[13Chad]

I'm a bit desperate here b/c I've been given some questions and cannot find the answers...I've tried I promise.

Some simple tests are needed to identify the following solutions (not labeled...as the question goes)

I have: Copper(II) Sulfate Cu(SO)₄, Aluminum(III) Sulfate Al₂(SO₄)₃, Lead(II) Nitrate Pb(NO₃)₂, and Iron(II)Sulfate FeSO₄

Now, I don't actually have to perform this experiment, just come up with some way of differentiating the samples.  Other than some color differences (CopperSulfate is cyan, LeadNitrate is clear, IronSulfate is blue-green which might be tough to tell from cyan) I'm pretty lost.

I know that if you add sugar to copper(II)sulfate it will be reduced to copper oxide and turn red but that's the only good one I can come up with.

I will ask one further question which is proving difficult to find.  I'm given the following question: in which member of the following pairs of complexes would 10Dq_oct. be the langer and why?
Cr(H₂O)₆²⁺ and Cr(H₂O)₆³⁺,
CrF₆^3- and  Cr(NH₃)₆^3-
MnF₆^2- and ReF₆^2-

I have no idea what 10Dqoct. Langer is supposed to mean.  This is and indendent study program and I've been through all my materials but cannot come up with these answeres.  Some help would.....help.

[Borek]

CopperSulfate is cyan, LeadNitrate is clear, IronSulfate

Copper sulfate, lead nitrate and iron sulfate - names are not one word.

Think what will happen if you add some other solutions, like - for example - NaOH or ammonia.

[13Chad]

Yes, I don't know why I chose one bit of the whole post to get lazy.

Well,

If I add ammonia to the Copper(II) Sulfate I would get a precipitate of copper(II) hydroxide at first but then a 4 ammonia molecules would replace 4 planar water molecules of the hexaaquacopper(II) ion

Pb(NO₃)₂ is soluble and if you heat it will crackle as it decomposes to lead(II) oxide but of course I could just add KI and look for some extremely bright orange precipitate (just remembered doing that experiment)

I don't know about ammonia and the others.

If I add NaOH to Cu(SO₄), insoluble Cu(OH)₂ should form...this wold differ from the same treatment of  Iron(II) sulfate because a white precipitate (light green due to oxidation) would form but it would dissolve in high concentration NaOH (heated too).

What about Aluminum(III) Sulfate? Add NH₃ and get a brown precipitate?

[LQ43]

just wondering,

do you have to identify the solutions as copper (II) sulfate, etc.

the tests mentioned would identify the cation, but not necessarily the anion.

you might have to do secondary tests to evaluate the anion

[13Chad]

I was pretty sure that I only needed to decide which sample was which.  Now you've got me thinking that maybe it's more difficult than that.  So far I've put together a pretty simply way to prove which is which...but not beyond that.

What about my second question...is this a term I should know?  What could the question be?

[Borek]

What happens to Al in NaOH?

No idea about second point. Could be it is nomenclature problem - ie I don't know nomenclature 

[13Chad]

adding NaOH to Al³⁺ precipitates aluminum hydroxide, Al(OH)₃,
this is white or even colorless with a refractive index similar to water so it would be barely visible.  This would be an easy distinction b/t it and ferrous hydroxide (it doesn't remain pure due to oxidation to Fe(OH)₃ which will be green)

[Borek]

Close, but that's not the whole truth. What happens to the white precipitate when you add excess NaOH?

[LQ43]

in which member of the following pairs of complexes would 10Dq_oct. be the langer and why?
Cr(H₂O)₆²⁺ and Cr(H₂O)₆³⁺,
CrF₆^3- and  Cr(NH₃)₆^3-
MnF₆^2- and ReF₆^2-

 10Dq_oct. be the langer

I think this is supposed to say "1.0 Delta(greek sign) oct  be the larger and why"

oct means octahedral

This looked familiar (I was an organometallic chemist a long time ago) and I'm pretty sure has to do with crystal field theory and crystal field stabilization energy (google this or find its Wiki)

Certain ligands cause the d orbitals of the transition metal to split in their energies (so they are not all the same) and some are higher than others. If its octahedral and the ligands are pointing on the xyz axes, the dxy, dyz and dxz orbitals (called "t2g" orbitals) are lower in energy because they point in between the axes, the dx2-y2 and dz2 (called "eg" orbitals) are raised in energy because they point directly at the ligand. The ligands themselves will determine how large this gap in energies between the t2g and eg orbital sets. Weak field ligands like Br- will cause a small splitting and strong field ligands like CN- and CO will cause a larger gap.

The oxidation number like +2 or +3 will determine how many electrons are in those d orbitals and the ligands will determine if those electrons will be low spin (paired up on the lower energy orbitals) or high spin (unpaired, some will go into the higher level orbitals)

Wow, that really took me back but this is really classical coordination chemistry.(Basolo and Johnson, wrote a great little book called "Coordination chem") - you might also find this topic in advanced inorganic text

To answer the question you need to determine if the ligands in the pair are strong or weak and what the CFSE would be for each, as well as how many electrons would be in the d orbitals

[13Chad]

Borek - I'm not sure what happens upon excess NaOH to Al(OH)₃ I suppose it dissolves.

Thanks for the help btw

Also, LQ43 - that's gotta be it.  I've worked it out as best I could.  I suppose I will find out if everything is accepted...but I wouldn't have known where to begin without your guess.

[Borek]

Borek - I'm not sure what happens upon excess NaOH to Al(OH)₃ I suppose it dissolves.

Exactly - that's characteristic of Al. Well, not only of Al, some other metals behave this way, like Zn, Sn, to some extent Pb and so on - but you have already identified Pb with iodide.