Timeline for Are these reaction equations for the formation of the brown ring complex correct?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2015 at 12:19 | comment | added | DavePhD | @Robin97 yes I added to my answer | |
| Jan 7, 2015 at 8:25 | comment | added | Rohinb97 | @DavePhD Please see the edit to my question and the comment above ^ | |
| Jan 6, 2015 at 20:43 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 6, 2015 at 20:23 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 6, 2015 at 20:07 | comment | added | Rohinb97 | So you are basically saying that $NO must$ attack as a neutral ligand only? But why? I guess I have convinced you why the hexaaqua ferrum(III) should be formed. Can you edit your answer and tell me why $NO^-$ doesn't attack it and make the ring complex? P.S. The comment section is getting too long. To make this question better and to make it useful please edit the answer to answer my question. | |
| Jan 6, 2015 at 19:49 | comment | added | DavePhD | @Rohinb97 that directly contradicts the mechanism described in the article cited in my answer. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ product". | |
| Jan 6, 2015 at 19:38 | comment | added | Rohinb97 | @DavePhD It should take the electrons from the formation of $Fe^{2+} \rightarrow Fe^{3+} +e^-$ and form $NO^-$ and then attack $[Fe(H_2O)_6]^{3+}$ and give our brown ring complex. | |
| Jan 6, 2015 at 18:14 | comment | added | DavePhD | @Rohinb97 I agree with "both the remaining $Fe^{3+}$ and $Fe^{2+}$ should form $[Fe(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{3+}$", but I don't follow your reasoning concerning NO- being freely present in the solution, how do you propose the NO- forms? | |
| Jan 6, 2015 at 18:01 | comment | added | Rohinb97 | @DavePhD I am saying that the first reaction is correct, and then $both$ the remaining $Fe^{3+}$ and $Fe^{2+}$ should form $[Fe(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{3+}$ and then NO should attack as a neutral ligand and as $NO^-$ respectively to give the final complex: $[Fe^{III}(H_2O)_5(NO^{-})]^{2+}$. Basically I just want to add the attack of $H_2O$ on $Fe^{3+}$ formed in the first reaction, and that too should give me the resulting complex. | |
| Jan 6, 2015 at 17:51 | comment | added | DavePhD | @Rohinb97 For Fe to provide an electron to reduce NO to NO-, Fe needs to be Fe2+. The book says it is Fe2+ that forms the brown complex. What's wrong with that? and what alternative are you proposing? | |
| Jan 6, 2015 at 17:35 | comment | added | Rohinb97 | This is what I was saying the whole time. The electrons from the oxidation of iron should also help NO to form $NO^-$ and then the reaction which I described above should happen. The reactions given in the book does not account for this part. That is why I feel the reactions given are wrong. | |
| Jan 6, 2015 at 17:28 | comment | added | DavePhD | @Rohinb97 It does form $[Fe^{III}(H_2O)_6]^{3+}$, Fe3+ in the first equation in the book should be construed as short-hand for $[Fe^{III}(H_2O)_6]^{3+}$, just as H+ might be considered as short hand for $H_3O+$. However, you are not accounting for gain of one electron in proposing that $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ is formed directly from $[Fe^{III}(H_2O)_6]^{3+}$ | |
| Jan 6, 2015 at 17:20 | comment | added | Rohinb97 | @DavePhD, I mean that why is that $Fe^{3+}$ in the solution just sitting there. It could have formed $Fe^{III}(H_2O)_6$ and then form $[Fe^{III}(H_2O)_5(NO^{-})]^{2+}$ by NO attacking and taking the place of $H_2O$. | |
| Jan 6, 2015 at 15:02 | comment | added | Lexicon | I suggest you also look up the concept of spectator ions. Although not applicable here, it is perfectly possible for a specie to be doing effectively nothing in the course of a reaction. | |
| Jan 5, 2015 at 21:24 | comment | added | DavePhD | @Robin97 The Fe3+ is hexaaquairon(III) and the Fe2+ is hexaaquairon(II). $\ce{[Fe(H2O)6NO]^{2+}}$ is a 7-coordinate transition state but it is not stable, one of the waters is displaced and $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ is formed. | |
| Jan 5, 2015 at 21:02 | comment | added | Rohinb97 | @DhananjayGupta, please see my edit and the comment above ^ | |
| Jan 5, 2015 at 21:02 | comment | added | Rohinb97 | @DavePhD, I have made an edit in my question which made me think why $Fe^{3+}$ ion shouldn't be there. Why isn't that being attacked by the water molecule? And now that you have clarified that Fe is in 3+ and not 1+, why should $Fe^{3+}$ be just $sitting$ there and do nothing? | |
| Jan 5, 2015 at 14:36 | comment | added | Lexicon | Aah i see. My bad then. Thanks a lot for the correction! | |
| Jan 5, 2015 at 12:29 | comment | added | DavePhD | @Martin I added to the answer now | |
| Jan 5, 2015 at 12:28 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 5, 2015 at 2:55 | comment | added | Martin - マーチン♦ | I see what you did here ;) However, it kind of does not answer the question. (Still upvoted though) | |
| Jan 5, 2015 at 2:03 | history | answered | DavePhD | CC BY-SA 3.0 |