# Bond energy and s-character [closed]

While studying the basics of chemical bonding, I've often come across the following statements: 1. The bond energy increases with the increase in the s-character in a hybrid orbital, i.e sp3 sp2-sp2, and so on.

I can't seem to correlate the two. If a hybrid has a higher bond energy, shouldn't it also form stronger bonds?

• "If a hybrid has a higher bond energy, shouldn't it also form stronger bonds?" Which statement contradicted this? – DHMO Oct 2 '16 at 10:35
• @downvoter: why the downvote? Because this question is too elementary? – DHMO Oct 2 '16 at 10:37
• I didn't downvote, but voted to close the question, since it is not quite clear to me what's it is all about. As @DHMO mentioned, there is no any contradiction out there. Still, I made an attempt to clear some doubts OP could possibly have, if I understand him/her correctly. – Wildcat Oct 2 '16 at 11:08

Let us start with the following most important part of the IUPAC definition for the bond energy.

bond energy (mean bond energy)

The average value of the gas-phase bond dissociation energies (usually at a temperature of 298 K) for all bonds of the same type within the same chemical species.

I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.

Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} < E_{\mathrm{sp^2}} < E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} > E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} > E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.

Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.

• @DHMO: if the sp hybrid orbital is stronger than the sp3 orbital, why is the sp-sp bond weaker than the sp3-sp3 bond? – user35546 Oct 2 '16 at 13:55
• @Kōhai, first note that orbitals aren't stronger/weaker relative to each other, but rather, they are lower/higher in energy. Now, the lower a bonding orbital is, the stronger is the corresponding bond. Finally, who told you that $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond is weaker than the $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ one? – Wildcat Oct 2 '16 at 14:25
• A quick Google search results in 347–356 kJ/mol for bond-dissociation energy (at 298 K) for $\sigma$ $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond in alkanes and 837-839 kJ/mol for the triple bond in alkynes, out of which the $\sigma$ $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ bond contributes 369 kJ/mol. – Wildcat Oct 2 '16 at 14:30
• Okay, just to ensure that I've understood this correctly, lower the hybrid orbital energy, higher is the dissociation energy of the bond it forms, right? I think I messed up between the energy and the bond energy. – user35546 Oct 2 '16 at 16:48
• @Kōhai, yes, the lower a bonding orbital is, the more energy is required to break the bond. In our case the bonding orbital is formed from a hybrid orbital of C atom and some orbital of another atom X. And for a given orbital of atom X, the lower the hybrid orbital of C atom is, the lower a bonding orbital will be. I think I can try to draw this... – Wildcat Oct 2 '16 at 16:52