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EDIT: I misquoted my teacher; he said something more along the lines of: in conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?


My teacher said that if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized.

1) Why is this true? 2) Is this even true?

I'm guessing this has to do with physical limitations. For example, consider a carbonyl oxygen. Oxygen is $\ce{sp^2}$ hybridized. The carbon the oxygen is bound to has already "used up" two of its four bonds. The carbon can only form two more sigma bonds in addition to the one sigma bond it already has in the $\ce{C=O}$ double bond.

But what about "hypervalent" (if there are such beasts) molecules? Here we come to a self-consistency issue. My teacher still believes in hypervalent molecules - i.e. he draws sulfuric acid with six bonds.

So let's consider sulfuric acid in its hypervalent form, with two double bonded $\ce{sp^2}$ oxygens. There are a total of six bonds and four sigma bonds, implying $\ce{sp^3}$ hybridization.

So my revised set of questions is as listed:

1) Is my teacher being inconsistent with himself here?

2) Does $\ce{sp^2}$ hybridization still imply $\ce{sp^2}$ hybridization for linked atoms? Note that sulfuric acid does not disprove my teacher's statement if we take sulfuric acid to have four bonds, not six (i.e. sulfuric acid is properly represented as charge separated due to the high ionic character of its $\ce{S-O}$ bonds. If I remember correctly, on a time-average basis, 1.7 electrons are in oxygen's valence and only 0.3 electrons are in sulfur's valence per bond.


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    $\begingroup$ To dispute the claim your teacher made, my example would be propene. There you have two $sp^2$ carbons, but bound to one of them is an $sp^3$ carbon. $\endgroup$ – tschoppi May 18 '14 at 9:50
  • $\begingroup$ and any alkyl carbonium ion where the carbonium ion carbon is $\ce{sp^2}$ hybridized as in t-butyl carbonium ion; and allene being the all carbon analogue of the $\ce{CO2}$ discussed in LDC3's answer $\endgroup$ – ron May 18 '14 at 13:28
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Acetone has the oxygen and the carbon $sp^2$ hybridized. In carbon dioxide, the oxygens are $sp^2$, but the carbon is $sp$ hybridized.

Added:
For $SO_4^{2-}$, the sulfur needs to utilize 2 of the d orbitals for the sigma bonds, so it would be spd^2 hybridized, and it uses the 2 p orbitals for the pi bonds. The sulfur is $sp^3$ hybridized. (According to Wikipedia).

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In conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

It can be seen as a general rule of thumb, but it is not accurate. The defining feature for conjugated systems is that they have (local) planarity. This is necessary in order to form delocalised $\pi$ orbitals. The $\pi$ orbitals are antisymmetric to that defining plane, hence have to be formed at least by $\ce{p}$ orbitals.

Since all $\ce{sp^2}$ orbitals are in a plane, there are available $\ce{p}$ orbitals perpendicular to that plane for $\pi$ orbitals. The best example for this is benzene and on a larger scale graphite.

The same concept can be applied to $\ce{sp}$ orbitals. The best example might be $\ce{HC#C-C#CH}$.

And therefore a combination of both. A fantastic example for this is phenylacetylene (BP86/def-SVP only orbitals of out of plane $\pi$ symmetry no 27, 26, 24, 21).

pi orbitals of phenylacetylene

An even much more fantastic example is the $D_{2h}$ symmetric $\ce{HC#C-CH=CH-C#CH}$ (BP86/cc-pVTZ only orbitals of out of plane $\pi$ symmetry no 20, 17, 16)

pi orbitals of (HCC-CH)2

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If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

Absolutely not, here are a few examples

enter image description here

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  • $\begingroup$ But only the middle molecule has a conjugated system, as his instructor mentioned. $\endgroup$ – LDC3 May 29 '14 at 2:34
  • $\begingroup$ Right ... that's what he mentioned I think. I don't recall and I haven't found the relevant 10 seconds in the many lecture videos he has made since. But in the middle atom that looks about right; we have a conjugated system and we have two sp2 atoms. Thinking about this further, this makes sense; with resonance and conjugation, we need free p-orbitals. This only happens with sp2 hybridization, not sp3 hybridization. Also we need parallel pi orbitals - again - sp2 hybridization enables this. $\endgroup$ – Dissenter May 29 '14 at 2:43

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