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Trying to prove that nitrogen doesn't split its lone pair to form 5 bonds, I thought of a situation that I couldn't rule out; the paired electron being excited to the 3s orbital, so that five bonds could be formed- in an "$\ce{sp3s}$" hybridisation.

I know hybridisation is just an observation-based explanation, but I'd like to know how to rule wacky situations like this out.

Can this - and all other nonstandard hybridisations you can come up with - be ruled out by reasoning that the electron excitations involved take up too much energy?

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    $\begingroup$ Yes, you are correct, the hybridisation scheme has to make sense in terms of energy. There is also a symmetry factor to consider. Without looking up a character table, my guess is that sp3s has the right symmetry to form a 5-coordinate system. However, it doesn't work as 3s is too high in energy in N. (Also when does N form 5 bonds??) $\endgroup$ – S R Maiti May 22 at 9:26
  • $\begingroup$ @ShoubhikRMaiti: haha, N doesn't form 5 bonds anywhere; I was just trying to rule out ways in which it could. Namely, this. $\endgroup$ – harry May 22 at 11:01
  • $\begingroup$ @ShoubhikRMaiti: wait, what? Sp3s can exist? (You said "too high in energy for N", are there cases where it can work out? ) $\endgroup$ – harry May 22 at 11:03
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    $\begingroup$ No sp3s cannot exist. Because once you go down the period you end up on P, where there is a 3d orbital between 3s,3p and 4s, so I doubt there is any case where sp3s exists. It has the right symmetry for N, but it doesn't work out due to energy differences. $\endgroup$ – S R Maiti May 22 at 11:07
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    $\begingroup$ @ShoubhikRMaiti: kay, guess it's a matter of experience to know if a hybridisation is invalid right off the bat. This answers it anyway, thanks. $\endgroup$ – harry May 22 at 14:05
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The counter question that might help you rule this out is: "Which 3s orbital?" Note that atomic Nitrogen doesn't have a 3s orbital as N itself belongs to the 2nd period(n=2).

enter image description here

Also note that in hybridization, Atomic orbitals of same or almost similar energies intermix to form same number of new Hybrid orbitals of same shap and equivalent energy.

That being said, you cant expect 2s or 2p orbitals to intermix with a 3s orbital

(if it was a case. Since this case isn't possible with N)

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    $\begingroup$ From what I understood from this answer, every atom does indeed have every orbital, it's just that they're unoccupied. Which would imply that there is an empty 3s orbital here. I just needed to know if the energy-not-enough explanation was correct, or if it's wrong, the correct explanation. $\endgroup$ – harry May 22 at 6:39
  • $\begingroup$ @harry You are correct, every atom has all the orbitals. But at the same time there are actually no orbitals. Orbitals aren't physical boxes in which we place electrons. But we tend to say that some orbitals aren't in an atom, what we mean by that is that a large amount of energy will be needed to move an electron to that orbitals. For a simple example we can excite the electron in hydrogen to higher states. But we never see this naturally because why would something like to horde energy when it has a way stabler option? $\endgroup$ – Nisarg Bhavsar May 22 at 7:29
  • $\begingroup$ @NisargBhavsar: right, thanks. So can every non-standard hybridisation you can think of be proven not to exist using this? $\endgroup$ – harry May 22 at 7:32
  • $\begingroup$ @harry What exactly do you mean by non-standard hybridizations? Does $\mathrm{d^3s}$ count as one? If yes, than I am sorry to say but it exists. $\endgroup$ – Nisarg Bhavsar May 22 at 7:37
  • $\begingroup$ @NisargBhavsar: I've heard of d3s. I meant ones like sp3s; hybridisations you can just make up using whatever explanation and that has never actually been used to describe a molecule before (so I guess they're "wrong" hybridisations). As far as I know, there isn't a comprehensive list of all the hybridisations that actually exist, so 'standard' and 'non-standard' are fuzzy terms. If anyone knows of such a list, linking it here would be great. $\endgroup$ – harry May 22 at 7:46

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