well hi everyone Mrs Hansen once again wrapping up some topics on resonance in Chapter 2 we've left off here uh with chapter 2 Section 10 which is called residence pattern recognition and we're going to learn that there are five general bonding patterns in Which Resident structures can occur and you must recognize these patterns in order to predict if residents will indeed be present so they're very critical uh and that's going to take some practice so we're going to start in and practice each of the five together the first of the five patterns is an alyc lone pair now remember what alyc Meant For Us alyc means that we have this double bond the two carbons in the double bond it's the position adjacent to it so this is the alyc position and if it has a lone pair you're looking at seeing a pair of electrons right here second one an alyc carbocation which would represent a structure in which in the alyc position there's a positive charge a lone pair of electrons adjacent to a carbocation so let's say we have a position where we have a set of dots next to a carbocation a pi bond between two atoms with different electr negativities that could be for instance a double bond which between a carbon and an oxygen for example that's a different electr negativities and the fifth is a conjugated Pi ring P so for example our aromatic compound which we're learning is Benzene where you have a cyclic structure that alternates double single double single double single that aromatic compound are called conjugated Pi Bonds in a ring so the only way to recognize these patterns is it's says is to do lots and lots and even more practice problems and examples and that's my goal today is to work some of these problems and provide examples with you so pattern number one talked about the ayic position and let's just remind ourselves of these two definitions the vinyl and alil refer to atoms of the pi Bond and next door to the double bond respectively so let's say the carbon involved in the actual double bond are in the vanilic positions these two carbons are involved in the bonds whereas the carbons adjacent to the carbons involved in the double bond are in the alyc positions so with this structure I can see that there would be four alyc positions we look for alyc lone pairs because they will exhibit resonance so remember rule one the pattern one said look for an alyc lone pair so the first of the five general pattern says if we find a double bond look in the position next door and see if they have a double bond or I'm sorry a lone pair of electrons we look for the alyc lone pairs because they will be in Residence they will be delocalized so for example I had written on this earlier let me erase that here is a carbon to carbon double bond all right so we know the carbons that are in the double bond are known as vanilic we're looking for the carbons that are adjacent to that so this is an alyc position and this is the alyc position directly next to the carbons in the double bond do you see any lone pair electrons the answer is yes right here notice that this lone pair is too far away so that's not the alyc position it's not directly next door but two steps down that's no good it has to be directly next door so we have a set of Lone pairs in the alic position which means it can exhibit resonance you know if I just back up what that means is that this lone pair can go here and these can go here and we can just start moving the electrons around all of the lone pairs drawn in the examples are alyc so let's just focus on that definition here's the two carbons in the double bond so the position that is one away is the ayic position here's the two carbons in the double bond one carbon or one position away is the alyc position that has two sets of electrons one would be delocalized one set would be Loc loc alized you're only allowed to move one set of electrons not both here we have a double bond the positions of the double bond one position away would be the alyc position again that meets the criteria it has two sets of electrons one would be localized one would be delocalized available for resonance here is a double bond the ayic position is one position away that carb anion contains a set of dots available for resonance and one more time here's a double bond one position away here's a set of Lone dots oops I missed the color coding and that would be available for residents and that's exactly what we've shown you one step at a time down here here is the ayic position of this carb anion the set of dots becomes the bomb Bond and the resulting structure ends up moving the positive charge to the left and the double bond to the right here the resulting resonance structure would have the oxygen losing a set of electrons forming a double bond the double bond forming a set of dots again remember to consider the formal charges as you develop your resonance structure in this structure the set of dots is formed from the pi Bond making the oxygen negative and this oxygen moved a set of dots to make a double bond making this guy 1 2 3 4 5 he ends up positive and here this double bond goes up to the nitrogen now is a set of dots two sets of dots it now is 1 2 3 4 5 6 so it will have a plus formal charge and this down here became a double bond 1 2 3 and this had a hydrogen so that's all set 1 2 3 the third one was a hydrogen so that hydrogen is still there so no formal charge and the resulting resonance structure here I have three sets of dots now on the oxygen making it a formal charge of minus1 the nitrogen had a set of dots oops let me clean that up the nitrogen now has a pi Bond 1 2 3 4 it would like five so that's now a positive formal charge all right so just remembering as we move electrons around forming the resonance structure always