effective nuclear charge

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caffeine jitters

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Hey all,
Quick question regarding the effective nuclear charge. Can't the equation for its calculation be simplified as "the number of valence electrons"? it seems that in all the examples I've seen, the effective nuclear charge is the same as the number of valence electrons.
Am I just jumping to conclusions?
Any help would be appreciated!

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Yes if you are calculating Zeff for a valence electron of a neutral atom, Zeff = Atomic number - shielding electrons which as you have noticed in that case is the same thing as the number of valence electrons. If given an ion it can be different. For instance Na+. The valence shell is now the second shell. so the number of shielding electrons is 2 instead of 10.

Zeff valence electron of Na = 11 - 10 = 1+
Zeff valence electron of Na+ = 11 - 2 - 9+
 
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Yes if you are calculating Zeff for a valence electron of a neutral atom, Zeff = Atomic number - shielding electrons which as you have noticed in that case is the same thing as the number of valence electrons. If given an ion it can be different. For instance Na+. The valence shell is now the second shell. so the number of shielding electrons is 2 instead of 10.

Zeff valence electron of Na = 11 - 10 = 1+
Zeff valence electron of Na+ = 11 - 2 - 9+
Awesome... I knew I was oversimplifying it.
Thanks for the explanation!
 
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