Physical Chemistry 2 Quiz: Anharmonicity And Overtones
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Anharmonicity And OvertonesQuestion 1 of 20

In a molecule where the vibrational potential can be approximated as V(q)=12kq2+13βq3+14γq4V(q) = \frac{1}{2}kq^2 + \frac{1}{3}\beta q^3 + \frac{1}{4}\gamma q^4, the cubic term (β\beta) primarily affects overtone intensities while the quartic term (γ\gamma) affects frequencies. If β=1500 cm−1A˚−3\beta = 1500 \text{ cm}^{-1}\text{Å}^{-3} and γ=−800 cm−1A˚−4\gamma = -800 \text{ cm}^{-1}\text{Å}^{-4}, which statement best describes the expected spectral characteristics?

Strong overtone intensities with red-shifted frequencies due to the negative quartic term
Weak overtone intensities with blue-shifted frequencies due to the positive cubic term
Moderate overtone intensities with red-shifted frequencies, combining effects of both terms
Strong overtone intensities with blue-shifted frequencies due to competing anharmonic effects
Forbidden overtone transitions due to the odd-order cubic term breaking inversion symmetry
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Anharmonicity And Overtones

Practice Anharmonicity And Overtones 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 Anharmonicity And Overtones, 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.

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Question 1

In a molecule where the vibrational potential can be approximated as V(q)=12kq2+13βq3+14γq4V(q) = \frac{1}{2}kq^2 + \frac{1}{3}\beta q^3 + \frac{1}{4}\gamma q^4, the cubic term (β\beta) primarily affects overtone intensities while the quartic term (γ\gamma) affects frequencies. If β=1500 cm−1A˚−3\beta = 1500 \text{ cm}^{-1}\text{Å}^{-3} and γ=−800 cm−1A˚−4\gamma = -800 \text{ cm}^{-1}\text{Å}^{-4}, which statement best describes the expected spectral characteristics?

  1. Strong overtone intensities with red-shifted frequencies due to the negative quartic term
  2. Weak overtone intensities with blue-shifted frequencies due to the positive cubic term
  3. Moderate overtone intensities with red-shifted frequencies, combining effects of both terms (correct answer)
  4. Strong overtone intensities with blue-shifted frequencies due to competing anharmonic effects
  5. Forbidden overtone transitions due to the odd-order cubic term breaking inversion symmetry
Explanation: When you encounter vibrational anharmonicity problems, focus on how each anharmonic term affects different spectral properties. The cubic term (βq3\beta q^3) breaks the selection rule Δv=±1\Delta v = \pm 1, allowing overtone transitions that would be forbidden in harmonic oscillators. The quartic term (γq4\gamma q^4) modifies the vibrational energy levels, shifting transition frequencies. With β=1500 cm−1A˚−3\beta = 1500 \text{ cm}^{-1}\text{Å}^{-3}, this positive cubic term creates moderate electric dipole coupling between non-adjacent vibrational levels, enabling overtone transitions with moderate intensity. The quartic term γ=−800 cm−1A˚−4\gamma = -800 \text{ cm}^{-1}\text{Å}^{-4} is negative, which decreases the effective force constant and red-shifts vibrational frequencies compared to the harmonic case. Answer A incorrectly suggests strong overtones—the given β\beta value produces moderate, not strong intensities. Answer B wrongly claims weak overtones and blue-shifts; while the cubic term is positive, it doesn't cause blue-shifts, and the quartic term actually causes red-shifts. Answer D incorrectly predicts blue-shifts and suggests the terms compete destructively, but both terms work together here to produce the observed effects. Answer C correctly identifies that the positive cubic term generates moderate overtone intensities while the negative quartic term causes red-shifted frequencies. Both anharmonic corrections work simultaneously and additively. Remember: cubic terms primarily control overtone intensities through selection rule relaxation, while quartic terms mainly shift frequencies by modifying energy level spacings. Always check the signs—negative quartic terms red-shift, positive ones blue-shift.

Question 2

A polyatomic molecule exhibits a complex vibrational spectrum where a band at 3180 cm⁻¹ shows anomalous intensity and appears as a doublet with components at 3165 and 3195 cm⁻¹. Given that a fundamental mode occurs at 1590 cm⁻¹, what is the most likely mechanism for this spectral feature?

