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Topic: Rh(n(ch)2n)3 complex  (Read 2986 times)

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Offline garnetwei

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Rh(n(ch)2n)3 complex
« on: November 02, 2012, 01:19:10 AM »
Is there anyone knows this complex? Rh(n(ch)2n)3. I need its name.
I cant find it on google.  Quite suffer when it took me quite few hours to search but no result...
Thanks.
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Offline vex

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Re: Rh(n(ch)2n)3 complex
« Reply #1 on: November 02, 2012, 12:57:59 PM »
Can you rewrite the formula using proper subscripts/superscripts and capitalization?
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Offline garnetwei

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Re: Rh(n(ch)2n)3 complex
« Reply #2 on: November 08, 2012, 10:25:39 AM »
The ligand is -n=ch-c=c-...I dont know how to write in as it is in structure form.
But, i guess, it is a pi acceptor ligand?
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Offline dipesh747

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Re: Rh(n(ch)2n)3 complex
« Reply #3 on: November 17, 2012, 08:37:02 PM »
There is the structure. Vex asked you to write the formula in the conventional (correct) manor, and your answer was very strange, you didn't even write N(CH)2N, you wrote down 3 carbons. Also, how could you think it was a pi acceptor ligand when you don't know how its bonded to the Rh?

The ligand is a 2 electron donor
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Offline dipesh747

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Re: Rh(n(ch)2n)3 complex
« Reply #4 on: November 17, 2012, 09:51:39 PM »
Apologies, the N atoms should have a double bond between them, not single. In which case, I assume that the Nitrogen-Nitrogen double bond could act as a donor atom too.
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Offline garnetwei

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Re: Rh(n(ch)2n)3 complex
« Reply #5 on: November 19, 2012, 05:39:20 AM »
Sorry. I think I have serious typo. The complex is Rh(N(CH)3)3. The ligand should be (N(CH)3) which is N=CH-CH=CH that attack to Rhodium.
I found that the oxidation of Rh is +3.
The Rh complex is a low spin complex because the ligands contain C=C and C=N
which are strong π acceptor. Strong π acceptor ligand can stabilize the t2g hence
increase the Δo between t2g and eg.

Is this correct?
Thanks for the comment. I appreciate it.
« Last Edit: November 19, 2012, 05:53:53 AM by garnetwei »
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