Answer the following questions for the molecular orbitals (MOs...

Question

Answer the following questions for the molecular orbitals (MOs) of 1,3,5,7-octatetraene:

f. What is the relationship between HOMO and LUMO and symmetric and antisymmetric orbitals?

Accepted Answer

Hello, friends. Let's work through this problem together. It says for the molecular orbitals of 135 heat trine determine how the Homo and Luo and the symmetric and anti metric orbitals are related. A, the Luo is antis me when the homo is symmetric B, the Luo is symmetric when the homo is antis me C the luo is symmetric when the homo is anti metric and vice versa. And D Homo and Luo are both symmetric answer choice C is interesting because it has that added vice versa statement which implies that the Luo and Homo symmetry can change for the same compound. So keeping that in mind, let's draw the molecular orbital diagram for 135 hex trine drawing the molecular orbitals label them as homo and Luo and also as symmetric and anti metric. And let's see that we can deduce. So 135 hex a trine that is a six carbon chain, 123456. And it is a conjugated poly with double bonds on carbons, 13 and five. So since we have six pi electrons, we are going to have six atomic orbitals, 123456 with one electron in each atomic orbital and orbitals are conserved. So when we draw molecular orbitals from atomic orbitals, we will also have six molecular orbitals. So let's draw 123, 456 molecular orbitals. And from bottom to top, we label them one si one side, two si three, the bonding orbitals in si four star si five star si six star, the anti boding orbitals in filling in our orbitals with our electrons, we use our electron principles. So pairing two electrons in si one opposite spins four and six electrons total. So that makes I three our homo our highest occupied molecular orbital and si four is going to be our Lomo our lowest unoccupied molecular orbital. Mhm So those are the two that we want to pay attention to their symmetry when we draw out the molecular orbitals, which we will do right now. So we draw six molecular orbitals for each energy level. And Johnny has a video on how to fill in these molecular orbitals. If the way I do it looks totally unfamiliar, make sure you go watch that video. It's the seven steps for filling in a molecular orbital diagram. It is super helpful. So starting with si one, we fill in the first molecular orbital, the bottom lobe and we do that the same for all of the energy levels. So the first molecular orbital, all six are shaded in the bottom lobe and then the last molecular orbital we alternate. So starting with the bottom lobe, the same as the first one. For si one, then for si two, we alternate top lobe, bottom lobe, top lobe, bottom lobe, top loop, then we will draw our nodes. So remember N minus one equals the number of nodes where N is the number of the energy level. So energy level one has no nodes, two has one node three will have two notes. Four, we have three nodes 123. And we always want to draw them as symmetric as possible. Sci five will have four nodes. Si six will have five nodes which is in between each molecular orbital. OK. So now we finish our shading every time we hit a node, we flip the shading starting from left to right. So si one, they will all be the same because there are no nodes side two, we have bottom bottom, bottom node, top, top, top side three, bottom, bottom node, top lobe, top lobe node, bottom, bottom node, si four, bottom lobe node, top lobe, top lobe node, bottom node or sorry, bottom lobe, bottom lobe node, top lobe size five, bottom lobe node, top lobe node, bottom bottom node, top node, bottom and size six were alternating for each molecular orbital. OK. So now let's look at the symmetry of each molecular orbital. So we want to let's forget the nodes that we just drew. We want a line. Let me choose a different color. Actually, we want a line down the center dividing our six molecular orbitals in half pretend like this is a mirror. And now we will determine if they are symmetric or anti symmetric. So if they match on either side of the mirror, they are symmetric if they don't, they are anti symmetric. So si one is symmetric si three is symmetric si four sorry si five is symmetric and SI two is anti metrics. Si four is antis metric in size six is anti symmetric. OK. So we see that our homo size three is symmetric in si four, our limo is anti symmetric. OK. So you might be thinking that answer choice A is the correct answer, right? But what about that vice versa statement. Remember I mentioned that and how it suggests that our molecular orbital diagram could have multiple options, multiple possibilities. This is the ground state molecular orbital diagram for 135 hex trine but this is a conjugated poly and conjugated polys can absorb light. And if they absorb light they become excited and the labels of homo and luo can change. OK. So let's look at the excited state of this compound. Luckily we can just copy and paste everything and change our electrons. So now this will be the excited state. So let's erase our homo and Lomo identification. So when we have electron excitation, we add either heat or light to one of these. Sorry to this compound, one of these electrons in si three is going to jump up to si four. So now let's identify the homo and the luo the highest occupied molecular orbital is going to be si four star and the highest or sorry, the lowest unoccupied molecular orbital is going to be five. So in this case, our homo is now anti symmetric in our limo is symmetric which is the opposite as the ground state, right. So the main principle here is that the homo is always going to be the opposite of the Lomo. So that means answer choice C is going to be the more accurate statement the Luo is symmetric when the Homo is anti metric and vice versa. So one is not always symmetric and one and the other is always anti symmetric they can change. But the relationship between the two states is always opposite, which may see the correct answer. That's it for this problem. I hope this was helpful. And thanks for watching.

Answer the following questions for the molecular orbitals (MOs) of 1,3,5,7-octatetraene:
f. What is the relationship between HOMO and LUMO and symmetric and antisymmetric orbitals?

Key Concepts

HOMO and LUMO

Symmetric and Antisymmetric Orbitals

Orbital Interactions

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