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if i have N would it be 7-3= zeff?
Ah, I think it's actually 7-2=5. If you go back to the s, p, d, f shells, there are only two inner electrons that shield the valence electrons from the 7 protons in the nucleus.
so Zeff = Z - S, with Z = 7 and S = 2
The simplest way is simply to assume all inner electrons shield perfectly, so Zeff = Z - n_ve, where n_ve = number of valence electrons. That gives a Zeff that begins with +1 in Group 1A and advances by 1 for each A group, ending in +8 for Group 8A. Zeff for the transition metals is hard to get by this method, and the method generally does poorly for Period 3 and below.
You can be more sophisticated by using "Slater's Rules" (which you can google), which assign a certain fraction of a +1 charge canceled by each electron, depending on its particular n and l values.
You can also look up sophisticated calculations on webelements.com.