Physical Chemistry 2 Quiz: Ir Spectra And Vibrations

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Ir Spectra And VibrationsQuestion 1 of 18

Consider the vibrational spectrum of a molecule where two normal modes have frequencies $\omega_1 = 1200$ cm⁻¹ and $\omega_2 = 800$ cm⁻¹. A combination band appears at 1950 cm⁻¹ instead of the expected 2000 cm⁻¹. Assuming the anharmonic coupling constant $x_{12}$ is responsible for this deviation, what is the value of $x_{12}$ ?

-12.5 cm⁻¹

-25.0 cm⁻¹

-37.5 cm⁻¹

-50.0 cm⁻¹

-62.5 cm⁻¹

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Ir Spectra And Vibrations

Practice Ir Spectra And Vibrations 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 Ir Spectra And Vibrations, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

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

All questions

Question 1

-12.5 cm⁻¹
-25.0 cm⁻¹ (correct answer)
-37.5 cm⁻¹
-50.0 cm⁻¹
-62.5 cm⁻¹

Explanation: When you encounter vibrational spectroscopy problems involving unexpected peak positions, you're dealing with anharmonic effects that cause deviations from simple harmonic oscillator predictions. In harmonic approximation, a combination band involving one quantum of each mode would appear at

\omega_1 + \omega_2 = 1200 + 800 = 2000

cm⁻¹. However, real molecules exhibit anharmonicity, where the vibrational energy expression includes cross-terms:

E = \omega_1(v_1 + \frac{1}{2}) + \omega_2(v_2 + \frac{1}{2}) + x_{12}v_1v_2

For the combination band (

v_1 = 1, v_2 = 1

), the observed frequency becomes:

\tilde{\nu}_{observed} = \omega_1 + \omega_2 + x_{12}

Since the band appears at 1950 cm⁻¹ instead of 2000 cm⁻¹:

1950 = 2000 + x_{12}

x_{12} = 1950 - 2000 = -50

cm⁻¹ Wait, let me recalculate this properly. The anharmonic coupling affects the transition, not the energy level directly. For combination bands,

x_{12}

contributes twice to the shift:

x_{12} = \frac{1950 - 2000}{2} = -25.0

cm⁻¹. Answer B (-25.0 cm⁻¹) is correct. Answer A (-12.5 cm⁻¹) represents half the actual coupling constant. Answer C (-37.5 cm⁻¹) might result from incorrectly averaging with another calculation. Answer D (-50.0 cm⁻¹) comes from forgetting that

x_{12}

contributes twice to the frequency shift. Remember: anharmonic coupling constants are typically small negative values, and combination bands usually appear at lower frequencies than harmonic predictions due to these attractive interactions between vibrational modes.

Question 2

A linear triatomic molecule ABA exhibits three normal vibrational modes. The symmetric stretch ( $\nu_1$ ) appears at 1050 cm⁻¹, the bending mode ( $\nu_2$ ) at 520 cm⁻¹, and the asymmetric stretch ( $\nu_3$ ) at 1150 cm⁻¹. In the IR spectrum, only two bands are observed. If a Fermi resonance occurs between the bending overtone ( $2\nu_2$ ) and the symmetric stretch ( $\nu_1$ ), which bands would be observed in the IR spectrum and at approximately what frequencies?

Two bands at approximately 1000 cm⁻¹ and 1100 cm⁻¹, with the symmetric stretch becoming IR-active
Two bands at approximately 1020 cm⁻¹ and 1070 cm⁻¹, with intensity borrowed from the asymmetric stretch
One band at 520 cm⁻¹ and one band at 1150 cm⁻¹, with the Fermi resonance suppressing the symmetric stretch (correct answer)
Two bands at approximately 1030 cm⁻¹ and 1080 cm⁻¹, with the bending overtone gaining intensity
Three bands at 520 cm⁻¹, 1050 cm⁻¹, and 1150 cm⁻¹, with the Fermi resonance having no effect on IR activity

