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I'm doing exercises on hybridisation, and I was given this molecule:

enter image description here

I'm wondering about this (electron deficient) oxygen atom. My intuition says it should be ${sp^2}$ like the answer says, but honestly I only see that it's $sp$ hybridised. Is it because you ask what it would look like if it was not electron deficient?


marked as duplicate by Todd Minehardt, Jan, Klaus-Dieter Warzecha, M.A.R., Wildcat Nov 24 '16 at 9:23

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  • $\begingroup$ I might add, that since oxygen has a lone-pair I somehow missed, that means it is of course not electron deficient afterall. $\endgroup$ – Brian Mar 17 '13 at 0:27

This would become much clearer if you show the lone pair on the oxygen atom.

One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized.

Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair.

  • 1
    $\begingroup$ Note: I always like to stress that the s and p orbitals don't actually combine in any way to form the $n$ $sp^n$ hybrid orbitals (This is a pretty common mistake). Hybridization is just an approximation used in quantum mechanics to explain the geometry of the compound better. $\endgroup$ – kaliaden Mar 16 '13 at 17:46
  • $\begingroup$ Ah of course. Somehow missed the lone pair. $\endgroup$ – Brian Mar 16 '13 at 17:56

You can find the hybridization of the atom by finding its steric number:

steric number = no of atoms bonded (to the atom you are finding the hyb. of) + lone pairs with that atom.

if Steric no comes to be 4, there is $\ce{sp^3}$ hybridization in the atom.

if steric no comes to be 3, there is $\ce{sp^2}$ hybridization in the atom.

if steric no comes to be 2, there is $\ce{sp}$ hybridization in the atom.

  • $\begingroup$ and this certainly is not true for any period different than 2nd. Try to use it on PH3 and you will end with sp3, however the free electron pair is s-type orbital. $\endgroup$ – Kris_R Apr 17 '13 at 6:32
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    $\begingroup$ Well, it helps in basic studies. $\endgroup$ – Rafique Apr 17 '13 at 8:22

+10 to kaliaden's comment. The "reason" is because linear molecules are $\ce{sp}$, trigonal-like molecules are $\ce{sp^2}$, and tetrahedral are $\ce{sp^3}$. Hybridization rules are completely bogus, and basically you know the answer by first observing the molecule experimentally, and post-hoc hammering it into these resonance structures.

This is nothing but a bunch of memorization. It is interesting to note that if you made $\ce{O}$ into $\ce{S}$, then it _may_not_ have the same shape. (Actually, it probably does, but where it is different, we sound off the d-orbital participation mumbo-jumbo)


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