Vibrational Energy Levels - Physical Chemistry 2

A molecule exhibits a vibrational progression in its electronic spectrum with bands at $25000$ , $25800$ , $26600$ , and $27400 \text{ cm}^{-1}$ . The spacing between consecutive bands is constant at $800 \text{ cm}^{-1}$ . If this represents transitions from $v''=0$ to $v'=0,1,2,3$ respectively, what can be concluded about the relative vibrational frequencies in the ground and excited electronic states?
Answer: The excited state vibrational frequency is $800 \text{ cm}^{-1}$ ; the ground state frequency cannot be determined When analyzing vibrational progressions in electronic spectra, you're observing transitions between vibronic levels (combined electronic-vibrational states). The key insight is understanding what the constant spacing between bands tells you about the vibrational frequencies in each electronic state. The observed bands represent transitions from the ground vibrational level of the ground electronic state ( $v''=0$ ) to different vibrational levels of the excited electronic state ( $v'=0,1,2,3$ ). The energy difference between consecutive bands in this progression equals the vibrational quantum spacing in the excited state: $\Delta E = h\nu' = 800 \text{ cm}^{-1}$ . This is because you're moving up one vibrational level at a time in the excited state while staying at $v''=0$ in the ground state. The correct answer is B because the constant $800 \text{ cm}^{-1}$ spacing directly gives you the excited state vibrational frequency, but this progression tells you nothing about the ground state frequency since all transitions originate from $v''=0$ . Answer A is wrong because you cannot determine the ground state frequency from this data. Answer C incorrectly assigns the $800 \text{ cm}^{-1}$ to the ground state when it actually reflects the excited state spacing. Answer D misinterprets what the $800 \text{ cm}^{-1}$ represents—it's not a difference between frequencies of the two states, but rather the vibrational quantum of the excited state. Study tip: In vibrational progressions, the constant spacing always corresponds to the vibrational frequency of whichever electronic state is changing its vibrational quantum number.
A polyatomic molecule undergoes a vibrational transition where two normal modes are simultaneously excited: mode A from $v_A = 0$ to $v_A = 1$ ( $\omega_A = 1500 \text{ cm}^{-1}$ ) and mode B from $v_B = 0$ to $v_B = 1$ ( $\omega_B = 800 \text{ cm}^{-1}$ ). Assuming the harmonic approximation and that both modes have the same infrared activity, what is the expected wavenumber of this combination band?
Answer: $2300 \text{ cm}^{-1}$ from the sum of both fundamental frequencies When you encounter questions about combination bands in vibrational spectroscopy, you're dealing with transitions where multiple normal modes are excited simultaneously. This is a fundamental concept in molecular spectroscopy that extends beyond simple single-mode excitations. In the harmonic approximation, vibrational energy levels are additive and independent. When two modes are excited together (mode A: 0→1 and mode B: 0→1), the total energy change equals the sum of individual energy changes. Since wavenumber is directly proportional to energy in spectroscopy ( $E = hc\tilde{\nu}$ ), the observed wavenumber for this combination band is simply $\omega_A + \omega_B = 1500 + 800 = 2300 \text{ cm}^{-1}$ . Answer D correctly gives $2300 \text{ cm}^{-1}$ from summing both fundamental frequencies. Answer A ( $1200 \text{ cm}^{-1}$ ) incorrectly applies a geometric mean, which has no physical basis in vibrational spectroscopy. Answer B ( $700 \text{ cm}^{-1}$ ) represents the difference between frequencies, which would correspond to a different type of transition (like a hot band or difference band), not a combination band. Answer C ( $1150 \text{ cm}^{-1}$ ) uses an arithmetic average, which also lacks physical justification for combination bands. Remember this key principle: combination bands always appear at the sum of the contributing fundamental frequencies in the harmonic approximation. This additive rule makes combination band identification straightforward—just add up the wavenumbers of all simultaneously excited modes.
In a two-level quantum system representing vibrational states $|0\rangle$ and $|1\rangle$ of a harmonic oscillator, a coherent superposition state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\phi}|1\rangle)$ is prepared. If the phase $\phi$ evolves as $\phi(t) = \omega t$ due to the energy difference between levels, what is the period of oscillation for the expectation value $\langle\hat{x}\rangle$ ?
