I was comparing the acidity of methane $\ce{CH4}$ and water $\ce{H2O}$, and a lot of things weren't making sense. Generally, the process of turning $\ce{HX}$ into $\ce{H+}$ and $\ce{X^-}$ can be broken into 2 hypothetical steps:
- Homolytic cleavage of the $\ce{H-X}$ bond, followed by
- Electron donation from $\ce{H}$ to $\ce{X}$.
The energy required for step 1 is correlated with bond length, while the energy released by step 2 is correlated with electronegativity of $\ce{X}$.
The way I see it, electronegativity comes as a direct consequence of atomic size. Adding an electron to smaller element means the electron is added to an orbital that's relatively close to the nucleus, which is more stable. Thus, smaller elements are generally more electronegative, particularly across a row in the periodic table.
Now we consider $\ce{CH4}$ and $\ce{H2O}$. Oxygen, being smaller than carbon, will be more electronegative; thus, $\ce{OH-}$ is more stable than $\ce{CH3-}$ and step 2 will be more favorable for $\ce{H2O}$. But oxygen being smaller than carbon also means the $\ce{O-H}$ bond length is smaller than the $\ce{C-H}$ bond length, and thus harder to homolytically break. Overall, I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark. Yet the $\mathrm{p}K_\mathrm{a}$ for $\ce{H2O}$ is 15.7, while the $\mathrm{p}K_\mathrm{a}$ for $\ce{CH4}$ is approximately 55!
How can I make sense of this? What are other factors that impact electronegativity and homolysis that I am not considering? Or are there other aspects to this problem that I am not considering entirely?