Crystal Field Theory Summary - Video Tutorials & Practice Problems
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1
concept
The greatest ligand-orbital interactions result in the greatest increase in energy.
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Now, when it comes to splitting patterns for de orbitals in complexes, it depends upon their geometries. We're gonna say orbitals with the strongest interactions with ligands have the greatest increase in energy. So here we have tetrahedral, optal and square planar. Now remember the difference in energy between our higher leveled orbitals and our lower leveled ones is defined as delta, our crystal field splitting energy. And we'd say that in relation to the three of them, the delta for tetrahedral would be the smallest octahedral is somewhere in the middle and square planar would have the highest delta. Now, if we look at tetrahedral here, remember the greatest interaction happens with the orbitals that lie in between our axes. That's why dxydyz and DXZ are at the top and then the ones on the bottom two on the bottom, those are the orbitals that lie on or along the axes, they have lower energy because they have lower interactions. So overall, we'd say that delta has low, well, tetrahedral has low delta for octahedral, for octahedral, the greatest interaction occurs with the orbitals that lie along the axes or on the axis. So those would be in the form of DX squared minus Y squared. And then DZ squared. And we'd say that the bottom three 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 delta one, as we can see that it lies somewhere in the middle, whether it's high or whether it's low, it really just depends. So we're gonna say it depends later on, we investigate more about octahedral species. We'll get into how do we determine if it's a low delta or high delta. But for right now, all we're gonna say it, 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 axis. So here we'd say that DX squared minus Y squared have the greatest interaction there. So it's at the top followed by DXY, which still has X and Y characteristics and then a little bit weird DZ squared, which is actually right in the middle of everything. And at the very bottom, Dyz and DXZ when it comes to square plane, and we'd say that it has a high delta, 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 is much greater, right? So just remember these key points when it comes to these three different complexes and the relationship with crystal field. Three.
2
example
Example
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2m
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Which one of the following complexes has 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 four ligands or four donor atoms. If we take a look at a, this can't be an answer because we have ethylene diamine three of them as are ligands, they are bent. So they have six donor atoms. This could not make a tetrahedral geometry. B is a possibility because in B, we have four ligands in the form of fluoride ions connected to the copper metal. So I will keep that for right now as a possibility C won't work either because just like a, we have six donor atoms in the form of six water molecules. D could also be a possibility because here we have four ligas, 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 would be three minus. Now, if we looked at the electron configuration of copper plus one, it would be argon. Remember here, neutral copper is argon four S 1 3d 10. It's an exception plus one means we've lost one electron. So we're gonna lose this four S one electron leaving us with 3d 10. 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 pt uh platinum two plus will be xenon 4 F-14 5 D eight. It loses its electrons in the six shell. So this is what we have left. So out of our two ions, one is ad 10 and one is ad eight. Remember if you're ad 10, then your geometry is tetrahedral. And if you're ad eight, 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.
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Problem
Problem
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
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