Answer: $T = \frac{2\pi}{\omega}$ When analyzing quantum superposition states, you need to track how the phase evolves over time and determine the periodicity of observable quantities like position expectation values. The time-evolved state is $|\psi(t)\rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\omega t}|1\rangle)$ . To find $\langle\hat{x}\rangle$ , you use the position operator in terms of ladder operators: $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger)$ . The key matrix elements are $\langle 0|\hat{a}|1\rangle = 1$ and $\langle 1|\hat{a}^\dagger|0\rangle = 1$ , while diagonal terms vanish. Computing the expectation value: $\langle\hat{x}\rangle = \sqrt{\frac{\hbar}{2m\omega}} \cdot \frac{1}{2}(e^{-i\omega t} + e^{i\omega t}) = \sqrt{\frac{\hbar}{2m\omega}} \cos(\omega t)$ Since $\cos(\omega t)$ has period $T = \frac{2\pi}{\omega}$ , answer B is correct. A ( $T = \frac{\pi}{\omega}$ ) would be half the actual period—this mistakes the angular frequency in the cosine for twice its actual value. C ( $T = \frac{4\pi}{\omega}$ ) represents twice the correct period, possibly from incorrectly handling the phase evolution. D ( $T = \frac{\pi}{2\omega}$ ) is one-fourth the correct period, likely from confusing the relationship between frequency and period. Study tip: For quantum oscillations involving superposition states, always write out the full time evolution and identify which trigonometric function emerges in your observable—the period directly follows from the argument of that function.
Two isotopomers of a diatomic molecule, $^{12}C^{16}O$ and $^{13}C^{16}O$ , have fundamental vibrational frequencies of $2143 \text{ cm}^{-1}$ and $2096 \text{ cm}^{-1}$ , respectively. What is the ratio of their zero-point energies?
Answer: $\frac{E_0(^{12}CO)}{E_0(^{13}CO)} = \frac{2143}{2096} = 1.022$ The zero-point energy is $E_0 = \frac{1}{2}h\nu = \frac{1}{2}hc\tilde{\nu}$ . Since the vibrational frequency is directly proportional to the wavenumber, the ratio of zero-point energies is simply the ratio of the fundamental frequencies: $\frac{E_0(^{12}CO)}{E_0(^{13}CO)} = \frac{\tilde{\nu}(^{12}CO)}{\tilde{\nu}(^{13}CO)} = \frac{2143}{2096} = 1.022$ . Choice B incorrectly uses the mass ratio, choice C inverts the ratio, and choice D uses an incorrect mass ratio formula. The key insight is that zero-point energy is directly proportional to frequency, not related to reduced mass through the square root relationship.
A diatomic molecule has a vibrational frequency of $\omega_e = 2990 \text{ cm}^{-1}$ and anharmonicity constant $x_e = 0.015$ . What is the maximum vibrational quantum number $v_{max}$ before the molecule dissociates, assuming dissociation occurs when the vibrational energy equals the dissociation energy?
Answer: $v_{max} = 32$ When you encounter vibrational spectroscopy problems involving dissociation, you're dealing with the anharmonic oscillator model, which accounts for the fact that real molecular vibrations deviate from perfect harmonic behavior at high energy levels. The key insight is that dissociation occurs when the vibrational energy reaches the dissociation energy $D_e$ . For an anharmonic oscillator, the relationship between these quantities is: $D_e = \frac{\omega_e^2}{4x_e\omega_e} = \frac{\omega_e}{4x_e}$ Substituting the given values: $D_e = \frac{2990 \text{ cm}^{-1}}{4 \times 0.015} = \frac{2990}{0.06} = 49,833 \text{ cm}^{-1}$ The maximum vibrational quantum number occurs when the vibrational energy equals $D_e$ . Using the anharmonic energy formula and solving for $v_{max}$ : $v_{max} = \frac{1}{2x_e} - \frac{1}{2} = \frac{1}{2 \times 0.015} - 0.5 = 33.33 - 0.5 = 32.83$ Since $v$ must be an integer, $v_{max} = 32$ . Answer A ( $v_{max} = 32$ ) is correct. Answer B (33) results from rounding 32.83 up instead of recognizing that $v = 33$ would exceed the dissociation energy. Answer C (31) comes from incorrectly rounding down too far. Answer D (34) likely results from computational errors or using an incorrect formula. Remember: when calculating maximum quantum numbers before dissociation, always round down to the nearest integer since exceeding the calculated value means the molecule has already dissociated.
