the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu if you haven't uh clicked in yet please pay attention to the clicker question and I'll give you a little bit more time to to click in and I'll give you a 10-second warning all right let's let's just go ahead and take 10 more seconds all right does someone want to say how they how they got this one right have anyone willing to say how they got it right we have a American Chemical Society whatever these are called that hang around your neck and clip things to them yes there's the word now oh there we go okay so this molecule is though phosphorus has five valence electrons and then each hydrogen has three so it's for a total of eight so it has three bonding and then one lone pair which makes it a tetrahedral but then the lone pair push the other bonds so it's less than 109.5 degrees excellent yeah so I see that some people just decided that there were three things bound to it and so then then they decided either 120 or less than 120 but you really need to do the lewis structure and see how many lone pairs there are first so once you do the lewis structure then you figure out the parent geometry s enforces some more tetrahedral system alright so more more practice on on these coming up all right so we're continuing on now with molecular orbital theory and we started the course talking about atomic structure and then we talked about atomic orbitals or wave functions and then we moved on to bonding that atoms can come together and bond and talked about the structure of molecules and now we're going to molecular orbitals so and then we'll also talk more about the structure of molecules based on those molecular orbitals on Wednesday so we're really coming all the way around we're using a lot of the ideas that you've seen before but now we're applying them to molecules and then to me this is really an exciting part I love getting up to the molecules and talking about structures and molecules and how orbitals really play a role in those properties that molecules have and then to me the really exciting I like reactions so after we finish with the structures we're going to talk more about how molecules react and so on Friday we're going to be starting that reactions and molecules and getting into thermodynamics so we're sort of winding our way away from orbitals for a while we will come back two D orbitals around Thanksgiving time but we'll have a long time before that where we're talking about reactions so we're about to kind of do a transition in the type of material but before we do that more orbitals but these are super cool because these are molecular orbitals so we're going to talk today about mo Theory mo for molecular orbitals so molecular orbital Theory presents the idea that these valence electrons are really going to be delocalized around these molecules and not just sitting on individual atoms so to think about this electron delocalization we need to think about molecular orbitals molecular orbitals as we've learned another word for orbitals also wave functions wave functions or orbitals orbitals or wave functions you need to consider the wave-like properties of electrons to think about where the electrons are going to be what is their probability density how are they going to be arranged with respect to the nucleus and so we take atomic orbitals and we bring them together as atoms come together to form bonds atomic orbitals come together to form molecular orbitals so we're going to be adding superimposing atomic orbitals to form these molecular orbitals and this is called a linear combination of atomic orbitals or lcao and so we're going to bring those atomic orbitals together and create molecular orbitals and we're going to create two types of molecular orbitals we're going to create bonding and antibonding and some basic math principles apply here and that is that you can create n atomic orbitals from n and molecular orbitals from n n atomic orbitals all right so that's really the basis of molecular orbital theory and now let's apply it to our friends the S orbitals all right so we're going to think about really simple molecules bringing together two atoms that are identical with each other and what happens to their s-orbitals when this happens so first we'll talk about bonding orbitals so bonding orbitals again rise from this linear combination of atomic orbitals the lcao and if it's a bonding orbital it's going to arise from constructive interference so we talked before about the properties of waves and one of the great properties of waves is that they can constructively interfere destructively interfere and orbitals are wave functions so they can constructively interfere and destructively interfere bonding orbitals are generated by the constructive interference so let's look at two atomic orbitals and so here we have an orbital the nucleus is in the middle so little thought is the nucleus and these two atomic orbitals are going to come together there's going to be a bonding event and so we have a 1s orbital from from atom a and a 1s orbital from atom B and they're going to come together and they're going to form a molecular orbital an a/b molecular orbital and this is called Sigma 1s so a sigma orbital is symmetric around the bond axis so the bond axis here would be just a vertical aligned between these two nuclei here and so this molecular orbital is symmetric around that bond axis there are no nodal planes for something that is symmetric there are no nodal planes for our