6
$\begingroup$

Ionization isomerism is defined as complexes which have same empirical formula but ionize to different ions in solution.

Consider the complex $\ce{[Pt(NH3)4Cl2]I2}$. The other possible ionization isomers are: $\ce{[Pt(NH3)4I2]Cl2}$ and $\ce{[Pt(NH3)4ICl]ICl}$.

But I have not come across the complexes similar to the last one with two different anions outside the coordination sphere. So, do these sort of complexes exist?

Moreover, what can be the IUPAC name for this compound?

tetraamminechloridoiodoplatinum(IV) chlorineiodine seems wrong.

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ -iodido-, -chloride iodide. $\endgroup$
    – Jan
    Nov 1, 2019 at 9:22

1 Answer 1

Reset to default
2
+25
$\begingroup$

I think it should be -

tetraamminechloridoiodoplatinum(IV) chloride iodide.

The rules of the naming of counter ions are pretty much the same as in general Inorganic Chemistry. I do know of a compound AgClI - named as Silver chloride iodide. So, I think the counter ions here - chloride and iodide should be named as by the nomenclature rules of anions i.e with an -ide suffix and alphabetically if there is the presence of more than one anion

Proof of Reference - Silver chloride iodide (NCBI PubChem)

EDIT: My teacher did give an example of it being a valid complex. I think it is not generally stated in the preliminary books of this topic to avoid confusion among students.I imagine it would be present in some more in-depth books of this topic.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ OK... but my original query was... Is this complex possible? Seems affirmative to me. So why there are no such examples in the books? $\endgroup$
    – Apurvium
    Nov 6, 2019 at 16:21
  • $\begingroup$ @Apurvium I think its a matter of the level of the books. The book I mentioned was of STD 12 level-which is the first time Coordination compound is taught.I think such examples would be found in some higher level books. $\endgroup$
    – Sristy
    Dec 3, 2019 at 19:10

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.