For b, why are s electrons better at shielding than p electrons?
For d, how come effective nuclear charge is lower for p electrons than s electrons in the same shell? Doesn't higher effective nuclear charge = more protons, and p-orbital elements have higher atomic numbers?
Question on 1E.5
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Re: Question on 1E.5
B) s electrons tend to be better at shielding because they are closer to the nucleus.
D) effective nuclear charge is the net positive charge experienced by VALENCE electrons. Think about it this way, if there are more shielding electrons between the nucleus and the outer shell, the net positive charge on the valence electrons is less than if the outer shell had less shielding between the nucleus and valence electrons. Since s is closer to the nucleus and has less shielding, the net positive charge on the valence electrons is greater than in the p block which has more shielding.
D) effective nuclear charge is the net positive charge experienced by VALENCE electrons. Think about it this way, if there are more shielding electrons between the nucleus and the outer shell, the net positive charge on the valence electrons is less than if the outer shell had less shielding between the nucleus and valence electrons. Since s is closer to the nucleus and has less shielding, the net positive charge on the valence electrons is greater than in the p block which has more shielding.
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Re: Question on 1E.5
b) To answer your question for b, s orbitals are more effective in shielding because remember that s orbitals are a sphere and have a higher probability of being closer to the nucleus whereas the p orbital has a lower probability (there is a node at the center). This indicates that effective nuclear charge is higher for electrons in s orbital.
d) For d, this builds off the reasoning in b that I gave. The greater the effective nuclear charge there will be less electron-electron repulsion and thus a smaller atom (you will find the electrons at a lower energy level). Remember that there is more electron-electron repulsion at higher orbitals. Hope this helps!
d) For d, this builds off the reasoning in b that I gave. The greater the effective nuclear charge there will be less electron-electron repulsion and thus a smaller atom (you will find the electrons at a lower energy level). Remember that there is more electron-electron repulsion at higher orbitals. Hope this helps!
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