Why doesn't carbon form 3 $\pi$ bonds? If s-s overlap is possible between 2 carbon atoms,then it will be possible for 3 p orbitals to overlap laterally .


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Okay so say you have two carbon atoms and you want to make a sigma bond with their s orbitals. They won't want to since their orbitals are filled but VSEPR theory tells us that carbon can rearrange and hybridize its orbitals to allow such bonds to happen. Normally you'd have your 3 pi orbitals along the 3 spatial axis, but the s orbital, holding 2 electrons, can hybridize with the p orbital that has no electrons to get two sp orbitals that have one electron each. This sp orbital will be along an axis we'll call z. This leaves us with two p orbitals orthogonal to our sp orbitals which may overlap laterally to give two pi bonds.

If we have two p orbitals overlapping to make two pi bonds, the third p orbital leaves us a problem. If we don't hybridize the orbitals, the third p orbital is empty and cannot make a bond. Even if you had a suitable lewis acid to try to give it some electrons, the bond would't be a pi bond since the orbital overlap would be along the inter nuclear axis. If we do hybridized then we end up with the situation described earlier.

Perhaps there could be a strange configuration where the inter nuclear axis is equiangular from the three base axis and the empty p orbital is filled by some electron donor and the bonding p orbitals overlap at right angles. I wouldn't really call it a pi bond at that point and anything in that position would likely collapse into a more stable state.

Basically the geometry of the p orbitals means that we can really only have two of them overlapping at any one time in the basic sense of a pi bond. This is assuming we are not counting the cases where a single p orbital participates in bonding with more than one atom such as in graphite, though I wouldn't really call that three distinct bonds.

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    $\begingroup$ VSEPR theory does not tell us that carbon hybridises its orbitals. VSEPR only lets us guess the molecular structure, then we would find the best fitting molecular orbitals. $\endgroup$ Aug 25, 2018 at 12:51

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