MO Theory: Homonuclear Diatomic Molecules Example 1

Jules Bruno
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using a molecular orbital diagram or M. O diagram. Right? The electron configuration for the dye forming an ion F to minus one. All right. So let's follow the steps step one. We need to determine the number of valence electrons that this particular an ion possesses Flooring is in group seven a. So it has seven valence electrons and there are two of them. So that's times two. So it's 14 valence electrons -1 means we've gained an additional electron. So in total we have 15 valence electrons. Step two, we're going to construct a molecular orbital diagram based on location of the valence electrons. Remember if we're dealing with period one elements, that means the electrons start in one S if we did what period two elements they start with to us and period three elements start with three us. Now we're going to use the molecular orbital diagrams we have here. Now remember we're dealing with two flooring, two foreigners involved because it's F two and it's minus one. So, remember we're going to say it has seven valence electrons. So, I'd be up down up down, Up, up up so that's 5, 6, 7 up up up and then there But remember we have one additional electron, you can put it on either one. So we'll put it here for that additional electron now we would just pull those electrons together into our molecular orbits. So here we have up down up down Up down. So that's 246, 10 11, 12, 13, 14, 15. So this would represent the molecular orbital diagram for the dive flora flooring and I are now here we filled in the total number of valence electrons for each element into the molecular orbital's. And now to fill in our electron configuration is a little bit different. So the way we do it is we show each one of our molecular orbital's as individual things here within parentheses. And up here we put the number of electrons found in each. So if we look we have two electrons within the sigma, two s orbital, two electrons and sigma star to us. Two electrons within sigma two P four total electrons within pi two P four, total electrons within pi star to pee. And then finally one electron within our sigma pop the sigma star to pee. So those are the numbers we place down here. Alright, we're gonna come back down here. We're just gonna fill them in. Right, so what do we say here? We said we had two electrons here. Two electrons here. Two electrons here. Four. And then we had if we come back up here we had four in pi two P four and P I start to pee and one in sigma star to pee. So come back down here make sure we fill them incorrectly. So four. And then finally here we have one. So this would represent the filled in molecular orbital electron configuration for our di flooring. And I'll so again this is the approach you would take in terms of filling out the molecular orbital for your designated diatonic molecule or ion, and then you would set up your molecular orbital electron configuration in this fashion and fill in the number of electrons found within each particular molecular orbital.