 So I guess the only thing this problem wants us to do, doesn't want us to build anything, but as of 6, it wants us to show the atomic orbitals, show them rehybridized into their molecular orbitals, their bonding orbitals. So hopefully already you can tell that, well, fluorine is not period 3 or below, so it cannot be hyperbabel. So hopefully you know that the sulfur is the central atom. Hopefully also you recognize that sulfur should normally only make two bonds, right? But of course it's making six bonds in this case, so it must have expanded its favor. And in order to do that, hopefully you recall that you've got to rehybridize your orbitals. So with all of that in mind, let's go ahead and write up the atomic orbitals for sulfur. So there's the 3S, the 3P, and there's three of them. And remember, when we rehybridize, we jump over that 4S and use those three Bs. The line's always too bad if you use whatever those charges. OK, so let's fill up according to our filling up rules, right? So 1, 2, 6, like that. 1, 2, 3, 4, 5, 6, OK? So of course we know it's not making just two bonds, and we know that those bonds are 90 degrees or 180 degrees or whatever part. So it must be rehybridizing just once again, another way to remember. And we know it's going to make six bonds, OK? So we need six half-filled orbitals in the sulfur's molecular orbital set. Is that OK? So in order to have six come out, we have to put six orbitals in, remember? So we're going to take the lowest energy orbitals, so the S orbitals, the 3, 3P orbitals, and in this case, two of the three Bs, right? Because 1 plus 3 plus 2 equals 6. Is everybody OK with that? We also, of course, need six electrons in order to do this. And we have six. So let's just kind of box in the orbitals that we're taking away. And so we're going to throw those into the blender and hybridize them. So six in, six out. What are these orbitals called? That's B3D2. That's B3D2, yep. So it's just S, P, 1, 2, 3, P, 1, 2, 2, 3. 5, 6, like that. And then we still have these three 3Ds like that. And we're just going to throw in our electrons, according to the rules that we already know, like that. In this case, we have 1, 2, 3, 4, 5, 6 half-filled orbitals, so we should be able to make six bonds. If you wanted to go further, right? Six bonds, or I guess we should start with the electronic geometry, right? So what would be the electronic geometry of something with six electron groups around it? Octahedral and molecular geometry? Same. It's the same. Why is that? Because in this case, all of those, all of them are bonds, right? Is there any questions on this one? So I think this is probably the hardest kind of one of these problems. But I guess we could do one other thing here. We could show the sulfur and its sp3d2 orbital. And this happens with all six of them, of course. I'm just going to show one of them, because it gets kind of an F with its 2p orbital. So there's your overlap. So that's the 2p orbital that's bonded with the sp3d2 orbital. sp3d2 orbital, sorry about that. So I don't know if there's anything else that we really should try on this problem. Is that good for you? Yep. OK. Any other questions? Sorry. No.