How can one predict whether a given complex ion will be square planar or tetrahedral when its coordination number is 4 using crystal field theory ?

Is it possible to theoretically predict this?


4 Answers 4


This is a really quick and dirty method since I don't know much about the calculations involved in crystal field theory. This method is based on some general observations I've made.

First of all, if your complex metal ion is not in group 8 ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) or it doesn't have a $d^8$ configuration (like $\ce{Rh(I)}$ or $\ce{Au(III)}$), it is usually tetrahedral (Assuming you already know that the coordination number is 4).

If your metal ion is in group 8 or has a $d^8$ configuration, look at the crystal field splitting diagram. Square planar complexes have a four tiered diagram (i.e. four different sets of orbitals with different energies).

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If it has a two tiered crystal field splitting diagram then it is tetrahedral.

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But this assumes you have the crystal field splitting diagram of the complex.

As to how you obtain these diagrams (the calculations involved), I don't know exactly how it's done for specific molecules.

Like I mentioned before, this is just a very basic way to distinguish between the two geometries. I'm sure the calculations involved to obtain these diagrams would be really complicated but would make more sense in some cases if you're looking for the actual reasons behind why some complexes are square planar while others are tetrahedral.


Predicting the structure of a complex depends on how much info you're given. If you only have the number of ligands in the complex or the molecular formula of the complex, then you can predict the structure based on this list here.

Then again, steric factors and electronic factors are also important in determining the structure and your prediction would be more accurate if you know what ligand(s) you're dealing with (for steric effects) and the electronic configuration and oxidation state of the central metal ion.

I don't know much about how exactly these factors influence the structure but in general, you only have to be worried about these if:

  1. the ligands are bulky or have some other special features
  2. certain special electronic configurations (I've read something about certain electronic configurations causing distortions in the structure especially in copper(II) complexes)
  • 6
    $\begingroup$ This is a good start. For the first row transition metals, you also need to know whether the electron configuration lends itself to high spin or low spin. The second and third row metals are always low spin. $\endgroup$
    – Ben Norris
    Apr 12, 2013 at 14:28
  • 1
    $\begingroup$ Supposedly , I am given a complex ion of the first period of transition elements and asked to predict its geometry , then how to do it ? Is there a way , to do this ? I am not being given experimental data or anything $\endgroup$
    – Sameer
    Apr 12, 2013 at 15:36
  • $\begingroup$ @BenNorris Thanks for mentioning that. I completely overlooked that point. $\endgroup$
    – kaliaden
    Apr 12, 2013 at 16:29
  • $\begingroup$ Seems there's a typo... Ni, Pd, Pt are in group 10. $\endgroup$
    – Shub
    Oct 3, 2022 at 13:42

There are all sorts of approximations available, but the short of it is that in general one cannot predict square planar vs tetrahedral without either doing experiments or doing some very detailed calculations.

Put another way, ligand field theories and the like are after the fact justifications of observations rather than predictions of observations.


Our teacher told us this trick to tell if complex is going to be square planar.
We are considering the fact that the coordination no. is 4.

If the metal has a $\ce{d^7}$, $\ce{d^8}$ or $\ce{d^9}$ configuration along with a strong field ligand or $\ce{d^4}$ with weak field ligand then complex will be square planar otherwise tetrahedral.

One exception i have come across for this is $\ce{[PtCl4]^2-}$ which is a square planar. You can read the explanation for it here.

  • 3
    $\begingroup$ How can pairing happen in $\ce{d^9}$ configuration? $\endgroup$
    – Yashas
    May 11, 2017 at 13:04

A method I use requires the memorization of the spectrochemical series, which arranges the ligands in increasing order of ligand strength. A strong ligand will almost always be able to cause pairing of electrons, producing a low spin, inner orbital complex, which in case of the coordination number 4, may correspond to the $dsp^2$ or $d^3s$ hybridization.

For instance,
$\ce{Ni^{+2}}$ produces a square planar complex with $\ce{CN-}$, as cyanide is a strong ligand. Nickel carbonyl, on the other hand, exists as a tetrahedral complex because nickel's oxidation is 0.

You shall be able to use this method most of the time, but in certain cases, there are some exceptions. For instance, Cobaltic ions behave differently than other $3d$ metal ions, and form a low spin complex with $\ce{H2O}$.

Good Luck!


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