Magnetic Properties of Complex Ions: Octahedral Complexes - Video Tutorials & Practice Problems

For octahedral complexes, Weak-Field Ligands create High-spin complexes and Strong-Field Ligands create Low-spin complexes.

For octahedral complexes. The type of Ln attached determines how electrons fill their de orbitals. We're going to say that if we have weak field ligands attached to our metal, this is gonna result in a small delta. If you have a small delta, that means that the orbitals are all going to be degenerate, they're gonna be considered being at the same energy. So electrons can fill the lower level of orbitals and then move up to the higher level. So this would give us a high spin complex. Typically high spin complexes result in paramagnetism for our species will be paramagnetic. If strong field ligands are attached, this would result in a large delta. So our orbitals are further apart. So there's more of an energy cost for our electrons to go from a lower energy state up to the higher energy orbitals. So they typically wouldn't do that. In that case, we create a low spin complex. You might also notice that I don't say anything about paramagnetism or diamagnetism because when we're dealing with strong field ligands attached and giving us an octal geometry, we actually need to check to see which type of magnetism our complex will possess, will it be diamagnetic or paramagnetic? So just keep this in mind when dealing with different types of lie Gans within octahedral geometries.

Example

Determine the spin and magnetism of the following complex ion in it, we have Cobalt connected to six ammonia. And the overall charge of a complex is two plus. Remember that ammonia is a neutral lien. Therefore, the charge is coming from the cobalt. That would mean that cobalt here is two plus. So step one, we find the number of D electrons in the transition metal cion Cobalt has an atomic number of 27. So when it's neutral, it has 27 electrons in its neutral state, its electron configuration will be argon four s 2 3d 7. But two plus means we've lost two electrons and we're gonna lose them from the highest shell number, meaning that we're gonna lose them from the four s orbital. So all we have now are seven electrons that are in the 3d orbitals. We have Phil according to Han's rule. Next, we identify the li in a strong field or weak field. It's a strong field leak ligand because the memory tool says that Larry cannot enter the neighborhood neighborhood stands for ammonia because it's a strong field lion. That means it's going to have a large delta. So there's a cost for electrons to try to go up to a higher tier in terms of the orbitals. So they're gonna try to avoid that this is gonna result in a low spin complex, meaning we fill in the lower energy orbitals first. So now that we know it's a low spin step three is we draw the octahedral crystal field splitting diagram. So here we have our three orbitals that lie um basically in between axis at the bottom and the ones that lie on or along the axis at the top, their difference is their delta. All right. Right. So I mean, actually this isn't drawn the best. Yeah. So now we have to fill in seven electrons and remember we're doing it by low spin. So we're gonna fill electrons in the split diagram and count the number of unpaired electrons. So that's step four low spin we fill in the lower energy orbitals first. So up up up come back around, down, down down at six. So far, remember we have seven electrons. So there's gonna be one more seven we can see here that there's one electron that's unpaired, meaning that this is a paramagnetic species. So we would say that this is a low spin and it is paramagnetic. These are the two things that we can say about this particular complex is

Determine the spin and number of unpaired electrons in the following complex ion: [Mn(en)₃]³⁺.

• A. high-spin; 1 unpaired electron
• B. low-spin; 1 unpaired electron
• C. high-spin; 2 unpaired electrons
• D. low-spin; 2 unpaired electrons

Determine the spin and number of unpaired electrons in the following complex ion: [Cd(H­₂O)₄]²⁺.

• A. high-spin; 2 unpaired electrons
• B. low-spin; 2 unpaired electrons
• C. high-spin; 0 unpaired electrons
• D. low-spin; 4 unpaired electrons

Which of the following complex ions is/are diamagnetic in nature?

I. [Mn(Br₄)]^2–                              II. [V(NO₂)₄]^4–                                 III. [Zn(NH₃)₄]²⁺                          IV. [Sc(H₂O)₆]³⁺

• A. I, II, and IV
• B. II and III
• C. I and II
• D. I, III, and IV
• E. iii and Iv