Factors affecting vibrational frequencies MCQs With Answer

Factors affecting vibrational frequencies MCQs With Answer

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Factors Affecting Vibrational Frequencies MCQs With Answer is designed for M. Pharm students focusing on Modern Pharmaceutical Analytical Techniques. Mastering why vibrational bands shift in IR/Raman spectra is essential for interpreting functional groups, intermolecular interactions, and structural changes in drug molecules, excipients, and biomacromolecules. This quiz covers core determinants such as force constant, reduced mass, hybridization, conjugation, ring strain, hydrogen bonding, solvent and phase effects, isotopic substitution, temperature and pressure, vibrational coupling, anharmonicity, and special phenomena like Fermi resonance and the vibrational Stark effect. Each question highlights practical analytical implications—from carbonyl and nitrile diagnostics to ionization and complexation effects—helping you translate spectra into reliable chemical and pharmaceutical insights.

Q1. Which single change most directly increases a bond’s IR stretching frequency according to the harmonic oscillator model?

  • Increasing the bond’s force constant (k)
  • Increasing the reduced mass (μ) of the vibrating pair
  • Strengthening intermolecular hydrogen bonding to that bond
  • Increasing ring flexibility (reducing ring strain)

Correct Answer: Increasing the bond’s force constant (k)

Q2. Replacing hydrogen with deuterium in a C–H bond typically causes which shift?

  • A shift from ~3000 cm−1 to ~2100–2200 cm−1 for the C–D stretch
  • A shift from ~3000 cm−1 to ~3600–3700 cm−1
  • No significant change in frequency
  • A shift into the fingerprint region (~1500–1600 cm−1)

Correct Answer: A shift from ~3000 cm−1 to ~2100–2200 cm−1 for the C–D stretch

Q3. Conjugation of a carbonyl (C=O) with a C=C system generally:

  • Lowers the C=O stretch by roughly 20–40 cm−1
  • Raises the C=O stretch by about 50–80 cm−1
  • Has no effect on the C=O frequency
  • Converts the C=O band into two distinct bands

Correct Answer: Lowers the C=O stretch by roughly 20–40 cm−1

Q4. Decreasing ring size (e.g., cyclohexanone → cyclobutanone) typically:

  • Raises the C=O frequency due to increased angular strain
  • Lowers the C=O frequency by enhancing conjugation
  • Only broadens the C=O band without shifting it
  • Eliminates IR activity of the C=O stretch

Correct Answer: Raises the C=O frequency due to increased angular strain

Q5. Intermolecular hydrogen bonding generally causes the O–H stretching band to:

  • A red shift (to lower wavenumber) and band broadening
  • A blue shift (to higher wavenumber) and band narrowing
  • Remain at the same frequency with increased intensity only
  • Split into an overtone doublet without shifting

Correct Answer: A red shift (to lower wavenumber) and band broadening

Q6. In aromatic aldehydes, electron-withdrawing substituents on the ring typically:

  • Electron-withdrawing substituents generally raise the C=O stretching frequency
  • Electron-donating substituents raise the C=O stretching frequency
  • Both types of substituents lower the C=O frequency equally
  • Show no predictable trend on the C=O frequency

Correct Answer: Electron-withdrawing substituents generally raise the C=O stretching frequency

Q7. Moving acetone from a nonpolar solvent (e.g., CCl4) to methanol typically leads to:

  • A red shift of the C=O stretch (lower wavenumber) due to H-bonding
  • A blue shift of the C=O stretch (higher wavenumber) with narrowing
  • Disappearance of the C=O band
  • No change because solvent polarity does not affect frequency

Correct Answer: A red shift of the C=O stretch (lower wavenumber) due to H-bonding

Q8. Fermi resonance most commonly manifests in IR spectra as:

  • Apparent splitting of a fundamental band due to coupling with an overtone/combination band
  • Replacement of all fundamentals with overtones
  • Only intensity changes without frequency shifts
  • A phenomenon exclusive to gas-phase spectra

Correct Answer: Apparent splitting of a fundamental band due to coupling with an overtone/combination band

Q9. Vibrational coupling in CH2 groups typically leads to:

  • Appearance of two distinct bands (symmetric and asymmetric) at different frequencies for the same group
  • Disappearance of all CH2 stretching bands
  • Shifting both symmetric and asymmetric stretches to identical frequency
  • Only affecting intensities without frequency changes

Correct Answer: Appearance of two distinct bands (symmetric and asymmetric) at different frequencies for the same group

Q10. Ionization of a carboxylic acid (COOH → COO−) is indicated in IR by:

  • Disappearance of ~1710 cm−1 band and appearance of asymmetric (~1550–1610 cm−1) and symmetric (~1360 cm−1) COO− stretches
  • Shift of the C=O to >1800 cm−1 without new bands
  • No change in the carbonyl region
  • Conversion to a single band at ~1650 cm−1

Correct Answer: Disappearance of ~1710 cm−1 band and appearance of asymmetric (~1550–1610 cm−1) and symmetric (~1360 cm−1) COO− stretches

Q11. Which C–H stretching mode appears at the highest wavenumber?

