Optical Isomerism of [Fe(H2O)6] Complex

[broegger]

Hi. I'm asked the following question:

What isomers does the complex $$[\text{Fe}(\text{H}_2\text{O})_6]$$ give rise to? Is it optically active?

My answer to the first question would be 'none': The 6 identical H2O molecules are arranged in an octahedral fashion around the central Fe-atom, so there are no asymmetries that could give rise to isomers. Is this correct? It seems to me that this question is phrased as if some isomers do exist (I'm not asked whether they exist or not).

The answer to the second question would of course be 'no', since there are also no optical isomerism.

Am I missing something?

 

[Gokul43201]

Your argument is insufficient. You could use essentially use the same argument to "prove" the optical inactivity of [Co(en)_3]^3+, but that would be wrong...wouldn't it?

While I'd still be inclined to agree with your result, I have never happened upon the bonding/geometry of aquo complexes myself, to say anything definitive. If there is some kind of hydrogen bonding between neighboring H2O ligands (which I think is unlikely), that might induce optical activity.

 

[broegger]

I see your point. If we assume that every H2O molecule occupies one ligand space and does not interact with it's neighbors -- would I then be correct? It's an introductory course, so I assume there's no pitfalls.

 

[Gokul43201]

I see your point. If we assume that every H2O molecule occupies one ligand space and does not interact with it's neighbors -- would I then be correct? It's an introductory course, so I assume there's no pitfalls.

I'm inclined to say yes, but I'd rather someone who's formally trained in this area weigh in. There may just happen to be some standard result (or exception to the rule) which is completely unobvious to deduce.

Do you have a link to a website talking about bonding/geometry in hexaquo complexes?

 

[broegger]

It's in the context of simple http://wwwchem.uwimona.edu.jm:1104/courses/CFT.html". I don't think there's any unobivious rules here, but I don't know...