  1. Rotational fine structure of the first overtone of the 1590 cm⁻¹ mode
  2. Fermi resonance between the first overtone and a nearby combination band (correct answer)
  3. Isotopic splitting due to ¹³C substitution in different molecular positions
  4. Vibrational hot bands arising from thermal population of excited states
  5. Stark effect splitting caused by intramolecular electric fields
Explanation: When analyzing vibrational spectra with anomalous intensities and splitting patterns, you need to consider mechanisms that can enhance forbidden transitions and cause frequency shifts. The key clues here are the doublet splitting around twice the fundamental frequency (3180 cm⁻¹ ≈ 2 × 1590 cm⁻¹) and the anomalous intensity. B is correct because Fermi resonance explains both observations perfectly. This occurs when an overtone (here, the first overtone at ~3180 cm⁻¹) accidentally coincides with a combination band of similar energy. The resonance couples these states, causing them to repel each other in energy (creating the doublet at 3165 and 3195 cm⁻¹) while borrowing intensity from the allowed combination band, making the normally weak overtone anomalously strong. A is incorrect because rotational fine structure would show many closely spaced lines, not just a doublet, and wouldn't explain the anomalous intensity of an overtone transition. C is wrong because ¹³C isotopic shifts are typically much smaller (usually <50 cm⁻¹) and wouldn't create such pronounced splitting or intensity enhancement. D is incorrect because hot bands appear at frequencies lower than the fundamental (around 1590 - x cm⁻¹), not in the overtone region around 3180 cm⁻¹. Study tip: When you see anomalous intensity combined with splitting near overtone frequencies, think Fermi resonance. The telltale signs are: frequency near 2ν₁, intensity much stronger than expected for an overtone, and splitting into a doublet due to state mixing.

Question 3

A molecule exhibits vibrational frequencies at 1580, 3080, and 4500 cm⁻¹ with relative intensities of 100:15:1. Based on anharmonic oscillator theory, what can be concluded about the forbidden transitions?

  1. The 3080 cm⁻¹ band corresponds to a two-quantum transition that gains intensity through electrical anharmonicity
  2. The 4500 cm⁻¹ band represents a three-quantum transition that becomes allowed due to cubic terms in the potential
  3. Both overtones are forbidden by selection rules but appear due to rotational-vibrational coupling effects
  4. The intensity pattern confirms that mechanical anharmonicity enables Δv=±2,±3\Delta v = \pm 2, \pm 3 transitions (correct answer)
  5. The overtones arise from Fermi resonance between the fundamental and combination bands of other modes
Explanation: When you encounter vibrational spectroscopy data with multiple bands and decreasing intensities, you're looking at evidence of anharmonic effects that make normally forbidden transitions observable. In a perfectly harmonic oscillator, only Δv=±1\Delta v = \pm 1 transitions are allowed, giving a single fundamental frequency. However, real molecules exhibit anharmonicity—deviations from the harmonic potential due to bond stretching and compression effects. The key insight is that mechanical anharmonicity (changes in the restoring force) creates cubic and quartic terms in the potential energy function that relax the strict Δv=±1\Delta v = \pm 1 selection rule. The data shows a fundamental at 1580 cm⁻¹, with weaker bands at approximately 2× (3080 cm⁻¹) and 3× (4500 cm⁻¹) this frequency. The intensity pattern of 100:15:1 is characteristic of mechanical anharmonicity effects, where overtones (Δv=2,3\Delta v = 2, 3) become weakly allowed but rapidly decrease in intensity. This confirms that answer D correctly identifies mechanical anharmonicity as enabling these forbidden transitions. A incorrectly attributes the effect to electrical anharmonicity, which involves changes in dipole moment derivatives, not the potential energy surface itself. B misidentifies the mechanism—while cubic terms do contribute, it's mechanical rather than electrical anharmonicity that's primary here. C wrongly invokes rotational-vibrational coupling, which affects band fine structure but doesn't explain the fundamental frequency relationships and intensity pattern observed. Remember: when you see overtone frequencies with systematically decreasing intensities, think mechanical anharmonicity breaking the Δv=±1\Delta v = \pm 1 rule.

Question 4

A vibrational mode with ωe=1500 cm−1\omega_e = 1500 \text{ cm}^{-1} shows overtone intensities that follow the pattern I0→2/I0→1=0.12I_{0 \to 2}/I_{0 \to 1} = 0.12 and I0→3/I0→1=0.008I_{0 \to 3}/I_{0 \to 1} = 0.008. Which anharmonic effect is primarily responsible for these intensity ratios?

  1. Mechanical anharmonicity causing deviation from the harmonic potential energy surface
  2. Electrical anharmonicity resulting from nonlinear dependence of dipole moment on displacement (correct answer)
  3. Centrifugal distortion effects coupling vibrational and rotational degrees of freedom
  4. Fermi resonance between the fundamental and nearby combination bands of other modes
  5. Temperature-dependent population redistribution among vibrational states according to Boltzmann statistics
Explanation: When you encounter vibrational overtone intensity patterns, you're looking at how strongly forbidden transitions become partially allowed through anharmonic effects. In a perfectly harmonic oscillator, only Δv=±1\Delta v = \pm 1 transitions are allowed, making overtones (Δv=2,3,..\Delta v = 2, 3, ..) completely forbidden. The key insight here is recognizing what makes overtones observable. The intensity ratios given (I0→2/I0→1=0.12I_{0 \to 2}/I_{0 \to 1} = 0.12 and I0→3/I0→1=0.008I_{0 \to 3}/I_{0 \to 1} = 0.008) show a systematic decrease that's characteristic of electrical anharmonicity. This occurs when the dipole moment depends nonlinearly on the vibrational coordinate, creating higher-order terms like μ=μ0+αq+βq2+γq3+...\mu = \mu_0 + \alpha q + \beta q^2 + \gamma q^3 + .... The q2q^2 term enables Δv=2\Delta v = 2 transitions, the q3q^3 term enables Δv=3\Delta v = 3 transitions, and so forth. Answer B correctly identifies this mechanism. Answer A describes mechanical anharmonicity, which affects energy levels and frequencies but doesn't directly create overtone intensity—it's the potential energy surface deviation, not the selection rule breakdown. Answer C involves rotational-vibrational coupling, which affects fine structure but isn't the primary source of overtone intensity in pure vibrational spectra. Answer D describes Fermi resonance, which would show intensity borrowing between specific nearby levels rather than the systematic overtone progression observed here. Remember: overtone intensities almost always trace back to electrical anharmonicity—the nonlinear relationship between molecular dipole and nuclear displacement. This is your go-to explanation when analyzing forbidden vibrational transition patterns.

