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Predict stretching frequency with a supplied bond model

Key idea: In the spring model, stretching frequency is proportional to √(force constant/reduced mass). Stiffer bonds absorb at higher wavenumber, while heavier isotopes shift absorption lower.

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  • GCE A-Level H3 Chemistry 2027
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Why does a C–H bond absorb at a higher wavenumber than a C–Cl bond, and a C≡C bond at a higher wavenumber than a C=C bond? A simple model of a bond as two masses joined by a spring explains these trends. You will be given the equations of the model if a question needs them; this lesson shows how to use them.

Check your understanding 1

A C≡C triple bond is stiffer than a C=C double bond, and both join two carbon atoms. Which do you expect to absorb at the higher wavenumber?

Higher wavenumber
Show answer

C≡C. With the same atoms the reduced mass is the same, so the stiffer bond vibrates faster and absorbs at a higher wavenumber. The model below explains why.

The bond as a spring

For two atoms joined by a bond that behaves like a spring, the model gives:

ω = square root of (k/μ), f = ω/2π, ν tilde = f/c, μ = m₁m₂/(m₁ + m₂)
  • k is the force constant: how stiff the bond is. Stronger bonds, such as multiple bonds, are stiffer.
  • μ is the reduced mass of the two atoms. It is not their total mass.
  • ω is the angular frequency, f the frequency and ν tilde the wavenumber.

So a stiffer bond absorbs at a higher wavenumber, and heavier atoms absorb at a lower wavenumber.

Check your understanding 2

The force constant of a bond increases by a factor of 1.44, and the reduced mass stays the same. By what factor does the stretching wavenumber change?

Show answer

square root of 1.44 = 1.20: the wavenumber increases by 20%, not by 44%, because it depends on the square root of k.

Units

With k in N m⁻¹ and masses in kg, square root of (k/μ) gives ω in rad s⁻¹. Divide by 2π to get f in Hz. Then divide by c = 3.00 × 10¹⁰ cm s⁻¹ to get ν tilde in cm⁻¹. (If you use c = 3.00 × 10⁸ m s⁻¹, the answer is in m⁻¹; divide it by 100 to get cm⁻¹.)

Check your understanding 3

The model gives ω = 6.00 × 10¹⁴ rad s⁻¹ for a bond. Calculate f and ν tilde, using c = 3.00 × 10¹⁰ cm s⁻¹.

Unit: Hz
Show the steps Optional working
Unit: Hz
Unit: cm⁻¹
Show answer

f = (6.00 × 10¹⁴)/2π = 9.55 × 10¹³ Hz

ν tilde = (9.55 × 10¹³)/(3.00 × 10¹⁰) = 3.18 × 10³ cm⁻¹

Dividing ω directly by c would miss the factor of 2π.

Comparing isotopes

Replacing an atom by a heavier isotope does not change the bond, so k stays the same and cancels in a ratio:

(ν tilde ₂)/(ν tilde ₁) = square root of (μ₁/μ₂)

Worked example 1

Replacing H by D

Problem

In the model, H³⁵Cl absorbs at 3000 cm⁻¹. Using masses of 1, 2 and 35 u for H, D and ³⁵Cl, and the same force constant, predict the wavenumber for D³⁵Cl.

Show full solution

  1. Reduced masses

    Method

    Calculate each reduced mass.

    Reason

    The model uses the reduced mass, not the total mass.

    Working

    Answer

    μ_HCl = (1 × 35)/(1 + 35) = 35/36 u, μ_DCl = (2 × 35)/(2 + 35) = 70/37 u

  2. Wavenumber ratio

    Method

    Take the square root of the inverse ratio of the reduced masses.

    Reason

    The force constant is the same, so it cancels.

    Working

    Answer

    (ν tilde _DCl)/(ν tilde _HCl) = square root of ((35/36)/(70/37)) = square root of (37/72) = 0.717

  3. Result

    Working

    Answer

    ν tilde _DCl = 3000 × 0.717 = 2150 cm⁻¹: lower, because D is heavier than H. These are model values, not measured wavenumbers.

Check your understanding 4

In the model, a C–H bond absorbs at 3000 cm⁻¹. Using masses of 12, 1 and 2 u for C, H and D, and the same force constant, predict the wavenumber for C–D.

Show the steps Optional working
Unit: u
Unit: u
Show the steps Optional working
Unit: cm⁻¹
Show answer

μ_CH = (12 × 1)/13 = 0.923 u and μ_CD = (12 × 2)/14 = 1.714 u. The ratio is square root of (0.923/1.714) = 0.734, so ν tilde _CD = 3000 × 0.734 ≈ 2200 cm⁻¹.

Limits of the model

The model is exact only for an ideal spring joining two atoms. Real bonds are not perfect springs, and in larger molecules the vibrations of neighbouring bonds affect each other. Use the model to explain trends; to identify bonds in a spectrum, use the data booklet values.

Check your understanding 5 Stiffness and mass changing together

In the spring model, one bond has a wavenumber of 1000 cm⁻¹. A second bond has 1.44 times its force constant and twice its reduced mass. Find the new-to-old wavenumber ratio and the second bond’s wavenumber.

Show the steps Optional working
Unit: cm⁻¹
Show answer

The ratio is square root of (1.44/2) = 0.849 (about 0.85). The new wavenumber is 1000 × square root of (1.44/2) = 849 cm⁻¹ (about 850 cm⁻¹). The greater reduced mass outweighs the greater stiffness, so the second wavenumber is lower.

Common mistakes

  • Using the total mass of the two atoms instead of the reduced mass.
  • Dividing ω by c without first dividing by 2π.
  • Mixing units: using c in m s⁻¹ and reporting the answer in cm⁻¹.
  • Forgetting the square root, so that doubling k is said to double the wavenumber.
  • Saying a heavier isotope changes the force constant. It changes only the reduced mass.

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