesn't change with isotopic substitution. Choice C incorrectly assumes no rotational frequency change. Choice D incorrectly predicts an increase in rotational frequencies.

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