The wave function for the $v = 1$ vibrational state of a harmonic oscillator is $\psi_1(x) = \sqrt{\frac{2\alpha^3}{\pi}} x e^{-\alpha x^2/2}$ , where $\alpha = \sqrt{\frac{m\omega}{\hbar}}$ . At what displacement $x$ from equilibrium does this wave function have its maximum amplitude?
Answer: $x = \pm \sqrt{\frac{\hbar}{m\omega}}$ at the maximum probability density To find the maximum of $|\psi_1(x)|$ , we take the derivative and set it to zero. $\frac{d}{dx}|\psi_1(x)| = \frac{d}{dx}\left[\sqrt{\frac{2\alpha^3}{\pi}} |x| e^{-\alpha x^2/2}\right] = 0$ when $1 - \alpha x^2 = 0$ , giving $x^2 = \frac{1}{\alpha} = \frac{\hbar}{m\omega}$ . Therefore $x = \pm\sqrt{\frac{\hbar}{m\omega}}$ . Choice A gives the classical turning points (where the probability density, not amplitude, is maximum for higher $v$ ), choice C uses an incorrect factor of 2, and choice D is incorrect since $\psi_1(0) = 0$ for the first excited state.
In the infrared spectrum of HCl, the fundamental band appears at $2886 \text{ cm}^{-1}$ . According to the harmonic oscillator selection rules, which of the following transitions would be strictly forbidden in the absence of anharmonicity?
Answer: The transition from $v = 0$ to $v = 2$ with absorption at $5772 \text{ cm}^{-1}$ The harmonic oscillator selection rule states that $\Delta v = \pm 1$ for allowed transitions. The $v = 0 \to v = 2$ transition has $\Delta v = 2$ , which violates this rule and is forbidden in the harmonic approximation. Such overtone transitions only become weakly allowed due to anharmonicity. Choice A has $\Delta v = 2$ but incorrectly suggests it's at twice the fundamental frequency. Choice C is allowed ( $\Delta v = 1$ ), and choice D is also allowed ( $\Delta v = -1$ , emission).
In the vibrational spectrum of a heteronuclear diatomic molecule, the spacing between adjacent vibrational levels decreases with increasing $v$ due to anharmonicity. If the energy difference between $v = 0$ and $v = 1$ is $4.28 \times 10^{-20} \text{ J}$ , and between $v = 1$ and $v = 2$ is $4.15 \times 10^{-20} \text{ J}$ , what is the anharmonicity constant $x_e$ ?
Answer: $x_e = 0.015$ based on the energy level spacing For an anharmonic oscillator, the energy difference between adjacent levels is $\Delta E_{v \to v+1} = h\omega_0 - 2h\omega_0 x_e(v + 1)$ . The difference between consecutive spacings gives: $\Delta E_{0 \to 1} - \Delta E_{1 \to 2} = 2h\omega_0 x_e$ . Therefore: $x_e = \frac{\Delta E_{0 \to 1} - \Delta E_{1 \to 2}}{2h\omega_0} = \frac{(4.28 - 4.15) \times 10^{-20}}{2 \times 4.28 \times 10^{-20}} = \frac{0.13}{8.56} = 0.015$ . Choice B uses the wrong denominator, choice C doubles the result incorrectly, and choice D uses an incorrect formula for the anharmonicity correction.
The Franck-Condon principle governs the relative intensities of vibronic transitions in electronic spectra. For a diatomic molecule where the excited electronic state has a significantly longer equilibrium bond length than the ground state, which vibrational progression would show the highest intensity in the absorption spectrum?