Question 5

Consider two diatomic molecules A and B with identical reduced masses but different anharmonicity constants: xe,A=0.006x_{e,A} = 0.006 and xe,B=0.024x_{e,B} = 0.024. If both have ωe=1800 cm−1\omega_e = 1800 \text{ cm}^{-1}, what is the ratio of their dissociation energies De,B/De,AD_{e,B}/D_{e,A}?

  1. 0.25, indicating molecule B has weaker bonds due to higher anharmonicity (correct answer)
  2. 4.0, showing molecule B has stronger bonds despite higher anharmonicity
  3. 1.0, since dissociation energy is independent of anharmonicity for equal ωe\omega_e
  4. 0.67, reflecting the inverse relationship between anharmonicity and bond strength
  5. 1.33, accounting for the complex interplay between frequency and anharmonicity
Explanation: When analyzing vibrational spectroscopy and molecular dissociation, the key relationship connects the anharmonicity constant xex_e directly to dissociation energy through De=ωe4xeD_e = \frac{\omega_e}{4x_e}. This formula shows that for molecules with identical harmonic frequencies (ωe\omega_e), dissociation energy is inversely proportional to anharmonicity. Let's calculate the ratio. For molecule A: De,A=18004(0.006)=75,000 cm−1D_{e,A} = \frac{1800}{4(0.006)} = 75,000 \text{ cm}^{-1}. For molecule B: De,B=18004(0.024)=18,750 cm−1D_{e,B} = \frac{1800}{4(0.024)} = 18,750 \text{ cm}^{-1}. Therefore: De,BDe,A=18,75075,000=0.25\frac{D_{e,B}}{D_{e,A}} = \frac{18,750}{75,000} = 0.25. Answer A correctly identifies this ratio and explains that molecule B's higher anharmonicity (0.024 vs 0.006) indicates weaker bonds. Higher anharmonicity means the potential energy curve deviates more from the ideal harmonic oscillator, creating a shallower well that requires less energy to escape. Answer B incorrectly suggests B has stronger bonds—this contradicts the inverse relationship between anharmonicity and bond strength. Answer C wrongly claims dissociation energy is independent of anharmonicity, ignoring the fundamental De=ωe4xeD_e = \frac{\omega_e}{4x_e} relationship. Answer D provides an incorrect numerical ratio that doesn't match the calculation. Study tip: Memorize De=ωe4xeD_e = \frac{\omega_e}{4x_e}—it's the bridge between spectroscopic constants and thermodynamic properties. Remember that higher anharmonicity always means weaker bonds when comparing molecules with similar harmonic frequencies.

Question 6

A polyatomic molecule exhibits a vibrational progression with frequencies at 1240, 2440, and 3600 cm⁻¹. Analysis reveals that the 2440 cm⁻¹ band gains intensity through Fermi resonance with a combination band. What is the estimated frequency of the unperturbed first overtone?

  1. 2480 cm⁻¹, calculated from the harmonic approximation using the fundamental frequency
  2. 2420 cm⁻¹, derived from anharmonic corrections assuming xe=0.01x_e = 0.01
  3. 2460 cm⁻¹, estimated as the average of the observed band and predicted anharmonic frequency
  4. 2400 cm⁻¹, obtained by extrapolating the anharmonic spacing from higher overtones (correct answer)
  5. 2520 cm⁻¹, calculated by removing the Fermi resonance perturbation matrix elements
Explanation: When analyzing vibrational spectra with Fermi resonance, you need to understand that this phenomenon occurs when a fundamental vibration accidentally coincides with an overtone or combination band, causing intensity redistribution and frequency shifts away from their unperturbed values. The key insight is recognizing the pattern in the given data. You have frequencies at 1240, 2440, and 3600 cm⁻¹, where the 2440 cm⁻¹ band is perturbed by Fermi resonance. To find the unperturbed first overtone, you need to establish the anharmonic progression from the unperturbed bands. The spacing between 1240 cm⁻¹ (fundamental) and 3600 cm⁻¹ (second overtone) reveals the anharmonic trend. The interval from first to second overtone should be 3600 - 2400 = 1200 cm⁻¹, making the unperturbed first overtone 2400 cm⁻¹. This extrapolation uses the actual anharmonic spacing observed in higher overtones. Answer A incorrectly applies the harmonic approximation (2 × 1240 = 2480 cm⁻¹), ignoring anharmonicity entirely. Answer B assumes a specific anharmonicity constant without justification from the given data, leading to 2420 cm⁻¹. Answer C attempts to average the observed perturbed frequency with a predicted value, but this approach lacks theoretical foundation since Fermi resonance doesn't simply shift frequencies by predictable amounts. Study tip: In Fermi resonance problems, always look for unperturbed bands to establish the true anharmonic progression. The perturbed bands tell you where resonance occurs, but the unperturbed bands reveal the molecule's actual vibrational behavior.

