b) Electrons in an s orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge because an electron in an s-orbial can penetrate to the nucleus of the atom. (true or false)
The answer to this is True which confused me because I thought the reason why this would be true is just that the s-orbital is closer to the nucleus compared to the other orbitals; therefore, they are more effective at shielding. Is the real reason because they can penetrate the nucleus of an atom??
1E.5 7th Edition
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Re: 1E.5 7th Edition
Hi Ricky!
You're partially correct in your thinking. The s-electrons are very close to the nucleus and this plays a role in the shielding that occurs. S-electrons penetrate through the inner shells of the atom. Because p-electrons, d-electrons and so on don't penetrate these inner shells like s- electrons, they are more effectively shielded from the nucleus and experience a smaller effective nuclear charge. The s-electrons close to the nucleus also repel the p-, d-, and f- electrons, which makes them less bound to the nucleus and less effected by the nuclear charge.
Hope this helps.
-Steven
You're partially correct in your thinking. The s-electrons are very close to the nucleus and this plays a role in the shielding that occurs. S-electrons penetrate through the inner shells of the atom. Because p-electrons, d-electrons and so on don't penetrate these inner shells like s- electrons, they are more effectively shielded from the nucleus and experience a smaller effective nuclear charge. The s-electrons close to the nucleus also repel the p-, d-, and f- electrons, which makes them less bound to the nucleus and less effected by the nuclear charge.
Hope this helps.
-Steven
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Re: 1E.5 7th Edition
I'm confused about this statement from the 6th edition in chapter 2:
In other words, an s-electron is bound more tightly than a p-electron and has a slightly lower (more negative) energy
Wouldn't an electron that is more tightly held have more positive energy because the nucleus "cancels out" more charge?
In other words, an s-electron is bound more tightly than a p-electron and has a slightly lower (more negative) energy
Wouldn't an electron that is more tightly held have more positive energy because the nucleus "cancels out" more charge?
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