(A wrong question?) Does Ni2+ have different t2g eg configuration for wfl and sfl?

  • #1
Gourab_chill
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Homework Statement:
Does Ni2+ have different t2g eg configuration for wfl and sfl?
Relevant Equations:
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I was expecting Ni2+ to be present also in the answer as it can give dsp2 and sp3 configuration.
[Ni(CN)4]2- and [Ni(NH3)4]2+ have dsp2 and sp3 respectively, right?
 
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  • #2
TeethWhitener
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Two things. First, the labels t2g and eg only apply to octahedral geometries. Are your nickel compounds octahedral?

Second, does [Ni(NH3)4]2+ exist on its own? I'm familiar with the hexaammine and the mixed ammine with other ligands (all octahedral). I can't say I've ever encountered a tetraammine nickel compound.
 
  • #3
Gourab_chill
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Two things. First, the labels t2g and eg only apply to octahedral geometries. Are your nickel compounds octahedral?

Second, does [Ni(NH3)4]2+ exist on its own? I'm familiar with the hexaammine and the mixed ammine with other ligands (all octahedral). I can't say I've ever encountered a tetraammine nickel compound.

Well so it means we are only considering octahedral compounds here? Then it is always sp3d2 for Ni2+ compounds.
Then what are the energy levels involved in tetrahedral geometry, if they are not t2g and eg?
I did search for [Ni(NH3)4]2+ and found that people asking questions on whether it is para or diamagnetic, it's hybridization etc. Are you sure it doesn't exist?
 
  • #4
TeethWhitener
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I did the same search and mostly found references to homework problems (not actual scientific papers). I did find one paper on a tetraammine nickel compound with a complicated molybdate counterion. The IR spectrum was interpreted as having two short and two long Ni-N bonds. I suspect most tetraammine nickel (II) compounds are probably hexacoordinate (octahedral/JT distorted) in reality.

Also, yes generally these discussions only happen in the context of octahedral complexes, since the ligand field splitting in tetrahedral complexes is always too small to affect spin states.
 
  • #5
TeethWhitener
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I should clarify: octahedral compounds will have different electronic structures from tetrahedral and square planar compounds. So a square planar nickel compound very well may have a different spin configuration from an octahedral compound. But for the purposes of this particular question, it’s pretty clear (at least to me) that they were asking about octahedral complexes alone.

Also, generally for tetrahedral complexes, t2 and e are more accurate labels reflecting the group symmetry (there is no gerade symmetry in the complex). This is moot for nickel square planar complexes, which do not have the 3-2/2-3 d-orbital splitting characteristic of octahedral or tetrahedral geometries.
 
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  • #6
Gourab_chill
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But for the purposes of this particular question, it’s pretty clear (at least to me) that they were asking about octahedral complexes alone.

Yes, I missed out that point earlier. I need to strengthen my concepts on CFT :)
 

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