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Topic: Cp Vs Cp*  (Read 5462 times)

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Offline phth

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Cp Vs Cp*
« on: February 17, 2016, 10:45:24 PM »
Hello,
I am sturggling to wrap my head around why Cp* is low spin compared to Cp: e.g. MnCp2 has 2 unpaired electrons and MnCp*2 only has 1.  I haven't really found a good explanation for this.  it's because the π* orbitals are higher up in energy?  Does anyone have a good reference or textbook to check out?

Offline clinz63

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Re: Cp Vs Cp*
« Reply #1 on: February 18, 2016, 12:01:13 AM »
Hmmm... Wouldn't Cp* have greater electron density due to the electron donating substituents causing greater stabilization of Mn orbitals? This high energy difference between bonding and antibonding orbitals of Mn leads to a low spin complex.

Offline Irlanur

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Re: Cp Vs Cp*
« Reply #2 on: February 18, 2016, 10:43:29 AM »
I would also say that Cp* is the better sigma-donor.

Quote
causing greater stabilization of Mn orbitals

I think better sigma donors actually increase the energy of the antibonding orbitals (which are the ones on the metal).

Offline clinz63

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Re: Cp Vs Cp*
« Reply #3 on: February 19, 2016, 12:41:32 AM »
Ah, yes. And pi donation by a ligand stabilizes the bonding orbitals?

Offline Irlanur

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Re: Cp Vs Cp*
« Reply #4 on: February 19, 2016, 01:11:45 PM »
Not really. Maybe have a look on your favourite MO-diagram of an octahedral complex.

Offline pm133

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Re: Cp Vs Cp*
« Reply #5 on: February 21, 2016, 07:04:29 AM »
Hello,
I am sturggling to wrap my head around why Cp* is low spin compared to Cp: e.g. MnCp2 has 2 unpaired electrons and MnCp*2 only has 1.  I haven't really found a good explanation for this.  it's because the π* orbitals are higher up in energy?  Does anyone have a good reference or textbook to check out?

Are you absolutely sure you are right about the numbers of unpaired electrons for MnCp2 ? High spin Mn d5 wouldnt give 2 unpaired electrons would it?

Offline phth

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Re: Cp Vs Cp*
« Reply #6 on: February 21, 2016, 09:05:55 PM »

Offline clinz63

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Re: Cp Vs Cp*
« Reply #7 on: February 22, 2016, 12:34:33 AM »
Wuuuut? Really? Wouldn't t2g and pi orbitals of ligands have the right symmetry to interact?

Offline pm133

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Re: Cp Vs Cp*
« Reply #8 on: February 22, 2016, 05:40:19 PM »
I mean 2 versus 4 unpaired electrons: 2 unpaired electrons in the upper 2 orbitals.  According to Crabtree (http://www.amazon.com/gp/product/1118138074/ref=pd_lpo_sbs_dp_ss_1/176-6000695-1376450?pf_rd_m=ATVPDKIKX0DER&pf_rd_s=lpo-top-stripe-1&pf_rd_r=1S5609HSYQPMPSZKWB72&pf_rd_t=201&pf_rd_p=1944687562&pf_rd_i=0470257628)

Sorry I am perhaps a bit tired tonight but I am confused. How do you get 4 unpaired electrons from d5?

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