How does the number of protons in the nucleus affect ionization energy trends?

[orgohell]

Hey guys, I am having trouble understanding/explaining some of the trends in ionization energy with calculated Zeff by slaters rules.
For example I was asked why is the 3rd ionization energy for Be greater then the 3rd ionization energy for O


My train of thought was to say its pulling an electron from a complete shell in Be (1s²) vs. pulling an electron from the 2p orbital...but how do I explain this with Zeff?

-Zeff of the 1s electron of Be²⁺ is=4-(1 x 0.35)=3.65
-Zeff of the 2p electron of O²⁺ is=8-(2 x .85)-(3 x 0.35)=5.25

so the p electron in the O still has a higher Zeff and wouldn't that mean its held tighter to the nucleus and more difficult to remove?

What am I missing here?

 

[jbsteele]

The key is to remember that it is not just the Zeff value that determines the ionization energy but also the number of protons in the nucleus. A higher Zeff value means that the electron is held more tightly to the nucleus, but if the nucleus has more protons then the extra positive charge will make it more difficult to remove the electron. In this case, Oxygen has 8 protons while Beryllium only has 4 protons, making it easier to remove electrons from Beryllium than Oxygen.