[5416339]

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,
-5416339

[BarbieCue]

omg! i really wish someone could answer thissss .. we share the same problem.

[cliverlong]

http://www.chemguide.co.uk/inorganic/transition/features.html

Transition metal ions

You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because in the empty atom, 4s orbitals have a lower energy than 3d orbitals.

However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital.

The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons.

Note:  This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

Remember this:
When d-block elements form ions, the 4s electrons are lost first.

[ardbeg]

Cu2+ is 3d9 BTW.

[opti384]

Note:  This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

I wonder how much further studies are required for the explanation. Any ideas?

[cliverlong]

Note:  This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

I wonder how much further studies are required for the explanation. Any ideas?

A lifetime maybe? Or at least university level study.

However, try these for a start

http://en.wikipedia.org/wiki/Electron_configuration#Aufbau_principle_and_Madelung_rule

And from that section

Ionization of the transition metals

The naive application of the Aufbau principle leads to a well-known paradox (or apparent paradox) in the basic chemistry of the transition metals. Potassium and calcium appear in the periodic table before the transition metals, and have electron configurations [Ar] 4s1 and [Ar] 4s2 respectively, i.e. the 4s-orbital is filled before the 3d-orbital. This is in line with Madelung's rule, as the 4s-orbital has n+l  = 4 (n = 4, l = 0) while the 3d-orbital has n+l  = 5 (n = 3, l = 2). However, chromium and copper have electron configurations [Ar] 3d5 4s1 and [Ar] 3d10 4s1 respectively, i.e. one electron has passed from the 4s-orbital to a 3d-orbital to generate a half-filled or filled subshell. In this case, the usual explanation is that "half-filled or completely-filled subshells are particularly stable arrangements of electrons".

The apparent paradox arises when electrons are removed from the transition metal atoms to form ions. . (read on)