s orbitals and so there's none for the molecular orbital either and we can also write this as 1sa plus 1sb so the atomic orbital from one si the atomic orbital one SP coming together to form Sigma orbital one s that is a bonding orbital because it's constructive interference it's a bonding mo or molecular orbital so now let's consider the wave-like properties and think about these atomic orbitals coming too as waves and what is happening so here we have the same equation up here we're bringing together one si and one SB but now let's think about this as a wave function so there is an amplitude associated with the wave function for one si and there is an amplitude associated with the wave function for one SB these wave functions come together here's one nuclei here's the other nuclei and for a bonding orbital it'll be constructive interference and so the amplitude where these wave functions atomic wave functions overlap will be increased when you have constructive interference so our Sigma 1s now has increased amplitude between these two nuclei due to that constructive interference so an increased amplitude between these two nuclei again this is the bonding axis here's one nuclei here's another so an increased amplitude here in create corresponds to an increased or enhanced probability density remember our wave function squared is probability density it suggests the likelihood of finding an electron in a certain region of volume and if it's enhanced by this constructive interference so if that density probability density is enhanced you have a greater chance of finding an electron between these two nuclei which will be attracted by both nuclei so now why don't you tell me what you think is going to happen to an electron that is in this region of constructive interference this increased area of probability density okay ten more seconds interesting okay so we should have probably not put up the answer there and repol so yes so here if we have an electron that's attracted to both nuclei then we want to think about whether that's going to be lower or higher in energy than something in an atomic orbital so if it's attracted to both it's going to be more stable II bound to those it'll be harder to remove that electron which means that it's lower down in energy and so we should look at that and we're going to so the answer is it should be lower in energy more stable harder to remove that electron it's in a sweet spot it has two positively charged nuclei and it's kind of hanging out right in the middle it's very happy it's gonna be more stable and that means lower in energy all right so let's take a look at that so the electron is lower in energy and the bonding orbital energy is also going to be decreased compared to the atomic orbitals and that has to be true if that's where the electron is so let's look at the atomic orbital from a and the atomic orbital from B and now I'm going to put the bonding orbital at a lower energy level first so I was going to put the electrons on and now I'm going to put the orbital and a lower at a lower energy so remember this is increasing energy here so the atomic orbitals would be up here whereas the molecular orbital is down here molecular bonding orbital will be lower in energy and now we'll put our electrons there so we have one electron up here and one here and so when they come together we're going to have two electrons in this molecular orbital so when you have these two electrons and they both occupy the bonding orbital and this is the case for h2 each H Adam is bringing one electron h2 has these two electrons and that's gonna make h2 more stable and we saw that before that you have to dissociate the h2 bonds you have to add energy into the system h2 is more stable than free h plus h so when you bring atomic orbitals together and you have constructive interference an increased probability of electrons between those two nuclei that's a sweet spot they're gonna those electrons are gonna be very happy there and that will result in a more stable lower energy structure that's bonding but whenever there is a positive event like this there's always a negative event cuz that's just how life works so we have bonding orbitals but we also have antibonding orbitals so antibonding orbitals arise from the linear combination of atomic orbitals lcao through destructive interference so here these are going to be destructively interfering and that will generate a molecular orbital that is an antibonding orbital so here our little nuclei again and this antibonding orbital is called Sigma 1s star so we can write an equation for this as one si minus one s B equals Sigma 1s star that is an antibonding molecular orbital and let's just think about how this kind of shape arises considering the wave-like properties of these atomic orbitals so I'm going to now move this equation up to the top and now I have my one at one si here my one s be here and now its destructive interference so we have opposite phase of the wavefunctions and when we'll put up a wave function there now the next one has the another the opposite phase so they're going to destructively interfere there is overlap over here but when you have destructive interference then the amplitude is going to decrease so now when we consider this destructive interference between these two orbitals you see that you have what arises between them instead of enhanced probability of finding an electron you actually get a node so you have decreased amplitude translates to decrease