Draw a molecular orbital energy diagram for ClF. (Assume that the sp orbitals are lower in energy than the p orbitals.) What is the bond order in ClF?

Video transcript

hey everyone in this example, we need to determine the bond order of nitrogen monoxide based on its molecular energy diagram. And we need to make note of the fact that our sigma bonding p orbital is going to be lower in energy than our pi orbital. So our first step is to analyze our total number of valence electrons in our molecule nitrogen monoxide. So looking at nitrogen first on our periodic tables, we would recall that nitrogen is located in group five a. And we would recall that oxygen on our periodic tables is located in group six a. So right next to our group number of nitrogen. And this is going to we should recall correspond to our number of valence electrons contributed by each of these atoms. So we would say that we have five valence electrons contributed from our nitrogen atom And then we would have six valence electrons contributed from our oxygen atom. And so to add up our total number of electrons in our molecule, we would have five plus six which would give us 11 total electrons for our molecular orbital diagram. And so we would recall that we begin our molecular orbital diagram with our two sigma s bonding molecular orbital. We then have higher energy from that molecular orbital, our sigma two a strict as anti bonding molecular orbital. And sorry, this should be sigma to us here, above. An energy from our anti bonding sigma two s molecular orbital. We would have our sigma two p bonding molecular orbital where we have just one slot here to fill in our electron pairs and then hire an energy above this orbital we have our pi two p bonding molecular orbital where we would have two slots to fill in our electron pairs. Then above an energy from that orbital we have our anti bonding pie sigma two P molecular orbital which also has two slots to fill in our electron pairs and then our last level which is highest in energy is going to be our sigma asterix two p anti bonding molecular orbital where we just go back to having one slot to fill in our electron pairs. So because we have a total of 11 electrons to fill in our orbital diagram, we're going to start off by honoring Hunt's rule by filling in the first orbital completely. First, that's lowest in energy. So down here we would fill in our first two electrons. Then we would move up in energy to the anti bonding sigma to us molecular orbital to fill in the next two electrons. Now we're at four total electrons. We want to continue on and fill in our sigma two p bonding molecular orbital with the next two electrons. And then moving on up. We want to go ahead and fill in our next four electrons by honoring huns rule first by placing the first electron in the first slot for our sigma. Or sorry, our pi two P bonding molecular orbital. Then we would go ahead and fill in our next electron in the second slot before pairing up our first slot And then pairing a bar 2nd slot. And so this is an honor of Hunts Hunts rule. And so far we would have a total of 2468, 10 electrons that we filled in. So now we just have one last electron to fill in and we can just place it in our first slot up here in our anti bonding pi two p molecular orbital. So our next step is to recall our formula for bond order, which we recall is found from taking one half multiplied by the difference between our bonding electrons subtracted from our anti bonding electrons. And so this is going to come straight from our diagram. So we would say that our bond order is equal to one half multiplied by the number of electrons that we filled in in every slot where we didn't have an asterisk which is our bonding molecular orbital. So we would count two electrons here plus another two to give us four bonding electrons. And then we have four more bonding electrons that we would add to our original for to give us a total of eight bonding electrons in our molecular orbital's. And so this is going to be subtracted from our anti bonding molecular sorry, our anti bonding electrons in our molecular orbital where that is at every position where we filled in electrons that had an asterix. So in our sigma to s asterix anti bonding molecular orbital, we have a total of two electrons and then we also have another anti bonding molecular orbital where we filled in one electron at pi sigma two P. So this would give us a total of three anti bonding electrons. And so we would have 8 -3 which would give us a total of five electrons. So that would give us one half times five, Giving us a bond order as the decimal 2.5. And so what we should recognize is that we have one unpaid electron in our diagram as well as a decimal for a bond order. And so we would say that therefore based on this one unpaid electron being present in our Diagram as well as our bond order being a decimal as 2.5. We would say our nitrogen monoxide is a para magnetic species and this is going to be due to the fact that it's going to be able to slightly attract and externally applied magnet because of this unpaid electron in its configuration. And so our final answer again is the bond order that we determined here, which was 2.5 for our nitrogen monoxide. 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