include form charge when the atom with the alyc lone pair as a negative charge the charge is delocalized with the lone pair as shown here and those my friends are the very things that we practiced on the previous slide we were just moving the electrons around from one position to the next so if a set of dots becomes a bond the bond becomes a set of dots the pi Bond becomes dots the lone pair becomes a pi Bond dots become a pi Bond Pi bonds become dots Pi bonds become dots loone paays become bonds and so forth It's just REM reminding you when you write out the new um structure to always always write the formal charge in the the new resonance structure so should we practice draw the appropriate resonance structure so let's put on the number of dots we know are there a pi Bond can become a setup electrons and when we're doing so we have whoops this Oxygen's in here a set of electrons forming up here now why didn't we do anything with this other oxygen this other oxygen has a set of Lone pairs that's also adjacent to that um to that Pi Bond so here's a set of lone pair electrons so I could move the pi Bond and create an O Negative up there but there's also another opportunity to move that pie Bond except to develop more noral charges let's kind of take a peak if I move the second loan pair we get an oxygen with a double bond and a positive charge so just wanted to kind of verify this also works this meets the criteria of having a lone set of electrons in the alyc position so this can also occur and when it does so you form a double bond here and now you have a positive formal charge so two arrows because there are two Al liic lone pair electrons here's a set and then the second that meets the same criteria would be here all right draw the appropriate resonance structure well all right this means that right here is a negative formal charge so that's 1 2 and we need a pair of electrons here 3 four and then there's a hydrogen even though it's not drawn we know it's there so our PI bond is the alyc position right here is the ayic position and it has a set of lone pair electrons even though they're not sh we know they were there cuz we see a negative formal charge that means that the electrons can become a bond and the bond becomes a pair of electrons and when we redraw that structure the pi Bond now is between those two carbons and the lone pair has moved out here and remember that would be a negative formal charge so we've moved a set of dots by moving the pi Bond and that's what we've shown in that particular structure you place all the lone pairs on the structure you locate any lone pair that's in the alyc position and you draw the arrows to convert from one structure to the next try another okay so I'm going to start by drawing all the lone pairs on the electrons here and the oxygens have I don't see any formal charges so which means um you know the formal charges they're all going to have two sets of dots on the oxygens all right let's find the double bonds here's one do you see any lone pairs in the alyc position and the answer to that is no so these are localized electrons they are not in resonance with anything all right so that's off the plate here we have I'll just pick a different color go back to highlighter here is our double bond do you see any lone pair electrons in the alyc position and the answer to that is yes right there I do so we have an an opportunity to exhibit resonance right all righty so we're going to delocalize that electron remember one of these is localized and one of them will be delocalized one set I'm referring to because you can only move one pair of electrons in resonance not both so when we move this this set of electrons becomes a pi Bond the pi Bond becomes a set of electrons and we get on the other side of that Arrow here's where we had the original oxygen that was not in position to exhibit resonance here the oxygen will now have a single Bond three sets of dots in the negative formal charge down here the O lost a set of dots and moved to a double bond resulting in one 2 3 4 five assigned electrons so that's a positive formal charge we got it that matches our screen oops oh the USA let me silence my phone here is the second of our five patterns the alyc carbo cat ion only one curved arrow is needed to show resonance for for the carbo cat ion and again we're just moving the pi bond to become in the other position so that the next uh positive charge moves to the other side of the alyc carbon and if there's multiple bonds which are called conjugated then multiple contributors are possible notice that conjugated means alternating single double single double single double that's conjugated but only move one set of arrows only move one set of electrons at a time only move one set of electrons and that way you'll remember to get all the possible resonance structures so here we have a just to highlight here is your double bond and here is the ayic position of the positive charge that means in the the first movement I've moved the pi bond between carbon 1 and two and I've moved the positive sign to position three now I can see that this next piie Bond meets the criteria to this alyc position and I can move that then one more time I only move one piie Bond at a time so draw the possible resonance structures here so the first Pi bond that is in in the correct proximity to the carbo cation are those two and so remember I can move one at a time so this Pi bond is going to move towards the positive and what happens when I redraw the pi bond is now here and the positive is now here and this Bond then is still located in the same position now I meet a new criteria that this Pi bond is in the correct position the alic position for a second resonance structure so I can add yet another