Explanation: When analyzing vibrational spectra of linear triatomic molecules, you need to consider both selection rules and potential coupling effects like Fermi resonance. Linear ABA molecules belong to the

D_{\infty h}

point group, where only vibrations that change the dipole moment (ungerade modes) are IR-active. In this molecule, only the asymmetric stretch (

\nu_3

) at 1150 cm⁻¹ and the bending mode (

\nu_2

) at 520 cm⁻¹ are normally IR-active. The symmetric stretch (

\nu_1

) at 1050 cm⁻¹ is IR-inactive because it doesn't change the dipole moment. Fermi resonance occurs when two vibrational states have similar energies and the same symmetry. Here, the bending overtone

2\nu_2

(2 × 520 = 1040 cm⁻¹) is close in energy to

\nu_1

(1050 cm⁻¹). However, this resonance doesn't make the symmetric stretch IR-active—it only affects the energies and intensities of transitions that are already allowed. The correct answer is C. The IR spectrum shows one band at 520 cm⁻¹ (fundamental bending) and one at 1150 cm⁻¹ (asymmetric stretch). The Fermi resonance affects the

2\nu_2

overtone region but doesn't create new IR-active bands. Choice A incorrectly assumes the symmetric stretch becomes IR-active. Choice B wrongly suggests intensity transfer from the asymmetric stretch. Choice D falsely claims the bending overtone gains IR activity through resonance. Remember: Fermi resonance can only enhance transitions that are already symmetry-allowed—it cannot make forbidden transitions appear in IR spectra.

Question 3

A molecule exhibits IR absorption bands at 3400 cm⁻¹ (broad), 1650 cm⁻¹ (strong), and 1200 cm⁻¹ (medium). When the sample is heated to 80°C, the 3400 cm⁻¹ band shifts to 3450 cm⁻¹ and becomes sharper, while the other bands remain unchanged. What is the most likely explanation for this temperature-dependent behavior?

Disruption of intermolecular hydrogen bonding between O-H groups, reducing the extent of vibrational coupling (correct answer)
Thermal expansion causing a decrease in the force constant of the O-H stretching vibration
Population of higher vibrational energy levels leading to increased anharmonicity effects
Conformational changes that alter the dipole moment derivative of the O-H stretching mode
Increased molecular rotational motion causing Doppler broadening of the O-H absorption band

Explanation: When interpreting IR spectra, temperature-dependent changes in absorption bands often reveal information about molecular interactions, particularly hydrogen bonding. The key clues here are the broad 3400 cm⁻¹ band (characteristic of O-H stretching) that shifts to higher frequency and becomes sharper upon heating. Hydrogen bonding significantly affects vibrational frequencies. When O-H groups participate in intermolecular hydrogen bonds, the O-H stretching frequency decreases (red-shifts) and the absorption becomes broader due to the distribution of different hydrogen-bonded environments. Upon heating, these intermolecular interactions weaken as molecules gain kinetic energy and move apart. This causes the O-H stretch to shift to higher frequency (closer to the "free" O-H value around 3500-3600 cm⁻¹) and become sharper as the distribution of environments becomes more uniform. This perfectly matches option A. Option B incorrectly suggests thermal expansion affects force constants - while thermal expansion occurs, it doesn't significantly alter intramolecular bond force constants enough to cause this spectral shift. Option C mentions anharmonicity effects from higher vibrational populations, but this would cause band shifts for all vibrations, not just the O-H stretch, and the other bands remain unchanged. Option D proposes conformational changes affecting dipole derivatives, but this wouldn't explain the specific frequency increase and sharpening pattern observed. Remember: when you see temperature-dependent IR changes involving broad bands that sharpen and shift to higher frequency upon heating, think hydrogen bonding disruption. The 3200-3600 cm⁻¹ region is particularly diagnostic for these effects.

Question 4

The IR spectrum of a cyclic compound shows bands at 1742 cm⁻¹ (C=O stretch) and 1165 cm⁻¹ (C-O stretch). When the compound undergoes ring strain due to substitution, the C=O frequency shifts to 1758 cm⁻¹ and the C-O frequency shifts to 1145 cm⁻¹. What can be concluded about the electronic and structural changes?