probability density between these two nuclei one here one here and that results in a node between the two nuclei so in the antibonding orbital there's even there's a lower probability that it will be in this sweet spot between these two between the two nuclei that have this nice positive charge for its little negative charge so there's pretty much no shot at being in that nice sweet spot and so what that ends up meaning is that an electron in an antibonding orbital is pretty much excluded from that inter nuclear region that region between those two nuclei and that results in a molecular orbital that has higher energy than the atomic orbital there's just no chance of being in that wonderful spot it's really very sad for the poor electron that has to occupy an antibonding orbital so now let's put this on our energy scale so we'll go back to our energy scale and we saw before that when we had one SI + 1 SP and you had a sigma one as a bonding orbital that was lower in energy electrons that occupy it are more stable compared to their positions in the atomic orbital but we also now have an antibonding orbital from destructive in farrance between the wave functions of the atomic orbitals and that's higher in energy and so that's up here so this is what our diagram is going to look like that brings two atomic orbitals together to create two molecular orbitals so the antibonding orbital up here is raised in energy by the same amount that the bonding energy is lowered and so that gives rise to this diagram and importantly as I mentioned we have n atomic orbitals forming and molecular orbitals so if you have two atomic orbitals we generate two molecular orbitals one is bonding lower in energy and one is anti-bonding higher in energy all right so let's take a look at some examples so always start with hydrogen so hydrogen has how many electrons one hydrogen hydrogen atom one so we have two hydrogen atoms and so we have two 1s orbitals 1s a 1s B and now we're going to where do we want to put our electrons and the highest energy possible lower energy where do we want to put them all we start with lower energy so we're gonna put them down here and so this is now the mo diagram for h2 and we can write the electron configuration for the mo diagram which you'll note is a different electron configuration than you were writing when you were looking at the periodic table because now we're not writing it in terms of 1s2 2s2 we're writing it in terms of molecular orbitals and there's some of these on the problem set and I tried to indicate example to make sure you know what kind of electron configurations we're talking about so for mo diagrams when it says electron configuration we're talking about where the electrons are with respect to the molecular orbitals so the answer to this would be Sigma 1 s 2 there are two electrons in the molecular orbital Sigma 1s good so that means that you can do the same thing for dye helium and that's a clicker question let's just take 10 more seconds okay yep all right so so someone want to tell me that is correct what what's number two what's number two four four hydrogen okay what about number three what was what was wrong there okay everyone's doing really well yeah so we had all of the electrons are were parallel in there and what what what is that violate what would be true in that case yup and so we they'd have they'd have the same four principal quantum numbers this is not allowed so that one's not good and what's wrong with four yes star was on the bottom so they were flipped around antibonding was lower in energy great okay so we'll just put those in over here and so we had two electrons helium brings two so two went into the bonding and two went into the antibonding with opposite spins and we have bonding lower-energy antibonding higher energy alright so then we can put in the electron configuration here and we have Sigma 1s 2 Sigma 1s star - so that's the electron configuration now interestingly you see you have two at lower energy and two at higher energy so there's no net loss or gain in energy of h2 compared to just two helium atoms and so that raises the question does helium 2 exist so what would molecular orbital Theory tell you about whether it exists and it would actually predict that it does not exist and the way that molecular orbital Theory gives you these predictions is through the calculation of something called bond order so bond order is half of the number of bonding electrons minus the number of antibonding electrons so let's just write out what the bond order for helium would be so helium 2 our die helium molecule so we have bond order equals 1/2 there's always a half and now the number of bonding electrons so how many bonding electrons do we have for helium - how many antibonding electrons do we have for die helium - and can someone do this math for me zero right so that would suggest a bond order of zero which means no bond and let's just compare that to hydrogen h2 which should have a bond order equal to 1/2 it's always a half in the hydrogen how many bonding electrons did we have we had two how many antibonding electrons do we have zero and the math bond order is 1 so that means one bond or a single bond so mo theory would predict that die helium has no bond ie it's not really a molecule without a bond but h2 should exist so let's take a look at what experiment tells us and it does exist but only really not that much so it was not discovered until 1993 which for some of you might seem like