Answer: $v' = 7 \leftarrow v'' = 0$ transition (7-0 band) When analyzing vibronic transitions, you need to understand how the Franck-Condon principle determines transition intensities based on vibrational wavefunction overlap. The key insight is that electronic transitions occur so rapidly that nuclei don't have time to move during the transition—they maintain their initial positions and momenta. For a molecule where the excited state has a significantly longer equilibrium bond length than the ground state, the potential energy curves are horizontally displaced. Most molecules at room temperature occupy the $v'' = 0$ vibrational level of the ground state. When vertical transitions occur from this level, the molecule finds itself at a bond length that's much shorter than the new equilibrium position in the excited state. The highest intensity transition occurs where there's maximum overlap between the ground state $v'' = 0$ wavefunction and an excited state vibrational wavefunction. Since the excited state equilibrium is shifted to longer bonds, this maximum overlap occurs with a higher vibrational level—often around $v' = 7$ for significant displacement. This makes answer D correct. Answer A (0-0 transition) would be strongest only if the equilibrium bond lengths were similar. Answer B (1-0) and answer C (3-0) represent intermediate cases that don't provide optimal overlap for the described large displacement scenario. These would show moderate intensities but not the maximum. Remember: when excited states have significantly different equilibrium geometries, the most intense vibronic transition isn't always the 0-0 band. Look for the vibrational level that best matches the vertical transition from the ground state geometry.
Consider a quantum harmonic oscillator where the probability of finding the particle in the classically forbidden region is calculated. For the ground state ( $v = 0$ ), this probability is approximately 16%. What physical insight does this provide about the relationship between quantum and classical mechanics?
Answer: The wave nature of matter creates probability distributions that extend beyond classical boundaries, even at the lowest energy state The key insight is that quantum mechanics describes particles as wave functions with probability distributions, not as classical point particles. Even in the ground state, the wave function has finite amplitude in regions where a classical particle with the same total energy could not exist (where kinetic energy would be negative). This is a fundamental difference between quantum and classical descriptions, not related to tunneling through barriers (choice A), violation of energy conservation (choice B), or modifications to the potential (choice D). The probability in the forbidden region reflects the inherent wave nature of quantum particles.
In a Raman spectrum of a diatomic molecule, the anti-Stokes line corresponding to the $v = 1 \to v = 0$ transition appears with much lower intensity than the Stokes line for the $v = 0 \to v = 1$ transition. If the vibrational frequency is $2000 \text{ cm}^{-1}$ and the intensity ratio $I_{\text{anti-Stokes}}/I_{\text{Stokes}} = 0.13$ , what is the approximate vibrational temperature of the sample?
Answer: $T \approx 1450 \text{ K}$ based on population differences When you encounter Raman spectroscopy problems involving intensity ratios between Stokes and anti-Stokes lines, you're dealing with population distributions governed by the Boltzmann equation. The key insight is that anti-Stokes scattering requires molecules to already be in excited vibrational states, while Stokes scattering starts from the ground state. The intensity ratio relates directly to the population ratio between vibrational levels: $\frac{I_{\text{anti-Stokes}}}{I_{\text{Stokes}}} = \frac{N_1}{N_0} = e^{-h\nu/k_BT}$ where $N_1$ and $N_0$ are populations of the $v=1$ and $v=0$ states respectively. Taking the natural logarithm: $\ln(0.13) = -\frac{h\nu}{k_BT}$ Converting the vibrational frequency to energy units: $h\nu = hc\tilde{\nu} = (6.626 \times 10^{-34})(3 \times 10^{10})(2000) = 3.98 \times 10^{-20} \text{ J}$ Solving for temperature: $T = -\frac{3.98 \times 10^{-20}}{(1.381 \times 10^{-23})(\ln(0.13))} = -\frac{3.98 \times 10^{-20}}{(1.381 \times 10^{-23})(-2.04)} \approx 1450 \text{ K}$ This confirms answer C is correct. A (1150 K) would give too high an intensity ratio. B (1820 K) and D (2100 K) would result in even higher ratios than observed, as higher temperatures increase the excited state population. Study tip: Remember that anti-Stokes intensity always decreases with lower temperature because fewer molecules occupy excited vibrational states. The exponential relationship makes temperature calculations straightforward once you set up the Boltzmann equation correctly.
