Posted by Kimora on Wednesday, December 10, 2008 at 4:51pm.
Confirm that the experimentally observed electronic configuration of K, 1s2 2s2 2p6 3s2 3p6 4s1 is energetically more stable than the configuration of 1s2 2s2 2p6 3s2 3p6 3d1
i know that it is energetically favorable for the electron to occupy the 4s state rather than the 3d orbital.but why?
The only thing I can tell you, without quantum mechanics and all the higher math that entails (which I can't do anyway), is that the 4s level is lower than the 3d when that electron goes into the 4s instead of the 3d.
i got it (thanks for the input)
i worked out the effective nuclear charge for both states and found out that for a 4s1 electron, Z* was 2 and for the 3d1, it was 1
thus, the 3d1 electron is shielded more than the 4s1 electron..so it could be said that the electron would occupy the 4s1 state than the 3d1 state because the electron is less shielded from the nucleus in this state thus it is easier for the electron to go there.
i'm a bit confused with the wording
I think an appropriate comment is that the effective nuclear charge is nothing more than a semi-empirical concept and empirical concepts actually don't explain anything. They are there and they work and they are based on experiment but by experiment we know where the electron goes anyway.
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