[vhpk]

Hi all,
I'm confused with Z_eff, can u help me with this?
Firstly, Z_eff depends on shielding electrons and the nuclear charge, right?
Secondly, in the period, when Z increases by one, 1 more proton is added to the nucleus and 1 more electron is added to the shell. But this electron isn't perfectly screened from the nucleus by the other electrons in the valence shell, therefore, Z_eff will increase ( since there are more electrons) but the Z_eff won't increase by one, but less than one .  Are all these right ?
Thirdly, down the group, I was taught that the Z_eff decreases because " orbitals of higher n values possess penetrating power toward the nucleus, and are thus not completely screened from the nuclear charge". However, when I surf the web, some say that it is almost unchanged. So which view point is correct ?
Finally, I can explain why Z_eff of AO2s of lithium is less than 1 because of the incomplete screening of the AO1s. However I can't explain why the Z_eff of AO2p of C is larger than 4.
Any help would be highly appreciated .
Thanks in advance

[vhpk]

Can anyone help me with this 

[renge ishyo]

Well, some of the wording in your first post was tricky but I will try to set it straight.

1. Yes, Z_eff is related to the shielding of outer electrons from inner electrons. It is always lower than the predicted nuclear charge "Z" that you would get from simply counting the protons because the effect of the shielding is to weaken the attraction for the electrons positioned on the outside shell of the atom.

2. When you move across the period the Z will increase by 1 since you are adding one proton as you move to the right from one element to the next one over. Z_eff will increase as well, but it will not increase by the full charge of 1 since the electron shielding will keep the newly added outer electron from feeling the full nuclear charge.

3. The third statement is somewhat confusing and the answer I think would depend on the orbital in question. For instance consider Lithium. The 2s electron is shielded from the inner electrons. So you would expect Z_eff to be less than the Z for lithium (which is 3), and it is. However, the decrease from Z to Z_eff is not as great as you might think because the 2s orbital can penetrate deeply towards the nucleus (see the picture on this site: http://www.wou.edu/las/physci/ch412/Periodic%20trends/periodic_trends.htm, notice that at least part of the blue 2s orbital still penetrates as close as the 1s orbital where it would feel more charge than the other part of the 2s orbital which is completely outside the 1s orbital). So when moving down the perodic table my advice is to repeat whatever your teacher says! The reason being that the issue is somewhat of a gray area which depends on the particular orbitals we are talking about.

[vhpk]

Thanks for your nice help. However, I still have no idea why Z_eff of AO2p of C is larger than 4 but not smaller than 4

[renge ishyo]

That one I dodged because the 2p orbital of carbon does indeed have a Z_eff less than 4...not sure what your source is, but here is mine: http://www.webelements.com/carbon/orbital_properties.html