 Okay, so this time we'll do one of these kind of anomalous atoms, so remember boron, carbon, and nitrogen have that weird feeling pattern, okay? So we're going to do the molecular orbital theory analysis of C2 2 plus. Okay, so in order to do this, we've got to write the electronic configuration of the atomic carbon, right? And, well in this case it's going to be carbon plus and carbon plus, right? Because if we add carbon plus and carbon plus, it's going to give us C2 plus. So at the bottom here, we've got the 2s orbital of the carbon and the 2s orbital of the other carbon. And then we show the bonding orbital, and again like I was saying, I don't have very much room on this board, so it's going to be really kind of flat. Anti-bonding orbital, so we call this the sigma 2s orbital, sigma star 2s orbital. Okay, up here we're going to have the 2p orbital, 1, 2, 3, 3, 2p. So remember in this case the pi orbitals are filled first, okay? So down here we're going to have 2 pi orbitals, then the sigma orbital, like that. Okay, so this is called pi 2p, and this is also a pi 2p. And this is sigma 2p, the pi star 2p, pi star 2p. The very top is going to be star, something like that. If you're not, you're going to have to memorize. Okay, so now let's fill up our electrons. Okay, so C plus has how many electrons? One, two, four, but minus that fourth one, so we only have three, right? So how do we fill that up? Well, one, two, like that. Is everybody okay with that? It's like you would normally do. C plus, same thing over here, one, two. Okay, so down here we're going to have 2 pi because we have four, so we have to have four. So here, what are we going to do? Well, one electron, so remember all of our rules, right? We've got to remember one's rule, we've got to remember the poly-exclusion principle, we've got to remember off-valve principle, okay? So if we remember all of those, these two, and I know all of my lines are kind of at an angle here, but these two are of equivalent energy, right? So what's going to happen? So we're going to have electrons looking like that. So now let's go about determining the bond order. So bond order is going to equal one-half bonding electrons minus the anti-bonding electron. So how many bonding electrons do we have? Help me out. Four, right? One, two, three, four. How many anti-bonding do we have? Two. So one-half, four minus two is two, so that's going to equal what? One. So does this molecule exist? Yes, it exists. Why? Because the bond order is greater than zero, okay? So let's draw the electron configuration for this thing, okay? So how will we do that? So what did we say? Well, we've got to put sigma 2s, and we're just doing the electron configuration for the valence show, okay? If we wanted to do the whole thing, it would be sigma 1s2, sigma star 1s2. Okay, so let's do the whole thing. So sigma 1s2, sigma star 1s2, right? That's the inner electrons. So sigma 2s, help me out. Two, good job. Sigma star 2s, what? Two, good job. And then here we say pi to e. So the last thing I want to ask is, would this be attracted to a molecule? What would you think? Yes. Why? Because we have these unfair colloquial systems, okay? So, uh...