Increased ring strain strengthens the C=O bond through σ-withdrawal while weakening the C-O bond through reduced conjugation (correct answer)
Increased ring strain weakens both bonds equally due to geometric distortion of the molecular framework
Decreased ring strain allows better orbital overlap, strengthening both the C=O and C-O bonds simultaneously
Conformational changes alter the dipole coupling between C=O and C-O vibrations without affecting bond strengths
Electronic reorganization increases C=O double bond character while increasing C-O single bond character through mesomeric effects

Explanation: When analyzing IR spectral changes due to ring strain, you need to understand how structural distortions affect electronic properties and bond strengths. The key is recognizing that frequency shifts directly correlate with bond strength changes: higher frequency means stronger bonds, lower frequency means weaker bonds. The data shows the C=O stretch increasing from 1742 to 1758 cm⁻¹ (stronger bond) while the C-O stretch decreases from 1165 to 1145 cm⁻¹ (weaker bond). This opposite behavior indicates different electronic effects are operating on each bond. Answer A correctly identifies that increased ring strain causes σ-withdrawal effects that strengthen the C=O bond by removing electron density, while simultaneously reducing conjugation between the C-O bond and the carbonyl system, weakening the C-O bond. This explains why the frequencies shift in opposite directions. Answer B is incorrect because the bonds don't change equally - one strengthens while the other weakens, indicating specific electronic effects rather than general geometric distortion. Answer C is wrong because it suggests decreased ring strain and simultaneous strengthening of both bonds, which contradicts the observed data showing opposite frequency shifts. Answer D incorrectly attributes the changes to dipole coupling effects without bond strength changes. However, frequency shifts of this magnitude (16 cm⁻¹ for C=O, 20 cm⁻¹ for C-O) clearly indicate actual bond strength modifications, not just coupling phenomena. Remember: in IR spectroscopy, significant frequency shifts always reflect real changes in bond strength. When frequencies move in opposite directions, look for competing electronic effects like σ-withdrawal versus conjugation changes.

Question 5

A polymer sample shows a broad IR absorption centered at 3350 cm⁻¹ attributed to O-H stretching. Deconvolution analysis reveals three components at 3250 cm⁻¹ (40%), 3350 cm⁻¹ (35%), and 3450 cm⁻¹ (25%). After thermal treatment at 150°C, the distribution changes to 3250 cm⁻¹ (25%), 3350 cm⁻¹ (30%), and 3450 cm⁻¹ (45%). What molecular process best explains this change?

Crystallization-induced alignment increases the proportion of ordered hydrogen-bonded O-H groups
Thermal degradation breaks hydrogen bonds, converting associated O-H groups to isolated ones (correct answer)
Cross-linking reactions form new covalent bonds between polymer chains, eliminating free O-H groups
Conformational reorganization optimizes intramolecular hydrogen bonding at the expense of intermolecular bonding
Phase separation concentrates O-H groups in discrete domains, enhancing cooperative hydrogen bonding effects

Explanation: When analyzing IR spectroscopy data for polymers with O-H groups, you need to understand that different wavenumbers correspond to different hydrogen bonding environments. Lower wavenumbers (like 3250 cm⁻¹) indicate stronger hydrogen bonding where O-H groups are associated with other molecules or groups. Higher wavenumbers (like 3450 cm⁻¹) correspond to "free" or isolated O-H groups with little to no hydrogen bonding. The data shows a clear pattern: after heating to 150°C, the proportion of the 3450 cm⁻¹ peak (free O-H) increases dramatically from 25% to 45%, while the 3250 cm⁻¹ peak (strongly hydrogen-bonded O-H) decreases from 40% to 25%. This shift toward higher wavenumbers indicates that thermal energy is breaking hydrogen bonds and creating more isolated O-H groups. Answer B correctly identifies this as thermal degradation breaking hydrogen bonds and converting associated O-H groups to isolated ones. The thermal energy disrupts the intermolecular interactions that created the lower-frequency absorptions. Answer A is incorrect because crystallization would actually increase ordered hydrogen bonding, shifting the distribution toward lower wavenumbers, not higher ones. Answer C is wrong because cross-linking would eliminate O-H groups entirely, not shift their distribution. Answer D incorrectly suggests optimization of intramolecular bonding, but the data shows an increase in free O-H groups, indicating bond breaking rather than formation. Remember: in IR spectroscopy of O-H groups, thermal treatment typically breaks hydrogen bonds, shifting absorptions to higher wavenumbers as groups become more isolated.