quite a long time ago but since we've been mostly talking about discoveries that were made in the 1800s it took a long time before someone could prove that da helium existed and if we look at the bond dissociation energy for h2 0.01 and compare that to h2 432 the helium really doesn't exist very much zero is a much better approximation of its bond than one would be it really is not a very good molecule so I would call this a win for molecular orbital theory it correctly predicts that h2 is not going to be a very good molecule our helium 2 is not going to be a very good molecule but h2 will be a good molecule and that that works all right so now let's consider 2 s so 2's orbitals are analogous to 1s except that you have to remember that they're bigger so we have our 1 s and r 2s but for the purposes of this it doesn't really matter so let's look at a diagram now that has 1s and 2s so we have lithium so how many electrons does lithium have yep so we'll put on lithium dilithium we'll put on lithium's it has 2 in 1s and 1 into s and so we have one lithium here and one lithium over here and our 1s orbitals are going to be lower in energy so they're down here our 2's orbitals are higher in energy so that goes for both the atomic and the molecular orbitals that are generated so bringing together 1s with 1s we get Sigma 1s and also Sigma star 1s and so we can start putting our electrons in we're going to start with the lowest energy and move up so we have 4 electrons that we need to put in so we fill up everything here and now we go up to our 2s the 2's orbitals will generate Sigma 2 s and Sigma 2 s star we have two electrons so they both can go down here so here is what our mo diagram looks like we can write out the electron configuration again based on this diagram so we have two electrons in Sigma 1s so we have a 2 there we have two electrons in Sigma 1s star or antibonding orbital and we have two electrons in Sigma 2's and now we can calculate the bond order which is another clicker question okay just ten more seconds excellent 84% so if we do out the math here we always have our 1/2 we have 4 bonding electrons we have 2 down here and 2 up here we have two antibonding electrons in our Sigma 1s star and so that gives us a bond order of 1 and in fact the dissociation energy does suggest there is a bond it's not a great bond it's a hundred and five kilojoules per mole not necessarily enough to power a starship but still this molecule does exist all right so let's look at beryllium now do another example so how many electrons is beryllium gonna have it'll have four so we'll put those up now we're going to start with our lowest energy orbital and we're gonna put some in the antibonding as well and then we'll do this and then we'll do that and we filled up everything and so we can write out our configuration we have two in our sigma 1s orbital we have two in our sigma 1s star we have two in our Sigma 2 s and 2 in our Sigma 2's star antibonding orbital there and now there are two different ways that I can calculate the bond order here and we'll do both of them and show you that they come out the same way one way that we can calculate this is to consider all electrons so if we consider all electrons our bond order and that's often just said be period o is 1/2 of our bonding and now we can count up both the 1s and the 2s so how many bonding electrons do we have we should have four so we have two here two here so for how many antibonding also four so that suggests a bond order of zero or we could just consider our valence electrons so that would be the the electrons in 2's and if we do that bond order equals 1/2 it's always a half so how many valence electrons do we have in bonding orbitals - how many an antibonding orbitals - and that gives the answer of zero so should always work unless you did something very strange you should be able to do it both ways and get the same answer so if this is a complicated problem you might want to only consider the valence electrons in fact on the test you may only be a asked to draw the molecular orbital diagrams for the valence electrons and you don't have to do the other the other ones ok so that should work both ways and we get a bond order of zero and in fact the dissociation energy is only nine kilojoules per mole it's a little stronger than dye helium but this is an exceedingly weak molecule so when you have the same number of electrons in bonding as anti-bonding doesn't lead to a very strong model molecule all right that's the S orbitals now it's time for molecules that have P orbitals as well so in this case now I have my p orbital and another p orbital and we have our nuclei in the middle and we're going to bring these together and we will have constructive interference and so here we're bringing together two P X of a + 2 P X B or 2 py a + 2 py B so just x + y we're can see right now and if we bring those together with constructive interference then we're going to form a bonding orbital that has enhanced probability density in both cases and a nodal plane along the bond axis because we had nodal planes along there to begin with so if we think about this and we have both of these or bringing them together and we're going to have they're going to interfere constructively enhance probability density here enhance probability density down here but we still have our nodal plane because we started out with one and so if we have a nodal plane this cannot be a sigma orbital it has to be a PI orbital because Sigma