Which of the following complexes are diamagnetic?
(a) [Ni(H₂O)₆]²⁺
(b) [Co(CN)₆]^3-
(c) [HgI₄]^2- (tetrahedral)
(d) [Cu(NH₃)₄]²⁺ (square planar)

Question

Which of the following complexes are diamagnetic?

(a) [Ni(H₂O)₆]²⁺

(b) [Co(CN)₆]^3-

(c) [HgI₄]^2- (tetrahedral)

(d) [Cu(NH₃)₄]²⁺ (square planar)

Accepted Answer

Welcome, everybody. Here's our next problem. Consider the complexes given below, identify which are diamagnetic. We have four different complexes. We have a zinc with four water ligands, which is two plus. And our problem tells us it's tetrahedral. Number two, we have rhodium with six NH three ligands, an overall charge on the complex of three plus. Number three, we have platinum with four cyanide ligands overall charge on the complex two minus and the geometry of square planar given to us. And then number four, we have palladium with six bromide ligands overall charged on the complex two minus. And our answer choices are a one and two B, 12 and three, C 12 and four or D three and four. So the main thing we need to identify here is which complexes are diamagnetic and diamagnetic would mean there are no unpaired electrons. So we need to look at how our G orbitals are filled and are they lead the electrons fill paired or unpaired? Which of course, we'll need to think about what kind of crystal field splitting we have. So let's start with our complex. And number one, our zinc complex figuring out the oxidation seed of zinc is pretty easy. In this one, we have water as our ligand. So water is a neutral ligand, of course. So it has no charge and the overall charge on our complex is two plus. So therefore, our zinc must be zinc two plus. So with that in mind, let's figure out how many d electrons we're looking at neutral zinc or elemental zinc as an electron configuration of argon in brackets and then four s 2 3d 10. And so therefore, our zinc two plus which has lost two electrons will be argon and then 3d 10, it will lose its four S two electrons. Well, we don't actually need to look at our orbi orbital diagram here. We know when we have tetrahedral or square planar complexes, the splitting is a little bit different from our octahedral complexes, but we have 10 D electrons. So all our D orbitals are filled in this case. And therefore, we have no unpaired electrons in this one. So we know that number one must be one of our diamagnetic. We're gonna go ahead and highlight it in blue to keep track of which ones are diamagnetic. We can look at our answer choices and cross out choice D since number one is not included in choice D sometimes especially in these long problems when you have to work your way through multiple in uh information or multiple sort of problems. If you keep an eye on your multiple choice options. You can manage to arrive at the correct answer by elimination without having to work your way through every single item in the problems. So something to keep in mind. Now, let's look at complex number two. So number two is our rhodium complex. First, let's determine the charge on our rhodium or the oxidation number. Our ligand is ammonia, which is also neutral. Our overall charge on our ion a complex, excuse me is three plus. So our rhodium must be three plus elemental rhodium as an electron configuration of its closest noble gasping krypton. So, Kr and brackets and then five s one 48 it has one of those irregular arrangements. So the regular configurations, so we're going to remove three electrons, we'll take away that s electron and it will remove two more of the D electrons. So the electron configuration for Rh three plus will be krypton in brackets and four D six. So we have six electrons to work with, we have six of those ammonia ligands. So our geometry is octahedral. We need to think about the splitting that's going to occur. Our ligand is ammonia and ammonia is a strong field ligand. This means we have a large delta and therefore, it's going to be a low spin complex. So let's look at our octahedral d orbital splitting. We have our two higher energy levels, our three lower, we know that our delta is going to be large now, it doesn't look very large in my drawing, but we know it's large. So we're going to fill in our six electrons knowing that we will completely fill in the lower orbitals before jumping up to the higher ones. So we fill in 1231 in each lower level and then go back for 56 again, one in each lower level, we have six electrons all paired up in the three lower energy levels. So we have no unpaired electrons. So we'll go ahead and highlight that in blue as well. Number two also will be diamagnetic. And that means we can eliminate choice c since it does not include number two, so we're down to just two answer choices left. Now let's move on to complex number three. So now we have this platinum complex, it has four cyanide ligands, cyanide has a charge of negative one. So that means our cyanide ligands are CN four must have a charge of negative four. So total