- #1
Mayhem
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I've been looking at trends in 1st row transition metals and trying to understand why some d-electron configurations are more common than others for each element, and I'm unable to find an easy pattern. It seems that getting rid of the high energy 4s electrons is an obvious pattern, but the resulting number of d-electrons isn't obvious to create a rule for.
Some thoughts:
Sc, Ti, V: d0 because it gives a [Ar] electron configuration. V can easily exist in four different oxidation states, though, as seen in vanadium flow batteries.
Cr: d0, see above. Cr(VI) is however a strong oxidizer. d3, removes high energy 4s electron, and three electrons can evenly distribute in the dxy, dxz, dyz (assuming octahedral field), minimizing correlation and exchange energy,
Mn: d5 high spin evenly distributes five electrons in all five d-orbitals, minimizing correlation and exchange energy.
Fe: Both d5 and d6, but often Fe(II) compounds will be prone to oxidizing to Fe(III) over time, depending on complex, possibly because there is a pairing energy in d6 high spin not present in d5 high spin.
Co: No particular thoughts as the above arguments do not hold for d6 (except 4s electrons are removed) and d7.
Ni: d8, the two 4s electrons are removed.
Cu, Zn: Not sure. Cu(I) does not tend to be stable in an oxygen atmosphere, and for some reason the d9 Cu(II) is most often found. This is interesting as, Zn(II) is d10.
My intuition is that the overall charge of the atom also plays a role. For (an extreme) example, d0
Zn12+ does not exist despite yielding the "stable" [Ar] electron configuration. Similar, less demonstrative, examples could be made for other elements.
There is obviously an underlying (probably quantum mechanical) mechanism that I'm not seeing. But what's the logic?
Some thoughts:
Sc, Ti, V: d0 because it gives a [Ar] electron configuration. V can easily exist in four different oxidation states, though, as seen in vanadium flow batteries.
Cr: d0, see above. Cr(VI) is however a strong oxidizer. d3, removes high energy 4s electron, and three electrons can evenly distribute in the dxy, dxz, dyz (assuming octahedral field), minimizing correlation and exchange energy,
Mn: d5 high spin evenly distributes five electrons in all five d-orbitals, minimizing correlation and exchange energy.
Fe: Both d5 and d6, but often Fe(II) compounds will be prone to oxidizing to Fe(III) over time, depending on complex, possibly because there is a pairing energy in d6 high spin not present in d5 high spin.
Co: No particular thoughts as the above arguments do not hold for d6 (except 4s electrons are removed) and d7.
Ni: d8, the two 4s electrons are removed.
Cu, Zn: Not sure. Cu(I) does not tend to be stable in an oxygen atmosphere, and for some reason the d9 Cu(II) is most often found. This is interesting as, Zn(II) is d10.
My intuition is that the overall charge of the atom also plays a role. For (an extreme) example, d0
Zn12+ does not exist despite yielding the "stable" [Ar] electron configuration. Similar, less demonstrative, examples could be made for other elements.
There is obviously an underlying (probably quantum mechanical) mechanism that I'm not seeing. But what's the logic?