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

Video transcript

hey everyone in this example, we are given the below molecules and we need to determine which of these molecules has the highest bond energy based on our molecular orbital theory. We want to recall that if we calculate bond order which can be found from taking one half multiplied by the difference between our bonding electrons minus our anti bonding electrons. That's going to give us a certain bond order. Where we should recall that the higher the value of our bond order that will correspond to a higher amount of energy in our bond. And we would also associate this with a shorter bond length. So beginning with our first example in our atom or sorry, our molecule F two. We want to first start off by finding our total number of valence electrons. So we would find flooring on our periodic table and we see that it's in group seven a where we recall that our group number tells us our number of valence electrons, meaning we have seven valence electrons. Now we have two molecules of flooring making up this molecule, two atoms of florian making up this molecule. So we're going to multiply this by two atoms To give us a total of 14 electrons. And so now we can go ahead and fill in our molecular orbital diagram. So starting off with the lowest energy level, we would have the sigma s bonding molecular orbital above that we want to fill in our sigma asterix s anti bonding molecular orbital above that we're going to have our sigma p molecular orbital here, which is a bonding molecular orbital, then we would have our pi P bonding molecular orbital where we would have two slots to fill in our electrons. Above that we would have our pi anti bonding molecular orbital here. And then lastly we would have our sigma anti bonding p molecular orbital. And so honoring our Hunt's role, we would fill in the first two electrons in our lowest energy level here at the sigma bonding molecular orbital. So we would fill in the first two here. Then we would fill in our next two electrons in our anti bonding molecular orbital. This leaves us with 10 more electrons to fill in our Diagram. So going on up to the sigma p bonding molecular orbital. We would fill in our two electrons and then moving to the pi bonding molecular orbital. We would honor Hunts Rule by first filling in one electron here and then the second electron in the second slot before pairing them up. So we have four electrons left to fill in and we're going to fill them in following our hands Rule one by one in each of these two slots and then pairing them up with our last two electrons. And so we can count and we would have a total of 2468, 10, 12, 14 of our 14 total valence electrons. I'm sorry, electrons total filled into our molecular orbital. And so now we want to go ahead and finally calculate our bond order. Where we're going to take one half multiplied by the difference between our bonding electrons, which are found in all of the molecular orbital's where we don't have an asterisk. So we would have to 5, 6, 7, 8. And that would complete our number of bonding electrons. So we would have eight bonding electrons subtracted from our anti bonding electrons which are all of the electrons where we filled in that have an asterisk. So we would have one too three, So that would be six total anti bonding electrons. And so 8 -6 gives us a difference of two. And when we take one half times two we have a bond order equal to one. And so we would go ahead and move on to our next Given molecule before we can determine an answer here. So our next molecule we have is the F two minus an eye on. Again. We recognize that because florian is in group seven a corresponding to seven valence electrons. We would have 14 electrons total for our two florin atoms. However, we do have a minus charge here. And we should recall that a negative charge for our ion corresponds to an added electron. So we would add one extra electron. And this would give us a total of 15 electrons that we would now fill in our molecular orbital above. So to make this simple, we're just going to copy our previous orbital here And paste this in so that we don't have to do all the work of drawing everything since this is almost the same molecule except for the fact that instead of 14 electrons, we now have 15 electrons where our 15th electron we can draw in our sigma anti bonding molecular orbital here. So we would go ahead and this is our sigma p anti bonding molecular orbital. We would go ahead and fill that in there and so now we can go ahead and move on to calculate our bond order, which again is one half multiplied by the difference between our bonding electrons. So this would be all of the slots where we don't have an asterix. So we would count to hear This would give us now four and then we would have A total of eight. And so we would have a total of eight bonding electrons subtracted from our number of anti bonding electrons. Where originally we have A total of six. And just to make sure let's count it through. We have all the spots where our asterisk is represents anti bonding electrons. So we would have 12, 34567. And so it does make sense that because we have this extra electron contributed from this negative chart here that we went from six anti bonding electrons to now seven anti bonding electrons. And so this would give us a difference of one electron. So we would have one half times one giving us a bond order of .5. And so so far we can say that because our bond order is decreasing based on this molecule, we have a lower bond energy here. Moving on to our third given molecule, we have the F 22 minus an eye on. And so beginning with our 14 total electrons in a neutral F two molecule, we would go ahead and because we have that two minus charge here, add two extra electrons. And so this would give us a total of now 16 electrons total. And so again, we're going to paste in our previous molecular orbital diagram here. So this is our orbital diagram for our third example and we want to go ahead and add in two more electrons. So right now we have a total of 14. We would fill in 15 here and then 16 in our second anti bonding p sigma orbital. So now we can go ahead and calculate our bond order for the last time by taking one half and multiplying it by the difference between our bonding electrons where we would have counted all of the positions where we don't have an asterix. So that's 12, 34 5, 6, 7, 8. So again we still have that eight of our bonding electrons maintained subtracted from. We went from seven anti bonding electrons to now we have that extra slot filled in for both positions of our anti bonding sigma p orbital. So this would give us a total of now eight anti bonding electrons. And so we would have a difference of 8 -8 being zero giving us a bond order of one half time zero. Which we would give us or sorry, which would give us zero as our result. And so overall because we have a bond order of zero associated with our third given molecule, we would say we have the low est bond energy contributed to this bond order. And so overall we would say that because our bond order for our first given molecule here, F two is equal to a value of one, we would say that therefore we have the highest bond energy out of our three given molecules. And so to complete this example, We would confirm that our F2 molecule with the bond order that equaled a value of one had the highest bond energy. So this is our final answer. To complete this example. If you have any questions, please leave them down below, but I hope I was able to help you understand. And I will see everyone in the next practice video