orbitals are not going to have a nodal plane along the bonding access all right so we could generate pi 2px or PI 2p why this way so a PI orbital is a molecular orbital that has a nodal plane through the bond axis or maybe I should say along the bond axis so here's our nodal plane right through the bond axis we can also have antibonding which means destructive interference so now I'm going to be subtracting one of these from one of these and I'm gonna get something that looks like this and it will be PI 2p x star or PI 2 P Y star and it will have two nodal planes so let's look at this so this is destructive interference I'm subtracting one of these from one of these so the phase has to change so I'm going to change the phase on one of them and then bring them together now we're not going to have that awesome constructive interference increased increased probability density the are negatively interacting with each other and this generates a nodal plane between the molecules they really look much more like this now so we have still our nodal plane through the bond we had that before we're always going to have that but now we have an additional nodal plane between the nuclei so in one case we have enhanced density probability density again and the other case in antibonding we have another nodal plane all right so now let's look at what happens to the energies of these PI orbitals and the diagrams that I'm about to show you we're only talking about px and py now we have for the moment forgotten - PZ so these diagrams are rated I for incomplete warning to the viewer people come to me and go where where are the two PZ yes they're not in these diagrams but when you're asked for a complete diagram you will always have to put those orbitals in and in fact completing these diagrams could be a question you get later okay but for now we're gonna have two PZ here but we don't have it's not forming a molecular orbital in this diagram this diagram is thus incomplete but we're going to start simple and build more complicated so only only two P orbitals first and then we'll add the third one because this compound doesn't actually need that orbital so we're good for now alright so we have moved on to the first element that has a P electron in AP orbital we have boron and we have two of them so now I dropped off one s to simplify this so we're starting just with now we just have our valence electrons so it's a good thing we can calculate bond order just using our valence electrons alright so we have two into s and 1 into P X or 2 py I could have put it either place so let's put in where they would go so we have two s orbitals they'll four go into bonding Sigma 2's first then into antibonding Sigma 2's star and now we have P P electrons and we're going to put them into our PI orbitals our molecular orbitals I'll put one in and where am I gonna put the other one am I gonna put it next to it or over here I think yeah so we're gonna do that because again when you're going to sit on a bus you want to have if they're degenerate in energy levels as they are here you're always going to put one electron in each orbital of the same energy first before you pair them up and they'll have parallel spins so we're reviewing things we learned before I love doing that ok so now let's see what our electron configuration is and this is just the valence electron configuration we're simplifying we're not going to consider the 1 the 1s orbital and we can write this down so we have two electrons in Sigma 2's and we have two electrons in Sigma 2's star antibonding orbital and we have one in pi 2px and 1 and PI 2 P Y and I don't put I can put a 1 or not put a 1 if I don't put anything 4 1 1 is assumed and we can calculate our bond order as well so we have 1/2 and again we're just using our valence electrons but that's ok we have 4 now who are bonding to down here to up here these are bonding orbitals and we have two that are antibonding and notice for our PI orbitals this is what we saw before bonding or lower in energy antibonding or higher in energy with the bonding orbitals we had constructive interference enhanced probability of the electrons near the nuclei and so that's lower in energy but in our anti-bonding ones we have a nodal plane in between our nuclei so we don't have any probability that electrons are right in between there because a nodal plane so those are higher in energy okay so here is our b2 diagram so now let's try the same thing for carbon c2 and that is a clicker question alright let's just take 10 more seconds okay so let's take a look at that over here the easiest thing to do to answer this question was to fill in the diagram in your handout and so if you did that you would have put two down here and you would have put two up here then you would have put one here one there another one there and another one there okay so now we have used these up and so our configuration is Sigma 2 s 2 Sigma star 2 pi 2px 2 pi 2py 2 and the bond order is 1/2 there are six bonding electrons 1 2 3 4 5 6 and two antibonding electrons and so that adds up to a bond order of two and so sometimes on a test there'll be a simple question what is the bond order but to get there you have to draw your whole molecular orbital diagram and figure out how many bonding and how many antibonding so these are not really that fast questions and it's nice sometimes we give you like a little space and you see this whole little molecular orbital diagram like fit in there to answer the question all