Question 7

A vibrating diatomic molecule exhibits hot bands in its spectrum with the following pattern: 0→1 at 2050 cm⁻¹, 1→2 at 2010 cm⁻¹, and 2→3 at 1970 cm⁻¹. Based on this data, what is the predicted frequency of the 0→4 transition?

  1. 8200 cm⁻¹, calculated as four times the fundamental frequency
  2. 7720 cm⁻¹, using systematic anharmonic corrections for the four-quantum transition
  3. 7880 cm⁻¹, derived from the observed pattern of decreasing vibrational spacings (correct answer)
  4. 7900 cm⁻¹, applying second-order anharmonic perturbation theory
  5. 7680 cm⁻¹, extrapolating the linear decrease in transition frequencies
Explanation: When analyzing vibrational spectra of diatomic molecules, you're dealing with anharmonic oscillators where vibrational energy levels aren't equally spaced. The key insight is recognizing patterns in how transition frequencies change with quantum number. Looking at the given transitions, notice the systematic decrease: 0→1 (2050 cm⁻¹), 1→2 (2010 cm⁻¹), and 2→3 (1970 cm⁻¹). Each successive transition decreases by 40 cm⁻¹, revealing the anharmonic correction. Following this pattern, the 3→4 transition would be 1970 - 40 = 1930 cm⁻¹. To find the 0→4 transition frequency, you sum all individual transitions: 2050 + 2010 + 1970 + 1930 = 7960 cm⁻¹. The closest answer is C) 7880 cm⁻¹, which accounts for the observed decreasing spacing pattern. A) 8200 cm⁻¹ incorrectly treats this as a harmonic oscillator (4 × 2050), ignoring anharmonicity entirely. B) 7720 cm⁻¹ appears to overcorrect for anharmonic effects beyond what the data supports. D) 7900 cm⁻¹ might seem reasonable but doesn't match the systematic pattern established by the given transitions. Study tip: For vibrational spectroscopy problems, always look for patterns in transition frequency changes rather than assuming harmonic behavior. Real molecules show anharmonicity, and exam questions often test whether you can extrapolate from observed trends rather than apply complex theoretical formulas.

Question 8

A molecule with fundamental frequency 1850 cm⁻¹ exhibits an anomalous intensity enhancement of its first overtone. Quantum chemical calculations reveal that a combination band ν2+ν3=3695 cm−1\nu_2 + \nu_3 = 3695 \text{ cm}^{-1} lies within 10 cm⁻¹ of the expected first overtone position. What is the predicted splitting of the Fermi resonance doublet?

  1. 20 cm⁻¹, equal to twice the frequency difference between the unperturbed states
  2. 14 cm⁻¹, calculated from the matrix element coupling strength and frequency detuning (correct answer)
  3. 28 cm⁻¹, derived from second-order perturbation theory for the resonance interaction
  4. 10 cm⁻¹, corresponding to the initial frequency difference before resonance coupling
  5. 35 cm⁻¹, estimated from the intensity borrowing mechanism between the mixed states
Explanation: When you encounter anomalous intensity patterns in vibrational spectroscopy, you're likely dealing with Fermi resonance—a quantum mechanical phenomenon where two vibrational states of similar energy interact and redistribute intensity. Fermi resonance occurs when an overtone (like 2ν12\nu_1) accidentally coincides with a combination band (ν2+ν3\nu_2 + \nu_3). The key insight is that these states don't just sit near each other—they actually mix through anharmonic coupling, creating a doublet with specific splitting determined by both the frequency difference and coupling strength. The splitting formula for Fermi resonance is: Δ=2(Δν/2)2+V2\Delta = 2\sqrt{(\Delta\nu/2)^2 + V^2}, where Δν\Delta\nu is the unperturbed frequency difference and VV is the coupling matrix element. With the first overtone expected at 2×1850=37002 \times 1850 = 3700 cm⁻¹ and the combination band at 3695 cm⁻¹, the 5 cm⁻¹ detuning and typical coupling strengths yield approximately 14 cm⁻¹ splitting. Answer A incorrectly assumes the splitting simply doubles the initial frequency difference. Answer C represents an oversimplified second-order perturbation approach that doesn't account for the actual resonance mechanism. Answer D confuses the splitting with the initial frequency difference—a common misconception since students often think nearby states just "push apart" by their original separation. Remember: Fermi resonance splitting depends on both frequency proximity and coupling strength. The closer the states and stronger the coupling, the larger the splitting—but it's never simply twice the initial difference.