For a harmonic oscillator with vibrational quantum number $v$ , the expectation value of the potential energy is $\langle V \rangle = \frac{1}{4}m\omega^2\langle x^2 \rangle$ , where $\langle x^2 \rangle = \frac{\hbar}{2m\omega}(2v + 1)$ . What fraction of the total energy is potential energy for the $v = 3$ state?
Answer: $\frac{1}{2}$ of the total energy is potential energy For any quantum state of the harmonic oscillator, the virial theorem applies: $\langle T \rangle = \langle V \rangle$ . Since the total energy is $E = \langle T \rangle + \langle V \rangle = 2\langle V \rangle$ , exactly half of the energy is potential and half is kinetic, regardless of the quantum number $v$ . This can be verified: $\langle V \rangle = \frac{1}{4}m\omega^2 \cdot \frac{\hbar}{2m\omega}(2v + 1) = \frac{\hbar\omega}{8}(2v + 1)$ , while $E_v = \hbar\omega(v + \frac{1}{2}) = \frac{\hbar\omega}{2}(2v + 1)$ , so $\langle V \rangle/E_v = 1/2$ . The other choices reflect common misconceptions about quantum number dependence.
A heteronuclear diatomic molecule has a fundamental vibrational frequency of $2890 \text{ cm}^{-1}$ . In the infrared spectrum, weak absorption bands are observed at $5780 \text{ cm}^{-1}$ and $8670 \text{ cm}^{-1}$ . What physical phenomenon primarily accounts for the appearance of these additional bands?
Answer: Anharmonic coupling allowing overtone transitions that violate the $\Delta v = \pm 1$ selection rule When you encounter infrared spectroscopy problems with additional bands appearing at specific frequency relationships, think about the selection rules governing vibrational transitions and what can cause deviations from ideal harmonic oscillator behavior. The key insight here is recognizing the mathematical relationship between the observed frequencies. The fundamental frequency is 2890 cm⁻¹, and the additional bands appear at 5780 cm⁻¹ (approximately 2 × 2890) and 8670 cm⁻¹ (approximately 3 × 2890). These are overtone transitions corresponding to $v = 0 \rightarrow 2$ and $v = 0 \rightarrow 3$ transitions, respectively. Answer C is correct because anharmonic coupling in real molecules allows these normally forbidden transitions to occur. In a perfect harmonic oscillator, the selection rule $\Delta v = \pm 1$ strictly applies, but real molecular vibrations exhibit anharmonicity that weakens this restriction, making overtone transitions weakly allowed. Answer A is incorrect because rotational fine structure would create many closely spaced lines around each vibrational band, not discrete bands at these specific multiples. Answer B is wrong because Fermi resonance requires two vibrational modes of the same symmetry with similar energies to mix—but this is a diatomic molecule with only one vibrational mode. Answer D is incorrect because isotope effects would shift the fundamental frequency itself and wouldn't create this specific 2:3 ratio pattern. Study tip: When you see weak IR bands at integer multiples of a fundamental frequency (2×, 3×, etc.), immediately think anharmonic overtones. The mathematical relationship is your strongest clue.
A molecule has a vibrational frequency of $1200 \text{ cm}^{-1}$ . At what temperature will the population ratio $N_1/N_0$ (where $N_1$ and $N_0$ are the populations of the $v = 1$ and $v = 0$ levels) equal 0.368?
Answer: $T = 1738 \text{ K}$ using the Boltzmann distribution The population ratio follows the Boltzmann distribution: $\frac{N_1}{N_0} = \exp\left(-\frac{h\nu}{k_BT}\right)$ . Given that $N_1/N_0 = 0.368 \approx e^{-1}$ , we need $\frac{h\nu}{k_BT} = 1$ . Using $h\nu = hc\tilde{\nu}$ : $T = \frac{hc\tilde{\nu}}{k_B} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{10})(1200)}{1.38 \times 10^{-23}} = \frac{2.385 \times 10^{-20}}{1.38 \times 10^{-23}} = 1728 \text{ K} \approx 1738 \text{ K}$ .
