Question

Zeff for the 4s electron in a copper atom, Cu

Answer 1

I'm assuming that you are using Slater's rules for effective nuclear charge. The general idea is to subtract the effect of shielding from the actual nuclear charge, Z. The formula is different for s electrons and d electrons, but the number of protons is the same for the same atom, copper, Z = 29. It is best if you write out the electron configurations out so you can see how many electrons are in each sublevel.

Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

For the 4s electron:

29 - [18(0.85) + 10(1.0)] = 3.7

For the 3d electron:

29 - [9(0.35) + 18(1.0)] = 7.85

You can conclude that the 4s electron experiences a lower effective nuclear charge and is therefore more likely to be lost than the 3d electron, should the copper atom become ionized.

Hope this helps!

Answer 2

I'm assuming that you are using Slater's rules for effective nuclear charge. The general idea is to subtract the effect of shielding from the actual nuclear charge, Z. The formula is different for s electrons and d electrons, but the number of protons is the same for the same atom, copper, Z = 29. It is best if you write out the electron configurations out so you can see how many electrons are in each sublevel.

Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

For the 4s electron:

29 - [18(0.85) + 10(1.0)] = 3.7

For the 3d electron:

29 - [9(0.35) + 18(1.0)] = 7.85

You can conclude that the 4s electron experiences a lower effective nuclear charge and is therefore more likely to be lost than the 3d electron, should the copper atom become ionized.