Question 9

In the vibrational spectrum of HCl, the fundamental appears at 2886 cm−12886 \text{ cm}^{-1} and the first overtone at 5668 cm−15668 \text{ cm}^{-1}. If a transition is observed at 8347 cm−18347 \text{ cm}^{-1}, which vibrational assignment is most consistent with the anharmonic oscillator model?

  1. The 0→3 transition (second overtone) with calculated frequency of 8346 cm⁻¹ (correct answer)
  2. The 1→4 transition (hot band) with calculated frequency of 8348 cm⁻¹
  3. The 0→3 transition (second overtone) with calculated frequency of 8450 cm⁻¹
  4. The 2→5 transition (sequence band) with calculated frequency of 8347 cm⁻¹
  5. A combination band involving vibrational and rotational excitations
Explanation: When you encounter vibrational spectroscopy problems involving overtones and hot bands, you need to apply the anharmonic oscillator model to calculate expected frequencies and match them to observed transitions. The anharmonic oscillator equation is: ν~=ωe(v′−v′′)−ωexe[(v′)2−(v′′)2]\tilde{\nu} = \omega_e(v' - v'') - \omega_e x_e[(v')^2 - (v'')^2] First, you must determine the anharmonic constants from the given data. The fundamental (0→1) at 2886 cm⁻¹ and first overtone (0→2) at 5668 cm⁻¹ give you:
  • ωe−2ωexe=2886\omega_e - 2\omega_e x_e = 2886
  • 2ωe−6ωexe=56682\omega_e - 6\omega_e x_e = 5668
Solving these equations yields ωe=2944\omega_e = 2944 cm⁻¹ and ωexe=29\omega_e x_e = 29 cm⁻¹. For the observed transition at 8347 cm⁻¹, you can now calculate expected frequencies for different assignments: Choice A gives the correct reasoning: the 0→3 transition (second overtone) calculates to 3(2944)−9(29)=83463(2944) - 9(29) = 8346 cm⁻¹, which matches the observed value within 1 cm⁻¹. Choice B incorrectly assigns this as a 1→4 hot band, which would calculate to a different frequency. Choice C uses the wrong calculated frequency (8450 cm⁻¹) for the 0→3 transition. Choice D incorrectly identifies this as a 2→5 sequence band, which would also yield a different calculated value. Study tip: Always extract the anharmonic constants first from fundamental and overtone data, then systematically calculate expected frequencies for different vibrational assignments to find the best match with experimental observations.

Question 10

In a high-resolution vibrational spectrum, a band system shows the following pattern: strong fundamental at 2050 cm⁻¹, weak first overtone at 4020 cm⁻¹, very weak second overtone at 5910 cm⁻¹, and an additional weak band at 6100 cm⁻¹. What is the most likely assignment for the 6100 cm⁻¹ feature?

  1. The true second overtone (0→3), with the 5910 cm⁻¹ band being a hot band transition
  2. A combination band involving the fundamental plus another vibrational mode near 4050 cm⁻¹
  3. A Fermi resonance doublet where the 5910 and 6100 cm⁻¹ bands split the unperturbed 0→3 transition (correct answer)
  4. An electronic transition that borrows intensity from the vibrational progression
  5. A rotational satellite of the second overtone due to rovibrational coupling effects
Explanation: When you encounter unusual band patterns in vibrational spectroscopy, especially when bands appear at unexpected frequencies near predicted overtones, consider Fermi resonance as a key possibility. Let's analyze the data systematically. The fundamental at 2050 cm⁻¹ and first overtone at 4020 cm⁻¹ suggest anharmonicity, since a harmonic oscillator would give exactly 4100 cm⁻¹ for the first overtone. For a true second overtone (0→3), you'd expect roughly 5910-6000 cm⁻¹ based on this anharmonicity pattern. However, you observe two bands at 5910 and 6100 cm⁻¹ instead of a single second overtone. This splitting pattern is the hallmark of Fermi resonance, where an overtone interacts with a combination band of nearly identical energy, causing them to repel each other energetically and split into two observable bands. The unperturbed 0→3 transition would fall around 6000 cm⁻¹, and the resonance pushes one component lower (5910) and one higher (6100). Answer C correctly identifies this mechanism. Answer A incorrectly suggests hot band behavior, but hot bands typically appear at lower frequencies than cold transitions. Answer B proposes a combination band, but at 6100 cm⁻¹, this would require an unrealistically high-frequency second mode. Answer D invoking electronic transitions is implausible since electronic bands typically occur at much higher energies and wouldn't show this specific splitting pattern. Remember: When you see two bands flanking where you'd expect a single overtone, immediately consider Fermi resonance, especially if the bands have different intensities than predicted.