right so let's just compare these two diagrams for a minute so in both cases we had two s orbitals two atomic orbitals four 2s and they both generated two molecular orbitals a bonding and in antibonding the bonding is lower in energy and the antibonding is higher in energy we also had two 2px atomic orbitals they generated two pi 2px orbitals one bonding one antibonding and the same for our two atomic orbitals for 2p why we had two of those and they Jen one lower-energy bonding PI bonding and one pi star antibonding so you always have an atomic orbitals generating and molecular orbitals so the stability of the resulting molecules in these cases depend on how many of the electrons are bonding how many are in lower energy as a result of formation of the molecule and how many are at higher energy as a result of formation of the molecule and if the net result are more electrons in lower energy more electrons in bonding orbitals then that molecule is more stable if there's a very slight or kind of no difference then that's not a very stable molecule so now let's just compare these two and think about which of these is going to be more stable so we have our configurations again so in the case of B 2 how many electrons are in lower energy or bonding orbitals yep we have 4 1 2 3 4 how many in higher to up here 4 carbon we had 6 1 2 3 4 5 6 2 and higher and so the bond order here was 1 the bond order here was 2 which is more stable higher dissociation energy which one do you think carbon right has a bond order of two it has more electrons in lower energy orbitals so it came out really well out of this bonding deal and the dissociation energy for B 2 289 whereas for C 2 599 so when molecules come together such that more of the electrons are in lower energy or bonding orbitals you form a nice stable molecule when molecules come together such that more of their electrons are an antibonding or higher energy that's not a happy molecule so I'll just end one way to think about this in this cartoon molecular breakup lines sometimes two atoms just have an incompatible number of valence electrons and they're just too many you just hear this molecule saying I'm sorry too many of your electrons are in my antibonding regions I don't know how many times we've all heard that but you know it's time to dissociate but our atomic orbitals well they can still be friends okay see you on Wednesday take a look at the clicker question all right let's let's just take 10 more seconds all right let's just go through this one so this is a review of where we were last time but so the correct answer is one so Sigma orbitals are cylindrical asymmetrical it's quiet down a minute you can hear hear the answer so this one is true the second one is not true a bond order of zero doesn't mean that you just have antibonding orbitals whenever you bring together two atomic orbitals you have to make two molecular orbitals so it isn't that sometimes you make bonding orbitals and sometimes you make antibonding orbitals every time you bring together two atomic orbitals you make two molecular orbitals one that's lower in energy and that's the bonding orbital and one that's higher in energy and that's the antibonding orbital so so that is not what that means a bond order of zero means that you have equal numbers of electrons in your bonding and antibonding orbitals so there's no net stabilization due to the formation of of these bonds so here bonding occurs when you bring together two atomic orbitals to make two molecular orbitals that are both of lower energy no every time you make the two orbitals one is lower in energy and one is higher than energy can't make two that are both lower in energy and a bond order of 1 means constructive interference is generated a one bonding orbital that's not what a bond order of one is and again every time you but generate a bonding orbital you generate an antibonding orbital and one means that you have twice as many electrons in your bonding orbitals as antibonding orbitals because the formula is half the number of bonding minus antibonding but that's good and it's important to remember that Sigma orbitals are symmetric around the bond axis alright so we had these diagrams for boron and carbon just talking about these the interactions of the PX and py and so we had forgotten about our PZ and you can't do that on a test to get into trouble so I'm always telling people for these two handouts you must include the molecular orbitals that are derived from P to Z so on a test you need to put them even if they're empty even if they don't have anything in them they need to be part of your molecular orbital diagram we didn't have them in the diagram because we hadn't talked about them yet so now we're going to talk about them so two PZ orbitals again this is an on Mondays handout you have this linear combination of atomic orbitals so now and you know all our P orbitals they all look the same as each other they're just different in orientation you have one along x1 along Y 1 along Z but they're the same so now we're going to bring our two PZ orbitals together and we're going to do it along the bonding axis so we've defined this as the bonding axis in the class so we'll bring them together and they'll be constructive interference with our bonding orbitals there's always constructive interference that generates bonding orbitals and so we're going to create a enhanced amplitude as the wave functions come together and it's going to be cylindrical symmetric so what type