In $\ce{[Cu(NH3)4]^2+}$, the $\ce{Cu^2+}$ ion has 9 electrons in the $\mathrm{3d}$ orbital with only one unpaired electron. How is it a square planar geometry? Where is that unpaired electron going?


3 Answers 3


It is very convenient to use crystal field theory to discuss this.

It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.

The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc.

Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate.

However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound.

This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.

  • $\begingroup$ Why should we use octahedral splitting system if its a square planar species ? $\endgroup$ Jul 23, 2015 at 13:44
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    $\begingroup$ @DhrubaBanerjee square planar species usually have enough place for two extra ligands. We do need to understand why it does not accept these ligands if available, so we have to consider octahedral splitting as well. It helps to draw the d-orbitals in octahedral splitting to understand why $d^8$ often (but not alwasy) square planar. $\endgroup$
    – permeakra
    Jul 23, 2015 at 14:14
  • $\begingroup$ Where does that unpaired lone electron get accomodated in such case ? $\endgroup$ Jul 23, 2015 at 14:20
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    $\begingroup$ @permeakra Okay, but where did that unpaired electron go? Because, if it remained in the $d_{x^2-y^2}$ orbital, then the ion would be $sp^3$ hybridised, but it is not. What is happening here? $\endgroup$
    – Aritra Das
    Dec 29, 2015 at 18:22
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    $\begingroup$ Since you are describing bonding in your answer, you are actually using ligand field theory as opposed to the older and a bit incomplete crystal field theory. $\endgroup$ Sep 8, 2016 at 7:26

The short answer is that $\ce{[Cu(NH3)4]^2+}$ does not exist; the compound you are observing is $\ce{[Cu(NH3)4(H2O)2]^2+}$. It is not square planar but a Jahn-Teller distorted octahedron. You can check out its structure in the image below.

tetraammin diaqua copper(II)

You can fill electrons into the energy diagram in a standard fashion. You will arrive at the $\mathrm{d}_{x^2-y^2}$ orbital for the unpaired electron — the one pointing towards the four ammine ligands.

Note that other copper(II) complexes may well be tetrahedral, as is the case for $\ce{[CuCl4]^2-}$. A combination of large ligand size around a small central metal and too much negative charge for a small species causes the tetrahedric structure to be preferred over $\ce{[CuCl6]^4-}$.

  • $\begingroup$ This is quite interesting. Can you tell me the source by any chance? Because I need to use this piece of information for a write-up. $\endgroup$
    – S R Maiti
    Feb 10, 2019 at 20:46

All Cupric complexes irrespective of the kind of ligands is always dsp2 square planar hybridization . Now the question is that why this is so? In tetrahedral complexes when Cu is in +2 state the one electron from 4s2 and one from 3d orbital gets removed and so it a 3d9 complex .Now from 3d9 complex one electron goes to 4p orbital and in this way it now becomes 3d8 and dsp2 complex.

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    $\begingroup$ Promotion from $3d$ to $4d$? Huh? $\endgroup$ Sep 6, 2016 at 10:08
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    $\begingroup$ Just don’t use hybridisation for transition metal complexes; it does more harm than good. $\endgroup$
    – Jan
    Sep 6, 2016 at 23:09
  • $\begingroup$ I have edited my answer and its 4p instead of 4d. $\endgroup$
    – Sommya
    Sep 8, 2016 at 5:27
  • $\begingroup$ Hybridisation theory for TM complexes is questionable at best. I will retract my downvote, but I really cannot recommend using hybridisation. $\endgroup$ Sep 8, 2016 at 5:34

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