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

Video transcript

welcome back everyone in this example, we need to determine which molecule between our sulfur molecule, our sulfide molecule and our and our per sulfide an ion with the highest spawned order based on our molecular orbital theory. So because this is based on molecular orbital theory, we want to recall how to draw our molecular orbital diagrams for each of our ions given in our molecules. So beginning with our sulfur diatonic molecule, we want to recall that sulfur according to a periodic table is across period three. And because we recall that our molecular or orbital diagrams should begin with the lowest energy level orbital first. According to Hunt's rule, we should recall that that begins at the sigma three us bonding molecular orbital, which is just one orbital. Now, before we fill things in, let's calculate our total valence electrons for our molecule here, we would recall that sulfur on our periodic table is located in group six a corresponding to six valence electrons. However, we are going to recognize that because we have the subscript of two representing the six valence electrons belonging to each sulfur atom in this molecule. That means we will multiply this by r two sulfur atoms to give us a total of 12 valence electrons for our molecular orbital diagram. And again, our molecular orbital diagram we recall only includes our outermost valence electrons, so filling in our first two with opposite spins according to our poly exclusion principle. That is out how we will begin our molecular orbital diagram. Now moving up higher and energy, we will recall that that is going to be our sigma anti bonding three S molecular orbital which is also just one orbital where we can fill in our next two electrons with opposite spins. Moving on up higher in energy, we would recall that we have our sigma three P bonding molecular orbital which is only one orbital where we fill in our next two electrons with opposite spins. Higher up in energy, we have our pi three p bonding molecular orbital where we recall consists of a total of two orbital's where we would follow our filling in our orbital's according to Pauli exclusion principles. So we would have our next four electrons filled in as follows. 123 and four and let's go ahead and make room because we have to keep going, we still have more electrons to fill in because we have a total of 12 electrons as we stated above. So higher up in energy we have to recall we have our pie anti bonding three p molecular orbital which also consists of a total of two orbital's and checking how many electrons we filled in so far. We count 2468, 10, meaning we have two more electrons left to fill in to meet our 12 total valence. So we would fill them in according to our poly exclusion principle one and two. To complete our 12 filled in valence electrons for our sulfur di atomic molecule here. Moving on to our next molecular orbital diagram, four hour s two minus an eye on. We are going to calculate total valence electrons. And we would recall that since our neutral molecule of sulfur consists of 12 valence electrons. And we recall that our an ion charge means we gain one electron. We would recognize that all we have to do is add one to our original total valence. So we would have 12 Plus one electron and that's going to give us a total of 13 valence electrons total for our molecular orbital diagram. So following the same process beginning with our filling of our lowest molecular orbital in energy. First we have our sigma three s bonding molecular orbital containing our first two electrons of opposite spins moving up higher in energy according to Hunt's rule, we have our sigma anti bonding three s molecular orbital which contains our next two electrons of opposite spins moving up higher in energy. We fill in our sigma three p bonding molecular orbital which consists of one orbital with our opposite spin electrons moving up higher in energy. We have our pi three p bonding molecular orbital which consists of a total of two orbital's where we fill in our electrons accordingly with 123 and four moving up higher in energy, We have our pie anti bonding 3P molecular orbital which consists of a total of two orbital's where we only need 13 valence electrons total. So we will only fill in as follows and three. And just to check we have 2468, 10, 11, 12, 13 fans electrons filled in. And just to expand on our molecular orbital diagram. So we know what is highest up in energy. We should recall that that is going to be our sigma anti bonding three P molecular orbital which is just one orbital which we leave unfilled. And this applies to both of our diagram so far. So sorry for leaving that one out but these remain unfilled for these first two ions and molecules. So moving on to our last given an ion, we have our per sulfide an ion and we would recall for its molecular orbital diagram, Us 2 to -. We recall that since our as 21 minus an ion has a total of 13 valence electrons. We can begin with that in our calculation and because we recognize that we have a two minus charge, that means we just need to recall that we gain two electrons. So actually let's start off with our neutral total valence electrons for a neutral molecule of sulfur, which we stated as 12 we would say plus two electrons from 12. And that would give us a total of 14 valence electrons. And sorry, 14 valence electrons for our molecular orbital diagram. And so beginning with the lowest energy molecular orbital first, we recall that again, that is our sigma three s bonding molecular orbital which contains our first two electrons of opposite spins moving up higher in energy, we have our sigma anti bonding three S molecular orbital which is one orbital containing our next two electrons of opposite spins moving up higher in energy. We have our sigma three P bonding molecular orbital which contains our next two electrons of opposite spins. Higher up in energy, we have our pi three P bonding molecular orbital which consists of a total of two orbital's with our electrons of opposite spins. Following our poly exclusion principle, we fill them in as 123 and four. Moving up higher in energy, we recall we have our pi anti bonding three P molecular orbital which consists of a total of two orbital's where we fill in our electrons as follows 123 and four. And to check our electrons so far we have 2468, 10, 12, 14 of our total 14 fields electrons that we must have filled in. That again leaves our sigma anti bonding three p molecular orbital empty as well. And so now that we have sketched out all of our molecular orbital diagrams, the next step is to calculate which has the highest bond order. So we must recall our formula for bond order, which we would recall is taking one half multiplied by the difference between our bonding electrons in our molecular orbital's subtracted from our anti bonding electrons in our molecular orbital diagram. And so calculating bond order for our sulfur di atomic neutral molecule. we would say that that is equal to one half multiplied by our bonding electrons. So according to our diagram that would be our electrons with the molecular orbital that don't have an asterix. So that's 2468 total electrons in our bonding molecular orbital here. So we would plug in eight electrons minus our anti bonding electrons according to our molecular orbital diagram. Those are the ones with the asterix, that's 24 Or sorry, 2, 3 and four. So we have a total of 8 -4 of our anti bonding electrons which is going to give us after. We multiply by one half a value equal to. So this yields a bond order equal to a value of two. So moving on to our next calculation of bond order for our us too one minus ion. We have our bond order calculation where we have one half multiplied by the difference between our bonding electrons according to our molecular orbital diagram, which are the ones without the asterix, so that's 2468 total bonding electrons plugging that. In our calculation we have eight electrons subtracted from our anti bonding electrons according to our molecular orbital diagram. That is the one with the asterix. So we have 24 and then five. That would be eight minus five times one half which will give us a bond order equal to a value. This gives us a bond order equal to three halfs which we understand as a decimal value, being 1.5. Moving on to our last bond order calculation for our per sulfide an ion, We have one half multiplied by our bonding electrons according to our molecular diagram. Again, the ones without the asterix. So that's 2468 total bonding electrons for our per sulfide. And sorry, wrong color subtracted from our non bonding electrons according to our molecular orbital diagram. Those are the ones with the asterix. So that would be 246 total anti bonding electrons. And sorry let's just check that one more time. That would be 24 and six. Yes. So we have 8 -6 non bonding electrons multiplied by one half which gives us a bond order equal to a value of one. And so according to our prompt, our objective for our final answer is to choose the species with the highest bond order. And so we would recognize that we have the highest bond order here for our sulfur di atomic molecules. So this is the highest bond order. And this is based on our molecular orbital theory. So for our final answer, we are going to highlight our sulfur di atomic molecule because it has the highest bond order with a value being too. So everything highlighted in yellow represents our final answer. To complete this example. I hope everything I reviewed was clear if you have any questions, leave them down below and I will see everyone in the next practice video.