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Below is a neat looking borax anion.

structure of the borax anion

While exploring the molecule in a computer simulation (Gaussian 16, level of theory: HF/3-21G), I found that the oxygen atoms in the 8-membered outer ring had sp2 hybrid orbitals. My curiosity led me to question the presented data. Using Natural Bond Orbital analysis, I found out that:

BD(1)B1(O2)
1s(20.94%)p3.78(79.06%)
BD(1)O2(B1)
2s(10.48%)p8.54(89.52%)

BD(1)O2(B3)
2s(23.60%)p3.24(76.40%)
BD1(1)B3(O2)
3s(36.38%)p1.75(63.62%)

LP(1)O2
s(17.50%)p4.71(82.50%)
LP(2)O2
s(48.39%)p1.07(51.61%)

Here BD denotes a 2-centered bond with atom(i)atom(j) and LP(i) denotes the lone pair and the s,p characters follow. My question is as follows:

Given the chaotic data, how does one analyse or interpret the harmony and claim that the oxygen[2] is indeed ~sp2 hybridised?

What is the correct hybridisation of the oxygen atoms not connected to a hydrogen?

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  • $\begingroup$ NBOA was done using Engine provided as Gaussian, theory: Hartree-Fock in basis set 3-21G $\endgroup$ – user96208 Aug 1 at 7:43
  • $\begingroup$ Oh, than that wouldn't be borax, because it's a salt of such anion and you made another mistake in simulation. not to say this title is particularly vague. $\endgroup$ – Mithoron Aug 13 at 19:18
  • $\begingroup$ @Mithoron Your suggested title was misleading, this atleast is in direct relation in what I want to know. And yes, Borax anion...that has been changed $\endgroup$ – user96208 Aug 13 at 19:26
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First, it's important to understand that the simple integer hybridization labels sp, sp2 and sp3 are gross approximations and only accurately describe the electron density of simple symmetric molecules like $\ce{BeH2}$ and $\ce{CH4}$. Atoms in more complex molecules (even something relatively simple like $\ce{H2O}$) cannot be described accurately using these labels. Instead, the contribution of atomic orbitals to localized bond orbitals must be described by non-integer combinations.

That said, the calculations you did are remarkably consistent with an sp2 hybridization of oxygen [2]. If I understand the notation correctly, it says that the bond from O[2] to B[1] is ~90% p (roughly a pure p orbital), while the other bond to B[3] is ~75% p (i.e. sp3). The two lone pairs on the oxygen average 67% p, perfectly sp2.

We are able to average the two lone pair orbitals because electrons are completely interchangeable and are not localized to a specific lone pair. Any hybrid orbital description that captures the combined electron density of the four electrons in the two non-bonding orbitals is acceptable.

So the oxygen can be described as having two sp2 nonbonding orbitals, an sp3 bonding orbital and a nearly pure p bonding orbital. That's pretty close to sp2 hybridization, where we expect three sp2 orbitals and one pure p. The only deviation here is the shift of ~ 10% s from one of the sp2 orbitals to the p orbital, so that we end up with an sp3 and a not-quite-pure p instead.

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  • $\begingroup$ Your interpretation was correct, Thanks for the explanation, can please explain how did you average?...say for lone pair, I did (5.78/2) ....and what you did was adding the percentages and obtaining a ratio. Also, last paragraph implies that as long as we donot have nearly 100\% p character, we cannot say the bond is $sp^2$ am i correct? $\endgroup$ – user96208 Aug 1 at 21:50
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    $\begingroup$ @AnindyaPrithvi - The reason you cannot do 5.78/2 is that the 4.7 and 1.1 p contributions aren't normalized. They are only relative to an s value of 1. To give them equal weight, you need to add 4.7/(4.7+1) + 1.1/(1.1+1), where the denominator represents the total s + p contribution. These normalized values are the percentages shown in parentheses in your calculation. For your second question, I guess you are talking about the B-O bond that is sp3? Yes, it's sp3, but the original statement is that the oxygen atom is approximately sp2 hybridized, which is roughly true. $\endgroup$ – Andrew Aug 1 at 22:26
  • $\begingroup$ This answer is extrapolating from incomplete data on a very, very crude level of theory. It's probably also using an NBO version years deprecated and the data which is actually not meant to be interpreted. It's not even clear that the structural data is sound. Sure, this is one way or interpreting it, but in my opinion, this is a more misleading interpretation than just guessing hybridisation from the structure with Coulson's theorem (given a crystal structure of something). $\endgroup$ – Martin - マーチン Aug 16 at 18:26
  • $\begingroup$ @Martin-マーチン Supposedly as the website states, Gaussian 16 is used, and Gaussian 16 uses relatively newer version of NBO as indicated in the answer. Structural data is sound? (structure of NBO or the compound)....and Apply coulson's theorem on the most optimised geometry of Borax( the B.angle is 109), you get sp3 as opposed all over the internet $\endgroup$ – user96208 Aug 16 at 19:07
  • $\begingroup$ @Martin-マーチン - fair point about the theory. I was approaching the question as being specifically about why the calculated result did not match OP's expectation and simply pointing out that it does match. Whether or not the expectation and the calculation correctly describe the situation is a different question. $\endgroup$ – Andrew Aug 16 at 20:19

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