Consider a quantum harmonic oscillator with vibrational states $|v\rangle$ . The transition dipole moment $\langle v'|\hat{\mu}|v\rangle$ is nonzero only when certain selection rules are satisfied. For an electric dipole transition in the presence of a weak static electric field that induces a small permanent dipole moment, which additional transitions become weakly allowed?
Answer: $\Delta v = 0$ transitions due to Stark shift-induced mixing of vibrational levels When you encounter quantum harmonic oscillator selection rules with external perturbations, think about how the perturbation changes the system's symmetry and mixes energy states. In an unperturbed harmonic oscillator, the electric dipole selection rule is strictly $\Delta v = \pm 1$ because the transition dipole moment $\langle v'|\hat{\mu}|v\rangle$ is only nonzero for adjacent vibrational levels. However, a weak static electric field creates a Stark effect that perturbs the system by mixing vibrational wavefunctions. This mixing means the pure vibrational states $|v\rangle$ are no longer exact eigenstates—they become linear combinations of the original states. The most significant mixing occurs between states of the same energy (or nearly the same energy), which allows $\Delta v = 0$ transitions to become weakly allowed. The static field induces a permanent dipole moment that enables these previously forbidden transitions through the mixed character of the perturbed wavefunctions. Looking at the incorrect options: Choice A is wrong because second-order perturbation effects don't specifically enable $\Delta v = \pm 2$ transitions in this context. Choice C incorrectly suggests all transitions become equally allowed, which violates the fundamental physics—the mixing is selective and intensity-dependent. Choice D is incorrect because the static field does create new allowed transitions beyond the original $\Delta v = \pm 1$ rule. Remember: external electric fields break symmetry in quantum systems. When you see Stark effect problems, focus on how field-induced mixing of wavefunctions creates new transition pathways, especially $\Delta v = 0$ transitions that probe the induced permanent dipole moment.
The vibrational partition function for a diatomic molecule is $q_{vib} = \frac{1}{1-e^{-\hbar\omega/k_BT}}$ . At what temperature would exactly 25% of molecules occupy vibrational states with $v \geq 1$ for $\text{HCl}$ with $\omega = 8.97 \times 10^{13} \text{ rad/s}$ ?
Answer: $T = 1840 \text{ K}$ When you encounter vibrational partition functions, you're dealing with the statistical distribution of molecules across different energy states. The key insight is that the fraction of molecules in excited states (v ≥ 1) equals 1 minus the fraction in the ground state (v = 0). The probability of being in the ground state is $P(v=0) = \frac{1}{q_{vib}}$ , so the fraction in excited states is: $P(v \geq 1) = 1 - \frac{1}{q_{vib}} = 1 - (1-e^{-\hbar\omega/k_BT}) = e^{-\hbar\omega/k_BT}$ Setting this equal to 0.25 (since we want 25% of molecules): $e^{-\hbar\omega/k_BT} = 0.25$ Taking the natural logarithm: $-\frac{\hbar\omega}{k_BT} = \ln(0.25) = -1.386$ Solving for T: $T = \frac{\hbar\omega}{1.386 \times k_B} = \frac{(1.055 \times 10^{-34})(8.97 \times 10^{13})}{1.386 \times (1.381 \times 10^{-23})} = 1840 \text{ K}$ This confirms answer A is correct. The other answers represent different calculation errors: B (2156 K) might result from using ln(0.75) instead of ln(0.25), C (1650 K) could come from arithmetic mistakes in the numerical calculation, and D (2890 K) likely stems from incorrectly using the full partition function instead of recognizing that excited state population equals the Boltzmann factor. Remember: for vibrational problems, the fraction in excited states always simplifies to the Boltzmann factor $e^{-\hbar\omega/k_BT}$ , making these calculations straightforward once you recognize the pattern.
Consider two diatomic molecules: $\text{H}_2$ ( $\omega_e = 4401 \text{ cm}^{-1}$ , $\omega_e x_e = 121 \text{ cm}^{-1}$ ) and $\text{I}_2$ ( $\omega_e = 214 \text{ cm}^{-1}$ , $\omega_e x_e = 0.6 \text{ cm}^{-1}$ ). For which molecule would the harmonic oscillator approximation be more accurate for calculating vibrational energy levels, and why?
