What Forms the Green-Blue Precipitate with Chromium (III) Ions and NaOH?

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Discussion Overview

The discussion revolves around identifying the green-blue precipitate formed when chromium (III) ions react with sodium hydroxide (NaOH). Participants explore various hypotheses regarding the nature of the precipitate, its color, and the underlying chemistry involved, including concepts of oxidation states and amphoteric behavior.

Discussion Character

  • Homework-related
  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant suggests that the green-blue precipitate may be [Cr(H2O)6]2+, noting that its solution is blue but expressing uncertainty about ion precipitation.
  • Another participant proposes that the precipitate could simply be a hydroxide, questioning the initial assumptions about the color and composition.
  • A participant expresses confusion over the color of the precipitate, indicating that sources disagree on whether it is green or blue and questioning the oxidation state of chromium in different scenarios.
  • One participant speculates that the precipitate is a chromium hydroxide compound, reasoning that the ion causing the green color has more hydroxide anions than the precipitate itself.
  • A later reply clarifies that oxidation states do not change in this reaction, emphasizing that chromium (III) remains unchanged and discussing the amphoteric nature of chromium.
  • Another participant suggests using the method of elimination to narrow down possible answers, arguing that NaOH cannot oxidize Cr(III) to higher oxidation states, thus ruling out certain options.
  • One participant challenges a previous statement about ions and anions, indicating a misunderstanding in the discussion and reinforcing the elimination approach to find the correct answer.
  • Finally, a participant concludes that the only plausible answer left is C, drawing parallels with aluminum compounds in qualitative analysis.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the identity and color of the precipitate, with no consensus reached on the correct answer. There is ongoing debate about the role of hydroxide ions and the oxidation state of chromium.

Contextual Notes

Participants acknowledge limitations in their reasoning, including confusion over color descriptions and the implications of amphoteric behavior. The discussion reflects varying interpretations of the chemistry involved, particularly concerning the formation and characteristics of the precipitate.

Who May Find This Useful

This discussion may be of interest to students and educators in chemistry, particularly those studying transition metal chemistry, precipitation reactions, and the behavior of amphoteric substances.

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Homework Statement


Screen Shot 2016-07-04 at 2.27.56 PM.png


Homework Equations

The Attempt at a Solution


I chose A as after googling, I learned it was green. My question is: what is the green-blue precipitate. I am inclined to say [Cr(H2O)6]2+ as it's solution is blue, but ions don't precipitate. I tried searching and could find nothing. Could someone tell me what it is?

Cheers!
 
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Why not just a hydroxide?
 
I just did further googling and it seems that A is not green according to some and green according to others. I am lost. Chromium (III) is green I think, but in A, C and D, the oxidation number of chromium is also 3. Could you give me some advice?

Why not just a hydroxide?
Is this for the blue-green precipitate or the green solution?

Thanks!
 
Sorry, that was for the precipitate.

Cr is amphoteric, dissolution involves OH-.
 
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So I guess the precipitate is some kind of chromium hydroxide compound (not sure what is the oxidation of chromium is). So I guess that means that the ion causing the green colour has more OH anions than the precipitate. That means A, B and E is not possible. Since NaOH is a base, it doesn't seem the formation of H3O+ is possible so I think the answer is C. My reasoning seems shaky so I could be very wrong. Is C the answer, if so, can you explain your logic so I can learn from it?

Cheers!
 
Oxidation doesn't change, there is no redox going on here.

As typical with most amphoteric elements, first hydroxide precipitates, then it dissolves in excess OH-. This is awalys a series of reactions of the

Me(OH)nm + OH- ↔ Me(OH)n+1m-1

type. One of them contains a neutral hydroxide, typically insoluble. Things get more complicated when you consider water ligands that are present as well (and replaced by OH-), but as a first approximation they can be ignored.
 
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So will the answer be C as it fits the formula you gave above? Thanks!
 
Point one I didnt get this statement from you.

TT0 said:
So I guess that means that the ion causing the green colour has more OH anions than the precipitate.

Ions can't have anions. I think you had a moment of confusion and wrote something else other than what you intended.

Coming to the question, why don't we use the method of elimination? Since the NaOH is (aq) it can't be a strong enough oxidising agent to oxidise Cr(III) to anything higher, especially Cr(VI). That leaves out B and E as answers.

Now because we are using basic medium H30+ ions can't originate. that leaves out D as an answer.

Googling tells me that Cr[(H2O)6]+3 is bluish in color. Puff goes A as an answer.

Leaves us only with C. And this also makes sense, because in qualitative analysis we place Chromium along with best buddy Aluminium, which forms a similar compound on addition of excess of base.
 

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