of orbital do you think this is going to be Sigma or pi it'll be Sigma because it's cylindrical asymmetric so we do not have any bonding plane along the bond axis and it's it's symmetric around and we have enhanced probability density and we have a wave function squared enhance probability of having an electron between the two nuclei and so this is a sigma 2p z so P orbitals can form Sigma molecular orbitals so we do have nodes passing through our nuclei here are our nuclei again we do have them they were here before in our P orbitals there's a nodal plane in our p-orbitals but we do not have in this case a node along the bond axis so that is our bonding so whenever you generate a bonding orbital which is lower going to be lower in energy we're gonna have our increased amplitude between the nuclei again our increased probability density and therefore lower energy so whenever you have constructive interference generating a molecular orbital of lower energy god creates something of higher energy that's just how life works so we also are going to have antibonding orbitals which are generated by destructive interference and again these orbitals can be thought about as wave functions in a property of waves is that they constructively interfere and destructively interfere so now we can subtract our two orbitals which base yours going to switch the sign and have they're gonna be out of phase so they'll destructively interfere and that's gonna look like this so now you're gonna generate a nodal plane between the two nuclei but it's still symmetric around the bond axis so this is a sigma 2pz star so it's an antibonding orbital so again it's Sigma so it's still cylindrically symmetrical with no nodal plane along that bond axis but you do have a new nodal plane that's generated so nodes pass again through the nuclei but also now between these two orbitals so we have a new nodal plane that's generated that's between these nuclei and that's a result of destructive interference generates that nodal plane you have decreased probability density for a nucleus via an electron being found there and so that poor electrons kind of shut out of that sweet spot the electrons like to be between those two nuclei where they have the two positive charges in the nucleus and then their little negative charge and they can sit right there and be very happy in a low energy state but here there's really lower probability density lower likelihood the electron will be found here and that generates the molecular orbital that's antibonding or higher in energy all right so now we have to go back to our mo diagrams and figure out where to put these new molecular orbitals on to our nice diagrams and it's not as simple as it was before because where we put these new sigma 2pz molecular orbitals depends on what Z is so it depends on the value of Z so if Z is less than the magic number of 8 then we have our pi 2px and 2py orbitals are lower in energy than our sigma 2pz molecular orbital but if we are equal to or greater than 8 then the sigma 2pz orbital is lower in energy than the pi 2px and 2py so less than 8 pi is first its lower in energy as you go up your energy scale and if Z is equal to or greater than 8 pi is second and Sigma is first so how are you going to remember this there could be many different ways one can remember if it I'll tell you how I remember it and I sort of my life revolves around my daughter and my dog and so at Thanksgiving we always have the question can I eat pie first so if you are under the age of eight you always want your pie first so if Z is less than eight pi comes first pi is lower in energy and Sigma is higher however when you mature to the grand age of nine say or ten if you were a kid ten is like the oldest you can possibly imagine being very mature and you can eat your Thanksgiving dinner and wait for pie so that is how I would remember it under eight pie is first equal to eight or greater you can wait till after dinner to have your pie pie comes second note that the ordering of the anti-bonding orbital's is the same so all you have to remember is down here depends on Z is pie first or is pie second so let's take a look at an example let's look at our friend molecular oxygen that has a Z equal to 8 so oxygen is at the old mature age of 8 and so it's going to have its sigma 2pz first it can wait for its PI orbitals until later so let's start putting in our electrons so we have each oxygen making up molecular oxygen or o2 brings two 2's orbitals to the Thanksgiving dinner table and two of them go down in energy into the bonding Sigma 2's orbital and to go into our antibonding Sigma star 2's orbital now we have 4 electrons in our PZ orbitals in our atomic PZ orbitals 4 from each molecule so we need to put all of those in so we always start with the lowest energy orbital so we'll put 2 in there then we'll go up we have 2 more here 2 more here we'll put them in singly with their spins parallel and then we'll pair them up in the lowest energy orbitals and then we have 2 more left so we're gonna have to put those up in our antibonding orbitals so they go into pi 2px and pi 2py star orbitals up here alright so now we can calculate the bond order for oxygen and that's a clicker question alright let's just take 10 more seconds all right so let's take a look over here so bond order again is half the number of bonding electrons minus the number of antibonding electrons and we have 8 bonding 1 2 3 4 5 6 7 8 and we have 4 antibonding 1 2 3 4 so that gives