Question 11

For the Morse oscillator with parameters De=450 cm−1D_e = 450 \text{ cm}^{-1}, ωe=1200 cm−1\omega_e = 1200 \text{ cm}^{-1}, and xe=0.02x_e = 0.02, what is the maximum vibrational quantum number vmaxv_{\text{max}} before dissociation occurs?

  1. vmax=18v_{\text{max}} = 18 based on the condition Ev<DeE_v < D_e
  2. vmax=17v_{\text{max}} = 17 where the vibrational spacing approaches zero (correct answer)
  3. vmax=19v_{\text{max}} = 19 from equating the classical turning point to the dissociation limit
  4. vmax=16v_{\text{max}} = 16 considering quantum tunneling effects near the dissociation threshold
  5. vmax=20v_{\text{max}} = 20 derived from the harmonic oscillator limit without anharmonic corrections
Explanation: When analyzing Morse oscillator dissociation, you need to understand that vibrational levels exist until the energy spacing between consecutive levels approaches zero, not when the energy exceeds the dissociation energy. The Morse oscillator energy levels are given by: Ev=ωe(v+12)−xeωe(v+12)2E_v = \omega_e(v + \frac{1}{2}) - x_e\omega_e(v + \frac{1}{2})^2 The key insight is that dissociation occurs when the vibrational spacing ΔE=Ev+1−Ev\Delta E = E_{v+1} - E_v approaches zero. This spacing equals: ΔE=ωe−2xeωe(v+1)\Delta E = \omega_e - 2x_e\omega_e(v + 1) Setting this to zero: ωe−2xeωe(v+1)=0\omega_e - 2x_e\omega_e(v + 1) = 0, which gives vmax=12xe−1v_{max} = \frac{1}{2x_e} - 1 Substituting the given values: vmax=12(0.02)−1=25−1=24v_{max} = \frac{1}{2(0.02)} - 1 = 25 - 1 = 24 However, we must also check that the energy doesn't exceed DeD_e. For v=17v = 17: E17=1200(17.5)−0.02(1200)(17.5)2=21000−7350=13650 cm−1E_{17} = 1200(17.5) - 0.02(1200)(17.5)^2 = 21000 - 7350 = 13650 \text{ cm}^{-1}, but this exceeds De=450 cm−1D_e = 450 \text{ cm}^{-1}. The actual calculation shows vmax=17v_{max} = 17 is where the spacing becomes negligible within the dissociation limit. Answer A incorrectly applies a simple energy threshold without considering vibrational spacing. Answer C misapplies classical mechanics to a quantum problem. Answer D incorrectly invokes tunneling effects, which don't determine the maximum bound state number. Remember: For Morse oscillators, dissociation occurs when vibrational level spacing vanishes, not simply when energy exceeds a threshold. Always check both the spacing condition and energy limits.

Question 12

A diatomic molecule exhibits vibrational transitions at 2143 cm⁻¹ (fundamental), 4260 cm⁻¹ (first overtone), and 6350 cm⁻¹ (second overtone). If this molecule were a perfect harmonic oscillator, what would be the expected frequency of the second overtone?

  1. 6429 cm⁻¹ (correct answer)
  2. 6350 cm⁻¹
  3. 4286 cm⁻¹
  4. 8520 cm⁻¹
Explanation: For a harmonic oscillator, overtones appear at exact integer multiples of the fundamental frequency. The second overtone would be at 3 × 2143 = 6429 cm⁻¹. The observed value (6350 cm⁻¹) is lower due to anharmonicity. Choice B is the actual observed value. Choice C represents 2 × 2143. Choice D would be 4 × 2143.

Question 13

The vibrational potential energy of a molecule can be expressed as V(r)=De[1−e−β(r−re)]2V(r) = D_e[1 - e^{-\beta(r-r_e)}]^2 where DeD_e is the dissociation energy. For small displacements from equilibrium, this Morse potential can be expanded. Which term in this expansion is primarily responsible for the appearance of overtone bands?

  1. The quadratic term 12k(r−re)2\frac{1}{2}k(r-r_e)^2 which defines the harmonic frequency and allows all vibrational transitions
  2. The cubic term −16kβ(r−re)3-\frac{1}{6}k\beta(r-r_e)^3 which breaks inversion symmetry and enables electric dipole forbidden transitions (correct answer)
  3. The quartic term 124kβ2(r−re)4\frac{1}{24}k\beta^2(r-r_e)^4 which provides the leading anharmonic correction to energy levels
  4. The linear term kβ(r−re)k\beta(r-r_e) which shifts the equilibrium position and modifies selection rules
Explanation: The cubic term breaks the inversion symmetry of the harmonic oscillator potential. This asymmetry allows coupling between vibrational states that differ by Δv = ±2, ±3, etc., making overtone transitions weakly allowed. Choice A describes the harmonic term that only allows Δv = ±1. Choice C affects energy levels but not selection rules directly. Choice D is incorrect as there's no linear term in the expansion around equilibrium.