arrow and I'll move the pi Bond and then the positive charge results on the next Carbon Let's see this guy stayed in the same place double B moved so now this carbon results in the positive charge we had three possible resonance structures in that conjugated double bond conjugated means alternating double single double we got it third pattern says Al lone pair adjacent to a carbo cation here is a carbo C notice that's an empty p orbital it's an SP2 hybridized carbon which means that empty orbital has room for a bond so I can turn it from SP2 to sp3 with a resonant structure here is a empty p orbital so this carbon is SP2 it has room to accept another set of electrons another bond to become an sp3 carbon so recognize that the formal charges are affected obviously but you can move electrons into a carbon that has an empty orbital and that's all it's saying carbocations have empty orbital so you can move a pair of electrons to give that carbon a fourth Bond so what might this look like here is a nitrogen with one two assigned electrons 34 so now look if I take this and I draw a circle here's 1 2 3 4 5 six assigned electrons nitrogen lives in group five so that's where the negative one formal charge comes up so remember even though they're not there I know there's two sets of dots on that nitrogen and this right here is an empty orbital on that carbon so what does the resonance look like let me clean that up I guess just put on the double the uh pairs of electrons and then you can see I'm going to move one set of electrons to make a pi Bond nitrogen still has one lone pair it's leading off to another carbon it now has a double bond leading to a carbon like this actually it should be like this the double bond is between carbon like this that's better all right so what about formal charges the nitrogen who still lives in group five has one two three four five assigned electrons no formal charge this carbon still lives in group four four assigned electrons no formal charge so there's no positives or negatives to write how about this guy here's the oxygens and it's next to a positive carbo ions remember this is an empty orbital SP2 that tells me that a pair of electrons can form a pi Bond and when it does so there's one set of electrons left on it and now a double bond satisfying the carbon's four bonds that it requires but now oxygen has 1 2 3 4 5 assigned electrons it would like to have six so it's left with a plus one charge we got it the fourth pattern is a pi bond between atoms of different electr negativity I mentioned for example between carbon and oxygen oxygen of course is much more electron Rich leaving the carbon electron deficient that's the definition of the polar bond the electron distribution is being pulled towards the top so we have really two extremes and this really occurs but causing the pi bond to be unequally shared that Pi Bond can be distributed to create formal charges and we have no formal charges I would say this is the major contributor but this is a valid resonance structure so for example remember there's a pair of electrons on the nitrogen they're not there but we know they are because nitrogen has no formal charge as have five electrons assigned to it here we have a double bond assigned around that nitrogen which one is the more electronegative element between carbon and nitrogen well clearly nitrogen is electron Rich right and that makes this guy electron deficient the polarity of the bond is in this direction so in terms of the arrow the arrow would show the pi Bond going to the nitrogen the more electronegative element that would give us a single Bond on the nitrogen two sets of dots and there's that Bond leading out to that uh other carbon this gives us a carbo cat ion and a negative formal charge on the nitrogen again I would say no formal charges is the major contributor this is valid but due to formal charges it's a minor contributor to the resonant structure how about one more doing your homework for you notice this oxygen has two sets of electrons making it very electron rich and the polarity of the bond pulling that group towards it so that means in terms of resonance the pi Bond can collapse and form uh sets of dots on the oxygen 1 2 3 four five six giving us a single bond to oxygen giving us a negative formal charge on o and a positive formal charge on carbon these two are in resonance this would be a major contributor no formal charge this would be a minor contributor due to the formal charge the fifth and final pattern of Resident pattern recognition is the conjugated Pi Bonds in a ring and we happen to show Benzene Benzene is a six cited 1 2 3 4 5 6 alternating double single double single double single we call that conjugated and really it's as easy as moving them between let me just say right now they're between carbon one and two three and four five and six the double bonds are located between 1 and two 3 and four four five and six but now not by twisting the molecule or anything now notice that they're between two and three four and five and 6 to 1 so we didn't just simply twist the molecule and kind of shift it clockwise or counterclockwise we actually broke bonds and move Bonds in order to create the resonance structure draw the resonant structures including all loone pairs well for this particular molecule the oxygen require lone pairs the nitrogen needs a lone pair and this is the aromatic functional group called Benzene and it is really the only alternating conjugated system so when I think about redrawing this all I'm really going to do is to to shift the positions of the double bonds and redraw the molecule just with the positions of the double bonds shifted by one position clockwise or counterclockwise doesn't matter you'll get the same picture