For a complex compound with coordination number 4, say having square planar geometry, and say all the ligands attached to the central metal atom are different, then we know that we will get 3 geometrical isomers.

Now my doubt is like in organic chemistry if we had 4 different groups around a carbon atom, we still named those isomers as Z and E isomers (by Cahn Ingold Prelog principle) . Similarly why can't we name the isomers as Z and E for the case of complex compound with square planar arrangement having four different ligands attached to it?

That is why can't we name those 3 geometrical isomers as Z and E isomers of the complex compound by Cahn Ingold Prelog principle?

  • 1
    $\begingroup$ You have three isomers, but only two designations, Z and E. Furthermore, what about this sort of complex where $\ce{L^1} > \ce{L^2} \cdots$ in the CIP system? I'm sure there is a IUPAC way to name such complexes, but it's not going to be the Z*/*E system. $\endgroup$ Jan 3, 2017 at 12:10
  • $\begingroup$ Presumably we do not use $E$ and $Z$ because there are three isomers so we would need three labels. $\endgroup$
    – Ben Norris
    Jan 3, 2017 at 12:15
  • 4
    $\begingroup$ According to IUPAC, the priorities of the ligands in such nontetrahedral compounds are still based on the CIP rules; however, additional rules may be needed in some cases. A general method for the notation is described in IR-9.3 of the Red Book (2005) and in P-93.3 of the Blue Book (2013). $\endgroup$
    – user7951
    Jan 3, 2017 at 12:44
  • $\begingroup$ @BenNorris Can these 3 isomers defined as cis- and trans- isomers? $\endgroup$
    – Apurvium
    Dec 11, 2019 at 9:32

1 Answer 1


Part I

The CIP rules in fact are used to determine the priority. However, a different notation system has been developed for inorganic nomenclature of coordination compounds.

For the square planar geometry, SP-4-x descriptor (an instance of ‘polyhedral symbol’) is used, where x is a CIP-based priority number of the ligand in trans (opposite) position relative to the ligand with the highest priority number (1).

In $\ce{[Mabcd]}$ complexes (ligand priority a > b > c > d, priority numbers 1, 2, 3, 4, respectively), for the 3 possible isomers, the descriptors are (forgive me the use of left/right superscript/subscript typography to depict square planar geometry, for the sake of simplicity):

  • $\ce{^\style{font-weight:bold}{a}_cM^b_\style{font-weight:bold}{d}}$    SP-4-4
  • $\ce{^\style{font-weight:bold}{a}_dM^b_\style{font-weight:bold}{c}}$    SP-4-3
  • $\ce{^\style{font-weight:bold}{a}_dM^c_\style{font-weight:bold}{b}}$    SP-4-2

For $\ce{[Ma2b2]}$ (priority numbers 1, 1, 2, 2) and $\ce{[Ma2bc]}$ (priority numbers 1, 1, 2, 3) complexes, where only 2 isomers are possible, the cis/trans notation can be used as well. The E/Z notation from organic chemistry is not used at all for coordination compounds central atom configuration.


  • $\ce{^{\style{color:grey}{(1)}{ }I}_{\style{color:grey}{(3)}{ }Cl}Pt^{Br{ }\style{color:grey}{(2)}}_{F{ }\style{color:grey}{(4)}}}$

  • $\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }NH_3}Pt^{NH_3{ }\style{color:grey}{(2)}}_{Cl{ }\style{color:grey}{(1)}}}$

  • $\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(1)}{ }Cl}Pt^{NH_3{ }\style{color:grey}{(2)}}_{NH_3{ }\style{color:grey}{(2)}}}$

  • $\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }Ph_{3}P}Ir^{PPh_{3}{ }\style{color:grey}{(2)}}_{CO{ }\style{color:grey}{(3)}}}$
    Vaska's complex (a different IUPAC name descriptor claimed there is disputed)

(References: see Loong's comment below the question.)

Part II

Now, your question is self-contradictory. You would like to use E/Z (two) descriptors for three different isomers.

Maybe you wanted to ask something like
Why are there 3 geometrical isomers of square planar [Mabcd] complexes, but only 2 isomers of tetrasubstituted ethene C2abcd?
(Note the latter statement is not true)

It's because for $\style{font-weight:bold}{\ce{[Mabcd]}}$, the four different ligands are at the vertices (corners) of a square. Which can be abstracted as a ‘closed loop’, i.e. a cycle graph. The combinatorics can be abstracted to free circular permutations, $P'_{n}={1 \over 2} (n-1)!$,
for $n=4$: $P'_{4}=3$.

For $\style{font-weight:bold}{\ce{C2abcd}}$, the situation is somewhat different. This is a combination of positional isomers $\ce{xyC=Czw}$, where the set $\{x,y\}$ is $\{\text{a}, \text{combination}_2(\{\text{b},\text{c},\text{d}\})\}$, and $\{z,w\}$ is a complement. There are (concidentally?) 3 of them (${n \choose k} = {{n!} \over {k! (n-k)!}}, {3 \choose 2} = 3$):

  • $\ce{abC=Ccd}$
  • $\ce{acC=Cbd}$
  • $\ce{adC=Cbc}$

(Here the different sortings of substituents at each carbon does not matter.)

However, the isomerism one is talking about when using the E/Z descriptors applies to geometrical isomers of a single positional isomer (take e.g. $\ce{abC=Ccd}$). Now, the sorting of substituents at each carbon does matter. What makes the isomers different is matching vs non-matching sort order at each carbon:

  • $\ce{abC=Ccd}$    ≡    $\ce{baC=Cdc}$    (ascending-ascending, descending-descending – matching)
  • $\ce{abC=Cdc}$    ≡    $\ce{baC=Ccd}$    (ascending-descending, descending-ascending – non‑matching)

which are 2 possibilities, and correspond to E and Z configurations, respectively (provided that the lexicographical order corresponds with the CIP priority order).

So, the total number of C2abcd isomers is 6 (3 positional isomers, each has 2 geometrical isomers). One think of the alkene as an object similar to the square planar metal complex, where the C=C can be arranged in the center in two orientations, horizontal and vertical, hence 2× as many possibilities than ${\ce{[Mabcd]}}$.

(I hope exactly zero mathematicians and chemists were harmed during the reading.)

  • 1
    $\begingroup$ Can you give any reference to the priority list followed during the numbering of the groups? Is it the same as the CIP rule followed in organic chemistry? Also is the same rule valid for octahedral geometry(take [M(ABCDEF)] for example) $\endgroup$ Dec 1, 2019 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.