Answer: $\text{I}_2$ , because the anharmonicity parameter $x_e$ represents a smaller fractional correction to $\omega_e$ When evaluating the accuracy of the harmonic oscillator approximation, you need to focus on how significant the anharmonic corrections are relative to the harmonic frequency. The harmonic oscillator treats the vibrational potential as a perfect parabola, but real molecules have anharmonic corrections that cause energy levels to deviate from this ideal. The key insight is comparing the anharmonicity parameter $\omega_e x_e$ as a fraction of the fundamental frequency $\omega_e$ . For H₂: $\frac{121}{4401} = 0.0275$ (2.75%). For I₂: $\frac{0.6}{214} = 0.0028$ (0.28%). Since I₂ has a much smaller fractional anharmonic correction, the harmonic approximation will be more accurate for this molecule. Choice A incorrectly assumes that higher frequency automatically means better harmonic approximation. While quantum effects are more pronounced in H₂, this doesn't make the harmonic model more accurate—it actually makes anharmonic corrections more significant. Choice C misinterprets the relationship between atomic mass and vibrational amplitude. While lighter atoms do have different vibrational characteristics, this doesn't directly translate to better harmonic approximation accuracy. Choice D incorrectly suggests that weaker quantum effects improve the harmonic approximation. The strength of quantum effects doesn't determine how well the harmonic model fits the actual potential energy surface. Remember: when comparing approximation accuracy, always look at relative corrections, not absolute values. The anharmonicity parameter as a percentage of the fundamental frequency tells you how much the real system deviates from the idealized harmonic model.
A molecule in its vibrational ground state ( $v = 0$ ) absorbs a photon and transitions to an excited vibrational state. The selection rule for electric dipole transitions in a harmonic oscillator is $\Delta v = \pm 1$ . However, if this molecule were placed in an intense infrared field where the classical oscillation amplitude becomes comparable to the molecular bond length, which statement best describes the expected changes in the vibrational spectrum?
Answer: Multiple-quantum transitions ( $\Delta v = \pm 2, \pm 3, ...$ ) become allowed due to field-induced mixing of vibrational wavefunctions When you encounter questions about vibrational spectroscopy in strong electromagnetic fields, you're dealing with nonlinear optics and field-induced perturbations to quantum mechanical selection rules. In weak fields, the harmonic oscillator model gives strict selection rules where only $\Delta v = \pm 1$ transitions are allowed for electric dipole absorption. This occurs because the transition dipole moment $\langle v'|\mu|v \rangle$ is only non-zero when vibrational states differ by exactly one quantum number. However, intense infrared fields create additional interaction terms in the Hamiltonian that couple non-adjacent vibrational levels. When the field amplitude becomes comparable to molecular dimensions, the electric field operator contains higher-order terms that mix vibrational wavefunctions. This mixing creates new pathways for transitions, allowing previously forbidden multi-quantum jumps ( $\Delta v = \pm 2, \pm 3, ...$ ). The intense field essentially "breaks" the harmonic approximation by introducing anharmonic coupling between states that were previously uncoupled. Option A is wrong because quantum mechanics doesn't prevent field-induced modifications to selection rules. Option C incorrectly focuses on frequency shifts rather than selection rule changes—while frequency shifts can occur, the primary spectroscopic change is the appearance of new transition pathways. Option D overstates the effect; while strong fields modify the system, discrete vibrational states don't completely disappear—they become coupled, creating new allowed transitions. Remember: intense electromagnetic fields can induce couplings between quantum states that are normally forbidden, leading to "forbidden" transitions becoming weakly allowed through field-mediated mixing.
The vibrational wave function for a quantum harmonic oscillator in state $v$ has $v$ nodes (excluding the boundaries at $x \rightarrow \pm\infty$ ). For a molecule with vibrational frequency $\omega_e = 3200 \text{ cm}^{-1}$ , if an electronic transition occurs from a vibrational level with 3 nodes to one with 1 node, what is the wavenumber change in the electronic absorption spectrum due to this vibrational contribution?
