Textbook problem 2.45 part b).

[Anna_Mohling_1D]

2.45 Consider the bonding in H2C=CHCHO. (a) Draw the most important Lewis structure. Include all nonzero formal charges. (b) Identify the composition of the bonds and the hybridization of each lone pair -- for example, by writing sigma(H1s, C2sp^2)

Hi! I am confused about how to write the composition of the bonds using the hybridizations of each lone pair (part b). Could anyone explain how to do this? Thanks in advance!

[Yu Jin Kwon 3L]

Hi Anna!

So part b is asking us to write the specific hybridization or composition of each lone pair or bond, so if you look at the structure of the H2C=CHCHO molecule, you'll see that there are 4 C-H bonds, 1 C=C bond, 1 C-C bond, and one C=O bond. We have to write the specific hybrid orbitals that are bonding together for the composition. For example, a C-H bond is a single bond (meaning sigma) and the specific hybrid orbitals that are bonding together are the 1s orbital of H and 2sp^2 hybrid orbital of C. Thus, this would be noted as σ(H 1s, C 2sp^2). This would be the label for all 4 of the C-H bonds!

Next, I'll explain the double bond situation! So for something like C=O, it has two components: the sigma bond and pi bond. The sigma one should be approached as the C-H example I talked about earlier. Thus, it is σ(O 2sp^2, C 2sp^2). But, for the pi bond, you don't look at the hybrid orbitals (basically shouldn't have something like 2sp^2 because that would be a hybrid or "mixture" of the s and p orbitals). So, you get π(O 2p, C 2p) because both O and C have an electron leftover in their p-orbital.

It's honestly pretty confusing, but I included like a visual aid (the answer to that problem), so hopefully that can help you!

[Sophia Stewart 3F]

So, you get π(O 2p, C 2p) because both O and C have an electron leftover in their p-orbital.

Could you say more about why the electron is left over in the p-orbital? I'm having trouble understanding why pi bonds are always only p.

[Yu Jin Kwon 3L]

So, you get π(O 2p, C 2p) because both O and C have an electron leftover in their p-orbital.

Could you say more about why the electron is left over in the p-orbital? I'm having trouble understanding why pi bonds are always only p.

Yeah for sure! So if you look at the hybridization of C and O in the lewis structure, you would see that they're both sp^2, meaning that their hybridized orbital has sp^2, but because p has 3 orbitals, you would say that there is an unhybridized p orbital (and that's why you have "an electron leftover in the p-orbital" for O and C.

As for the reasoning why the pi bond has to be the leftover electron in the p-orbital, here is my guess:
We already know that the O and C atoms can only have 3 regions of electron density (because of the hybridization of sp^2), meaning they can only have 3 hybridized orbitals (as mentioned before). In each orbital (as shown in the picture I attached), there is one unpaired electron (which is what gets paired when that atom bonds with another atom), and so essentially, all of the orbitals in the atom have to be bonded with another. Thus, the three hybridized orbitals all bond with another atom, following the hybridization we created based on the Lewis structure (sp^2 = 3 regions of electron density). That means that the remaining unhybridized orbital (which is the p as explained above) has to join one of the hybridized orbitals (because the Lewis structure only shows three regions of electron density), so the unhybridized orbital has no choice but to "join" one of the other single sigma bonds, which means it becomes the pi bond of the created double bond.
^^That is honestly just how I like to think about it, but I'm unsure if it's completely right. At this point, I just basically accepted that the remaining electrons in the unhybridized p orbitals will be the pi bond in the double bond.

I really hope this helps, and let me know if you have any other questions!

[Sophia Stewart 3F]

That helps a lot! I think I pretty much get it now. However, I do have one follow up question: If The hybridization were something like sp^3, would the leftover orbital still be a p? In other words, since there are only 3 orbitals, even if hybridization were using all three, or even into a d orbital, would pi bonds always use the p?

[Yu Jin Kwon 3L]

That helps a lot! I think I pretty much get it now. However, I do have one follow up question: If The hybridization were something like sp^3, would the leftover orbital still be a p? In other words, since there are only 3 orbitals, even if hybridization were using all three, or even into a d orbital, would pi bonds always use the p?

Hi sorry I just saw this! If the hybridization was sp^3, that would mean the atom has 4 regions of electron density, and if the atom was something like C (typically has 4 single bonds = sp^3) or let's say F (one single bond and three lone pairs = sp^3), it wouldn't require a leftover orbital because they only need that many orbitals to fulfill how many bonds they need. But, let's take this a step further (like you mentioned with a d orbital) and think about situations in which we see atoms that "have access" to a d-orbital (like n=3). Then, there would be a leftover d-orbital, and if the atom requires that orbital (for a double bond), then the pi bond would use the d-orbital (because the hybridized orbital may use up all of the 3sp^3 orbitals).

In short, I do not think pi bonds would always use p (it can use d or maybe even f), BUT for the context of CHEM 14A, we only need to think about hybridizations up to sp^3 up to n=2 I believe. Thus, within those restrictions, the pi bond would be found in the p-orbital.

Hope this helps :))