 Okay, so let's do this problem. It's very similar to the last problem that we did. We're going to take the atomic orbitals from the atoms that make up this polyatomic ion and make the molecule, or the polyatomic ion from it. Okay, so molecular orbitals. So remember what we do. We have to build an atomic orbital valence shell, electron diagram. So in this case it's the 2s and 2p. Let's go ahead and fill these orbitals before we do anything else. Okay? So we just count 1, 2, 3, 4, 5, 6, 7. 1, 2, 3, 4, 5, 6, 7. Then f plus is going to have one glass than that, so it must be 4, 5. Everybody got that? So now let's build the molecular orbital diagram. Remember for each bonding orbital we have an antibonding orbital. This one's going to be the sigma 2s orbital. This one's going to be the sigma star 2s orbital. And then remember for these small atoms the pi and sigma orbitals will repel each other. So we're going to have that normal, what you would think of as the normal electron theory. So the sigma 2p comes before the 2. This is where you're getting confused or whatever, but they label this just 1. And then the antibonding orbitals would be pi star 2p. Bonding sigma orbital and sigma star. So this question asked for both to figure out what the bond order is and the electron configuration. So in order to do both of those we're going to have to fill our molecular orbital diagram. So remember you're filling roles. So 1, 2, 1, 2. So that's all four of those electrons. So how many electrons do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. So 1, 2, 3, 4, 5, 6, 7, 8, 9. Okay so let's do the electron configuration first since it's right up there shown to us. So what's the electron configuration? So we just start from the bottom and remember put it in those parentheses. So sigma 2 star 2. Sigma 2s, sorry, sigma 2s 2. Of course it would be something I would say. Sigma star 2s 2. Sigma 2p 2. 2p 4. Hopefully you can still see this on the camera. Pi star 2p 3. You can see all that. Well at least you can see it for a second. And then the bond order. One half of the bonding electrons minus the anti-bonding electrons. So 1, 2, 4, 6, 8. Right? The bonding electrons. So anti-bonding 1, 2, 3, 4, 5. So 3 halves. So it's just telling you how strong the bond is. So anything that gets higher and higher and higher. As long as you're getting higher you're getting a stronger bond. A bond order of a half is not as strong as a bond order of 1. So if you want to think of a single bond that's a bond order of 1. A double bond would be a bond order of 2. I mean it's between a single and a double bond. So I guess we've already answered that question. But I guess the next question would be does this molecule exist? Or would this ion exist? Why? Because it's above 0. Even a bond order of a half exists, but it's very weak bond. But stronger than a non-no bond. Is everybody okay with that one? I think we captured it in a bottle there.