Which of the following complexes are paramagnetic?

(a) [Mn(CN)₆]^3-

(b) [Zn(NH₃)₄]²⁺ (tetrahedral)

(c) [Fe(CN)₆]^4-

(d) [FeF₆]^4-

Video transcript

Welcome back everyone. Let's look at our next problem. It says among the given complexes below predict which will be paramagnetic. And then we have four different complexes listed by formula and we have answer choices. A one and two B one and three C two and three or D three and four. So let's recall that a paramagnetic com complex will be one which has unpaired electrons. The other option is diamagnetic in which all the electrons are paired up. So we need to look at our structures and determine how many electrons are present in that outer valence shell of our central metal atom and then determine how they fill in the D orbitals. Remembering that we're going to have our crystal field splitting. So we need to look at our ligands to determine will this be a weak or strong field. So let's start with number one in brackets with an overall two plus charge we have cr and then in parentheses, NH three subscript six. So our central metal atom is cr or chromium. So we need to determine how many D electrons this chromium has, it's a transition element. So it will have D electrons in its outer shell. So the just neutral chromium, we'll use our periodic table here has we have brackets argo and then it has a four s 1 3d 5 configuration. So a non uh regular configuration and we need to then figure out what is the charge on our chromium atom to find out how many electrons we have. So we know that the charge on ammonia is zero and our overall charge is plus two. So the charge on chromium must be plus two to give it that overall charge. So if we have chromium plus two, we're going to remove two electrons. So we'll have in brackets are gone, we'll remove that four s one and then one of our 3d electrons and get 3d 4. So we've got four D electrons to think about filling in our orbitals. So we have our split D orbitals with a two higher energy and a three lower energy. So we need to look at ammonia to see what kind of crystal field splitting will have. And ammonia is a strong field ligand. So we have a strong field splitting. That means we have high spin hi spin here. There's a larger separation between those orbitals or excuse me, I misspoke there, we have a strong field splitting. So we have low spin, there's a larger separation between our orbitals. So we will completely fill in the lower energy orbitals before adding any electrons to the upper orbitals. So strong field, low spin. As you can see, it's easy to get a little mixed up there. So we have our four D electrons to fill in, we're calling that we will completely fill in a lower energy. So 1231 in each orbital and the fourth electron makes a pair in the first of the de orbitals. So you can see we have two unpaired electrons here. So we have a paramagnetic complex there, we're supposed to predict which ones are paramagnetic. So number one is our answer choice has to have number one in it. So we're going to eliminate C which is two and three and eliminate D which is three and four A and B both have one in them. So wh we're through with our first complex. Now we move on to our second, I'm going to draw a little line to separate here. So our second complex, we have an overall charge of four minus within the brackets. We have N I and then CL six, we have a central nickel atom, we need to determine what charge it has. So we have overall charge of four minus, we have cl six, each chloride has a minus one charge. So that would be minus six coming from the chlorides overall charge of minus four. So our nickel must be plus two. So let's think about the structure of just plain old nickel that's going to be argon in brackets is our nearest noble gas and it will be four S two and 3d 8. And then we have nickel two plus. So we're going to remove two of those electrons, remove the s orbital electrons and have argon and brackets 3d 8. So we have eight D electrons to fill in chloride is a weak field ligand. So we will have high spin, there's not as much of a separation between those orbital levels. So we will fill in all of our orbitals, all five with a single electron before we start pairing them up. So we have eight D electrons. So we fill in 12345. And now we add the 2nd 6783 pairs on the bottom two unpaired on the top. And actually, if you remember, you can recall that from D eight to D 10, it doesn't matter what the field splitting is, it'll look the same. So we have two unpaired electrons. This one is paramagnetic as well. So number two must be paramagnetic and that's going to lead me to answer choice. A one and two, answer choice B is one and three, but we only have two paramagnetic complexes in all of our answer choices. So we know two should have been in there. So B can be eliminated. So just by elimination, we've got our answer. But we're going to be thorough and go through each one here. Let's move on to structure number three. And I'll draw a line to separate them. This one has an overall charge of four plus within the brackets we have PD. And then in parentheses, NH three subscript six ed is Paladin. Let's determine the charge on that. So our overall charge is four plus our ligands are ammonia, which is a neutral ligand. So all of the charge must come from the palladium. So it must be plus four. So let's think about our number of electrons uncharged palladium. We look at our periodic table has a structure of in brackets krypton. And then this has a non-standard electron configuration. So it's krypton and then four D 10 with no S electrons, but we have PD four plus. So we remove four electrons to get in brackets krypton and four D six. So we have six of the electrons to fill in it's jar D orbitals. Our ligand is ammonia. And as we saw in our first structure, this means we have a strong field splitting. So low spin, we're going to fill in the lower energy ones completely first six electrons. So we're going to put 123 in the bottom low spin. So the second set of three go in on the bottom as well. So we have six electrons all on the bottom 3d orbitals. So they're all paired up. So zero unpaired structure number three is diamagnetic, which we already knew since we've already eliminated our other answer choices. But again, we're just an answer of being thorough here. Finally, we have structure four, which is within brackets, an overall charge of two minus we have CD and then in parentheses, CN subscript four. And it says we have a tetrahedral structure. Uh That's because in general tetrahedral structures are high spin complexes. Let's find the charge on our a cadmium. Here, our overall charge is two minus. Our ligand is cyanide which has a negative one charge, there's four of them. So that's negative four from our four cyanide. So our academia must be positive two to give an overall charge of negative two, the uncharged cadmium from the periodic table, we have brackets crypton and then five S two, four D 10. So this has completely filled the orbitals. So cadmium two plus also krypton in brackets five S two electrons are taken away and we have four D 10. So we actually don't need to know if this is high spin or low spin because we have 10 D electrons. So we are filling in all of our D orbitals with zero unpaired electrons as we expect. So once again, the complexes which will be paramagnetic will be the two that have unpaired electrons. Just choice A one and two. See you in the next video.