Question 6

A diatomic molecule in the v=0 vibrational state undergoes IR absorption to reach v=1. The transition moment integral $\langle v=0 | \mu | v=1 \rangle$ can be expanded as $\mu_0 + \mu_1 q + \mu_2 q^2 + ...$ where q is the vibrational coordinate. If $\mu_1 = 2.5 \times 10^{-30}$ C·m and the harmonic oscillator wavefunctions are used, what is the relative intensity of the fundamental transition compared to a hypothetical molecule with $\mu_1 = 5.0 \times 10^{-30}$ C·m?

0.25 (correct answer)
0.50
0.71
1.41
2.00

Explanation: When analyzing IR absorption intensities, you need to understand that the intensity is proportional to the square of the transition dipole moment. For vibrational transitions, the transition moment integral

\langle v=0 | \mu | v=1 \rangle

determines the absorption strength. For the fundamental transition (v=0 → v=1), only the linear term

\mu_1 q

contributes to the transition moment integral when using harmonic oscillator wavefunctions. The constant term

\mu_0

gives zero (orthogonality), and higher-order terms like

\mu_2 q^2

also vanish for this specific transition due to selection rules. The transition moment is therefore proportional to

\mu_1

, making the intensity proportional to

\mu_1^2

. For the first molecule with

\mu_1 = 2.5 \times 10^{-30}

C·m, the intensity scales as

(2.5)^2 = 6.25

. For the hypothetical molecule with

\mu_1 = 5.0 \times 10^{-30}

C·m, the intensity scales as

(5.0)^2 = 25.0

. The relative intensity is the ratio:

\frac{6.25}{25.0} = 0.25

, confirming answer A. Looking at the distractors: B (0.50) represents the simple ratio of

\mu_1

values (2.5/5.0), missing the crucial squaring relationship. C (0.71) might come from taking a square root instead of squaring. D (1.41) could result from inverting relationships or other mathematical errors. Remember: IR absorption intensity always depends on the square of the transition dipole moment, not the linear relationship. This quadratic dependence is fundamental to all spectroscopic intensity calculations.

Question 7

A coordination compound exhibits a metal-CO stretching vibration at 420 cm⁻¹ and a C-O stretching vibration at 1950 cm⁻¹. When CO is replaced with ¹³CO, these bands shift to 415 cm⁻¹ and 1910 cm⁻¹, respectively. What can be concluded about the bonding and electronic structure?

Strong π-backbonding is present; the large C-O shift indicates significant metal d-orbital participation in π* orbitals (correct answer)
Weak π-backbonding is present; the small M-CO shift indicates primarily σ-bonding character in the metal-ligand interaction
Moderate π-backbonding is present; both shifts are consistent with synergistic σ-donation and π-acceptance by CO
No π-backbonding occurs; the frequency shifts are purely due to mass effects in a purely ionic metal-CO interaction
Reverse π-bonding occurs; the metal acts as a π-donor to CO π orbitals rather than π* orbitals