  • sp C–H (alkynyl) near ~3300 cm−1
  • sp2 C–H (alkenyl/aryl) near ~3100 cm−1
  • sp3 C–H (alkyl) near ~2900 cm−1
  • All three appear at the same frequency

Correct Answer: sp C–H (alkynyl) near ~3300 cm−1

Q12. How do substituents influence the –C≡N (nitrile) stretching frequency?

  • Electron-withdrawing substituents increase the –C≡N stretching frequency
  • Electron-donating substituents increase the –C≡N stretching frequency
  • Substituents do not influence nitrile frequencies
  • Hydrogen bonding always blue-shifts nitrile bands regardless of context

Correct Answer: Electron-withdrawing substituents increase the –C≡N stretching frequency

Q13. What is the typical temperature effect on a strongly hydrogen-bonded O–H stretching band upon heating?

  • Heating a strongly H-bonded O–H band causes a blue shift and narrowing as H-bonds break
  • Heating invariably red-shifts all stretching modes
  • Temperature has no effect up to ~80 °C
  • Heating converts the O–H fundamental into an overtone-only feature

Correct Answer: Heating a strongly H-bonded O–H band causes a blue shift and narrowing as H-bonds break

Q14. How does pressure generally influence vibrational frequencies in condensed phases?

  • Increasing pressure generally blue-shifts stretching frequencies by stiffening intermolecular environments
  • Increasing pressure red-shifts all fundamental bands
  • Pressure only changes band intensity, not frequency
  • Solid-state bands are unaffected by pressure

Correct Answer: Increasing pressure generally blue-shifts stretching frequencies by stiffening intermolecular environments

Q15. What is commonly observed when moving from gas phase to condensed phase for polar functional groups (e.g., OH, C=O)?

  • Moving from gas to condensed phase often produces small red shifts and broadening due to intermolecular interactions
  • Always a blue shift and narrowing of the bands
  • No change in frequency or band shape
  • Elimination of combination and overtone bands

Correct Answer: Moving from gas to condensed phase often produces small red shifts and broadening due to intermolecular interactions

Q16. Replacing 16O with 18O in a carbonyl group is expected to cause:

  • A shift of the C=O stretch to lower wavenumber by roughly 2–4% (≈30–40 cm−1 for a ketone)
  • A shift of the C=O stretch to higher wavenumber by ~2–4%
  • No measurable change in the C=O frequency
  • Replacement of the IR band with a Raman-only feature

Correct Answer: A shift of the C=O stretch to lower wavenumber by roughly 2–4% (≈30–40 cm−1 for a ketone)

Q17. Anharmonicity in molecular vibrations primarily:

  • Lowers fundamental frequencies relative to the harmonic model and allows overtones/combination bands
  • Raises fundamental frequencies and suppresses overtone formation
  • Only affects band intensities without changing frequencies
  • Occurs exclusively in the gas phase

Correct Answer: Lowers fundamental frequencies relative to the harmonic model and allows overtones/combination bands

Q18. Which factor most reliably lowers the Amide I (mainly C=O stretch) frequency in peptides/proteins?

  • Strengthening hydrogen bonding (e.g., in β-sheets) lowers the Amide I frequency
  • Increasing temperature lowers the Amide I frequency
  • Deuteration of the backbone nitrogen markedly lowers Amide I
  • Electron-withdrawing substituents on side chains primarily raise only Amide II

Correct Answer: Strengthening hydrogen bonding (e.g., in β-sheets) lowers the Amide I frequency

Q19. Regarding the vibrational Stark effect, nitrile (–C≡N) probes commonly show:

  • Stronger local electric fields (more polar environments) often blue-shift the –C≡N stretch
  • Stronger fields red-shift the –C≡N stretch in all cases
  • No measurable sensitivity of –C≡N frequency to local fields
  • Stark effects are limited to O–H stretching modes

Correct Answer: Stronger local electric fields (more polar environments) often blue-shift the –C≡N stretch

Q20. How does halogen identity affect the C–X stretching frequency?

  • Replacing chlorine with fluorine in a C–X bond shifts the stretching band to higher wavenumber because of lower mass and a stronger bond
  • Replacing chlorine with bromine shifts the stretching band to higher wavenumber
  • Halogen identity does not affect C–X stretching frequencies
  • All C–X stretches appear at ~1000 cm−1 irrespective of halogen

Correct Answer: Replacing chlorine with fluorine in a C–X bond shifts the stretching band to higher wavenumber because of lower mass and a stronger bond

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