Vibrational frequency

Question

Rank the C – H bonds from strong to weak.

¹²C-¹H : ¹³C-¹H : ¹²C-²H : ¹²C-H

Explain reasoning using stretching/vibrational frequency (in the content of isotope effects), and frequency connection to bond strength.

Answer 1

This is an odd question... Because nominally, the differences in bond strengths would be pretty stinking small, so asking about very small differences in bond strengths seems a little pointless... These are all C-H bonds, and without additional information (such as the hybridization of the carbon atom, which would have a much larger affect on bond strength than isotope) they would all be extremely similar.

In any case, the basic idea which can be applied to all four of these cases (well, kinda -- one is a little bit of a stretch) is that bonds between heavier atoms are slightly weaker than bonds between lighter atoms. This is because the "reduced mass" of the pair of atoms is in the denominator of Hooke's Law, which is used to calculate the frequency of a vibration. You can look up the formula for those (both "reduced mass" and Hooke's law), but the idea is that if either of the bonded atoms is present as its heavier isotope, the reduced mass will be higher, which will make the frequency lower (since reduced mass is in the denominator). Also, changing the isotope of the lighter atom (in this case, H) has a MUCH larger effect, since the proportional change is much larger -- changing 1H to 2H doubles the mass of that atom, while changing 12C to 13C only increases its mass by ~8%.

SO! If heavier atoms have lower frequencies, and the effect is MUCH more pronounced when changing the isotope of the lighter atom, the first three situations are clear. The strongest bond is between 12C and 1H, since both atoms are in their lower isotope. The next strongest is 13C-1H, since changing the isotope of C only has a small effect on the frequency. And 12C-2H is the weakest of the three, since changing the isotope of the hydrogen weakens the bond the most.

The fourth case (12C-H) is odd, since the isotope of H is not specified. I suspect that when the original problem was transcribed, perhaps that isotopic information was accidently missed...? If it were, say, 12C-3H (the only other isotope of hydrogen), it would be weaker again than 12C-2H. But if the absence of isotope for H is intentional, I can only assume it would be referring to naturally abundant hydrogen. Naturally abundant hydrogen is 99.97% 1H, so that would make this functionally identical to 12C-1H. And claiming the bond is 0.03% weaker because of the 0.03% of 2H is a little disingenuous -- it's not exactly an average like that... It would be better to think of it as 99.97% made up of bonds identical to 12C-1H (because they are) with a vanishingly small portion of the bonds (0.03%) the same as the 12C-2H ones, rather than all the bonds being 0.03% weaker.

In any case, it's a little bit of a weird question, I gotta say. And I wonder if the last one was really 12C-3H.