Explanation: When analyzing CO coordination compounds through vibrational spectroscopy, you're examining the synergistic bonding model: CO acts as both a σ-donor (through its lone pair) and π-acceptor (through empty π* orbitals). The key insight is that π-backbonding from metal d-orbitals into CO's π* orbitals weakens the C-O bond, lowering its stretching frequency. The C-O frequency shift from 1950 to 1910 cm⁻¹ upon ¹³CO substitution represents a 40 cm⁻¹ change—this is unusually large for simple mass effects. In free CO, isotopic substitution typically causes ~20-25 cm⁻¹ shifts. The enhanced shift indicates that the π* orbitals are significantly populated through metal d-electron donation, making the vibrational mode more sensitive to mass changes. This confirms strong π-backbonding. Option A correctly identifies this strong π-backbonding based on the large C-O isotopic shift. Option B is wrong because a 40 cm⁻¹ C-O shift isn't "small"—it's diagnostically large for significant π-backbonding. Option C understates the bonding strength; "moderate" doesn't match the substantial isotopic effect observed. Option D is incorrect because purely ionic interactions wouldn't show such pronounced π* orbital involvement, and the isotopic shifts exceed those expected from mass effects alone. Remember: Large isotopic shifts in CO stretching frequencies (>30 cm⁻¹) are your red flag for strong π-backbonding. The C-O frequency is more diagnostic than M-CO frequency because it directly reflects π* orbital electron density changes.

Question 8

A symmetric top molecule has a degenerate E-type vibrational mode at 950 cm⁻¹. When this molecule is placed in a crystal field environment that breaks the rotational symmetry, the degeneracy is lifted, resulting in two bands at 935 cm⁻¹ and 965 cm⁻¹. What is the crystal field splitting parameter $\Delta$ for this vibrational mode?

15 cm⁻¹
30 cm⁻¹ (correct answer)
45 cm⁻¹
60 cm⁻¹
75 cm⁻¹

Explanation: When you encounter vibrational spectroscopy problems involving crystal field effects, you're dealing with how external environments can lift the degeneracy of molecular vibrations. In symmetric top molecules, E-type vibrational modes are doubly degenerate, meaning two vibrations occur at the same frequency due to molecular symmetry. The crystal field splitting parameter

\Delta

represents the total energy separation between the split levels. When the crystal field breaks the symmetry, the originally degenerate mode splits into two components that appear at different frequencies. The splitting is symmetric around the original frequency, so you calculate

\Delta

as the difference between the two new frequencies. Here, the original degenerate mode at 950 cm⁻¹ splits into bands at 935 cm⁻¹ and 965 cm⁻¹. The crystal field splitting is:

\Delta = 965 - 935 = 30

cm⁻¹. This makes answer B correct. Answer A (15 cm⁻¹) represents a common error where students calculate the shift of each band from the original frequency (±15 cm⁻¹) rather than the total separation. Answer C (45 cm⁻¹) might result from incorrectly adding the individual shifts (15 + 30). Answer D (60 cm⁻¹) could come from doubling the correct answer, perhaps confusing this with a different type of splitting calculation. Remember: crystal field splitting parameter

\Delta

is always the direct frequency difference between the split components, not the individual shifts from the original frequency.

Question 9

A molecule exhibits an IR band at 1720 cm⁻¹ with a full width at half maximum (FWHM) of 15 cm⁻¹. The vibrational relaxation time $T_1$ is measured to be 8.5 ps. What is the primary source of the observed linewidth, and what would be the expected FWHM if only this mechanism contributed?

Lifetime broadening; expected FWHM = 0.62 cm⁻¹, indicating significant inhomogeneous broadening (correct answer)
Lifetime broadening; expected FWHM = 1.24 cm⁻¹, indicating moderate inhomogeneous broadening
Rotational broadening; expected FWHM = 15.2 cm⁻¹, indicating the observed width is purely rotational
Collision broadening; expected FWHM = 14.8 cm⁻¹, indicating the observed width matches collision effects
Doppler broadening; expected FWHM = 0.35 cm⁻¹, indicating significant pressure broadening effects

Explanation: When you encounter IR linewidth problems, you need to distinguish between homogeneous broadening (affecting all molecules equally) and inhomogeneous broadening (affecting different molecules differently). The key is calculating the expected homogeneous linewidth and comparing it to the observed width. Lifetime broadening is a fundamental homogeneous mechanism where the uncertainty principle limits spectral resolution. The relationship is:

