the color and paramagnetism of complex ions and coordination compounds gonna be the topic of this lesson my name is chad and welcome to chad's prep where my goal is to take the stress out of learning science now in addition to high school and college science prep we also do mcat dat and oat prep as well i'll be sure to leave a link in the description below for where you can find those courses now this lesson's part of my new general chemistry playlist which is just about finished but if you'd still like to be notified every time i post these last few lessons then subscribe to the channel click the bell notification so let's start with color here and it turns out that complex ions and coordination compounds are very often associated with very bright vivid colors so and not all of them but the vast majority that you come across you know usually we associate these transition metals and transition metal complexes with these bright colors there's a very good reason for that now you've got to understand kind of the nature of color in a compound and what's really going on here is you're going to have an orbital and it's got to have an electron in it here down here a low energy orbital and then you're going to have a higher energy orbital so and it turns out when you absorb light of just the right energy you might end up promoting an electron from this lower energy orbital to this higher energy orbital well it turns out that the energy of that light is going to correspond perfectly to this gap in energy between these two orbitals well if you take a look at say the oxygen in this room well the oxygen in this room i say take a look at it but it's invisible and it's invisible for a reason if you look at the highest energy orbital that has electrons and oxygen and then the lowest energy orbital above that the gap is huge and it's so huge that the energy difference corresponds to the energy of ultraviolet light which your eyes and my eyes cannot see which is why we can't see oxygen in this case so if a compound is going to appear colored then it needs to absorb light that you and i can see and that's colored light visible light it needs to absorb light in the visible spectrum and unfortunately again this gap compared to oxygen is going to have to come down in a significant way that gap is just too big in say oxygen things of a sort well it turns out comparatively speaking then if you take a look at the gaps in energy in the d orbital splitting whether that be in an octahedral complex whether it be low spin or high spin in a tetrahedral complex or in a square planar complex if you look at some of these d orbital splittings that energy gap is much smaller than the one i just referenced in say oxygen or something like this and as a result it often corresponds to visible light and so visible light might get absorbed now one thing you have to realize though is that even though a certain color of visible light is going to get maximally absorbed that's not the color it's actually going to appear to you and i or something called the color wheel and if you absorb say you know red light well red light is on one side of the color wheel if you're absorbing that red light then you're going to reflect back most of everything else and the average of everything else would be on the opposite side of what we call the color wheel and i'll put it up on the board here an opposite of red typically something going to be more in the green spectrum so if it absorbs red light it's probably going to appear green in color to you and i and so i just want to make sure you realize that the color it appears to your eyes and the color of light that's actually being absorbed are not the same color we call them complementary colors and they're on opposite sides of that color wheel and so uh we will probably exclusively talk about the color of light being absorbed not the color of light at which it appears to your eye and i just want to make sure you realize that distinction here so all right so it turns out a larger gap in energy in these d orbital splittings is probably going to be shifted towards the violet end of the spectrum the highest energy visible light and the smaller gaps will be shifted towards the red end red end of the spectrum and i say violet but most of the time we just actually kind of call one end of the spectrum red and the other one blue which notice it goes blue indigo violet so we really should call it the violet end of the spectrum but for whatever reason we often refer to red shifted and blue shifted to you know depending on which end of the spectrum it shifted towards and things of a sort so just keep that in mind so but larger gaps uh absorbing light more towards the blue violet region so and then smaller gaps in energy between the d orbital splittings gonna be absorbing light closer to the red end of the spectrum all right so we may not know the exact color but at least we might be able to comparatively say it's more one way versus another more towards the red or more towards the blue violet if you will okay so way this works then requirement for being colored then so is you've got to be one of these deoval complexes and it doesn't matter if your octahedral tetrahedral are square planar but the key is you have to have an electron down in one of the lower sets and then you have to have an empty place for it to go one of these four spots in the octahedral or one of these six spots in the tetrahedral and the square plane is a little bit special if you recall for a square planar they only exist for d8 so these are always going to have eight electrons and so square planar we expect them to be colored all the time because they're always going to have at least one electron in this case eight electrons down here in one of these lower sets and they're always going to have an empty spot in this case it's got two empty spots in that highest energy orbital in the orbital splittings and so for a square planer complex yes we expect them to be colored now for octahedral so here with one electron in the d orbital yeah i've got an energy uh electron in one of these low energy orbitals and i've got four empty spots up here to which i could go to and so if it absorbs light of just this energy difference this same energy as the crystal field splitting energy well then it's going to absorb visible light in all likelihood and appear colored to the naked eye so when