Use the MO energy diagram in Figure 8.22b to describe the
bonding...

Question

Use the MO energy diagram in Figure 8.22b to describe the
bonding in O2+, O2, and O2-. Which of the three is likely to
be stable? What is the bond order of each? Which contain
unpaired electrons?

Accepted Answer

Well everyone's in this video. We're being asked to draw the diagram. So the molecular orbital diagram where the following mall kills an ion. So we have F two plus F two as a neutral state and F to minus. We also have to go ahead and identify which of the three is going to be the most stable. Where they determine the bond order for each. As well as determining which one of these three are going to be para magnetic. So let's first go ahead and draw out the molecular orbital diagrams. So we start with the neutral one. So that's going to be f. two. So we know that's going to have our um container two S and two P. So start with the s orbital's recall that central is going to be our M. O. So the molecular orbital to the right, we'll have our atomic orbital's and because we're dealing with just the same atoms here just flooring we're going to have the same energy levels for each. So we have a two S. Here. A two S here. Then we'll have to orbital's one more higher and one lower. So the lower one is going to be our sigma. The one that's higher is going to be our sigma star for our M. O. Diagram. This for just the s orbital interactions and then it's these connecting here. Alright, let's deal with our p orbital next So maybe we'll maybe put this s just a little lower. So because we need that space. So of course we have the three orbital's for two p same thing for the other side to pee. All right. So now we'll have antibiotic bonding for our sigma and pi So we have are sigma all the way at the bottom that we have two pi bonds here or not pi bonds but Pie orbital's and then on top same thing. So, we'll have our pi star and then we'll have our sigma star again. We'll just put this a little bit lower. All right. So you're dealing with again, F two. Let's go ahead and draw these bonds are connected, apologize for the messiness here. Alright, so basically these are just connecting here like this. All right. So let's go ahead and fill up the orbital. So we know that we're dealing with F two. So that's flooring for the group seven A. So we have seven electrons from each. So we'll go ahead and add the electrons in and our atomic orbital's first. It doesn't matter if we do the right or left. They're going to be the same. So, I'm gonna go ahead and start with the left. So again we're riding in seven electrons because seven with Florence in Group seven A. Which has seven valence electrons. So we have 12. So that's two in the s orbital so continue on We have 3456 and seven. So same thing on the right side we have 123456 and seven. So because we have seven from each atomic orbital the M. O. Is going to have a total of 14 seven from each of our florins. Alright so we have 14 total. So 10 11 12 13 and 14. Alright this again just for our F to our neutral compound. Next what we can do is draw the diagram for an F. Two plus. So we're going ahead to remove this electron So it's the same exact or are format here for our diagram. The same exact thing here. So starting off again central with being M. O. To the right we have our atomic orbital's to the left. We have so have the atomic orbital's again we're dealing with the same atoms here so our energy levels are going to be the same. So we have our to us and to us then we'll have a higher one and lower one. Lower one will be our sigma. Higher one will be our sigma. Start and then we have our interactions here like So alright we're dealing with the p orbital's so we'll have 12 and three. Move this over to the right a little so again this is going to be two P. And then over to the left 123. We have to pee now for our demo we'll have are sigma and pi sigma and pi put this a little lower. Alright same thing for what we have on top of the anti bonding. We have to for our pie star and one for a six month star of course, drawing in the interactions. All right. The same exact thing for atomic orbital. We have seven valence electrons from each flooring atom. So 123456 and seven. Same thing for the right? We have 123456 and seven. Let's go ahead and specify for this one here we're dealing with F two plus. Alright. And now for a central So again we have seven electrons each from each atom. So that's 14. But because we're going ahead to take away one electron for t minus one is 13. We'll be adding 13 electrons into our M O orbital's. We've won 123456789, 10, 11, 12 and 13. Alright, so that's 13 electrons. So go ahead and actually take a little pause here and going head to calculate the bonds orders of age. So let's recall that the formula for bond order will just put in maybe purple. So, B O, which stands for bond orbital is equal to half, multiply the sum of our electrons. So we have electrons in our bonding subtracted from the electrons in antibiotic orbital's. Alright, so we're going to go ahead and calculate the bond order for the F two neutral molecule first I'll do in blue. So again we have the B O. Then half. So counting the number of electrons in our bonding orbital. So anything without the star which is these three here and this one here. So that would be total of eight electrons in our bonding And then in our anti bonding we have six so eight minus six is 22 divided by two is equal to one. So I bought orbital for our F two neutral is going to be equal to one over to the right, we have F two plus get out of the students in blue. Probably move this to the bottom here. Alright, so bond order is equal to the number of electrons in our bonding here. Let's go ahead and count. So we have 12345678. So we have eight in the bonding and team bonding. Let's see we have 1234 and five. So eight minus five is equal to three. Dirty divided by two is equal to 1.5. So that's going to be the bond order of F two plus the last M. O diagram that we have to go ahead and figure out and drought is going to be F to minus. So we're adding an electron into this molecule to make it a an ion. So again with our demo diagram, we have central being our M. O. And then we have our atomic orbital's over to the right and left. So again we're only dealing with one type of item here, we're just flooring so they will have the same energy levels. So here have to us another to us and then we have our bottom and top bottom is going to be our sigma, top is going to be our sigma star. Of course we have these interactions going on. Alright, we're dealing with RP orbital's same exact thing. So we have our three orbital's here for the two p same thing over to the left, we have 123 so we have to pee but for our M. O. We have our one down here and two above. We will put this down a little bit. Alright, so for our one orbital here at the bottom, that's going to be sigma. The one with two is going to be our pie. All right, and the same thing on top. We have our pi star now are anti bonding and then our sigma star again we have those interactions going on so we can go ahead and connect it with a dotted line or try to at least. Alright, alright, one more. Maybe this one will go ahead and fix a little. All right now go ahead and add the electrons the same thing on the atomic orbital side for the right and left. We have Florence and that's in group 78. So we have seven valence electrons each starting with the left, I'll do 123456 and seven. Same thing on the right, we have 123456 and seven now from the middle we have a total of 14. Just as if it's neutral because we have seven from each floor in or sometimes two is 14. But here let's not forget that we're drawing the diagram for F two minus. So it has this negative charge because we're adding one extra electron. So instead of 14 14 plus, one is 15. So we'll have 15 electrons that will add in this central M. O diagram. So we have 123456789, 10, 11, 12, 13, 14 and 15. Now go ahead, calculate for our bond order. So again be oh you're going to have the number of electrons in our bonding. Let's go ahead and count. We have 1234567 and eight. So that's eight. Now for anti bonding we have 123456 and seven. So then eight minus seven is 11 times half is equal to half. So the bond order for our f two minus is equal to half. So the highest bond order is going to be the most stable molecule. So since F E two plus, which if we scroll up, we'll see Over to the right here. So this one here we have a bond order of 1.5 since FE two is the highest bond order, then it's going to be the most stable. So maybe I'll write this answer down below The F two plus is the most stable I am. That's whenever answers. And then another answer is going to be if it's para magnetic or which one is param eidetic. So para magnetic means that there's going to be unparalleled in this demo diagram. But you can look quickly. We see here for the first one, the neutral F two. We have no unparalleled trains in the middle for the right, we see one for our F two plus because this one is a new paired. Going down below. We see that this one is a paired. So our para magnetic Is going to be our F two plus And our F 2 -. So basically all of our ions. So this is going to be my final answer for this problem.

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