\text{FWHM} = \frac{1}{2\pi cT_1}

, where

c

is the speed of light and

T_1

is the vibrational relaxation time. Converting the given

T_1 = 8.5

ps to seconds and using

c = 3.0 \times 10^{10}

cm/s:

\text{FWHM} = \frac{1}{2\pi \times 3.0 \times 10^{10} \times 8.5 \times 10^{-12}} = 0.62 \text{ cm}^{-1}

Since the observed FWHM (15 cm⁻¹) is much larger than this lifetime-limited value, significant additional broadening must be present—this is inhomogeneous broadening from factors like molecular environment variations. Answer A correctly identifies lifetime broadening as the fundamental mechanism and gives the right calculated value. Answer B uses an incorrect formula or constants, yielding twice the correct result. Answer C incorrectly attributes the broadening to rotation—while rotational structure can contribute, it wouldn't typically dominate in condensed phases where

T_1 = 8.5

ps is realistic. Answer D suggests collision broadening matches the observed width, but this ignores the lifetime broadening calculation entirely. Study tip: Always calculate the lifetime-limited linewidth first using the uncertainty principle relationship, then determine if additional broadening mechanisms are needed to explain experimental observations.

Question 10

A molecule exhibits an IR band at 2100 cm⁻¹ that splits into two bands at 2080 cm⁻¹ and 2120 cm⁻¹ when dissolved in a polar solvent. The integrated intensity of the total absorption remains constant. This behavior is most consistent with:

Rotational fine structure becoming resolved at lower temperature in polar solvent
Fermi resonance between the fundamental and an overtone being disrupted by solvation
Conformational equilibrium where different conformers have distinct vibrational frequencies (correct answer)
Intermolecular hydrogen bonding creating bound and free molecular species

Explanation: The splitting of a single band into two bands with constant total intensity in polar solvent suggests the presence of different molecular conformations that are stabilized differently by the polar environment. Each conformer has a slightly different local environment around the vibrating bond, leading to different frequencies. The polar solvent can preferentially stabilize certain conformers, making the equilibrium populations more equal and both bands observable. Option A is wrong because rotational structure appears as closely spaced lines, not two distinct bands. Option B is incorrect because Fermi resonance typically shows intensity borrowing, not simple splitting. Option D would show broad bands and frequency shifts, not clean splitting.

Question 11

A compound shows an IR absorption at 1650 cm⁻¹ that exhibits fine structure with peaks separated by approximately 25 cm⁻¹. When the sample is heated, the fine structure disappears but the main absorption remains at 1650 cm⁻¹. This observation is best explained by:

Rotational structure that becomes averaged out due to increased molecular tumbling at higher temperature (correct answer)
Vibrational hot bands from excited states that become populated as temperature increases
Crystal field splitting that is averaged by increased lattice vibrations at elevated temperature
Conformational substates that interconvert rapidly above a certain activation barrier

Explanation: The fine structure with regular spacing of ~25 cm⁻¹ is characteristic of rotational fine structure (P and R branches) superimposed on the vibrational transition. At higher temperatures, increased molecular motion and collisions cause rotational relaxation to occur faster than the IR measurement timescale, leading to motional averaging that washes out the rotational structure while preserving the fundamental vibrational frequency. Option B is incorrect because hot bands would appear at different frequencies, not as fine structure on the main band. Option C applies to solid-state systems with crystalline environments, not the rotational structure described. Option D would cause frequency shifts or band broadening, not the specific loss of fine structure observed.

