Crystal Field Theory Summary - Online Tutor, Practice Problems & Exam Prep

• Crystal field splitting energy (\
• ) varies based on the geometry of the complex: tetrahedral has the lowest, octahedral is intermediate, and square planar has the highest.
• In tetrahedral complexes, the orbitals dxy, dyz, and dxz have higher energy due to greater interaction with ligands, while orbitals on or along the axis have lower energy.
• For octahedral complexes, orbitals along the axes (dx^2-y^2, dz^2) have higher energy, and those between the axes have lower energy; the exact magnitude of depends on additional factors.
• Square planar complexes exhibit the most complex splitting, with dx^2-y^2 and dxy having the highest energy due to strong interactions on and between the x and y axes, while dyz and dxz have the lowest energy.
• The difference in energy levels in square planar complexes is significantly greater compared to tetrahedral and octahedral, indicating a high .

The greatest ligand-orbital interactions result in the greatest increase in energy.

Now, when it comes to splitting patterns for D orbitals in complexes, it depends upon their geometries. We're going to say orbitals with the strongest interactions with ligands have the greatest increase in energy. So here we have tetrahedral, octahedral, and square planar. Now remember, the difference in energy between our higher leveled orbitals and our lower leveled ones is defined as Δ, our crystal field splitting energy. And we'd say that in relation to the three of them, the Δ for tetrahedral would be the smallest, octahedral lies somewhere in the middle, and square planar would have the highest Δ.

Now if we look at tetrahedral here, remember the greatest interaction happens with the orbitals that lie in between our axes. That's why d_xy, d_yz, and d_xz are at the top and then the ones on the bottom of the two on the bottom. Those are the orbitals that lie on or along the axis. They have lower energy because they have lower interactions. So overall we'd say that Δ has low well. Tetrahedral has low Δ for octahedral.

For octahedral, the greatest interaction occurs with the orbitals that lie along the axis or on the axis. So those would be in the form of d_x²-y² and then d_z². And we'd say that the bottom 3 again are the ones that lie in between the axes. They have less interaction and therefore have lower energy. And when it comes to the D1, as we can see that it lies somewhere in the middle. Whether it's high or that it's low, it really just depends. So we're going to say it depends.

Later on, we investigate more about octahedral species we'll get into how do we determine if it's a low Δ or a high Δ? But for right now, all we're going to say that it just depends. For square planar, this one has the most complex, difficult splitting pattern. Remember here the interactions are greatest on and in between axes that involve the X and Y axes. Here we'd say that d_x²-y² have the greatest interaction there, so it's at the top, followed by d_xy, which still has X and Y characteristics, and then a little bit weird d_z², which is actually right in the middle of everything and at the very bottom d_yz and d_xz.

When it comes to square planar, we'd say that it has a high Δ. We can see how this stacks up much higher and longer than the other ones, so the difference between the higher energy levels and the lower ones we can see as much greater, right? So just remember these key points when it comes to these three different complexes and the relationship with Crystal Field.

Example

Which one of the following complexes have the smallest crystal field splitting energy? Now remember that belongs to tetrahedral complexes, so out of the choices we have to identify which one has this type of geometry. So in order to be a tetrahedral geometry, you need to have 4 ligands or 4 donor atoms.

If we take a look at a this can't be an answer because we have ethylenediamine, three of them, as our ligands. They are bidented, so they have 6 donor atoms. This could not make a tetrahedral geometry. B is a possibility because in B we have 4 ligands in the form of fluoride ions connected to the copper metal O. We'll keep that for right now as a possibility.

C won't work either because just like A we have 6 donor atoms in the form of 6 water molecules. D could also be a possibility because here we have 4 ligands. We have two ammonia molecules and two chloride ions. So let's figure out which one is it B or D?

For B we'd have copper one ion because the copper one ion is combating these four fluoride ions. But overall your charge will be 3 minus. Now if we looked at the electron configuration of copper plus one, it would be argon. Remember here neutral copper is argon 4S¹3 D¹⁰. It's an exception. Plus one means we've lost one electron, so we're going to lose this 4S¹ electron, leaving us with 3D¹⁰.

If we go to D, we have platinum here. Platinum would have to be two plus because the chlorides together are two minus. Ammonia is neutral so it doesn't contribute to charge. That's why overall this has no charge. So here platinum, its electron configuration will platinum 2 plus will be xenon 4, F¹⁴-5 D⁸. It loses its electrons in the 6th shell, so this is what we have left.

So out of our two ions, one is AD¹⁰ and one is AD⁸. Remember, if you're AD¹⁰ then your geometry is tetrahedral, and if you're AD⁸ your geometry is a square planar. Remember, square planar has the highest crystal field splitting energy, so this can't be the answer we're looking for, the smallest which was tetrahedral geometry. So B is our final answer.

What is the correct order when the following complexes are arranged in ascending order of the Δ values?

i) [Mn(NO₂)₆]^4–               ii) [Ni(CN)₄]^2–                 iii) [Zn(OH)₄]^2–

• A. i < ii < iii
• B. ii < i < iii
• C. ii < iii < i
• D. iii < i < ii