would we expect maybe an octahedral or tetrahedral complex not to be colored well i'll give you an example so if you have a metal ion that is d0 no d electrons well then there's no electrons down here or here and as a result you don't have any electrons that could get promoted up to the higher set of d orbitals so if you're d zero no guarantee you're going to be colored we don't necessarily expect to be colored so other side of the coin is going to be d 10. and so d10 whether you fill it in low spin or high spin in this case i'll do it with the octahedral here and then the tetrahedral here and notice in either case i do have electrons now down in these lower energy sets of orbitals but i don't have a place for it to get promoted in the upper set because they're full as well and so for d10 no reason it should be absorbing light corresponding to the crystal field splitting energy no reason we'd expect it to be colored now it could be colored for other reasons it turns out this d orbital splitting is not the only reason so that compounds can be colored and if you take organic chemistry you'll learn about pi electrons and things of this sort so however in this chapter this is the principal reason why things might appear to be colored and so if you have d one two three four five six seven eight or nine we should expect that compound to be colored is it guaranteed no is it pretty likely yes but if you're d0 or t10 we probably are going to expect it to be colorless and not be colored is that guaranteed well no but again it's just part of the trend really high likelihood that something with no d electrons or 10 d electrons is going to end up being colorless all right so that's color now we've got to talk about what referred to as paramagnetism and diamagnetism empty these out a little bit and so these are going to be different than ferromagnetism kind of the magnetism we associate with iron so and that deals with you know little areas of iron where you've got spins kind of aligned and things of this sort and and i'm not going to go there but this is related to spin but it's going to be the spins typically of unpaired electrons so it turns out electrons have a spin associated with them so and we like to say either spin up or spin down and that's when we write electrons you know we like to represent that with an up arrow or a down arrow but there are two opposite spins so that and they are exactly opposite in magnitude as well and so it turns out when you pair up electrons their spins cancel each other out and so to have an overall spin associated with an atom here you've got to have unpaired electrons and it turns out that leads to a state we call paramagnetism we say that that substance is paramagnetic and that's that's the requirement you just gotta have unpaired electrons and it turns out the more unpaired electrons you have the more paramagnetic it becomes what does this ultimately means in terms of magnetism well it means that if you have one of these lovely complexes or metal ions or whatever moving through a magnetic field it's going to experience a slight attraction as a result and the more paramagnetic it is the more of that attraction it's going to feel now on the other hand what if all the electrons are all paired up well if the electrons are all paired up we refer to as being diamagnetic instead and in this case instead of experiencing a slight attraction it's going to experience a very very very very very slight repulsion instead so but because all the electrons are paired up there's no net spin and that's why it's going to end up not having this paramagnetism associated with it and so you're going to have to look at the number of d electrons and you have to fill them into one of these diagrams depending on what kind of complex you're told you have and just figure out do i have unpaired electrons yes or no if yes well then that's going to be a paramagnetic complex and if no well then that's going to be a diamagnetic complex and so by simply looking at the number of d electrons now and how they might fill in on the type of complex you have we now have a way of evaluating whether it's colored or whether it's going to be paramagnetic or diamagnetic and those are the kind of questions we're going to answer here let's take a look all right so the first we're going to take a look at here is sc3 plus and first order of businesses you just have to figure out how many d electrons we're dealing with and so scandium is going to look like argon 4s to 3d1 so if you recall in the crystal field theory lesson we went right back over we started off by figuring out electron configurations reviewing it a little bit so we could figure out these number of d electrons and this is definitely one of the examples we went over so this is scandium but scandium three plus we're going to lose three electrons well you lose the s's before the d's but that's two that we have to lose for the s's and then 1d and all we're going to have left with is just plain old argon and what this ultimately means is that scandium does not have any d electrons at all and so in this case if you had to choose would you expect it more likely to be colored or colorless well with no d electrons no electrons that get promoted from the lower set to the upper set we'd expect it to be colorless and then paramagnetic or diamagnetic well in this case if you're isoelectronic to argon a noble gas where you've got a filled octet in that last shell and stuff like that all the electrons are going to be paired and so you're going to be i can spell this diamagnetic all right so that's sc3 plus moving on to copper plus and recall that copper is one of those lovely exceptions and if we get the electron configuration now for copper argon and instead of 4s2 you've got to remember that argon i'm sorry that copper is argon 4s13d10 but again we don't have copper we have copper with a plus one charge we've got to remove one electron and in this case you gotta remember that you remove electrons from the 4s before the 3d and so we're going to lose that one electron right there and we're just going to have argon 3d 10 but the big deal here is that we have 10 d electrons that's what we really need to figure out well with 10 d electrons these are all full so notice i didn't tell you if it was octahedral or tetrahedral with 10 d electrons you'd know it's not square planar because this only exists