us a bond order of 2 and the bond order equation is one that you do have to memorize that will not be given to you on an exam and with a bond order of two we have a pretty big number for dissociation energy again that's the energy you have to put into a bond to break it to dissociate it and that means it's a pretty strong bond if it's a big number if you need a lot of energy that's a strong bond another thing that you can see from this diagram is that Oh 2 is a bi radical it has two lone pair electrons so two unpaired electrons which also makes it paramagnetic or attracted to a magnetic field so whenever you have unpaired electrons you will have a paramagnetic species and diamagnetic means they're all paired and so there's some questions on problem sets and on exams so you need to know the definitions of those alright so we talked about this when we're doing Lewis structures if you recall and we drew a beautiful lewis structure of molecular oxygen that had two lone pairs on each oxygen and a double bond but i told you that that was not really a complete description of molecular oxygen that molecular oxygen was actually a by radical but you would not get a hint of that from the Lewis structure so if you draw the Lewis structure as a by radical then you have single electron here and a single electron there and a single bond but we also see the bond order as 2 so this one does describes the by radical in the but doesn't really describe that double-bond character that it has so neither of these Lewis structures really completely describe molecular orbital a molecular oxygen and we need a molecular orbital diagram to really help us understand the properties of molecular oxygen that it's pretty strong bond between it and has double bond character but it also is very reactive it's a by radical so this is just a bizarre molecule and really that is why this diagram right here really tells us why our life our planet is what it is right here that describes it it's because of oxygen that we have the life forms that we did life was very different on this planet before molecular oxygen came about you did have microbes that lived anaerobically without oxygen but when oxygen cane everything changed and so here this if you said can you explain life to me you can draw this and though there it is this explains life as we know it because life as we know it exists because of this crazy molecule that is o2 nothing else really like it it is an amazing molecule that allows us to live so there you go so I say you don't learn anything in chemistry this diagram I just explained life as we know it to you right there all right we also it's not just odd there's a few other elements that yeah are pretty important and one of them is nitrogen we wouldn't really be matched anywhere without nitrogen either oxygen is no nitrogen is pretty special too okay this this also helps maybe these two molecular orbital diagrams really sum up life as we know it alright so let's look at molecular nitrogen n2 so we have two electrons in our two s orbitals so we're going to bring them together we're gonna put two in the lower energy orbital our Sigma 2 s + 2 and our antibonding orbital and then we have three in our p orbitals from this nitrogen 3 electrons from this nitrogen and so we'll put them down here we'll put in two three four five six so we didn't need to use any of these anti-bonding orbital's and this is a Z less than eight case so here we have our PI orbitals first because it is Z less than eight so let's look at the bond order here and the bond order is half of eight one two three four five six seven eight eight bonding electrons and just these two antibonding electrons so 1/2 of 8 minus 2 is 3 and you have a really big number for your dissociation energy 941 this is a very stable molecule and you can draw the lewis structure of this without much difficulty this one works quite well and you get a triple bond and two lone pairs so this would be a diamagnetic molecule no unpaired electrons no no radicals here but this is a crazy strong bond it's really hard to split nitrogen because of this triple bond and molecular orbital Theory just tells you it should be a triple bond it should be really really strong interactions between these nitrogen's and so this is a very hard thing we want to do this industrially and we'll talk more about this when we get to chemical equilibrium how do you split the nitrogen bond this is actually currently a big area of research how you break that bond because we need nitrogen for life and so how do you split it there's lots of nitrogen and two in the atmosphere but we need it here and we need it usable so we need to break that bond to use it so we'll come back to this in chemical equilibrium and how people are able to split the nitrogen bond so forget Man of Steel we should have Man of nitrogen that's strong nitrogen strong forget steel Man of nitrogen okay one more thing to do before we move on to today's handout and this is really fast you can also be asked to draw molecular orbital diagrams where both atoms are not the same so if you are asked to do that you can use the following rules if Z is less than 8 for both atoms pi is first if Z is not less than 8 for both atoms you don't really know what it's hard to know what to do so you don't need to know what to do and that's it ok so we might tell you something about it and then you can do it but otherwise just just just worry about things when Z is less than 8 so that's molecular orbital Theory