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I have basic knowledge of crystal field theory (CFT) and how to predict geometries in complexes with a coordination number of 4.

For $\ce{[Ni(NH3)4]^2+}$, we find that the metal center, $\ce{Ni^2+}$, has a $\ce{d^8}$ system. I believe that $\ce{NH3}$ should behave as a weak field ligand here so a high spin complex should be formed with tetrahedral geometry.

I am making this assumption because, to my knowledge, $\ce{NH3}$ behaves like a weak field ligand in hexaammine complexes of $\ce{Cr^2+}$, $\ce{Mn^2+}$, $\ce{Fe^2+}$ and $\ce{Co^2+}$ and high spin complexes are formed.

I am not sure about this, since $\ce{NH3}$ lies in the middle of the spectrochemical series (its a middle field ligand?). Is my analysis correct?

sushant_padha
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  • What shape do you think it would be if it were low spin? What shape is the platinum analogue? Why is it different? – Ian Bush Jul 18 '23 at 15:01
  • I think square planar for both? Because $\ce{Pt}$ belongs to the $\ce{5d}$ series so pairing of electrons in lower orbitals is energetically favorable in most cases. But how is that related to the complex in question? – sushant_padha Jul 18 '23 at 18:11
  • In both complexes the metals are d8. In both complexes they have 4 ammonia ligands. But the structure is different. So maybe the fact that the ligand is ammonia is not the main deciding factor between the two structures? Maybe it is the metal? – Ian Bush Jul 18 '23 at 18:15
  • You might also want to look at https://chemistry.stackexchange.com/questions/40880/why-is-pdcl42-square-planar-whereas-nicl42-is-tetrahedral – Ian Bush Jul 19 '23 at 09:34

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