Bonding, Structure, and Resonance
Sigma bonds come in six varieties: Pi bonds come in one
Last updated: December 13th, 2022 |
You may recall from Gen chem (and no doubt your first week of o-chem as well), that orbitals on carbon come in two flavors: s and p.
s orbitals should be familiar as the spherical-shaped orbitals. The electrons of hydrogen, for instance, are in a 1s orbital.
p orbitals are shaped like figure-eights, or loops. The electrons in p orbitals are slightly farther away from the nucleus than those in s orbitals, so they are a little bit higher in energy. The p orbitals therefore fill with electrons only after the s orbitals are filled.
Hybridization is a concept that we address here.
What’s observed from analyzing the structure of molecules such as CH4 is that the shapes cannot result from the electrons being in s or p orbitals alone, but instead are a consequence of the electrons in s and p orbitals mixing to form hybrid orbitals. If you draw an analogy to how we could make a “hybrid” soft drink by mixing different proportions of Sprite and Pepsi, these new orbitals aren’t fully s or fully p, but are a combination of both. The “flavor” of each bond depends on the relative proportions of s orbital and p orbital content:
- sp3 = 25% s character, 75% p character
- sp2 = 33% s character, 66% p character
- sp = 50% s character, 50% p character.
It is these hybrid orbitals that form sigma bonds (σ bonds). Sigma bonds are created from the head-on overlap of orbitals. Those orbitals will be some combination of s orbitals and p orbitals.
[What’s happened to the missing p orbitals for sp2 and sp hybridized carbon? Well, these p orbitals aren’t involved in hybridization or in sigma bonding – instead, they maintain 100% p character, are available to form π bonds, which are created by the side-by-side overlap of orbitals. π bonding involves p orbitals exclusively. ]
Since carbon can exist in one of these three hybridization states, we can therefore have six varieties of carbon-carbon sigma bonds:
Now, recall that for any given quantum number, s orbitals are lower in energy than p orbitals. The electrons in s orbitals are held more closely to the nucleus than electrons in p orbitals. Take a look at the relative bond lengths and strengths for each of these six situations.
General principle – the more s character the bond has, the more tightly held the electrons will be. [Note 1]
Now, the number of π bonds that can form will be dependent on the number of unhybridized p orbitals available – 1 for sp2 hybridized carbons, 2 for sp hybridized carbons (the two π bonds will be at right angles to each other in the latter case).
In contrast to sigma orbitals, there is only one way to form a C-C π bond – from the overlap of two p orbitals.
Question: Relative to sigma bonds, do you think p-p bonds (π bonds) would be stronger or weaker?
1) What’s their % s-character?
2) Energy required to break C-C bond in ethane:
Energy required to break apart the C=C bond in ethene:
Notes
Note 1. This is also why the protons in acetylene are more acidic than those in ethene and ethane – the electrons in the conjugate base are more tightly held, making them more stable. If it seems contradictory that the bond is both stronger and yet more easily broken, remember that bond energies measure homolytic cleavage (See article: Bond Energies Measure Homolytic Cleavage)
(Advanced) References And Further Reading
Tables of bond lengths determined by X-ray and neutron diffraction. Part 1. Bond lengths in organic compounds. Allen, F. H.; Kennard, O.; Watson, D. G. ; Brammer, L.; Orpen, A. G.; Taylor, R. J. Chem. Soc., Perkin Trans. 2, 1987, S1-S19
DOI: 10.1039/P298700000S1
A useful reference, this paper collects bond lengths of various organic compounds determined from X-ray crystal diffraction data.
This is awesome, helped me understand the concept better
Excellent, Ramone. Glad it was useful.
thank you! good explanation of the hybridization blending s and p orbitals.
Thank you, that was really helpful. Could you explain the diplole moment, its calculation, bent, linear shape, trigonal, and tetrahedral shape effect on dipole moment. Also, could please an example lets say why does Acetonitrile is more polar than Methanol? From electronegativity perspective, Oxygen has a higher electronegativity than Nitrogen but why in case of Acentonitrile it’s having higher electronegativity than Methanol? Does the triple bond has an effect due shortness of its bond compared to single bond?
Thank you ver much
I could not find this information anywhere else! Thank you
Thanks for such a clear explanation…can u tell plz s character increase or decrease bond length with reason???
sp3 bond lengths are longest, sp2 bond lengths are shorter, and sp bond lengths are the shortest.
what is meant by “s” character???
Hi Alina – “s” character refers to hybrid orbitals. An sp3 hybrid orbital is made of 1 s orbital and 3 p orbitals, so it has 25% s character and 75% p character.
An sp2 hybrid orbital is made of one s orbital and 2 p orbitals so it has 33% s character and 66% p character.
An sp hybrid orbital is made from 1 s orbital and 1 p orbital so it has 50% s character and 50% p character.
Is it the C=C bond or C-C is stronger in 2-methylprop-2-ene ?
The bond dissociation energy of a C-C single bond is generally about 80 kcal/mol. The bond dissociation of a C-C pi bond is generally about 60 kcal/mol. The pi bond is weaker than the sigma bond.
Exam in a few, this was exactly what I needed to hear
Quiet clear.. ..But could you please tell the reason why sp3 bond length are greater than sp2 and sp
More s-character means the electrons are held tighter to the nucleus, which means a shorter bond length. In an sp-hybridized C-H bond the electrons have 50% s-character and will be held closer to the nucleus than in sp2 C-H (33%) and sp3 C-H (25%).
This is an eye opener to me! I learned so much and I wish I discovered this ages ago.
how does the C-H bond length change due to %s character
C-H (sp3) : 1.50 Angstroms. C-H (sp2) 1.40 Angstroms. C-H (sp) 1.30 Angstroms.
I will take that Sprite/Pepsi analogy with me to the grave, made it so easy for me to understand!
Awesome, thanks for letting me know Jesse!
Please tell me how to compare bond lengths in conjugated and aromatic compounds?
Thanks for this amazing explanation…😊
I am a bit confused by the representative bond length table, as it is listed in almost increasing bond length with the exception of the swap of sp-sp3 and sp2-sp2. Although the different is only 1pm I presume there are other factors affecting the bond length between 2 specific elements, hence the “representative bond length table”. Given 2 arbitrary elements that can take the either the sp, sp2, sp3 hybridization will the ranking of their bond lengths always follow table’s representative bond lengths?
The reason I ask this is Quiz #679 as the ranking of the bond lengths would follow this table if it were in order, neglecting the actual numerical values, otherwise would the solution not be incorrect?
The key is to keep one variable constant while the other changes.
For example in the first 3 cases, one side is sp3, while the other side goes from sp3 to sp2 to sp. With increasing s-character we should expect (and we do see) a strengthening of the bond as shown by the decreasing bond length.
For the next two examples we are keeping the right-hand side constant (sp2) while increasing the s-character of the left hand side (sp2 to sp) and we again see a decrease.
It would be unfair to be asked to guess whether sp3-sp is stronger/shorter than sp2-sp2 because we are changing both sides at the same time and it is difficult to predict the outcome.
A fair question would instead keep one side constant and change the other side in a linear fashion.