At my university, we had an organic chemistry exam question that stated,"The trend in the values of the ir absorbtions for the c-h bonds of methane ($\ce{CH4}$: $2995\,\mathrm{cm}^{-1}$), (ethylene: $3080\,\mathrm{cm}^{-1}$), and (acetylene: $3305\,\mathrm{cm}^{-1}$) can be explained by...

The answer was " by the increasing bond strength of the carbon-carbon single, double, and triple bonds. I thought the correct answer was "increasing s character of the carbon atoms". Could I get an explanation as to why the first response is valid?


  • 2
    $\begingroup$ I think that you're correct. The C-C bond strength should only explain the C-C wavenumber. The C-H bond strength is correlated with hybridisation of the carbon in question and as you suggested an increasing s character leads to a stronger bond. Sorry if you saw my earlier answer (now deleted), I entirely misread the question in my haste. $\endgroup$ – orthocresol Nov 7 '16 at 19:52
  • $\begingroup$ But could the first answer technically explain the reason for the C-H trend? For example, If the C=C bond is stronger than the C-C bond, then that's an indication that the H bond to the C will be stronger if attached to the C=C? I'm just playing devils advocate. $\endgroup$ – user34748 Nov 7 '16 at 19:59
  • $\begingroup$ Methane has two vibrational normal modes with C-H stretches, $3026 , 3156 \pu{ cm^{-1}} $ and the other molecules similarly have different modes in which CH bonds stretch. (In a normal mode all atoms move in synchrony but with different phases depending on symmetry). Thus its not as simple as suggesting amount of s character in a CH bond as all such bonds move together in such a way that there is a fixed 'in and out of phase' relationship with one another. The correlation more a sort of 'rule of thumb' than a strict scientific explanation. I feel sure the 's' character is the answer expected. $\endgroup$ – porphyrin Nov 8 '16 at 11:15
  • $\begingroup$ There is no such thing as increasing or decreasing s character of a carbon atom. See, carbon atom has but one s-orbital; it can't have more, it can't have less. You may, though, speak of a bond having more or less s character. $\endgroup$ – Ivan Neretin Dec 8 '16 at 6:04

As orthocresol suggested in his comment, you are correct. First of all, methane does not contain a $\ce{C-C}$ bond, so that answer could not explain the wave number form methane. In addition, you can think of bonds as tiny harmonic oscillators. Specifically, they behave like a mass on a spring, absorbing frequencies that match the compound's frequency of oscillation. Remember that for a mass on a spring, $\omega = \sqrt{\frac{k}{m}}$, where k is the spring's stiffness (related to the strength of the bond) and m is its mass. This is why stronger bonds (such as a $\ce{C=C}$ bond over a $\ce{C-C}$ bond) have a higher frequency, and therefore wavenumber (these two quantities are directly proportional). Likewise, the $sp$ orbitals in acetylene are stiffer than the $sp^2$ orbitals in ethylene, which are in turn stiffer than the $sp^3$ orbitals in methane. This explains why methane has a lower wave number than ethylene, which in turn has a lower wave number than acetylene.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.