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I was solving a question where I was asked to identify the reagents given in the question which could react with a compound provided.

The first compound was methanoic acid. One of the reagents given was $\ce{FeCl_3}$. According to the answer key, these 2 could react. On googling the reaction, I found that methanoic acid loses its acidic proton and is attacked by $\ce{FeCl_3}$ and the proton released then removes chlorine to form the complex. It all made sense because $\ce{CH_3COOH}$ was an acid.

But the next compound was 3-carboxy-4-nitro-benzene sulfonic acid. And I wanted to check its reactivity with $\ce{FeCl3}$. So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$. But I couldn't think of a reason why.

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  • $\begingroup$ Looks like carboxylates make stronger complex with Fe than sulfonates. $\endgroup$ Jun 17 '20 at 8:38
  • $\begingroup$ Methanoic (=derived from methane ) acid is formic acid $\ce{HCOOH}$, while $\ce{CH3COOH}$ is ethanoic ( = derived from ethane ) acid aka acetic acid. $\endgroup$
    – Poutnik
    Jun 17 '20 at 8:52
  • $\begingroup$ Acetic Acid reacts with $\ce{FeCl3}$ to form Iron (III) acetate complex. I think the sulfonate with it's large benzene ring won't be able to form a similar complex. $\endgroup$ Jun 17 '20 at 13:33
  • $\begingroup$ @AniruddhaDeb What could have prevented it? $\endgroup$ Jun 17 '20 at 13:37
  • $\begingroup$ @AniruddhaDeb iron does form Sulfonate complex. See this: nrcresearchpress.com/doi/pdfplus/10.1139/v81-098. $\endgroup$ Jun 18 '20 at 4:32
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It all made sense because $\ce{CH_3COOH}$ was an acid

$\ce{CH_3COOH}$ being an acid facilitates the complex formation as the chlorido ligands can be easily removed as $\ce{HCl}$. But, that's not the only reason why the red colored complex $\ce{[Fe(OAc)_3]}$ was formed.

You have to understand that the stability of the complexes is also determined by how well the ligand interacts with the central metal ion.Here, the ligand forming the coordination complex is $\ce{OAc^-}$ and this is associated with enough negative charge density to stabilize the complex formed.

So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$

This is because although as you have said has a sulfonic acid group, it is not the acidic strength of this ( or acidic strength of $\ce{CH_3COOH}$ in the first question) that drives the complex formation.It mainly depends upon the strength of the ligand and effective nuclear charge of the central metal ion.The ligand in this case, is associated with lot of electron withdrawing groups like $\ce{-NO_2}$ and that makes this a poor ligand.Also, the steric hindrance due to the bulky groups also has a role to play for the complex formation not to take place.

NOTE:

All my arguments made regarding stability of complexes is as per the Crystal Field Theory which assumes the interaction between metal ion and ligand to be purely ionic.That's why, I was comparing the charge density of the ligands in the 2 cases.

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  • $\begingroup$ @VamsiKrishna Where did you find this question btw? And I hope you understood the reason :) $\endgroup$ Jul 4 '20 at 17:44
  • $\begingroup$ Amazing I am finally relieved! It was in a worksheet that my sir had provided, but he left shortly so I couldn't get the solution. Thanks a lot fpr your answer! $\endgroup$ Jul 4 '20 at 18:07
  • $\begingroup$ Happy to know that my answer helped you! :) $\endgroup$ Jul 4 '20 at 18:20

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