for d8 complexes so but with 10 d electrons it didn't matter if you fill in the octahedral the tetrahedral they're going to be full either way and if they're full either way well then there's no empty spots up here for the electrons down here to get promoted to and so if i depict if it's more likely to be colored or colorless i'm going to pick colorless and also if we fill these in again you'll see that regardless of whether it's octahedral or tetrahedral they're all paired up all of them and if there's no unpaired electrons then we're going to expect it to be diamagnetic all right moving on to fe 2 plus here and so fe2 plus and notice here i specified its octahedral low spin and notice i didn't specify that on the last ones here because with either zero electrons or 10 electrons it wouldn't make a difference but fe2 plus it turns out is going to end up being d6 and it's going to totally make a difference how you fill them in whether it's tetrahedral or whether it's octahedral and low spin or octahedral and high spin but with 60 electrons we'll find out again it can't be square planar so it but three options so i had to tell you which of those three it was going to be so let's figure this out though we've got again iron is argon 4s2 3d6 but we need to lose two electrons and so again you remove the s's before the ds it's just going to be argon 3d6 and so once again the key here is that we've got six d electrons that's the important part and now we've got to look and say okay because there's 60 electrons it's low spin octahedral let's fill those in let's take a look at what that looks like here let's empty these back up okay so in this case six electrons low spin means we fill up everything down low before we ever go up high so this means we have a very large crystal field splitting energy and so it's easier to pair the electrons up down low than to bump them up up high so in this case we filled in those 60 electrons and we can see that yeah we'd totally expect this thing to be colored but again typically d1 to d9 you expect those things to be colored now some of you might learn an exception for things that have five d electrons and they're all unpaired and so for one to promote it would have to switch its spin which is a rare event and if you learn that well great then you're probably on the hook of that for that but most of the rest of you you're probably just going to learn about the the two extremes where if you're d0 or d10 probably colorless d1 through d9 colored if you've learned that d5 exception i'll leave that to you but not something most students are normally presented with at this level okay so in this case because it's d6 if i had to pick color to colorless i'm definitely picking colored got electrons down there that could get promoted up with the absorption of visible light so we'd expect to be colored and then as far as paramagnetic dye magnetic i can see that all the electrons are paired all of them and so it's going to end up being dia magnetic all right so our next example here is fe2 plus so but instead of being octahedral and low spin like the last example it's now going to be octahedral and high spin well we already figured out that fe2 plus was argon 3 d6 we lost the two 4s electrons to become 2 plus so same as the last example we did so big thing again here is that it's 60 electrons and having 60 electrons so we'd expect it to be colored rather than colorless so if we fill in those in a high spin fashion here we'll see why so filling in 60 electrons so one two three and if we fill them in high spin you go up high before pairing up anything down low so four five and then six and so now we've got four electrons down here and i still have two empty spots up here to which they could go and so yes we'd expect it to be colored and again usually d1 to d9 we expect to be colored now as far as paramagnetic or diamagnetic we can see that we've got one two three four unpaired electrons this thing is very paramagnetic as a result so fe2 plus makes a really good comparison in this regard because as we saw with the fe2 plus low spin all the electrons repaired and it was diamagnetic but with the fe2 plus high spin so we had four unpaired electrons and ended up being paramagnetic instead all right cr3 plus and so if we take a look at cr3 plus now and erase our electrons here from the last example before we get started so but again you gotta remember that chromium is one of your exceptions here and so chromium is electron configuration is argon 4 s1 3 d5 so instead of s2 d4 it's s1 d5 promotes one from the 4s to the 3d to get them both half filled here and now we got to lose three electrons well we're going to lose that 4s electron first and then we've got to take away two more to become three d three and notice this didn't say if it was low spin or high spin it's because it didn't have to you've got to have between four and seven uh inclusive d electrons to have distinct low and high spin cases so complexes and so this case with d3 whether you filled it in the low spin away or hyphen weight it's going to look exactly the same and so it doesn't make a difference so i didn't have to specify and so in this case we fill in 3d electrons one two three we see what that's going to look like and whether this had been filled in the low spin way or the high spin weight looks exactly the same for those first three electrons again that's why it doesn't matter but again having d electrons but not being full we expect it to be colored we definitely have electrons in that lower set that can be promoted up to the higher set empty spots so and then we see that we've got three unpaired electrons and so we'd expect it to be paramagnetic all right so last couple examples here on this question uh so in this case we're giving an entire complex here and you've got to figure out well okay there's six fluorines there they're monodentate ligands this has got a coordination number of six so it's got to be octahedral didn't tell you that it was octahedral but from the formula of the complex you can deduce that it's octahedral but in this case instead of saying high spinner low spin i told you that f minus is weak field okay so this is a little more challenging question a little more things to figure out here well first off we gotta figure out what's the oxidation state of fe here well in this case the fluorine the fluoro ligands are minus one