Help the chem teacher - hybrid orbitals

General Questions
#1

I’ve been researching this, but I still have some questions. I got betrayed by another teacher who said that the hybridization is based on the number of bonds, but I find that it’s the bonds (not counting double or triple bonds) plus the unshared pairs that creates the orbitals: water has sp3 hybridization because of the 2 bonds and the 2 unshared pairs on the oxygen. Right?

I need to know if something like F2 has hybridization because it’s bonding once. I know the unshared pairs want to get away from each other and the bonding pair, but does this mean that it has sp3 hybrids in the valance shell? I can’t find even my college-level book addressing this directly.

Oh dear God, I just found in this book that H2S is thought not to have hybrids, because the bond angles are at 92 degrees, close to 90 degrees, which is not an sp3 hybrid-type angle. It says that unhybridized angles occur in “hydrides of other 6A elements and larger 5A elements.” I guess "larger means “larger than oxygen”? Are they saying that HCl doesn’t have hybrids because Cl is from 6A?

#2

You are right. The orbitals on water are sp3 hybridized. The lone pairs occupy the other two orbitals. Likewise, the bond in fluorine gas is sp3 hybridized.

The hybridization of orbitals starts to fall apart a bit with heavier elements, particularly when they are bonding to much smaller elements like hydrogen. This is because getting a favorable interaction requires that the orbitals overlap well. It’s my guess, that in the case of H2S, the s orbital just doesn’t interact well with the hydrogen s orbital. As a result, most of the interaction is with the p orbitals which are at 90 degrees. The way organic chemists draw bonds to sulfur is not necessarily the most descriptive way.

If you want to get really technical, you’ll see that hybridization has less to do with the number of bonds and more to do with bond angles. The bonds in water are only nominally sp3. If they were purely sp3, the bond angle would be 109.5, but it is 104.45, which is slightly closer to the 90 expected of a p orbital. So it’s probably closer to s.9p3.1. Meanwhile the lone pairs will have a higher angle because they have more S character. Their orbitals are probably closer to s1.1p2.9. The numbers are really just proportionally related to coefficients applied to a linear combination atomic orbitals.

The partial hybridization becomes more interesting with strained molecules where your forced geometry creates hybridizations that change the character of the bonds. Acetylenic hydrogens are more acidic than vinyl hydrogens, but the proton on a cyclopropene lies someplace in between because the constrained geometry has forced more s character into the bond.

#3

I was ready to agree with you, WarmNPrickly. My first thought was that the orbitals of F[sub]2[/sub] are indeed sp[sup]3[/sup] hybridized, with all of the lone pairs occupying the hybrid orbitals not involved in bonding.

However, I also seem to recall that forming hybrid orbitals takes energy, which is usually paid back by the formation of strong covalent bonds. In the case of fluorine and other period-2 elements, because the energy of the 2s orbital is so much lower than that of the 2p orbitals, it takes an even greater amount of energy (relatively speaking) to form hybrid orbitals. In addition, in the case of F[sub]2[/sub], the bonding between the two fluorine atoms can very easily be explained by the orbital overlap of the unfilled, unhybridized p orbitals. Therefore I’m not sure that F[sub]2[/sub] does in fact contain hybrid orbitals.

(BTW, I’m very pleased with myself that I easily understood both the question as well as the answer, after more than a decade out of the classroom, and some 12 years since I last taught chemistry. Where does the time go?)

#4

The s orbital isn’t that much lower than the p orbital. It is lower, obviously or else it wouldn’t fill first. But, if you were dealing with the idealized case of the hydrogen atom, then the only relevant quantum number is the principle quantum number, and the 2s has the same energy as the 2p. It’s only electron-electron repulsion that brings the 2p orbital to higher energy and having another cloud of fluorine electrons nearby quickly changes that situation.

#5

Could you explain more about “quickly changes that situation”?

#6

robbie said

(BTW, I’m very pleased with myself that I easily understood both the question as well as the answer, after more than a decade out of the classroom, and some 12 years since I last taught chemistry. Where does the time go?)

Ya did good, Robbie! :slight_smile:

#7

The only reason that the 2p orbitals are higher in energy than the 2s orbitals is due to electron electron repulsion. The second fluorine atom puts another cloud of electrons near the first fluorine atom and this is enough to change the energy of the atomic orbitals. Fluorine is why all of the period two elements hybridize their orbitals the way that they do.

#8

Also remember that hybridization is a qualitative approximation that helps make bonding easier to understand for 15-year-olds. It’s great for many simple molecules, but I recommend not putting too much effort into trying to rationalize the exceptions.

#9

True enough. What I was trying to say was that for multi-electron atoms, the higher the principal quantum number, the closer in energy the subshells are, and therefore the easier it is to hybridize. In other words, the difference in energy levels between 2s and 2p is greater than the the difference between 3s and 3p.

See here for a visualization.

Thanks! :slight_smile:

#10

pssht. Just solve the wave function for all electrons involved, and the hybrid orbitals will fall into place.

:wink:

/barely made it through p-chem in college, glad I’m doing my chemistry in a lab, not trying to explain it to students. Because I’ve forgotten nearly everything outside my field.