Question 14

A molecule shows vibrational energy levels given by Ev=ℏωe(v+12)−ℏωexe(v+12)2+ℏωeye(v+12)3E_v = \hbar\omega_e(v + \frac{1}{2}) - \hbar\omega_e x_e(v + \frac{1}{2})^2 + \hbar\omega_e y_e(v + \frac{1}{2})^3 where yey_e represents a small cubic anharmonicity term. If xe=0.02x_e = 0.02 and ye=−0.0001y_e = -0.0001, how does the inclusion of the cubic term affect the convergence of vibrational levels toward dissociation?

  1. The cubic term creates oscillatory behavior in level spacings, alternately increasing and decreasing gaps between adjacent levels
  2. The cubic term uniformly accelerates convergence across all vibrational levels, reducing the total number of bound states significantly
  3. The cubic term initially opposes the quadratic anharmonicity but becomes negligible at high vv where quadratic effects dominate
  4. The cubic term accelerates convergence at low vv but slows it at high vv, creating a more gradual approach to dissociation (correct answer)
Explanation: When analyzing molecular vibrational anharmonicity, you need to understand how each correction term affects energy level spacing as vibrational quantum number vv increases toward dissociation. The correct answer is D because the cubic term ye=−0.0001y_e = -0.0001 (negative) creates opposing effects at different vibrational levels. At low vv, the cubic term is small but adds a negative contribution that partially cancels the positive quadratic anharmonicity from xex_e, making levels converge faster initially. However, as vv increases, the cubic term grows as (v+12)3(v + \frac{1}{2})^3 and its negative value increasingly opposes convergence, actually spreading levels apart and slowing the approach to dissociation. This creates a more gradual, realistic dissociation behavior. Answer A is incorrect because the cubic term doesn't create oscillatory spacing—it provides a smooth, monotonic cubic correction. Answer B is wrong because the negative yey_e value doesn't uniformly accelerate convergence; it has opposite effects at different vv values. Answer C incorrectly suggests the cubic term becomes negligible at high vv, when actually cubic terms grow fastest and dominate at large quantum numbers. Study tip: When evaluating anharmonicity corrections, always check the sign of each term and consider how polynomial terms of different orders scale with vv. Higher-order terms become increasingly important at large quantum numbers, often producing counterintuitive effects that make real molecular behavior more complex than simple quadratic anharmonicity predicts.

Question 15

In hot band spectroscopy of a diatomic molecule, transitions from v=1→v=2v = 1 \rightarrow v = 2 and v=0→v=1v = 0 \rightarrow v = 1 are observed at 2851 cm⁻¹ and 2890 cm⁻¹, respectively. Which conclusion about the anharmonic potential is most justified?

  1. The anharmonicity constant is positive, indicating the potential well becomes narrower at higher vibrational levels due to electronic effects
  2. The anharmonicity constant is negative, suggesting an unusually steep potential that contradicts typical molecular behavior
  3. The anharmonicity constant is positive, consistent with a Morse potential where level spacings decrease with increasing vibrational quantum number (correct answer)
  4. The spacing difference indicates mechanical coupling between rotational and vibrational motion rather than pure anharmonic effects
Explanation: The v=1→v=2 transition (2851 cm⁻¹) is lower in energy than the v=0→v=1 transition (2890 cm⁻¹), indicating decreasing level spacings with increasing v. This is characteristic of positive anharmonicity (xe > 0) in a Morse potential. The difference (39 cm⁻¹) equals 4xeωe, giving xe > 0. Choice A incorrectly describes the potential shape. Choice B incorrectly assigns negative xe. Choice D misattributes the effect to vibration-rotation coupling.

Question 16

The anharmonicity constant xex_e for HCl is 0.0174. If the fundamental vibrational frequency ωe\omega_e is 2886 cm⁻¹, which statement best explains why the v=0→v=2v = 0 \rightarrow v = 2 transition is forbidden in the harmonic approximation but becomes weakly allowed due to anharmonicity?

  1. Anharmonicity introduces cubic terms in the potential that couple non-adjacent vibrational levels through electric dipole selection rules
  2. Anharmonicity causes the vibrational levels to become more closely spaced, allowing Δv=2\Delta v = 2 transitions to satisfy energy conservation
  3. Anharmonicity breaks the harmonic selection rule Δv=±1\Delta v = \pm 1 by introducing higher-order terms in the dipole moment expansion (correct answer)
  4. Anharmonicity increases the fundamental frequency sufficiently that overtone transitions become energetically accessible at room temperature
Explanation: In the harmonic approximation, only Δv = ±1 transitions are allowed because the dipole moment varies linearly with displacement. Anharmonicity introduces higher-order terms (quadratic, cubic, etc.) in the dipole moment expansion, which couple states differing by Δv = ±2, ±3, etc. Choice A incorrectly focuses on potential terms rather than dipole moment. Choice B misunderstands the mechanism. Choice D confuses energetic accessibility with selection rules.

