Magnetic Properties of Complex Ions: Octahedral Complexes - Online Tutor, Practice Problems & Exam Prep

• In octahedral complexes, the type of ligand influences electron distribution in d orbitals.
• Weak field ligands cause a small splitting of d orbitals (small delta), leading to degenerate orbitals at similar energy levels.
• With small delta, electrons fill lower energy orbitals first, resulting in high spin complexes which are typically paramagnetic.
• Strong field ligands result in a large delta, creating a significant energy gap between lower and higher d orbitals.
• Due to the large delta, electrons are less likely to occupy higher energy orbitals, forming low spin complexes.
• The magnetism (paramagnetic or diamagnetic) of complexes with strong field ligands and octahedral geometry needs specific examination.

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

For octahedral complexes, the type of ligand attached determines how electrons fill their D orbitals. We're going to say that if we have weak field ligands attached to our metal, this is going to result in a small Δ. If you have a small Δ, that means that the orbitals are all going to be degenerate. They're going to 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. Tor species will be paramagnetic if strong field ligands are attached. This would result in a large Δ. 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'd 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 octahedral 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 ligands within octahedral geometries.

Example

Determine the spin and magnetism of the following complex ion. In it we have cobalt connected to six ammonias and the overall charge of the complex is 2 plus. Remember that ammonia is a neutral ligand, therefore the charge is coming from the cobalt. That would mean that cobalt here is 2 plus. So step one we find the number of D electrons in the transition metal cation.

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 4S²3D⁷. But two plus means we've lost two electrons and we're going to lose them from the highest shell number, meaning that we're going to lose them from the 4S orbital. So we have now are 7 electrons that are in the 3D orbitals we have filled according to Hun's rule.

Next we identified A ligand as 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 log in. 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 going to try to avoid that. This is going to 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 as we draw the octahedral crystal field splitting diagram. So here we have our three orbitals that lie basically in between axes at the bottom and the ones that lie on or along the axis at the top. Their difference is their delta. Alright, alright, so actually this isn't drawn the best. So now we have to fill in seven electrons. And remember we're doing it by low spin O. We're going to fill electrons in the split diagram and count the number of unpaired electrons.

So that's Step 4. 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 7 electrons, so there's going to be one more 7. We can see here that there's one electron that's unpaired, meaning that this is a pair of magnetic 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 ion.

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