Why does the weaker C–H bond have a higher wavenumber than the C=O bond?

Question

My understanding is that a stronger bond has a higher wavenumber in IR spectrum. But why does the C–H vibration have a higher wavenumber than the C=O vibration? The latter is a double bond, so I think it should be a stronger bond than the C–H single bond.

Answer 1

A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object:

$$\omega = \sqrt{\frac{k}{m}}$$

We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $\mu$, defined by:

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$

where $m_1$ and $m_2$ are the masses of the two bodies. So, for a C–H bond, we have

$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$

and for a C=O bond, we have

$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$

From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $k$ for the C=O bond in formaldehyde is approximately $2.46$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).

So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $\bar{\nu}$, is directly proportional to the (angular) frequency $\omega$.

$$\begin{align} \frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\ &= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\ &= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\ &= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\ &= 1.74 \end{align}$$

This is not too far from the experimental value of $\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$ (using values in the middle of the typical ranges for alkanes and aldehydes).

Answer 2

According to Skoog, Analytical Chemistry:

Using classical mechanics, assuming a diatomic molecule, the frequency of vibration $\nu$ may be described by $$\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} $$

where $k$ represents the force constant of this bond, and $\mu$ the reduced mass of the particles bond together, defined as

$$\mu = \frac{m_1 \cdot m_2} {m_1 + m_2}$$

An equivalent quantum chemical / mechanial description includes $c$, the speed of light, yielding a result expressing the radiation in wavenumbers: $$\bar{\nu} = \frac{1}{2\pi \cdot c} \sqrt{\frac{k}{\mu}} $$

So (1), with the same reduced mass $\mu$, the observed frequency increases with increasing force constant $k$. And (2) $\nu$ equally increases while lowering the reduced mass $\mu$.

Skoog further mentions as typical force constant of single bonds the range of $3 \times 10^2 \ldots 8 \times 10^2\,\pu{N m}^{-1} $ with an average of $5 \times 10^2\,\pu{N m}^{-1}$; while stating $1 \times 10^3\,\pu{N m}^{-1}$ and $1.5 \times 10^3\,\pu{N m}^{-1}$ for typical double, and triple bonds, respecitively.

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Consequently, for a carbonyl $\ce{C=O}$ double bond, assuming as the mass $m_1$ of a carbon atom of

$$m_1 = \frac{12 \times 10^{-3}\,\pu{kg / mol} } {6.0 \times 10^{23}\,\pu{atoms / mol}} \times 1\,\pu{atom} = 2.0 \times 10^{-26}\,\pu{kg}$$

and for oxygen a mass $m_2 = 2.7 \times 10^{-26}\,\pu{kg / atom}$, yields a reduced mass $\mu = 1.1 \times 10^{-26}\,\pu{kg}$. Using the above mentioned equation determines as vibration frequency (in wavenumbers) a value of

$$\bar{\nu} = 5.3 \times 10^{-12}\,\pu{s \cdot cm}^{-1} \sqrt{\frac{1 \times 10^3\,\pu{N \cdot m^{-1}}}{1.1 \times 10^{-26}\,\pu{kg}}} = 1.6 \times 10^3\pu{cm}^{-1}$$.