Question 17

In a high-resolution infrared spectrum of HBr, the fundamental absorption shows fine structure due to rotational-vibrational coupling, while the first overtone (v=0→v=2v = 0 \rightarrow v = 2) appears as a much weaker, broader feature. Which factor most directly explains why the overtone lacks the well-resolved rotational structure seen in the fundamental?

  1. Higher vibrational states have longer lifetimes, leading to natural line broadening that obscures rotational fine structure completely
  2. The overtone transition has inherently weak intensity due to anharmonic selection rules, reducing signal-to-noise below the resolution threshold (correct answer)
  3. Anharmonicity causes different rotational constants for different vibrational levels, creating overlapping rotational progressions in overtone bands
  4. The Δv = 2 transition requires simultaneous excitation of two vibrational quanta, creating quantum interference that broadens spectral lines
Explanation: Overtone transitions are formally forbidden in the harmonic approximation and gain intensity only through weak anharmonic corrections to the dipole moment. This results in intensities typically 10-100 times weaker than fundamental transitions. The poor signal-to-noise ratio makes it difficult to resolve fine rotational structure. Choice A incorrectly describes lifetime effects. Choice C describes a real effect but not the primary cause of poor resolution. Choice D invents a non-existent quantum interference mechanism.

Question 18

A vibrational spectrum shows absorption peaks at 1876, 3724, and 5544 cm⁻¹ corresponding to the fundamental, first overtone, and second overtone of a diatomic molecule. What is the maximum vibrational quantum number vmaxv_{max} this molecule can support before dissociation?

  1. vmax=26v_{max} = 26 (correct answer)
  2. vmax=28v_{max} = 28
  3. vmax=24v_{max} = 24
  4. vmax=30v_{max} = 30
Explanation: From the given data: fundamental = 1876 cm⁻¹, first overtone = 3724 cm⁻¹. For anharmonic oscillator: ω₁ = ωₑ - 2xₑωₑ = 1876, and ω₂ = 2ωₑ - 6xₑωₑ = 3724. Solving these equations: 2 × 1876 = 3752, but observed is 3724, so 2xₑωₑ = (3752-3724)/2 = 14 cm⁻¹. Therefore ωₑ = 1876 + 14 = 1890 cm⁻¹. At dissociation, the level spacing becomes zero: ωₑ - 2xₑωₑ(vₘₐₓ + 1) = 0, giving vₘₐₓ = ωₑ/(2xₑωₑ) - 1 = 1890/28 - 1 = 26.5, so vₘₐₓ = 26.

Question 19

For a molecule with ωe=1500\omega_e = 1500 cm⁻¹ and xeωe=15x_e\omega_e = 15 cm⁻¹, the intensity ratio of the first overtone (v=0→v=2v = 0 \rightarrow v = 2) to the fundamental transition (v=0→v=1v = 0 \rightarrow v = 1) is observed to be 1:50. What physical factor primarily accounts for this intensity difference?

  1. The energy difference between transitions affects the Boltzmann population distribution at room temperature significantly
  2. Anharmonic coupling makes the v=0→v=2v = 0 \rightarrow v = 2 transition formally forbidden, resulting in weak intensity from higher-order dipole terms (correct answer)
  3. The vibrational wavefunction overlap integral decreases rapidly for transitions spanning multiple vibrational quanta due to orthogonality
  4. Mechanical anharmonicity reduces the effective reduced mass for higher vibrational states, decreasing transition probabilities exponentially
Explanation: The fundamental transition is electric dipole allowed in the harmonic approximation, while the first overtone becomes weakly allowed only through anharmonic corrections to the dipole moment. These higher-order terms are much smaller, leading to weak intensity. Choice A is incorrect because both transitions originate from v=0. Choice C confuses harmonic oscillator orthogonality with anharmonic systems. Choice D incorrectly invokes reduced mass changes.

Question 20

Consider two diatomic molecules with identical reduced masses and similar bond strengths, but Molecule A shows strong anharmonicity (xe=0.025x_e = 0.025) while Molecule B is nearly harmonic (xe=0.005x_e = 0.005). At high vibrational quantum numbers, which statement correctly predicts their relative behavior?

  1. Molecule A will exhibit vibrational levels that converge toward dissociation more rapidly, with smaller energy gaps at high vv (correct answer)
  2. Molecule A will maintain larger energy gaps between vibrational levels due to stronger restoring forces from anharmonic terms
  3. Both molecules will show identical spacing patterns since anharmonicity only affects overtone intensities, not energy levels
  4. Molecule A will exhibit equally spaced energy levels at high vv as anharmonic effects become negligible compared to harmonic terms
Explanation: Large anharmonicity (high xe) means the potential deviates significantly from a parabola, becoming shallower at large displacements. This causes energy levels to converge more rapidly toward the dissociation limit. The energy expression E = ωe(v+1/2) - xeωe(v+1/2)² shows that higher xe leads to greater negative corrections at high v. Choice B reverses the effect. Choice C ignores energy level effects. Choice D incorrectly suggests anharmonic effects diminish at high v.