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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?

Thanks!

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    $\begingroup$ 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
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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.

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