According to MO theory, which molecule or ion has the shortest bond length? O₂, O₂^- , O₂^2-

Video transcript

hey everyone in this example, we need to determine among the following molecules which has the shortest bond length based on molecular orbital theory. So beginning with our first molecule, we have B2 And we want to go ahead and first calculate our total number of valence electrons. So we would find boron on the periodic table. And we would see that it's in group three a recalling that our group number corresponds to the number of valence electrons. We would determine that we have three valence electrons. Now we're going to multiply these three valence electrons by the two atoms that make up our molecule to give us a total of six electrons total For our neutral B two molecule. And so now we can go ahead and get into drawing our molecular orbital diagram which beginning at our lowest energy sub level. We would have our bonding molecular orbital sigma s. And then above that we want to go ahead and fill in our anti bonding molecular orbital sigma asterix us. And then because we still have two more electrons that we would have to fill in, we're going to extend our molecular orbital diagram so that we include our pi bonding molecular orbital here. And this should actually have two slots to fill in our electrons. So we want to recall our hunts rule which states that we would fill in our electrons according to the lowest energy level first, before proceeding to the next energy level. So we're sorry our orbital, our first orbital first before proceeding to the next bonding orbital. So we would go ahead and Draw in our 1st 2 electrons where we have one downward arrow representing the first electron and then the second electron represented by an upward arrow. So we've used a total of two of our six electrons and now we're going to fill in the next two electrons in our anti bonding molecular orbital following Hunt's rule yet again. Now we have two electrons left because we have a total of six, we said where we will fill them in our bonding pi molecular orbital. And so following Hunt's role, we would fill in the first orbital with one electron and then the second orbital with the second electron. And so this would give us a total of all of our six electrons in our molecular orbital diagram. And so now in order to determine our bond length, we want to recall our formula for bond order, where we would recall that it can be found by taking one half, multiplied by the difference between our bonding electrons in our diagram minus our anti bonding electrons in our diagram. And so we can find our bond order here by taking one half and multiplying by our bonding electrons which we would count in our sigma s molecular orbital here we have two electrons and then we have another two bonding pi molecular orbital's sorry, molecular orbital electrons up here. So this would give us a total of 12, 3, 4 bonding electrons. So we would plug that in as four electrons that we're bonding. And then in our anti bonding molecular orbital we have a total of two electrons that we filled in our sigma asterix s anti bonding molecular orbital. So we would say minus two electrons. And so that would be one half Times The Difference of two Electrons. And this is going to give us a value where our bond order is equal to one. And so because our bond order is equal to one, we would say that therefore our bond length we can just say is moderate for now because we need to compare it to the rest of our given molecules. So let's go ahead and move on to the second given molecule where we have a B two minus an ion. So again we want to look to boron on our periodic tables, we see it in group three a corresponding to three valence electrons which we will multiply by our two atoms that make up our molecule giving us a total of six electrons. However, we do have a minus charge here, meaning that we have an an ion and we should recall that a negative charge means that we will gain one electron. And so this extra electron gives us a total of seven electrons total that we would fill in our molecular orbital diagram. And so we would again begin our molecular orbital diagram with our sigma bonding molecular orbital first for sigma s above that we want to fill in our anti bonding molecular orbital represented by sigma asterix s here. Then we would continue on where we would have our bonding pi molecular orbital here where we have two slots to fill in our electrons and we can stop our diagram here and start filling in our electrons. So following Hunt's rule, we would fill in our bonding molecular orbital completely first, then moving on up, we would fill in our anti bonding molecular orbital with the next two electrons and then continuing on to the bonding pi molecular orbital. We would fill in according to Hunt's rule, first the first orbital then our second orbital and then we would pair up the first orbital leaving us with one unpaid aired electron in our pi bonding molecular orbital. So overall we can go ahead and calculate our bond order and so we would say it's one half multiplied by the difference between our bonding electrons, where we would count two in our sigma s bonding molecular orbital and then a total of three bonding electrons in our pi bonding molecular orbital. So that would give us three plus two which would give us a total of five bonding electrons. And we would take this from the difference of our anti bonding electrons which we have in our sigma asterix Anti bonding molecular orbital where we filled in two electrons. And so that would give us 5 -2, which would leave us with three. And so we would have one minus or sorry, one half times three which would give us a result equal to 1.5 as our bond order. And so therefore we would say that so far our bond length of our B two minus an ion is greater than our bond length of our B two neutral adam or sorry molecule. And so we need to go ahead and consider our last given molecule where we have the B two minus and ion. So again we know that we have three valence electrons because we're dealing with the boron atom which we will multiply by the two atoms that make up our molecule giving us a total of six electrons. However we recall just like we did above that the two minus negative charges, telling us that we will Gain two electrons extra. And so this would give us a total of eight electrons total. And so to draw in our molecular diagram, we again want to start with the sigma s bonding molecular orbital. Then we would proceed to the sigma asterix as anti bonding molecular orbital. After that we want to go ahead and fill in our pie bonding molecular orbital and we can stop our diagram here because we only have eight electrons to fill in. So beginning with our principle we start off with the bonding sigma s molecular orbital and fill in our first two electrons. We would then fill in our Next two electrons in our anti bonding molecular orbital. And then moving on to the pi bonding molecular orbital, we have four more electrons to fill in. So we'll follow hun's rule by filling in our first two electrons like this and then our last two electrons with the downward arrow to pair up our electrons in the pi bonding molecular orbital. And so this gives us a total of 246, 8 of our total electrons. So now we can go ahead and calculate our bond order here, which is again equal to one half, multiplied by the difference between our bonding electrons which we determined according to our diagram is going to be a total of two from sigma s and then a total of four from our pi bonding molecular orbital. So that's four plus two to give us a total of six bonding electrons minus our anti bonding electrons. Again located in our sigma asterix s anti bonding molecular orbital where we just have two electrons there. And so this would give us a total of 6 -2 which would give us four electrons. And so this would be one half times four which would give us a bond order equal to a value Giving us a bond order equal to a value of two. And so just to make a correction here, this should be less than Our first bond length from our first given example B two. Now that we know all bond orders, let's go ahead and just make our ranking for a final answer. So we would say that We have for our B2 to minus molecule a bond length of two which is going to be a shorter bond compared to our second molecule. Where we have a bond length for the B two minus an ion of 1. and then our longest bond length will be associated with the Bond order. That was the smallest number coming from our b. two molecule. And so we want to recall that bond order, the higher the bond order that would correspond to a lower bond length or we would say shorter bond length. So we would say that we have the shortest. I'm sorry that would be with the B 22 minus an eye on we would have the shortest bond length From our b. 2 to minus an ion which had the bond order Equal to a value of two. And let's go ahead and make sure that that's eligible. So our final answer to complete this example is that Our B 2 to minus molecule which had the bond length or the bond order equal to a value of two corresponds to the shortest bond length of our given molecules. So this will be our final answer corresponding to answer choice see and our multiple choice. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.