Using the molecular orbital energy ordering for second-row homonuclear diatomic molecules in which the p2p orbitals lie at lower energy than the s2p, draw MO energy diagrams and predict the bond order in a molecule or ion with each number of total valence electrons. Will the molecule or ion be diamagnetic or paramagnetic? d. 9

Video transcript

Hey everyone in this example, we're given an unknown species with 14 valence electrons and we need to draw in its molecular orbital energy diagram to determine the bond order as well as whether our species will be dying, magnetic or para magnetic. We're told to use the molecular orbital energy diagram where the pi two P orbital is at a lower energy than our pie to pee. Or sorry our sigma two p orbital. So we should go ahead and use the information that we have 14 electrons total to fill in our given unknown species here for our molecular orbital diagram. So we would recall that the lowest energy level in our molecular orbital diagram starts at sigma two S, which is our bonding molecular orbital. Moving on up higher in energy, we would have our anti bonding sigma asterix two S molecular orbital higher in energy than that orbital. We would have our pie two P molecular orbital where we would have two slots to fill in our electron pairs and then above energy in that orbital we would have our sigma to p molecular orbital where we again have just one slot to fill in our electron pair. Next we have higher in energy our pie asterix two p anti bonding molecular orbital where we have two slots to fill in our electron pairs and then the highest in energy would be our sigma asterix two p anti bonding molecular orbital where we just have one slot to fill in our electron pairs. So we would go ahead and fill in our 14 electrons by honoring our hunts rule by fully filling in our first orbital first which is our bonding to us molecular orbital. Next we would continue up in energy to fill in our anti bonding two s sigma molecular orbital above. An energy from the sigma to us. Anti bonding molecular orbital. We have our pi two p molecular orbital where we would honor Hunts Rule by first filling in our next electron here in the first slot and then in the second slot like this before pairing them up. So now we have a total of eight electrons and we need to go ahead and fill in six more electrons in our diagram. So we would go ahead and move up an energy to the sigma two p orbital where we would fully fill in our next two electrons and now we have four more electrons to fill in where we will land at our anti bonding pi two p molecular orbital. Where to follow Hunt's rule. We would fill in our first electron here, our second electron in the second slot. Before we would then pair our last two electrons up to complete our 14 total electrons for our diagram here. And so our next step is to recall our formula for bond order, which we recall is calculated by taking one half multiplied by the difference between our bonding electrons subtracted from our anti bonding electrons. And so what we would have for bond order is a value of one half multiplied by the difference between our bonding electrons which is every slot that did not have an asterisk in our molecular diagram. So we have to we can count here two electrons. We have another four electrons to add to that giving us six bonding electrons And then we have another two electrons to add to our original six bonding electrons to give us a total of eight bonding electrons based on our orbital diagram. So that would be eight electrons minus the number of anti bonding electrons which is every slot where we filled in electrons that had an asterisk in our diagram. So here we would have two anti bonding electrons and then here we would add four anti bonding electrons to the two that we just counted which would give us a total of six anti bonding electrons. And so that would give us 8 -6 which would give us a difference of two. And so we would have one half times two which would give us a bond order equal to a value of one. And so because we have a bond order equal to a value of one and we have zero um paired electrons in our diagram we would say therefore our species is die a magnetic and this is due to the fact that our unknown species is going to be able to slightly repel magnets because it has all of its electrons and its configuration paired up together. And so this would complete this example as our final answers, which would correspond to choice A in our multiple choice. So I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.