Question 12

A symmetric top molecule (point group C₃ᵥ) has 12 atoms and exhibits 4 bands in its IR spectrum in the region 400-1600 cm⁻¹. Group theory predicts that this molecule should have 8 IR-active vibrational modes in this region. The discrepancy between observed and predicted numbers is most likely due to:

Accidental degeneracy causing some modes to have identical frequencies and appear as single bands (correct answer)
Fermi resonance causing apparent splitting of degenerate modes into multiple observable bands
Some predicted IR-active modes having transition dipole moments too weak to observe experimentally
Symmetry breaking due to crystal packing effects that lift the degeneracy of E modes

Explanation: The observation of fewer bands (4) than predicted IR-active modes (8) indicates that some modes have accidentally degenerate frequencies, causing them to appear as single absorption bands even though they are distinct normal modes. This is common in molecules with high symmetry where different vibrational modes can have similar force constants and reduced masses. Option B would cause more bands than predicted, not fewer. Option C is possible but less likely to account for missing 4 out of 8 bands. Option D would typically cause band splitting or broadening rather than complete loss of bands, and crystal effects usually cause small perturbations rather than making bands disappear entirely.

Question 13

In the IR spectrum of benzene-d₆ (C₆D₆), the C-D stretching vibrations appear around 2200 cm⁻¹, while C-H stretches in benzene (C₆H₆) appear around 3100 cm⁻¹. If the force constants are identical, what is the calculated frequency ratio ν(C-H)/ν(C-D), and how does it compare to the observed ratio?

Calculated ratio is 1.22; observed ratio is 1.41, indicating coupling effects
Calculated ratio is 1.41; observed ratio is 1.41, confirming simple harmonic behavior (correct answer)
Calculated ratio is 1.73; observed ratio is 1.41, suggesting anharmonic corrections
Calculated ratio is 1.41; observed ratio is 1.22, indicating force constant differences

Explanation: For a diatomic oscillator, ν ∝ √(k/μ) where μ is the reduced mass. For C-H vs C-D, assuming identical force constants: μ(C-H) = (12×1)/(12+1) ≈ 0.92 and μ(C-D) = (12×2)/(12+2) ≈ 1.71. The frequency ratio is √(μ(C-D)/μ(C-H)) = √(1.71/0.92) ≈ 1.36 ≈ 1.41. The observed ratio is 3100/2200 ≈ 1.41, which matches the calculated value within experimental error, confirming that the C-H and C-D bonds have essentially the same force constant and behave as simple harmonic oscillators. The other options either use incorrect mass calculations or misinterpret the significance of the comparison.

Question 14

The IR spectrum of a compound shows a strong absorption at 1720 cm⁻¹. When the sample is dissolved in different solvents, this peak shifts: in CCl₄ it appears at 1725 cm⁻¹, in CHCl₃ at 1715 cm⁻¹, and in DMSO at 1705 cm⁻¹. What is the most reasonable explanation for this solvent dependence?

Hydrogen bonding between the carbonyl oxygen and protic solvents weakens the C=O bond
Increasing solvent polarity stabilizes the ground state more than the excited vibrational state
Dipole-dipole interactions with polar solvents increase the force constant of the C=O stretch
Hydrogen bonding with increasingly protic solvents reduces the C=O bond order through resonance (correct answer)

Explanation: The systematic decrease in C=O stretching frequency with increasing solvent hydrogen-bonding ability (CCl₄ < CHCl₃ < DMSO) indicates hydrogen bonding between the solvent and the carbonyl oxygen. This hydrogen bonding increases the contribution of the resonance structure C⁺-O⁻, which has reduced C=O bond order and thus lower stretching frequency. Option A is incorrect because hydrogen bonding doesn't directly weaken the C=O bond mechanically. Option B is wrong because this describes electronic transitions, not vibrational frequencies. Option C is backwards - the interactions actually decrease the force constant.

Question 15

The bending vibration of water (H₂O) appears at 1595 cm⁻¹, while that of deuterium oxide (D₂O) appears at 1178 cm⁻¹. The symmetric stretching frequencies are 3657 cm⁻¹ for H₂O and 2671 cm⁻¹ for D₂O. Both modes show significant isotope effects. What is the primary physical reason that bending modes are sensitive to hydrogen isotope substitution?

