Look at the MO diagrams of corresponding neutral diatomic
specie...

Question

Look at the MO diagrams of corresponding neutral diatomic 
species in Figure 8.22, and predict whether each of the following 
ions is diamagnetic or paramagnetic. Diagrams for 
Li2 and C2 are similar to N2; Cl2 is similar to F2. 
(c) F2-

Accepted Answer

Hello. Everyone in this video. We want to go ahead and determine whether our C two minus ion is going to be para magnetic or die magnetic. How can you determine this is by drawing out the molecular orbital diagram. So M. O. Diagram let's go ahead and do so. So here we know we have a sigma three S. And of course we need the anti bonding. So we have sigma star three S. Then we'll work our way up to the p orbital. So then we'll have our sigma for three P. And then we'll have our pi three P. Now that we have our bonding. Let's go ahead and draw also the anti bonding. So have to for our pi star three P. And then sigma star threepeat. Alright so here again we're doing cl two minus. So chlorine is in group 78. It means we have seven valence electrons. So seven times two because we have two items of chlorine is equal to 14 electrons. And then of course we have our minus one charge. That means we have one extra electron. So we'll add one here. That gives us a total of 15 electrons. To go ahead and fill in. Let's go and do so in our M. O diagram. So we have 123456789, 10, 11, 12, 13, 14 and 15. Let's go ahead and call what diane magnetic and para magnetic differences are. So for dia magnetic it's going to be when all the electrons are paired. And then for paramedic that's going to be when one of our electrons are paired. So let's apply that and just say UNP aired electron. So we see on electron that's para magnetic and all of them are paired. That's diabetic. So looking at the freshly drawn diagram we see here at the top orbital the sigma star three P. We have an unpaid electron and we're gonna have an unpaid electron. That's a paramedic. So this ion here, r c L two minus is para magnetic. And this right here is going to be my final answer for this problem.

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