Why a higher s character increases a carbon atom's electronegativity?

To make us understand certain phenomenon like the higher acidity of alkynes than alkanes and alkenes, our teacher told us to learn this as a general rule:

Higher s character leads to a more electronegative carbon atom.

Note: By s character I mean the percentage of s in the hybridisation of the carbon atom(eg $25\%$ s character in $\ce{sp^3}$ hybrid carbon)

Though this well explains the above mentioned phenomenon (and certain others like the decreased $\ce{C-Cl}$ bond length in chlorobenzene), but I was unable to figure out why this thing(increased electronegativity) happens after all.

This did not provide me a satisfactory answer. Electronegativity is after all the ability of an atom to attract the shared pair of electrons towards itself. So, how would one explain this using the concept of orbitals or molecular orbital theory?

I would also like to know if the same argument is valid for other p-block elements?

• higher s character -> lower potential energy -> stabler electrons – DHMO Oct 2 '16 at 14:22
• @DHMO then this should be : stabler electrons->more electron attracting power. Right?But why? – Amritansh Singhal Oct 2 '16 at 14:28
• because electrons want to lower their potential energy. – DHMO Oct 2 '16 at 14:29
• @DHMO then for a C-H bond, electrons should prefer to move towards the H atom(100% s character). Is it something that I'm missing? – Amritansh Singhal Oct 2 '16 at 14:31
• But you are comparing with another C-H bond where the C is sp2 hybridized. Comparatively more electrons go to C. Also, although you are right this time, you cannot compare different orbitals of different atoms. For example, the 2s orbital of N has a lower energy than the 2s orbital of C; the 1s orbital of H has a higher energy than the 2s orbital of F. – DHMO Oct 2 '16 at 14:32

Electronegativity is the power of an atom to attract bonding pairs of electrons to itself. It clearly depends on the nuclear charge: the larger it is, the more strongly the nucleus attracts electrons and the closer to the nucleus they stay, so the larger is the electronegativity. Besides, electronegativity also depends on the number of other electrons present in the atomic shells ahead of valence electrons participating in chemical bonding. These electrons shield the valence electrons from the positively charged nucleus decreasing its effective charge and consequently lowering the electronegativity of the atom.

Now, back to the question. $\mathrm{2s}$ orbitals are lower in energy than $\mathrm{2p}$ ones, so that electrons occupying the former are (at average) closer to the nucleus than that occupying the later. Thus, the more $\mathrm{s}$ character a hybrid orbital has, the lower its energy is and the closer (at average) electrons occupying it are to the nucleus. The lower the energy of a hybrid orbital of an $\ce{X}$ atom is, the lower is the energy of the molecular orbital in formation of which it participates, and consequently, the closer (at average) the bonding pair of electrons stay to that $\ce{X}$ atom.

So, with respect to a bonding pair of electrons that occupy a molecular orbital to which a hybrid orbital of an $\ce{X}$ atom contributes, the $\ce{X}$ atom indeed appears to be more electronegative as the $\mathrm{s}$ character increases. Specifically for $\ce{C}$ atoms, the larger the $\mathrm{s}$ character of a hybrid orbital is, the lower in energy is a bonding $\sigma_{\ce{C-C}}$ orbital formed from the hybrid ones: $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$. So that electrons occupying the $\sigma_{\ce{C-C}}$ orbital stay closer to $\ce{C}$ atoms (at average) as the $\mathrm{s}$ character of hybrid orbitals increases.

But if we compare between $\ce{C}$ and $\ce{H}$, is there some other factor to be accounted for because clearly H atom due to its very lower potential energy electrons should attract the shared electrons more.
the factor OP is looking for is the nuclear charge, which is 6(!) times larger for $\ce{C}$ atom comparing to the $\ce{H}$ one.