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Regarding the above picture, why is the C-C $\sigma$ bond composed of two $\ce{sp^2}$ orbitals while the C-H $\sigma$ bonds composed of $\ce{sp^3}$ hybrid orbitals? I don't see why; I thought the geometry around the carbons was trigonal planar. Why does hybridization differ across atoms in the molecule?

I'm thinking the textbook authors made a mistake; yes, there are four C-H bonds, but this does not mean there is $\ce{sp^3}$ hybridization because there are two carbon atoms, each with two C-H bonds and one C-C bond. Perhaps the authors mistakenly took there to be one carbon and four C-H bond (i.e. methane.) That would be $\ce{sp^3}$ hybridization.

Or perhaps I'm wrong altogether. Please enlighten me!

EDIT: Saw this in another text:

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3 Answers 3

up vote 6 down vote accepted

You are not mistaken. Your book is wrong. If one desperately wants to involve the hybridisation concept (which makes sense for carbon), then the description should be in accordance with the molecular geometry of ethene:


So the carbons can be described as $\ce{sp_{x}p_{y}=sp^2}$ hybridised, hence there are four $\ce{H_{s}-C_{sp^2}}$ $\sigma$ bonds, one $\ce{C_{sp^2}-C_{sp^2}}$ $\sigma$ bond and one $\ce{C_{p_{z}}-C_{p_{z}}}$ $\pi$ bond.

However the concept of hybridisation is no law, but yet another tool to describe bonding. In case of methane, (ethane,) ethene and ethyne it is quite simple to come to certain $\ce{sp^{x}}$ hybrid orbitals where $\ce{x}\in\mathbb{N}$ - different molecule may lead to fractional $\ce{x}$. (Not a rule, just a guideline)

In response to Dissenters comment:

The coordinate system is chosen arbitrarily to indicate the molecular plane to be $xy$, as indicated in the picture taken from the german article of wikipedia:

ethene plane

As the π orbitals have to be perpendicular to this plane they only can be composed of $\ce{p_{z}}$ orbitals.

The molecular plane can be chosen different, e.g. as the $xz$ plane, the the $\ce{sp^2}$ orbitals would be formed from $\ce{sp_{x}p_{z}}$ and the $\pi$ bond would consist of $\ce{p_{y}}$ orbitals.

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Thank you. Do we have to use the px and py orbitals in that order in hybridization (you wrote spxpy). –  Dissenter May 19 at 5:18
Okay! And if I understand correctly the z-axis is taken to be up/down (not up/down like the y-axis in 2D) but up and down like standing up and sitting down. –  Dissenter May 19 at 5:21
And yes, I have to involve the hybridization concept since it is emphasized in all the introductory level organic textbooks. You might be appalled at the level of qualitative analysis these organic textbooks start out with (having us draw cartoons of molecular bonding; mixing and matching aspects of valence bond theory with molecular orbital theory, etc.) I guess authors are just trying not to let the math take away from the chemistry at this point. –  Dissenter May 19 at 5:26
@Dissenter The hybridisation concept makes a lot of sense for carbon in most cases. This is talking about simple organic molecules - just be a little critic about the concept (your question shows that you are) and learn where there might be the limits of that concept. I am not opposed to using hybridisation as a tool - however, the bonding description could be accurately obtained without it (It would not be such a nice picture though). I have added some more information to the answer, I hope that further clarifies the plane I have chosen. –  Martin May 19 at 5:36
Thank you. What other ways to describe bonding are there? –  Dissenter May 19 at 5:37

Martin has provided an excellent answer to your question. And although I've arrived late at the party I'd like to comment on the hybridization in ethylene. All of the introductory texts I've come across use the same bonding description as your text: the carbons in ethylene are hybridized such that we have 6 carbon $\ce{sp^2}$ bonds (4 $\ce{C-H}$, 2 $\ce{C-C}$) and 1 pi bond based on overlap between the unhybridized p porbital remaining on each carbon. Actually, if we refine this analysis to take into account the known $\ce{H-C-H}$ angle of 117 degrees, we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bonding are $\ce{sp^{2.2}}$ hybridized and the carbon orbitals involved in the $\ce{C-C}$ bond are $\ce{sp^{1.6}}$ hybridized.

I want to point out that, while staying within the MO framework where we only consider the s and p orbitals on carbon, there is an alternate way of describing the hybridization in ethylene - and these two descriptions predict the same physical reality, so they are equivalent descriptions. The texts arbitrarily choose to hybridize only 2 of the carbon p orbitals along with the carbon s orbital. It is just as valid to mix all 3 p orbitals with the s orbital. If we constrain 2 of the resultant orbitals just as we did above (those that will bond to hydrogen) to produce an $\ce{H-C-H}$ angle of 117 degrees, then these two orbitals will be $\ce{sp^{2.2}}$ hybridized, just as the earlier analysis concluded. The remaining two orbitals on carbon, however, can consequently be described as $\ce{sp^{4.3}}$ hybridized. The picture that emerges from this alternate hybridization scheme is one of bent bonds - a 2-membered ring if you will! Something to think about.

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Thank you, are you saying there is a easily derived relationship between s character and bond angle? –  Dissenter May 19 at 14:34
That's a part of what I'm saying. See the following link for the 2 relevant formulas (Coulson's theorem and sum of the s character=1) chemistry.stackexchange.com/questions/10653/… –  ron May 19 at 14:53
(five sp2 bonds not six) The representation of $\ce{sp^{2.2}}$ is highly inaccurate as it only gives the relation between s- and p- character. Symmetry in ethylene dictates that the surrounding of carbon has to be planar, so it is only possible to use in plane p-orbitals, and there are only two. Hence the hybridisation is better described as $\ce{s^{0.91}p^{2}}$. And a $\ce{sp^{4.3}}$ orbital is just not possible. Also molecular geometry (and symmetry) dictates a $\pi$ bond and this is only possible by combining the p-orbitals. –  Martin May 27 at 4:43
@Martin 1) I should have said 6 orbitals emanationg from the 2 carbons; 2) $\ce{s^{0.91}p^2}$ is the same as $\ce{s^{1}p^{2.2}}$, just divide 2 by 0.91, the "s" portion is always normalized to 1; 3) Do you agree that $\ce{s^{1}p^{2.2}}$ is possoible (I assume yes)? Then why not $\ce{sp^{4.3}}$? You seem to accept a continuity of hybridization indices between 1 and 3, truth is it runs from 0 (pure s) to infinite (pure p). 4) Google bent bonds in ethylene, Linus Pauling and Bent were proponents. –  ron May 27 at 14:30
I am generally not a big fan of hybridisation, as I believe it oversimplifies most bonding situations. It is basically only another term for a certain type of linear combination. Most bonding situations are perfectly fine described without any hybridisation. 2+3) In a hybrid orbital $\ce{s^{$x$}p^{$y$}}$ I will accept, that $x\in\mathbb{R}\in(0,1)$ and $y\in\mathbb{R}\in(0,3)$, everything else is overcomplicating an oversimplified concept. 4) I love bent bonds! –  Martin May 28 at 3:46

The bonding model commonly accepted is neither "right" or "wrong" - it is a model, i.e. an approximation of reality using simple concepts (sort of like describing movement of trains as being linear and at constant speed, and ignoring wobbles, curves, speeding up and slowing down). There is an alternate model, although not popular, that postulates $sp^3$ hybridization only. Try to search for "banana bond theory".

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