Questions: Vibrational Frequency and Force Constant

Question 1

An O−H bond absorbs near 3500 cm⁻¹ in the IR spectrum. When hydrogen is replaced by deuterium to form O−D, where would you expect the new absorption?

• A. Near 3500 cm⁻¹ — the force constant increases to compensate for heavier deuterium
• B. Near 7000 cm⁻¹ — deuterium is twice as heavy, so the frequency doubles
• C. Near 2600 cm⁻¹ — the heavier reduced mass lowers the frequency while the force constant stays nearly the same
• D. Near 1750 cm⁻¹ — deuterium substitution halves the vibrational frequency

Answer: (C)

Replacing H (mass 1) with D (mass 2) roughly doubles the reduced mass μ. Since ν ∝ 1/√μ, the frequency drops by a factor of √2 ≈ 1.41. Starting at 3500 cm⁻¹, the O−D stretch appears near 3500/1.41 ≈ 2480–2600 cm⁻¹ — consistent with observed values. Crucially, the force constant k is essentially unchanged because it reflects bond stiffness (electronic structure), not atomic mass. Isotope substitution is a clean probe of reduced mass effects precisely because it changes μ without changing k.

Question 2

A C≡N triple bond absorbs at a higher wavenumber than a C=N double bond, which absorbs higher than a C−N single bond. What is the primary reason for this trend?

• A. Carbon-nitrogen bonds with more electrons have smaller reduced masses, increasing frequency
• B. Triple bonds have larger force constants due to greater bond stiffness, and higher k means higher vibrational frequency
• C. The reduced mass decreases with bond order because electrons contribute to molecular mass
• D. Triple bonds are shorter, and shorter path length means higher frequency

Answer: (B)

The force constant k measures bond stiffness — how strongly the bond resists stretching. A triple bond involves three shared electron pairs, making it much stiffer than a double bond (two pairs) or single bond (one pair). Since ν ∝ √k, higher k gives higher frequency. The reduced mass is essentially the same for C≡N, C=N, and C−N (the atomic masses don't change with bond order), so the frequency trend is driven entirely by the increasing force constant. This is why C≡C absorbs near 2100 cm⁻¹, C=C near 1650 cm⁻¹, and C−C near 1000 cm⁻¹.

Question 3

Deuterium substitution (replacing H with D) changes the force constant of the bond, which is why the vibrational frequency shifts to lower wavenumber.

• T. True
• F. False

Answer: False

The force constant k reflects the bond's electronic structure — how stiff the bond is — which is determined by the number and type of electrons shared between the atoms. Replacing H with D does not change the electronic structure of the bond, so k remains essentially unchanged. What changes is the reduced mass μ: deuterium is twice as heavy as hydrogen, increasing μ and thereby lowering the vibrational frequency according to ν = (1/2π)√(k/μ). This is precisely what makes isotope substitution a clean diagnostic: it selectively probes the reduced mass without perturbing the force constant.

Question 4

A bond with a larger force constant will absorb at a higher wavenumber in the IR spectrum, all else being equal.

• T. True
• F. False

Answer: True

The vibrational frequency ν = (1/2π)√(k/μ). For fixed reduced mass μ, a larger force constant k means a stiffer bond that vibrates faster, appearing at higher wavenumber (cm⁻¹ = ν/c). This is why multiple bonds absorb at higher wavenumbers than single bonds between the same atoms: C≡O (triple bond character, large k) absorbs near 2143 cm⁻¹ in CO gas, while C=O (double bond, smaller k) absorbs near 1700–1750 cm⁻¹, and C−O (single bond, smallest k) absorbs near 1000–1200 cm⁻¹.

Question 5

A chemist observes an IR absorption at 2100 cm⁻¹ and suspects it is either a C≡C or a C≡N stretch. Explain how the concept of reduced mass could help distinguish between them, and what isotopic experiment would test the assignment.

Model answer: The reduced mass for a C≡C vibration (both atoms are carbon, mass 12) differs from that for C≡N (nitrogen mass 14 vs carbon mass 12). For C≡C, μ = (12×12)/(12+12) = 6 amu. For C≡N, μ = (12×14)/(12+14) ≈ 6.46 amu. Since ν ∝ 1/√μ, the C≡N stretch should appear at slightly lower wavenumber than C≡C with the same force constant. To test the assignment, you could perform isotopic labeling: replace ¹²C with ¹³C (mass 13) in the triple bond. For C≡C, substituting one carbon shifts μ to (12×13)/(12+13) = 6.24 amu, predicting a specific frequency shift. For ¹³C≡N, μ = (13×14)/(13+14) ≈ 6.74 amu. The magnitude and direction of the observed shift would confirm which assignment is correct.

This illustrates the power of isotope labeling in spectroscopic assignment. Because isotopic substitution changes μ but not k, the predicted frequency shift is precisely calculable from the mass ratio. If the observed shift matches the prediction for ¹³C≡C, the original assignment is confirmed; if it matches ¹³C≡N, the other is. This approach is widely used in inorganic and biochemistry to assign IR bands in complex molecules.