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Electron stability and Z effective question.


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I learned that electron stability depends on the Zeff value because it takes into account shielding. However, the Zeff of, say a 2s He electron is less than the Zeff of a 2p Carbon electron. BUT the He electron is obviously a lot more stable, as it is a noble gas and doesn't bond easily. I know that the He electron is closer to the nucleus where the positive charge is, but the total charge on it is STILL lower than the charge felt by a 2p carbon electron. It seems like it shouldn't matter how close it is, because in the end it is effected by a lower charge.


Would I be right in saying then that the fact that the He has a full 2s shell which is the lowest energy state is more important in terms of stability than the fact that its Zeff is lower? Or does it really just have to do with how close it is to the nucleus?

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You must consider not only the size of the atom as well as the distance these differentiating electrons are from the nucleus, but also the fact that carbon simply has more protons than a helium atom. As you go across any given row, you observe the trend that Zeff increases by 0.65. This is due to the addition of 1 proton and the addition of 1 'same-group' shielding electron, 1.00 - 0.35 = 0.65. There is no real defined 'trend' for descending a column, but Zeff does increase.


In terms of stability, all you are really looking for, at least as an undergrad, would be having a full orbital, or half-filled, and in some cases, addition to the d-group is favored over addition to the s. Obtaining the full orbital raises the ionization energies of these atoms considerably--this rise is peaked at noble gases.

Edited by aeontide
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