but everything else is identical and so when you do that you're just going to draw in those arrows as I had shown this is the answer key to your homework make sure you remember that the oxygen gets two sets of Lone pairs the nitrogen gets one set of lone pair and that you've just shifted the position of the double bonds they were between carbon 1 and two between carbon 3 and four and between carbon five and six now notice where they are they're between Carbon 2 and three four and five 6 to one and so you're just moving the position clockwise or counterclockwise and you can shift these with your arrows it's okay to go this direction and it's absolutely okay to go this direction Direction so just pick clockwise or counterclockwise when you're drawing your arrows in your homework so here's a summary of those five patterns that you need to work to memorize to recognize and be able to U manipulate in order to practice resonance structures we talked about an alyc lone pair the alyc position these are the two carbons in the double bond the lone pair must be one position away remember that this is a d localized pair of electrons and two sets would be localized you're only allowed to move one set of electrons we had an alyc carbocation remember this is an SP2 empty orbital so it will happily accept a pair of electrons a lone pair adjacent to a carbocation again this is an empty orbital it will happily accept a pair of electrons notice this I should stress two sets of arrows one Arrow one Arrow one Arrow if we have an electr negativity difference based on polarity and here in a cyclic structure such as Benzene to move all three of those bonds that are conjugated we ended up using three curved arrows and that's going to take a little bit of practice this is what I mean a lot of the homework here in Chapter 2 is about resonance structures so draw resonance structures for each of the following compounds so what pattern do you recognize here I have a a double bond and adjacent to it is another double bond so that double bond really has no place to go does it it has no positive no pair of electrons it doesn't have an empty orbital so this right here is a very localized double bond however this double bond based on polarity we recognize a pattern that says due to the electron distribution based on electro negativity difference I'm able to move this set of electrons which I put a Green Dot there and create a set of Dots and a negative formal charge on the oxygen so Pi bond is delocalized here I'm noticing a conjugated set of bonds and I notice it next to a positive SP2 carbo cat I so this is going to have a first set of arrows in which the the piie bond can shift over and when it does so this Bond stayed the same the pi Bond shifted here and the carbo cat I is now at Carbon 3 we now recognize that this Pi bond is then able to do the same thing it can shift forming a third resonance structure first Bond here next Bond whoops next bond is now here and the positive ended up at the very end so there were three structures where the alic position was a carbo cat ion again you're going to have lots of practice no worries well you know what it looks like I missed this position this double bond does indeed exhibit resonance structures for each of the following compounds I'm so glad I spun ahead because look now ah got it you probably are coming out of your seats ready this carbon I neglected to put a positive carbocation there it only has three bonds I so sorry I missed that oh my goodness embarrassing because now this alyc position this double bond does indeed meet the criteria of being next to a positive carbocation duh which means the pi Bond can then move and we would have here's the negative the pi Bond now is here and the positive charge has moved over here on that particular carbocation come on up Bubby so now there are indeed three structures for this one as well ah y so um the first structure didn't meet the criteria until I put a carbocation next to it and now it delocalized this piie Bond now meets the criteria of being delocalized because it's got a neighbor of a positive carbocation and therefore can shift positions so all three are valid resonance structures for both A and B having three valid structures when we assess resonance structures we look at the hybrid between them and decide which one of those really is the major contributor and which will be the minor contributing structures when multiple resonance structures can be drawn we know a blend of them all occurs is called a hybrid but typically not all the Resident structures contribute equally so let's go through some following rules and kind of list in order of importance to give us you know how to determine the most significant which I've been calling the major contributor which are less significant which we call the minor contributors rule one when assessing the best structure for residents the the one that is the major contributor to the blend of the hybrid the most significant resonance form have the greatest number of filled octets we want all atoms if possible to have an octet notice here for instance this carbon has an SP2 hybridization it has an empty orbital that empty orbital means that it only has six veence electrons or three bonds if I were to move that and as it did here move a pair of electrons look now this carbon has four bonds this oxygen has three Bonds in a lone pair that means all of them are obeying the octet rule that is a better structure if you obey the octet rule that's a major contributor rule one rule two in determining the major contributors of the hybrid structure of a resonance says the structure with fewer formal charges is more significant and I might have been saying that all along it says if I have zero on All Atoms