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Cr and Cu question relating to electron configuration


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Hi there everyone!


Why do Cr and Cu not follow the other d-block metals in electron configurations?


We have this:


[math]1s^22s^22p^63s^23p^63d^34s^2[/math] (Sc)

[math]1s^22s^22p^63s^23p^63d^54s^1[/math] (Cr)


[math]1s^22s^22p^63s^23p^63d^84s^2[/math] (Ni)

[math]1s^22s^22p^63s^23p^63d^{10}4s^1[/math] (Cu)


My question is this, I can see why Cr forms the structure it does - the electron arrangement is more stable as it only has one electron per orbital (In the d and s shells) so there is no electron replusion and hence more stable but Cu makes no sence.


Can someone explain why these elements behave the way they do? I have been trying to think of reasons for this apparent break in the trend but I cna find no reason for it so maybe you cna help!




Ryan Jones

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There is an extra stability associated with having a half-filled shell or completely full shell. Since the 3d and 4s orbitals are comprable in energy, removing an electron from the 4s orbital to make the 3d subshell completely full or half full makes the atom more stable.

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Remember that the 4s shell fills in before the 3d shell does, because as Yggdrasil pointed out, their energy levels are pretty much the same. When you move from Nickel to Copper, you would logically think that you'd go from 4s2-3d8 to 4s2-3d9. In Nickel (4s2-3d8) you have four totally filled 3d subshells and one empty one. If you just add one electron to that as you move to copper, then you'd have four totally filled shells and one half-filled shell. Since the 3d shell can hold a total of ten electrons, it would wind up being 9/10'th full which is an odd fraction. Shells want to be either half full, or totally full. Anything in between isn't good and if it's possible to rearrange the electrons to accomodate that then you'll probably see it happen. So in copper, an electron is taken from the 4s shell and added to the 3d shell along with the extra electron that copper has over nickel. As a result, you get a completely filled 3d shell and a half-filled 4s shell. This is nice and stable and everybody's happening.


For fun, pick any random d or f block element and write out it's electron configuration before you look it up. Keeping in mind what has been said in this thread, you should even be able to figure out the 'odd' ones. :D

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It might be that all this reasoning about shells want to be half full and so on is correct, but there are counter examples. In the lanthanides there is even more complicated behavior with oddities in between (the f orbitals come in play then).


So, I agree with the answers, because they make it easy to remember which are the oddities for the transition metals. But now look at the second row of transition metals (elements Y ... Cd), or look at Pt and Au, over there this reasoning does not work anymore.


There only is one way to really understand all these oddities and that is numerically solving the quantum mechanical wave equations for all electrons in an atom, also taking into account relativistic effects. This is a hell of a job and requires a lot of computational power, but it has been done for us already. Only then you see the very subtle differences in energy levels. Probably this is not a nice answer in terms of easy understanding, but sometimes you have to take things as they are. Simple underlying physical laws can give rise to really complex behavior, which only can be understood by performing massive computations.


Another nice example of a very simple system, following very simple laws is a three-body mechanical system with Newtonian laws of gravity. Such a simple system exhibits amazingly complex behavior, which only can be understood by simply running the simulation for a long time.

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I have a question here... if Cu wants to have a full shell why can it sometimes form Cu2+ when Cu+ is much more stable?


Again, it is a matter of quantum mechanics. The energy levels are so close to each other that sometimes Cu(+) is more stable and another time Cu(2+) is more stable.


The stability strongly depends on the ligands which are coordinated to the copper core:


Water ligands: Cu(2+) is stable, Cu(+) disproportionates to Cu and Cu(2+)

Ammonia ligands: Both Cu(2+) and Cu(+) are stable.

Chloride ligands: Cu(+) is more stable, in the presence of both Cu and Cu(2+) they comproportionate to Cu(+)


So you cannot say that Cu(+) is more stable, it really depends on the compounds in which it is incorporated.

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  • 3 years later...

Is someone going to answer the question that was asked?

Simply stating because "half-filled" and "completely filled" is more stable is not even an answer.

Atoms can't desire or want to be "filled" - that is anthropomorphistic (as stated by Ferdinand in the post above).


The "half filled" 3d orbital of Cr is more stable because the 4s energy level is close to the 3d energy levels - therefore moving electrons from the 4s energy level into the 3d minimizes the repulsion between the electrons, and entropy is favored (if the two electrons stay in the 4s there is more order).


I do not understand the "completely filled" state. I thought that the "+" charged nucleus of an atom (z effective) for a 4s is more "penetrating", which means that further away electrons (such as the 3d) are more shielded (by the spherical s shell electrons which block the "+" charge) and therefore "feels" less attraction towards the nucleus.



Why does Cu obtain an [Ar]4s^(1)3d^(10)? You can't use the entropy argument (or electron repulsion argument) any more because the electron would not favor either 4s or 3d more... and it seems that due to the 4s being more penetrating (closer to the "+" nucleus with less sheilding) that the electron would be more stablized to be in the 4s energy level?


Confused. ):

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With regards to wht you fill Cu in that order, its a result of the effective nuclear charge increasing going across the period. As you increase the nuclear charge, Zeff goes up and so the orbitals are contracted to the nucleus and are more stable...however the s and d orbtials are contracted by different amounts. Generally, the 4s is lower in energy than the 3d so you fill the 4s before you fill 3d. However, at copper, the effective nucealr charge has increased so muh that the d-orbitals are now more stable than the 4s (3d orbtials are affected more than the 4s) and so now you just follow the normal rules and fill the lowest energy orbitals first.

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  • 1 month later...

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