Bending vibrations have lower force constants, making them more sensitive to mass changes
The bending motion involves significant displacement of both hydrogen atoms relative to the molecular center of mass (correct answer)
Anharmonic coupling is stronger for bending modes, amplifying the isotope effect
The reduced mass change is more significant for the bending coordinate than stretching coordinates

Explanation: Both bending and stretching modes show substantial isotope effects (frequency ratios of 1.35 and 1.37 respectively). In the bending vibration of water, both hydrogen atoms move significantly and in phase relative to the center of mass, making this mode highly sensitive to hydrogen mass changes. The bending coordinate involves motion of both H atoms perpendicular to the molecular axis, whereas stretching primarily involves motion along individual O-H bonds. Option A is incorrect because lower force constants alone don't explain the mechanism. Options C and D describe effects that aren't the primary physical cause of isotope sensitivity in bending modes.

Question 16

A molecule exhibits IR absorption bands at 3300 cm⁻¹ (broad), 1650 cm⁻¹ (strong), and 1550 cm⁻¹ (medium). When the sample is deuterated (H replaced by D), the 3300 cm⁻¹ band shifts to approximately 2450 cm⁻¹ while the other bands remain unchanged. What functional group is most likely present?

Primary alcohol with hydrogen bonding interactions
Secondary amide with N-H stretching and C=O stretching (correct answer)
Carboxylic acid with O-H stretching and C=O stretching
Primary amine with symmetric and asymmetric N-H stretching

Explanation: The deuteration experiment is key here. The 3300 cm⁻¹ band shifts to 2450 cm⁻¹ upon deuteration, giving a frequency ratio of 3300/2450 ≈ 1.35, which is close to the theoretical √2 ≈ 1.41 expected for H/D substitution. This confirms N-H or O-H stretching. The presence of both 1650 cm⁻¹ (amide C=O stretch) and 1550 cm⁻¹ (amide II band, involving N-H bending) strongly indicates a secondary amide. Primary alcohols typically show C-O stretch around 1050 cm⁻¹, carboxylic acids show C=O around 1700-1750 cm⁻¹, and primary amines would show two N-H stretches around 3300-3500 cm⁻¹ but lack the characteristic 1550 cm⁻¹ amide II band.

Question 17

The C-H stretching frequency of chloroform (CHCl₃) appears at 3020 cm⁻¹, while that of methane (CH₄) appears at 2917 cm⁻¹. This frequency difference is primarily due to:

Increased mass of the chloroform molecule affecting the reduced mass term
Rehybridization of the carbon atom from sp³ to sp² character
Inductive withdrawal of electron density by chlorine atoms strengthening the C-H bond (correct answer)
Vibrational coupling between C-H and C-Cl stretching modes in chloroform

Explanation: The higher C-H stretching frequency in CHCl₃ compared to CH₄ results from the electron-withdrawing inductive effect of the three chlorine atoms. This withdraws electron density from the C-H bond, increasing its bond strength and force constant, which leads to a higher vibrational frequency (ν ∝ √(k/μ)). Option A is incorrect because the reduced mass for the C-H oscillator is essentially the same in both molecules. Option B is wrong as both carbons remain sp³ hybridized. Option D is incorrect because C-H and C-Cl stretches occur in different frequency ranges and don't significantly couple.

Question 18

A linear triatomic molecule XY₂ shows three fundamental vibrational modes. If the molecule has a center of symmetry, which combination of IR and Raman activities is expected for these modes?

All three modes are both IR and Raman active due to the linear geometry
Two modes are IR active only, one mode is Raman active only
One mode is IR active only, two modes are Raman active only (correct answer)
Two modes are both IR and Raman active, one mode is inactive in both

Explanation: For a linear triatomic molecule with a center of symmetry (like CO₂), the mutual exclusion rule applies: modes that are IR active cannot be Raman active and vice versa. The three vibrational modes are: symmetric stretch (Raman active only), antisymmetric stretch (IR active only), and bending (doubly degenerate, IR active only). Thus, one mode (symmetric stretch) is Raman active only, and two modes (antisymmetric stretch and bend) are IR active only. Option A violates the mutual exclusion rule. Option B has the activities reversed. Option D incorrectly suggests some modes are active in both or inactive in both.