Using the molecular orbital energy ordering for second-row homonu...

Question

Using the molecular orbital energy ordering for second-row homonuclear 
diatomic molecules in which the p2p orbitals lie at 
higher 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. 14

Accepted Answer

hey everyone in this example, we're told that we have a molecular species with 11 valence electrons. We need to determine whether it's going to be dia magnetic or para magnetic. After we sketch its molecular orbital diagram and that we need to use the molecular orbital diagram where the pie to be orbital is at a higher energy than the sigma two P orbital. So beginning off with our 11 valence electrons from our molecular species, we're going to use this as information on how to fill in our molecular orbital diagram. So we would say that because we have 11 valence electrons, we're going to start out by honoring our Hunt's rule and filling in our lowest energy level first where we would start off at sigma one s our bonding molecular orbital above that we want to go ahead and fill in our sigma asterisk one us for our anti bonding molecular orbital and then above our sigma asterix as molecular orbital. We would have the next lowest in energy which would be the sigma one Or sorry, Sigma two p bonding molecular orbital above our sigma two p bonding molecular orbital. We're going to fill in our pi two P bonding molecular orbital which has two slots where we can fill in our electrons And then we would above that have our pie anti bonding with an asterisk two P molecular orbital where we also again have two slots to fill in and then above our pi two p anti bonding molecular orbital, we would have our sigma asterix anti bonding two P molecular orbital and this would actually only have just one slot to fill in. So again, honoring our Hunt's role we would fill in the lowest energy level or the lowest orbital first. And so that begins at the one sigma s bonding molecular orbital where we would have one electron here and then the second here. Then we would go ahead and move up to our anti bonding sigma one s molecular orbital where we would fill in our next two electrons. Then we would go ahead and move up to the bonding to p molecular orbital where we would fill in our next two electrons. And then when we get to the pi two p molecular orbital, we want to go ahead and fill in by honoring Hunt's rule where we place the first electron here in the first slot and then the second one in the second slot. And then we would go ahead and continue to pair our electrons. So we would pair the next one and then our second slot we would pair this electron here. So so far we can count a total of 2468, 10 electrons. And we have a total of 11. So we want to go ahead and fill in our last electron in our next energy level up which is our anti bonding pi two p molecular orbital. So we would fill it in here. And so this would complete our molecular orbital energy diagram for our valence electron species. Now we want to go ahead and find our bond order and we should recall that our bond order is going to be equal to one half multiplied by our bonding electrons and subtracted from our anti bonding electrons. And so what we would have is that our bond order is equal to one half multiplied by our bonding electrons which are each slot which doesn't have an asterisk in our molecular orbital diagram so that we would have to hear. Then we would have another two giving us four and then we would have four added to the original four be counted which gives us a total of eight bonding electrons. And so this is subtracted from our number of anti bonding electrons where we filled in in each slot of our orbital diagram that had an abstract So here we would count two And then we would count a single anti bonding electron here which 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 and we would have one half times five which is going to give us a bond order equal to a value of 2.5. Now, based on this pond order, we would also correspond it with our molecular orbital diagram and recognize that we have the presence of one unpaid electron. And so because we do have one unpaid electron as well as our bond order of .5 which is a decimal here. we would say. Therefore we have a para magnetic species. And so this would be our final answer to complete this example based on a species with 11 total valence electrons. So I hope that everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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