for formal charge that would be the major contributor if I have formal charges on the nitrogen here is A+ one over here the nitrogen is a negative one it is a valid structure because all are obeying the octet rule and the nitrogens kind of balance one another out I have a plus one on one and the negative on the other so overall it's zero um and that's okay that's a major contributor that's a valid structure but what makes this one not valid is that it does not have a full full octet on this carbon it's an SP2 carbon so the first two structures have full octets the first one of course has no formal charge so bang that's the largest contributor the second obeys the octet with formal charges so okay that's that's good but the last one is a very minor contributor even though it's a valid resonance structure because it's obeying the way to move electrons it's going to be the few you know the minor contri distributor because it has an empty orbital on the carbon if the compound has an overall formal charge Focus only on drawing resonance that form and show the delocalization of the charge so I'm allowed to move a pair of electrons into a pi Bond I'm allowed to move a pi Bond and form a pair of electrons R I'm delocalizing the negative charge but what have I done to move to this this being a nonvalid structure again I'm seeing that it's an SP2 hybridized I'm seeing here that I have -2 overall on this oxygen here so I really Haven an accomplish much with this this last move by taking the set of electrons and moving this when I'm done is move this back up that's how I got this third structure um and doing so really accomplished breaking the octet rule for the carbon and I wouldn't want to do that so only two really major contributors and the following is insignificant because moving that last set of electrons back kind of made a a positive charge on the carbon bad move a structure with a negative charge on the more electron negative element will be significant the structure with a positive charge on the least electr negative is also valid so in other words if I had carbon as a negative formal charge compared to oxygen with a negative formal charge clearly oxygen is more electr negative and therefore it should exhibit the negative charge not the carbon here we have a positive sign on the oxygen and here we have a positive sign on the nitrogen now who's more electr negative between n and O remember in our periodic table c n o f as we move across Period 2 the closer you get to flor in the more electronegative you are and so we'd want the positive on the nitrogen Which is less Electro negative than the oxygen which is more Electro negative so the positive sign belongs on the least Electro negative the negative sign belongs on the more electronegative element that creates a more major contributor to the resonant structure if you had to rank the resonance structures from most to least significant in this example what might you say all right so they're all valid resonance structures as you're examining them I notice letter A places a negative formal charge on oxygen and letter B places a formal charge of positive on the nitrogen which is good because this is less electr netive this is more Electro negative in terms of oxygen compared to nitrogen nitrogen is less electr negative than oxygen so it should bear the formal charge of positive oxygen should bear the formal charge of negative how about letter B well I'm upset about letter B right off the bat cuz I have an empty orbital so I'm breaking the octet rule how about letter c now you know C looks nice doesn't it all octet rules I mean everybody has eight electrons that's good carbon has four bonds oxygen has two bonds two sets hydrogen Yep this is perfect no formal charge on anybody and the octet rule is being followed excuse me so I like C as the most followed by a because it it has good octet rule the octet rule is being followed except it has formal charges but that's okay because the negative is on the more Electro negative element so that's good and then the least would be letter B with that empty orbital and that's what we verified here the major contributor is letter C filled octet on all no formal charges at all number two would be letter the structure a filled octet on all formal charges negative on most electronegative element and the least minor contributor would be Part B because of that SP2 carbon only three bonds there and so that's kind of the homework part here is drawing the significant resonance forms for each compound below and indicate the major resonance so if I have a negative on the nitrogen means it has two sets of dots on it I can move a set of dots to become a pi Bond and when I do so I form one hydrogen leading out here it now has a double bond 1 2 3 4 five and then of course I can move these guys around and I can have the double bonds appearing this way so which is the major contributor 1 2 3 4 1 2 3 4 5 that's not a valid how did I know that's not valid because this here gives me five Bonds on this carbon so that's not valid what if we only moved the first Pi Bond and left the resonance structure as is here right so instead of these arrows did that accomplish anything the answer is no cuz this carbon still has five bonds so that's no good so the pair of electrons on the nitrogen really don't move they don't produce a valid structure so the only type of resonance structure that could occur for this is really just moving the conjugated pairs such as this and when we move the conjugated pairs we still end up with the same structure so those are equal valid structures how about the letter or the next example here 1 2 3 4 5 six double bond here's an N to an H all right so we have 1 two three I know that there's a lone pair on the nitrogen