# Why does methane have 4-different bond dissociation energies?

I will elaborate on my question here.

In my Lecture Notes it is mentioned that for methane, there are 4 different $$\text{C-H}$$ bond energies. I plotted a quick graph of the bond energies to analyse the trend.

Here the vertical axis represents the bond dissociation energy while the horizontal axis represents the bond dissociation number. Looking at the graph, there is no clear trend so I decided to turn my attention to the structure.

$$\text{CH_4}$$ has the following structure

It is from here that I am unable to proceed. Here are my questions.

1. Am I barking up the wrong tree by looking at the structure and how do I account for the changing bond energies
2. Is there a VSEPR structure for molecules like $$\text{CH_3}$$ $$\text{CH_2}$$ $$\text{CH}$$? If so how do I figure it out.
• All bonds in methane are not broken together. Say one C-H bond is broken by heteroolysis, it is no longer methane. It becomes a methyl cation, where C-H bond enthalpy is different. Likewise, the pKa values of the two hydrogen atoms in sulphuric acid are not same. Apr 7 '21 at 12:36
• But why? It is still a bond between C-H @AdityaRoychowdhury Apr 7 '21 at 12:49
• @SanjayAdhith because molecules are more complicated than localized bonds between parts of them. And even if everithing would be fully localized, which is quite true for molecules as CH4, CH3 ions or radical, and so on, the potential felt by electrons isn't / can't be the same. Apr 7 '21 at 13:17
• VSEPR is consistent with the geometries of the various radicals, but the energetics are more difficult to unravel. The following article gives this a shot: sciencedirect.com/science/article/abs/pii/S2210271X14001285 Apr 13 '21 at 10:59
• Also, it looks like the 2nd and 3rd bond dissociation energy values are switched (see for instance the article referenced in the answer). Apr 13 '21 at 19:15

The dissociation energies are different because bonds don't all break at once

Most chemical processes happen step by step. Even in simple molecules like methane where there are only 4 bonds, bond dissociation almost always occurs stepwise.

The implication of this is that you would not expect each step to involve the same energy (and this applies whatever sort of bond-breaking occurs; some might create a $$\ce{H+}$$ and a $$\ce{CH3-}$$ others a $$\ce{H.}$$ and a $$\ce{CH3.}$$ ).

The issue is, in general, that the second bond-breaking step is happening to a different molecule to an isolated methane. Whatever the second step is, the starting point is not a complete methane. And, since the structure of a molecule affects the properties of all the bonds in it (though the effect might be small in a large molecule where the bonds are far apart), the second dissociation will, in general, be different from the first (and the bond energy will vary as a consequence).

While, hypothetically, a single concerted breaking of all 4 bonds could happen, this is likely to be vanishing rare or extraordinarily hard to achieve. If you could somehow do that reaction, then you would measure a different result. But that isn't what chemists are usually talking about when measuring or reporting bond strengths so any such result would be a rare oddity unrelated to most achievable results.

• Ah okay that makes sense. Am I right to say that the MO will change every time a bond is broken? Nov 28 '21 at 8:34
• @SanjayAdhith yes. The bonding structure of a molecule changes when atoms are removed. Nov 28 '21 at 11:48

Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.

Now let's get into a bit more interesting compound: ethane. Very interesting!

The table is only part of the information on ethane given in an open access article (Ref 1). Note that the first C-H bond dissociation requires 422 kJ/mol, while the second requires 150 to 465 kJ/mol, depending on which bond you are breaking, because of what product you are making. Each line in the table is worth several paragraphs of explanation.

I admire your desire to tie up your knowledge neatly here - get a lot of string, because it's a very big package!

Ref 1. https://pubs.acs.org/doi/pdf/10.1021/acs.jpca.5b01346 Branko Ruscic, J. Phys. Chem. A 2015, 119, 28, 7810–7837

• This is a well put together answer but I am afraid to say that I don't think that it answers OP's question. Apr 7 '21 at 18:20
• @Nisarg: Thanks, you're right. I was answering the question so it satisfied me. In particular, I left out VSEPR calculations because the methane data really just say that the bonding between C and H is pretty uninteresting as far as CH4 goes. CH4 is too simple. BUT, bond energies help us to understand more complex molecules, and the bigger picture involves looking not just at the original molecule, but also at the final product. Apr 7 '21 at 21:52