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Topic: CFSE and spin complexes  (Read 3138 times)

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Offline Firehchicken

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CFSE and spin complexes
« on: February 07, 2018, 04:24:43 AM »
Hi,

In question 33 the answer is B). I can figure out that Ni2+ has 8 3d electrons. Pairing all 8 electrons in low spin would give 6 paired electrons in t2g and 2 unpaired in eg. So, how do I obtain zero unpaired electrons?

In question 34, the answer is D. Fe2+ has 6 3d electrons. Pairing them in low spin would give 3 electron pairs in t2g, with the eg orbital left empty. So my question is, why is it +2P and not +3P? Surely, there are 3 paired electrons so it should be 3P right?

Offline mjc123

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Re: CFSE and spin complexes
« Reply #1 on: February 07, 2018, 04:58:08 AM »
33) Are all Ni2+ complexes octahedral?
34) What are you comparing this state with? What would be the energy of a spherical, isolated Fe2+ ion?

Offline Firehchicken

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Re: CFSE and spin complexes
« Reply #2 on: February 07, 2018, 05:27:54 AM »
33) Are all Ni2+ complexes octahedral?


34) What are you comparing this state with? What would be the energy of a spherical, isolated Fe2+ ion?

33)Umm yeah, as far as I have been taught. I havent been taught any other geometries??

34)I'm sorry, I don't quite follow what you're trying to say?

Offline mjc123

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Re: CFSE and spin complexes
« Reply #3 on: February 07, 2018, 07:33:15 AM »
Have you not been taught that square planar MX4 geometry is common for d8 ions? If not, you probably don't know how the d orbitals split in this geometry. There are no triple degeneracies; you get four sets of orbitals; eg, a1g, b2g, b1g (in order of increasing energy). How would 8 electrons go into these orbitals?

An isolated ion would have all five 3d orbitals degenerate. How do 6 electrons go into these orbitals? What is the pairing energy?

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