Solution Chemistry

[SaintJude]

Passage-based question from...Kaplan MCAT Qbank

Passage excerpt:
" Copper (II) is generally stable in aqueous solutions, and many copper (II) compounds are well-known. However, copper (I) cation is not stable in aqueous solutions.

Initially unaware of this fact, a student planned to prepare copper (I) iodide by treating copper (II) with a mild reducing agent in the presence of iodide. The student was surprised when when adding a soluble copper (II) salt to the solution of potassium iodide when a precipitate formed immediately: revealed to be Cu2I2 ("copper (I) iodide"). Blah blah blah.

Question:

What is the most likely reason for the apparent stability of Cu2I2 ?

A. The copper is actually in the +2 oxidation state.
B. Cu2I2 contains copper in both the 0 and +2 oxidation state.
C. The potassium from the KI keeps it from being oxidized.
D. It does not react because it is insoluble.



Can someone explain this? The answer is D. How was I supposed to know that B is incorrect? There are many examples of compounds, usually oxides, containing elements in two oxidation states. Help!

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[Pisiform]

hmm i dunno if this helps but the same way you found out A is incorrect by finding out the oxidation number of Cu in Cu2I2

2x - 2 = 0
x = +1 .... is the oxidation no. of copper in Cu2I2 .... A & B are wrong ... is this what you are asking?

 

[MedPR]

Passage-based question from...Kaplan MCAT Qbank

Passage excerpt:
" Copper (II) is generally stable in aqueous solutions, and many copper (II) compounds are well-known. However, copper (I) cation is not stable in aqueous solutions.

Initially unaware of this fact, a student planned to prepare copper (I) iodide by treating copper (II) with a mild reducing agent in the presence of iodide. The student was surprised when when adding a soluble copper (II) salt to the solution of potassium iodide when a precipitate formed immediately: revealed to be Cu2I2 ("copper (I) iodide"). Blah blah blah.

Question:

What is the most likely reason for the apparent stability of Cu2I2 ?

A. The copper is actually in the +2 oxidation state.
B. Cu2I2 contains copper in both the 0 and +2 oxidation state.
C. The potassium from the KI keeps it from being oxidized.
D. It does not react because it is insoluble.



Can someone explain this? The answer is D. How was I supposed to know that B is incorrect? There are many examples of compounds, usually oxides, containing elements in two oxidation states. Help!

I've never heard of Copper having a 0 oxidation state. Maybe I'm missing some huge chunk of chemistry, but the only things I've heard of having an oxidation state of 0 are S8 and O2.

Even if Copper does have a 0 oxidation state, aren't halides always -1? So the Cu in Cu2I2 would have to be +2/2 = +1?

 

[Ssina]

B to me says Cu will have 0 and +2 oxidation state in the same time in Cu2I2, which I can't see how you would have 2 different oxidation states... and since it'll have +1 oxidation state and passage says it's "not stable at +1" I kinda assumed it'll precipitate out

 

[SaintJude]

So can only answer this by POE.

Choice A seems to suggest that iodine has an oxidation state of -2, which is impossible.

Choice B is plausible for some compounds but not for this Copper compounds specifically, because an oxidation state of 0 is copper in its metal form and it's unlikely that a copper metal would be a constituent of a compound.

Choice C specifically mentions K ion, when its known that potassium ion is a spectator ion because it's impossible to oxidize or reduce it in aqueous solution (like Na+, Li+,, etc..) And since it won't react in water, we can ignore this spectator ion. So there is no reason to think K could actually prevent any reaction, let alone oxidation.

So..we're left with Choice D I guess

 

[MedPR]

So can only answer this by POE.

Choice A seems to suggest that iodine has an oxidation state of -2, which is impossible.

Choice B is plausible for some compounds but not for this Copper compounds specifically, because an oxidation state of 0 is copper in its metal form and it's unlikely that a copper metal would be a constituent of a compound.

Choice C specifically mentions K ion, when its known that potassium ion is a spectator ion because it's impossible to oxidize or reduce it in aqueous solution (like Na+, Li+,, etc..) And since it won't react in water, we can ignore this spectator ion. So there is no reason to think K could actually prevent any reaction, let alone oxidation.

So..we're left with Choice D I guess

When I read precipitate formed immediately: revealed to be Cu2I2 and then saw answer D, I knew that was it. Precipitate formed immediately = none of it is soluble.

 

[dranet]

Passage-based question from...Kaplan MCAT Qbank

Passage excerpt:
" Copper (II) is generally stable in aqueous solutions, and many copper (II) compounds are well-known. However, copper (I) cation is not stable in aqueous solutions.

Initially unaware of this fact, a student planned to prepare copper (I) iodide by treating copper (II) with a mild reducing agent in the presence of iodide. The student was surprised when when adding a soluble copper (II) salt to the solution of potassium iodide when a precipitate formed immediately: revealed to be Cu2I2 ("copper (I) iodide"). Blah blah blah.

Question:

What is the most likely reason for the apparent stability of Cu2I2 ?

A. The copper is actually in the +2 oxidation state.
B. Cu2I2 contains copper in both the 0 and +2 oxidation state.
C. The potassium from the KI keeps it from being oxidized.
D. It does not react because it is insoluble.



Can someone explain this? The answer is D. How was I supposed to know that B is incorrect? There are many examples of compounds, usually oxides, containing elements in two oxidation states. Help!

Its got to do with solubility rules. MCAT like these they come up a lot!
"3. The chloride (Cl-), bromide (Br-), and iodide (I-) ions generally form soluble salts. Exceptions to this rule include salts of the Pb2+, Hg22+, Ag+, and Cu+ ions. ZnCl2 is soluble, but CuBr is not".
The answer isn't there so it has to be D. It was a solution chem q. so solubility may be the issue.