each and there's six of them for a total of minus six so if the whole complex is minus three then the iron here has to be in the plus three oxidation state and so if we take a look at fe plus three recall that iron again is argon 4s2 3d6 but if we're at fe3 plus we got to lose three electrons the first two we lose are the four s's and then we'll lose one of the three d's and so in this case the big key is that this has five d electrons okay now again instead of being told it's low spin or high spin you're told that it is weak field so you got to recall that weak field corresponds to a small crystal field splitting energy which corresponds to a high spin so weak field goes with high spin strong field goes with low spin so when you see weak field you're supposed to think oh that's going to be high spin and so high spin again means we go up high before we pair anything up and so there's your d5 and we got five unpaired electrons and we can definitely see in this case that that it's likely to be colored for most of you however it's not a given here that it's going to be colored it turns out so but i'm going to say colored because that's how most of you are going to have this presented so and in this case though with five unpaired electrons it's definitely going to be highly paramagnetic all right last example here is nickel two plus in a square planar complex and in this case if they tell you square planar you should automatically know that it's going to be d8 so but if we figure that out so nickel is usually argon 4s2 3d8 and then we lose two electrons to become nickel two plus and lose the four s's and yeah it's just argon 3d8 so but square planar has to be a d8 and it has to fill in just like this for a square planar and so yes you're going to expect every square planar complex to be colored because you have electrons down here and you have an empty spot right up here it could go to so but also if you fill in eight electrons on the square planar splitting pattern they're all going to get paired up every time and so this guy's going to be diamagnetic all right so the last question we're going to deal with is which of these two lovely complexes more likely to absorb blue light all right so if we take a look here uh in this case with fecn64 minus you gotta realize that you've got iron in the plus two oxidation state so each of the cyanides are minus one and so iron must be plus two to get an overall minus four charge and then fe h2o6 the aquas are neutral so if it's going to be plus two it's all due to the iron he's plus two so we're dealing with fe2 plus in both cases and so in this case the only way you can answer this question is if you are familiar with the spectrochemical series and in this case it's on the handout i'm expecting you to look at this i would just not ask this blindly i would provide you with a brief spectrochemical series before asking you this question and so the thing you got to look at here is that cyanide is the highest ligand on the the brief spectrochemical series i give you which means it's going to result in the largest splitting crystal field splitting energy here between those whereas water is lower and so this crystal field splitting energy associated with this complex is going to be smaller now in this case i never said if it's high spin or low spin so because i'm largely not going to care so if we just fill this in say one two three four five six one two three four five six so whether i fill them both in low spin so well then the energy that's going to get absorbed is again the crystal field splitting energy the energy of the light that's going to absorb it's going to correspond to that energy in either case so and had i filled say this one in like well maybe this one's high spin because it's a smaller crystal field splitting energy maybe that's true and it's still not going to matter because whether i go one two three four five six so the energy of the light that's gonna get absorbed is still going to correspond to exactly the same crystal field splitting energy and so whether these are low spin or high spin in this example does not matter it's totally irrelevant so the big thing you're supposed to take away is if one of these is going to absorb blue light well blue again is at the high end of the visible light spectrum so i want to pick the largest crystal field splitting energy the energy of the light absorbs corresponds to that crystal field splitting energy so and in that case therefore i would pick fecn64 minus to be closer to the blue end of the spectrum for the light absorbs i'd expect feh 206 to be closer to the red end of the spectrum for the light being absorbed again i don't know that this is actually going to be blue light being absorbed and i don't know that this is actually going to be red light being absorbed it's just this one's closer to being at the blue end of the spectrum and this one should be closer to being at the red end of the spectrum and so oftentimes we just define the the visible spectrum by its extremes the the blue end at the higher energy and the red end at the lower energy that sort of thing cool and that's exactly how that works with using that spectrochemical series now if you found this lesson helpful a like and a comment let me know are pretty much the best things you can do to support the channel and i've got one chapter left in this playlist it's going to be an introduction to organic chemistry and not all of you are going to get there so a lot of classes get to the end of semester and they just don't have time and some just even preemptively cut it off before we can get there they don't even plan to cover it so but for some of you that are going to include that chapter i've covered it uh next week so if you want to be notified still subscribe to the channel and hit that bell notification so uh but i'm not gonna wait too long before getting going on my next playlist so if you have again a choice between biochemistry redoing physics or doing a freshman bio course let me know in the comment section i'll take your vote quite seriously i'm not just trying to get comments here and if you're studying for your final exams take a look at my general chemistry master course that includes final exam rapid reviews practice final exams if you're trying to do things quickly and efficiently free trial is available i'll leave a link in the description below happy studying