Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Shadow on December 06, 2014, 04:25:43 AM
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Are Cu2+ ion octahedral complexes low spin or high spin? They are paramagnetic, but I am unsure.
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How many electrons are in D shell of Cu2+? Could they be arranged in both "configurations"?
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Are Cu2+ ion octahedral complexes low spin or high spin? They are paramagnetic, but I am unsure.
what are the ligands present around Cu2+ ...it is the nature of the ligand which determines if it is high or low spin complex.Strong ligands like CN- gives high spin complexes and weak ligands like Cl- gives low spin complex.
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so what would be the difference in HS and LS Cu(II) ?
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I believe that I have already answered this in another forum...
Just in case it was just a coincidence:
There are not two possibilities for low spin and high spin complex for a d9 metal ion. The only cases where you can have both is for d4, d5, d6 and d7.
Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy. You need extra energy for doing two things:
Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, Δ_o
Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)
In the complexes where Δ_o is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite. But this only applies for d4, d5, d7 and d8.
Now think about d1 d2 and d3: It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each T2g orbital.
Now d8 d9 and d10. There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals. You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital? This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin. The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.
Btw, this should be on the inorganic chemistry board, not organic.
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it's nice that you know the answer. but I guess he could have figured that out himself...
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In the case of a Cu2+ isolated in vacuum, I expect the lowest energy configuration to be that of neutral Co, that is [Ar]3d74s2. Or?
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Who knows :-D
It could also be d8s1 so you would have half-filled s orbital and eg while having full t2g
edit: never mind, its in vacuum -> no crystal field splitting
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it's nice that you know the answer. but I guess he could have figured that out himself...
Yeah, I never post answers directly on this forum, but he asked the exact same question in a question-answer chemistry directory and I had already responded it there, so I've just copy/pasted it here too just in case someone finds the thread in the future and doesn't know the answer aswell.
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Are Cu2+ ion octahedral complexes low spin or high spin? They are paramagnetic, but I am unsure.
what are the ligands present around Cu2+ ...it is the nature of the ligand which determines if it is high or low spin complex.Strong ligands like CN- gives high spin complexes and weak ligands like Cl- gives low spin complex.
Nnnno, opposite, ligands high up in the spectrochemical serires, as CN-, will give low spin complexes, and vice versa
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Yeah, I never post answers directly on this forum, but he asked the exact same question in a question-answer chemistry directory and I had already responded it there, so I've just copy/pasted it here too just in case someone finds the thread in the future and doesn't know the answer aswell.
check.
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Why does the configuration in vacuum resemble Co?
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it depends on what type of complex it forms i.e the ligands attached to it. to strong field ligands as CN ,it forms [Cu(CN)4]2- , a square planar complex and all d-electrons paired, gives diamagnetic complex. but as for Oh complexes, it is paramagnetic ,a J-T distortion occurs. d-(x2-y2) orbital accommodating the odd electron.
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But JT distortion won't change the spin moment/
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I'm sorry , made a mistake in the last post , square planar complexes are paramagnetic for Cu(II) also.