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Topic: Cr(III) amine complex  (Read 8204 times)

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Offline Rutherford

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Cr(III) amine complex
« on: February 15, 2013, 08:22:20 AM »
The synthesis of chromium(3+) ammine complexes usually starts from a freshly prepared in situ solution of a chromium(2+) salt. To the solution of a chromium(2+) salt, the solution of ammonia and a solid ammonium chloride are added. Then a stream of air is passed through the solution. The red precipitate is formed that contains 28.75 wt.% of N. Determine the composition of the precipitate and give the reaction equation.

I wrote the possible formula: Cr(NH2)xCly(OH)z. There is the proportion: 14x:M=28.75:100, M=48.696x g/mol. xmax=6. If x=1, the molar mass is to small, for x=2 its the same. For 3 or 4 no compound fits. For 5 and 6 the mass gets too big. I am stuck here now.

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Re: Cr(III) amine complex
« Reply #1 on: February 15, 2013, 08:31:21 AM »
Ignoring everything else for a moment - NH2? 14?
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Offline Rutherford

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Re: Cr(III) amine complex
« Reply #2 on: February 15, 2013, 08:46:36 AM »
It is the weight percent of nitrogen only, so it should be 14x.

Offline Rutherford

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Re: Cr(III) amine complex
« Reply #3 on: February 15, 2013, 01:29:51 PM »
I even assumed that it is an anionic complex, with NH4+, but still couldn't get a reasonable solution  ???.

Offline AWK

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Re: Cr(III) amine complex
« Reply #4 on: February 15, 2013, 06:41:28 PM »
Chromium complex contains NH3 not NH2
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Re: Cr(III) amine complex
« Reply #5 on: February 15, 2013, 06:42:48 PM »
For 5 and 6 the mass gets too big.

Too big for what?
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Offline Rutherford

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Re: Cr(III) amine complex
« Reply #6 on: February 16, 2013, 05:14:17 AM »
Thanks for the replies.
It is Cr(NH3)xCly(OH)z
Again, from the proportion 14x:M=28.75:100, M=48.696x g/mol.
Now, NH3 makes a difference. If x=5, M=243.5g/mol, this would mean 1 or 3 chlorine atoms. As Cr has the charge 3+, I need 3 negative ions, so y+z=3. The mass without chromine and amino groups is 106.5 which corresponds to 3 chlorine atoms, so the formula is [Cr(NH3)5Cl]Cl2. Is this good?

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Offline Rutherford

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Re: Cr(III) amine complex
« Reply #8 on: February 16, 2013, 12:36:18 PM »
Good. Now I have another question: How can one prepare a solution containing Cr2+ ions using metallic chrome? Specify the conditions.

When it gets dissolved in acid it will be Cr3+, but what to use to reduce it to Cr2+? Or is there a better way?

Offline AWK

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Re: Cr(III) amine complex
« Reply #9 on: February 16, 2013, 02:21:40 PM »
read
http://en.wikipedia.org/wiki/Chromium(II)_chloride

or buy (relatively inexpensive)
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Offline Rutherford

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Re: Cr(III) amine complex
« Reply #10 on: February 16, 2013, 02:29:28 PM »
Got this, thanks.
« Last Edit: February 16, 2013, 02:40:49 PM by Raderford »

Offline Rutherford

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Re: Cr(III) amine complex
« Reply #11 on: February 16, 2013, 02:57:39 PM »
Arrange the hexammine complexes of iron(2+), chromium(3+) and ruthenium(2+) in a row of increasing stability towards the acidic water solutions. Explain your choice.

How can this be determined? What property do I need to know for this?

Offline AWK

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Re: Cr(III) amine complex
« Reply #12 on: February 16, 2013, 05:22:00 PM »
Hint - stability constants
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Offline Rutherford

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Re: Cr(III) amine complex
« Reply #13 on: February 17, 2013, 04:33:33 AM »
No stability constant is given, so I have to deduce the answer logically  :(.

Maybe, the bigger the cation, the repulsion of donated electrons from NH3 will be smaller as the distance between them increases, so it would be: iron(2+)<chromium(3+)<ruthenium(2+).
Is my reasoning correct?

Offline XGen

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Re: Cr(III) amine complex
« Reply #14 on: February 25, 2013, 12:13:33 PM »
3. What oxidizer can be used instead of oxygen to obtain the same product?

4. What product will be formed if the experiment described above is performed under inert atmosphere without oxygen?

These are another couple questions that I've been having trouble with.

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