4
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The molecule is shown in the textbook as:

enter image description here

I can rationalise that the N is sp2 hybridised with a single electron in the p-orbital perpendicular to the plane of the S/N ring. (The sp2 lone pair is omitted for clarity) The S atom is thought to be sp3 hybridised, but how can this explain the double bond formed with N? Can an sp2 and sp3 orbital overlap??

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    $\begingroup$ I think that bond is d pi- p pi. $\endgroup$ – jatin Mar 20 '16 at 16:34
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    $\begingroup$ S has 1 electron in 3d, N has 3 electrons in p orbitals as usual, 2 of which are used in 2 sigma bonds while the 3rd is participating in d pi-p pi bonding. $\endgroup$ – jatin Mar 20 '16 at 17:10
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    $\begingroup$ @jatin I don't think p pi - d pi is really widely accepted, anymore. $\endgroup$ – SendersReagent Mar 20 '16 at 19:00
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    $\begingroup$ You can see in this image the bond angles at each atom in the crystal structure. You can calculate hybridization mathematically from the bond angles. I haven't done so yet, but when I get a chance, I'll look into it. You can definitely have at least some bonding between $sp^2$ and $sp^3$ orbitals: the classic example is carbocation hyperconjugation. But between nitrogen and sulfur? Not much. Image from Acta Cryst, Sec. C, 1991, C47, 1784. $\endgroup$ – SendersReagent Mar 21 '16 at 0:12
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    $\begingroup$ How long is it going to take for people to accept that d orbitals are never involved in main group bonding? The 4s orbital is lower energy than 3d, so why is that not invoked? I located the origin of that diagram as an Indian textbook first published in 1986 so that diagram is quite suspect. $\endgroup$ – gsurfer04 Jan 14 '17 at 22:06

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