here's a positive sign here all right so I better give myself yeah I'm going to just redraw that so I have some room cyclic NH and just to remind ourselves this is where the double bond was and we had a positive sign directly here okay I just wanted to redraw and work this through we had a Dots here so notice that we have a double bond and the alyc position is a carbocation which means I can move this and the position then changes where the double or where the positive sign becomes comes so now the double bond is here and this becomes here now all of a sudden I have a double bond adjacent to a lone pair of electrons so we can move again and this particular structure says the Dots here this is an H there we go all right so we have an adjacent pair so that means this particular structure can move those dots to become a pi bond this is just where that is now notice these dots will never go this direction never ever because hydrogen will never form a double bond so this here will come in as a double bomb and this then will form a lone pair so the double bond now is here the double bond loan is here and that positive charge has been removed because we have filled the empty orbital this carbon who only had three bonds before now has four bonds remember there's a hydrogen there even though it's not drawn we know it's there so that carbon went from SP2 to sp3 therefore we got rid of the positive charge nitrogen has 1 2 3 four assigned electrons so the nitrogen should have a + one charge what's the most significant resonance structure and that's really what you're asking yourself so the first structure had a positive charge on the carbon the second structure has a positive charge on the carbon and no formal charge on the N or the H right that's just all zero so I'm going to say these two are equal contributors because all we really did was move the positive charge from one carbon to the next however in this structure what happened is that we moved the positive formal charge which was on the carbon right so this is on the carbon on structure one and two but carbon over here it's now on the nitrogen and that's the more electronegative element not the least so that would be the least contributing factor even though we ended up having an sp3 carbon we had a hydro we still had a nitrogen with a formal charge we'd rather have the positive formal charge on a carbon than we would on a nitrogen when we think about resonance and hybrid structures we're mentioning that they're really a hybrid of both so this is just a practice to really think about what ises it mean to be a blend of both structures draw a resonance hybrid for example for the following compound so this is really just challenging us to say I know that I can draw a new structure by moving the electrons and in doing so I've moved them I didn't just rotate the circle I moved them but as a hybrid I want to show them as a combination of both structures and that's where we're used these Dash lines to represent a delocalized bond so off often times for Benzene to represent the hybrid structure you'll just see it with a circle inside and that's telling you I want you to Invision this structure equal Bond length in all directions and the hybrid structure for the caroan ion would show an equal distance from Carbon 2 to carbon 3 as well as carbon uh 2: one so again you're just using a dash line to represent equal distribution we're heading to the home stretch looking at delocalized loan pairs recognize when it means to be localize which is not involved in Residence and the term delocalize which means they are involved in resonance and remember to be delocalized a lone pair of electrons must be adjacent to an atom with an unhybridized p orbital the lone pair on the nitrogen is delocalized because it's next door to a carbon with a p orbital these sets of electrons this carbon is SP2 hybridized right it has three electron domains three are Bond bonded Z unbonded 330 is trigonal planer that meets the criteria of being delocalized which means we can really set in these arrows as saying this Pi Bond becomes a set the lone pair can become a pi Bond and that resonance structure would give us a carbon leading up to an oxygen now with a negative formal charge leading down to the nitrogen which then donated a pi Bond leading r r and now of course um nitrogen as 1 2 3 4 this would become a positive charge now of course this would be the major contributor with no formal charges obeying the octet rule this would be a minor contributor but still a valid resonance structure so the lone pair of electrons on nitrogen are D localized remember these two sets on the oxygen they did not move so they're considered to be localized on oxygen it would only be the third set of electrons that could be delocalized as we move back to the original structure so in one structure that nitrogen is sp3 hybridized and the other indicates that nitrogen is an SP2 hybridized right so here is SP2 here would be sp3 which one is it and the answer is yes it's SP2 it is a delocalized set of electrons and therefore can be involved in resonance structures so to be delocalized the lone pair is involved in resonance if it is considered to be localized it is not involved in resonance so let's say for example here is a lone pair of of nitrogen on the nitrogen atom and we see that it is in the alyc position because here's a double bond and yet I really can't draw a valid structure because this is already an aromat compound it has that conjugated positive negative so this is not a valid structure since this ring is already aromatic the lone pair in this molecule cannot overlap to the adjacent so this lone pair ends up to be parallel to the axes instead of perpendicular to the axes so these are localized electrons