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Doubt regarding removing of electrons from shells !

by 5416339
Tags: doubt, electrons, removing, shells
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5416339
#1
Jul14-10, 10:36 AM
P: 18
Cu - 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Cu2+ - 1s2 2s2 2p6 3s2 3p6 4s0 3d7

This is the configuration of the respective atoms and ions..

According to Aufbau's principle.The subshells with lesser energies get filled first !

Which means 4s sub shell is filled and only after that 3d10 sub shell is filled because the energy of 3d subshell is be greater than 4s subshell ie.4s < 3d energy.

Now if we are removing electrons and converting Cu to Cu2+ it means that energy should be provided to remove the electrons,If that is the case then.

Since, 3d subshell has more enery than 4s orbital,therefore less energy will be required for removing electrons from 3dorbital because they already have greater energy than 4s orbital..

But 4s electrons are removed while ionizing ..How ?

I hope you understood my question...Please giveme a rpoper reason for this..

Thanks,
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alxm
#2
Jul14-10, 12:38 PM
Sci Advisor
P: 1,866
Quote Quote by 5416339 View Post
According to Aufbau's principle.The subshells with lesser energies get filled first !
Correct. (it's the aufbau principle BTW, 'aufbau' means 'building up' in German)

Which means 4s sub shell is filled and only after that 3d10 sub shell is filled because the energy of 3d subshell is be greater than 4s subshell ie.4s < 3d energy.
But this is at odds with what you just said. The 4s shell should be filled before the 3d shell.

But 4s electrons are removed while ionizing ..How ?
Well I'm not 100% sure of the configuration you gave. But more importantly: It's not really important.
See, Madelung's rule and such are only approximate rules. There are many exceptions. (the mentioned [Ar]4s13d10 configuration for neutral copper being one. Filling the final 'd' orbital gives a reduction in energy which makes it more stable than having [Ar]4s23d9. Filled and half-filled orbitals have additional stability due to Unsöld's theorem) The ordering of the levels isn't constant at all, change the number of electrons and their internal order changes. (e.g. K has 4s1 but Ar-, with the same number of electrons puts the last one in the p orbital!)

So it's not very predictable, except for these general rules that don't always hold. The reason why it's not so important is that the electron configuration is for a single atom in vacuum. In practice, you rarely have a Cu2+ ion in vacuum (and if you do, it's not really chemistry you're doing anyway). Once you have ligands surrounding the copper ion, the levels will change again depending on the coordination environment (ligand-field splitting). In the case of Cu, you have a Jahn-Teller axis as well.

In short, there's no easy way to predict the electronic structure of a transition-metal complex. But on the up side, you'll seldom be required to do so unless you're studying it at the more advanced level.
5416339
#3
Jul15-10, 07:32 AM
P: 18
Thats fine ! But i don't think you have answered my question and the Configurations are right ! Don't know how !

alxm
#4
Jul15-10, 11:17 AM
Sci Advisor
P: 1,866
Doubt regarding removing of electrons from shells !

Quote Quote by 5416339 View Post
Thats fine ! But i don't think you have answered my question and the Configurations are right ! Don't know how !
Your question was fundamentally wrong. There's no reason to assume the ordering of the orbitals will be the same when you ionize the atom. Period.


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