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:

c. Which MOs are symmetric, and which are antisymmetric?

Accepted Answer

Hello everyone. Let's do this problem together. It says for the molecular orbitals of 135 heat trine determine whether it is symmetric or anti symmetric. And we have four answer choices differing in which three molecular orbitals they've labeled as symmetric and which three they have labeled as anti metric. So from the answer choices, we can see that there are a total of six molecular orbitals. And this makes sense, right? Because if we draw out 135 hex a trine, which hex tells us we have six carbons, 135, trying tells us we have three conjugated double bonds. Since we have those three conjugated double bonds, we are going to have six atomic orbitals, which in turn equals the number of molecular orbitals which will be six. So we'll start with high energy and draw our six molecular orbitals and recall that the lowest energy molecular orbitals, the bottom half are the bonding molecular orbitals and the higher energy half of the molecular orbitals are anti boding molecular orbitals. So in our atomic orbitals, we have six electrons, 123456, right two from each pi bond. So we fill our molecular orbitals using our electron principles like hunt's rule off ball principle, right? So we start at the lowest molecular orbital which is P I, one with two electrons fill si two with two electrons fill si three with two electrons and that is six electrons. So we see that our bonding molecular orbitals are all filled which leaves si 45 and six. And those should be si four star si five star si six star because they are anti bonding orbitals, they do not have electrons in them. OK. So we are almost done drawing our molecular orbitals to tell if they are symmetric or anti symmetric. We want to draw out all of the orbitals in that energy level. So since we have six electrons, we will have six lobes and I will draw this for each molecular orbital and we want to fill in the phases or shade in the phases of these orbitals in the molecular orbitals. And we do so according to Johnny's seven rules of how to fill in molecular orbitals. So if that doesn't sound familiar, go back and make sure you watch that video, it's very helpful. So we start by filling in the first lobe with the top phase shaded and we do this for all of the molecular orbitals. So this one stays the same in all of the molecular orbitals. And then going back to the first molecular orbital si one we fill in the top phase as well. But now when we move up to the different molecular orbitals, we are going to alternate the phase. So filling in side two on the bottom lobe, si three on the top lobe, si four on the bottom. So five on the top in size six on the bottom. OK. So now to fill in the rest of the orbitals, we should draw our nodes and recall that the number of nodes is the energy level minus one. So the first molecular orbital molecular orbital, 11 minus one is zero. So we have zero nodes, then we have one, two nodes for si 33 nodes for si four and so on. And we always want our nodes to be as symmetric as possible. So let's finish drawing the nodes for side two, our one node will be directly down the center between orbitals three and four, SI three. We have two nodes. So we'll have two orbitals in each chunk or each section size four, we have three nodes. Oops, that's not correct. I five, we have four nodes in size six, we have five nodes. So between every orbital OK. So now we shade from left to right orbital all the way to the left to the orbital on the right. And every time we hit a node we flip the phase. So for si one, there are no nodes. So all of the top phases are going to be shaded for side two, the first three orbitals will have the top phase filled. Then we hit a node. So we flip to the bottom phase. Si three, the first two are shaded in the top phase. Then we hit a node. The next two are filled in on the bottom phase. Last two are shaded on the top phase. Then si four, we have the first node shaded or sorry, first phase shaded on the top. Then we hit a node. So we switched to the bottom phase for the next two orbitals. Then we hit another note. So we switch to the top again for the next two and then we hit a node. So the last one is filled on the bottom which we already filled that in size five, we have the top phase a node, bottom phase, a node, top phase, top phase node, bottom phase node, top phase and lastly size six alternating phases for every orbital. So top phase, bottom phase, top, bottom, top and bottom. OK. So now the question is asking us to determine which molecular orbitals are symmetric or anti symmetric. So we want to draw a plane of symmetry in the middle of these molecular orbitals. So there are six of them. So we draw between the three orbitals or having three on each side of the plane and we see which phases are symmetrical. So basically we can act as if it's a mirror, right, the plane of symmetry is a mirror. So in si one, this is symmetric right, they look the same on each side of the mirror. So that is symmetric SI two, they do not mirror each other. So that is asymmetric si three, they mirror each other. So that is symmetric si four, they do not mirror each other. So that is asymmetric si five, they mirror each other. So that is symmetric and size six does not mirror each other. So they are asymmetric. So notice how we alternate symmetric or anti symmetric based on the molecular orbital, right, every other molecular orbital is symmetric or anti symmetric. So that makes c the correct answer for this problem because C says that si 13 and five are symmetric and si 24 and six are antis metric. That's it for this problem. I hope this was helpful and I'll see you next time.

Answer the following questions for the molecular orbitals (MOs) of 1,3,5,7-octatetraene:
c. Which MOs are symmetric, and which are antisymmetric?

Key Concepts

Molecular Orbitals (MOs)

Symmetry in Molecular Orbitals

Conjugation and its Effects

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