Answer: $\Delta\tilde{\nu} = -6400 \text{ cm}^{-1}$ When analyzing vibrational transitions in electronic spectra, you need to understand how vibrational quantum numbers relate to energy changes and spectral shifts. The key insight is connecting nodes to vibrational quantum numbers: a vibrational state $v$ has $v$ nodes. So the initial state has $v'' = 3$ and the final state has $v' = 1$ . The vibrational energy levels are given by $E_v = \hbar\omega_e(v + \frac{1}{2})$ , where the energy difference between adjacent levels is $\hbar\omega_e$ . The change in vibrational quantum number is $\Delta v = v' - v'' = 1 - 3 = -2$ . This means we're transitioning to a vibrational state that is 2 levels lower in energy. The wavenumber change due to this vibrational contribution is: $\Delta\tilde{\nu} = \omega_e \times \Delta v = 3200 \times (-2) = -6400 \text{ cm}^{-1}$ The negative sign indicates the transition involves less vibrational energy in the final state, shifting the absorption to lower wavenumber (red shift). Choice A gives $+6400 \text{ cm}^{-1}$ , which incorrectly uses a positive sign—this would apply if transitioning from $v=1$ to $v=3$ . Choice C gives $+3200 \text{ cm}^{-1}$ , using the wrong magnitude and sign—this represents a single vibrational quantum increase. Choice D gives $-3200 \text{ cm}^{-1}$ , which has the correct sign but wrong magnitude—this represents only a single vibrational quantum decrease. Study tip: Always track the direction of vibrational transitions carefully: $\Delta v = v_{final} - v_{initial}$ , and remember that negative $\Delta v$ values give red shifts in spectra.
A diatomic molecule in a $^1\Sigma$ electronic state undergoes a vibrational transition. The electric dipole moment operator $\hat{\mu}$ can be expanded as $\hat{\mu}(r) = \mu_0 + \left(\frac{d\mu}{dr}\right)_e (r - r_e) + ...$ where $r$ is the internuclear distance. For this molecule, $\mu_0 = 0.5 D$ and $\left(\frac{d\mu}{dr}\right)_e = 2.0 \text{ D/Å}$ . Which statement about the vibrational transition intensities is most accurate?
Answer: $\Delta v = 0$ transitions are forbidden, but $\Delta v = \pm 1$ transitions have intensity proportional to $\left(\frac{d\mu}{dr}\right)_e^2$ When analyzing vibrational transition intensities, you need to consider how the electric dipole moment varies with internuclear distance and apply selection rules. The key insight is that different terms in the dipole expansion give rise to different types of transitions. The transition intensity depends on the transition dipole moment $\langle v'|\hat{\mu}|v\rangle$ . Using the given expansion, this becomes two separate contributions: $\mu_0\langle v'|v\rangle$ from the constant term and $\left(\frac{d\mu}{dr}\right)_e\langle v'|(r-r_e)|v\rangle$ from the linear term. For the constant term $\mu_0$ : The integral $\langle v'|v\rangle$ equals 1 only when $v' = v$ (orthogonality of vibrational wavefunctions), making $\Delta v = 0$ transitions allowed with intensity proportional to $\mu_0^2$ . For the linear term: The integral $\langle v'|(r-r_e)|v\rangle$ is non-zero only when $v' = v \pm 1$ due to the properties of harmonic oscillator wavefunctions, making $\Delta v = \pm 1$ transitions allowed with intensity proportional to $\left(\frac{d\mu}{dr}\right)_e^2$ . However, $\Delta v = 0$ transitions (pure rotational transitions within the same vibrational level) are typically forbidden in vibrational spectroscopy because we're specifically looking at vibrational changes. Answer C correctly identifies this. Answer A incorrectly attributes $\Delta v = \pm 1$ intensity to $\mu_0^2$ instead of $\left(\frac{d\mu}{dr}\right)_e^2$ . Answer B ignores the $\mu_0$ contribution entirely. Answer D incorrectly suggests $\Delta v = 0$ transitions are allowed in vibrational spectroscopy. Study tip: Remember that the constant term in dipole expansion gives $\Delta v = 0$ transitions, while the linear term gives $\Delta v = \pm 1$ transitions—but focus on which transitions are actually observed in vibrational spectroscopy.

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