charge of negative four, our overall charge of our complex is negative two. So therefore our platinum must be platinum two plus. So let's look at our electron configuration for platinum or elemental platinum. We have xenon being the closest noble gas and we have six S one 4 F-14, we're up into our f orbitals and then five D nine. Again, that non or irregular electron configuration and we have lost two electrons to become PT two P. So that electron configuration will be xenon in brackets. And we're going to lose that one s electron and one of our d electrons to give a xenon 4 F-14, 5 D eight, we have only four ligands in this complex. And our problem tells us that we have a square planar geometry. So square planar has a very unusual splitting that we have to keep in mind different from the octahedral splitting. It has a large delta, but that delta is between the one highest level orbital, the X squared plus y squared orbital and then the four lower levels which we have two, one above the other and then the lowest two at the same level. But the key thing to remember is that our delta here is between the highest orbital energy energy level orbital and four lower ones. And that's where our splitting will occur. So we fill in the lower four before we jump up to the highest, we have eight electrons. So that means one electron in each of these lower four. And then we go back and put it in the second electron before we even look at the highest energy level. Well, since we have four lower level electrons, eight electrons total, we have all eight fitting in the lower levels. And so we have and all eight are paired. So once again, we have no unpaired electrons and we have a diamagnetic complex. And in fact, square planar geometry is quite common for complexes where the metal has ad eight electron configuration since it does then completely fill those four lower energy levels. So it's quite a stable arrangement. So again, now we have answer choice three, having no unpaired electrons. And just as I mentioned at the beginning, we can eliminate choice A because three is missing by process of elimination, we only have one answer choice left. And that's choice B 12 and three with no unpaired electrons. So if I run a test, I'm done, I'm picking that and moving on, we're on this tutorial video. So we'll be thorough and look at our fourth complex here, which we will assume is going to be paramagnetic. So let's just confirm that to be thorough, I will need to scroll down a bit just to have some space. My complex number four, we have this overall charge on our complex of two minus our ligand is six bromides which have a charge of negative one. So since we have six of them, the total charge is negative six from the ligands and we know our complex charge is negative two. So therefore, our palladium PD must be plus four. So it's lost four electrons. So let's look at elemental palladium, it's electron configuration, which is krypton as the nearest noble gas. And it has one of the exceptional configuration and just has four D 10, no S electrons. So we now have lost four electrons. They'll come out of those D orbitals. And so palladium plus four or I should say four plus is going to be krypton with four D six as our electron configuration. So let's think about our geometry and our splitting, we have six ligands. So we have octahedral geometry. We have to look at our ligand to determine what kind of splitting we have. And bromide is a weak filled ligand. So it will have a small delta and therefore a high spin. So we just look at our actual orbitals are two higher energy orbitals, our three lower ones and fill in our six electrons, remembering that we have a small delta. So 123 and then we jump up to the higher level because of that small delta 45 and then we have 1/6 electron. So a second electron goes down in one of the lower energy levels. So as you can see, we have four unpaired electrons, this is most definitely paramagnetic, not diamagnetic. So we're going to I like that or actually, we won't highlight it to avoid confusion. But we see as we expected that is paramagnetic, not diamagnetic, scroll back up. And we have confirmed that our correct answer. Choice is indeed choice. B the complexes that are diamagnetic. Our choice B 12 and three. See you in the next video.

Which of the following complexes are diamagnetic?
(a) [Ni(H₂O)₆]²⁺
(b) [Co(CN)₆]^3-
(c) [HgI₄]^2- (tetrahedral)
(d) [Cu(NH₃)₄]²⁺ (square planar)

Verified Solution

Key Concepts

Magnetism in Coordination Complexes

Crystal Field Theory

Ligand Field Strength

Which of the following complexes are diamagnetic?(a) [Ni(H2O)6]2+(b) [Co(CN)6]3-(c) [HgI4]2- (tetrahedral) (d) [Cu(NH3)4]2+ (square planar)

Verified Solution

Key Concepts

Magnetism in Coordination Complexes

Crystal Field Theory

Ligand Field Strength

Which of the following complexes are diamagnetic?
(a) [Ni(H₂O)₆]²⁺
(b) [Co(CN)₆]^3-
(c) [HgI₄]^2- (tetrahedral)
(d) [Cu(NH₃)₄]²⁺ (square planar)