Draw the molecular orbital picture of octa-1,3,5,7-tetraene, ...

Question

Draw the molecular orbital picture of octa-1,3,5,7-tetraene, indicating which is the HOMO and which is the LUMO.

Accepted Answer

All right. Hi everyone. So for this question, lets go ahead and provide the molecular orbital diagram for Decco 13579 Penta eat Lapel. It's Lumo and Homa now recall first and foremost, right, that the lumo stands for lowest unoccupied molecular orbital, whereas Homo stands for highest occupied molecular orbital. Now, before we go ahead and decide what the diagram actually looks like, we have to figure out how many molecular orbitals there are going to be in this case, right? Because the Decco 13579 P A in has five py bos, right, each with two carbons, there are going to be a total of 10 P orbitals in the actual diagram, right. So 10 P orbitals overlap to form the 10 molecular orbitals necessary for our five pi bots. Now, half of them, in this case, five are going to be bonding orbitals or P orbitals, whereas the other five are going to be anti bonding orbitals or pie star. So essentially, right, the lumo is the lowest energy anti bonding orbital, whereas the Homo is the highest energy bonding orbital. So let's go ahead and fill out this diagram that I went ahead and actually prefilled myself to save time during this video, right. So as you can see here, I went ahead and I showcased all 10 orbitals here. The first five being bonding and the second five being anti bonding as denoted with the star. So I also went ahead and true the 10 P orbitals in each level. So the reason why I went ahead and showcased the P orbitals here is because we have to showcase the number of notes, no recall that the number of nodes is going to depend on the energy level of that particular orbital. So for example, the second orbital should have one node, whereas the first orbital should have zero nodes and so on. Now because the first orbital should have zero nodes, all of my P orbitals should be aligned in the same direction, which means that the same side of them should be shaded in like. So now the 10th orbital, the very highest one should have nine nodes in total because the rule is N minus one in which end is which molecular orbital you're dealing with, right. So 10 minus one means that I should have nine nodes in the very last one, which means that in this case, my P orbital should alternate and be in different directions. Now, because my second molecular orbital should only have one node. There's going to be only one region in which the orientation of my P orbitals is going to change, right. So the first half of them are going to be pointing in one direction in this case upwards and then the other half are going to be pointing in the opposite direction in this case downwards. And so from this point forward, right, the trick to deciding the orientation of the individual P orbitals is to keep your nodes symmetrical relative to the center. So for example, our third molecular orbital here should have a total of two nodes which I must keep symmetrical relative to the center. Right. So I can divide this into two nodes by having the four and the center pointing in one direction and then the three on either side pointing in the other. So the outer six I'm going to depict pointing upwards, whereas the inner four are going to be pointing downwards. So now we can go ahead and do the same thing with the fourth molecular orbital that should have three notes, right. So I can separate by keeping into consideration the fact that my three nodes should be symmetrical relative to the center. So here, what I've done is fill out the rest of the molecular orbital by making sure to switch my direction going from left to right, right. So the first two here in the fourth molecular orbital are going to be pointing upwards. But then the next three, because there was a node pointing downwards until I got to the next node. And so on. So that in my fifth molecular orbital having a total of four nodes, I can go ahead and change the direction after every two individual P orbitals that I go ahead and shade in. And so from this point forward, right, for the remaining molecular orbitals, you have left to shade the number of nodes is going to be one less than the number of the orbital itself. And so the number of nodes should be kept symmetrical relative to the center. And so here is what the completed diagram should look like after all orbitals have been shaded in or individual p orbitals have been shaded in. So now let's go ahead and add in our pi electrons, right? Recall that we've got 10 pi electrons in the compound given to us Decca 13579 Penta inc. So here recall that each pi orbital can be occupied by a maximum of two electrons and you start adding electrons from the lowest energy level, which in this case is the first one. And so I'll be adding two electrons each going from bottom to top and both electrons should have opposite spins inside of each orbital, right? So now our highest occupied molecular orbital is in this case, going to be the fifth bonding orbital, whereas the lowest unoccupied molecular orbital is going to be the first anti bonding orbital immediately above si five. So are lumo is going to be thy six star and there you have it. Here is your completed molecular orbital diagram with the homo and lumo label and I know it was a lot, right. But here is your final answer. And if you stuck around to the end, I thank you so much for watching and I hope you found this helpful.

Draw the molecular orbital picture of octa-1,3,5,7-tetraene, indicating which is the HOMO and which is the LUMO.

Key Concepts

Molecular Orbitals

HOMO and LUMO

Conjugation

Watch next

Draw the molecular orbital picture of octa-1,3,5,7-­tetraene, indicating which is the HOMO and which is the LUMO.

Key Concepts

Molecular Orbitals

HOMO and LUMO

Conjugation

Watch next

Draw the molecular orbital picture of octa-1,3,5,7-tetraene, indicating which is the HOMO and which is the LUMO.