Quantum numbers of 20th electron of mn, M=0,n=4,l=0,s=+/-1/2 Is it correct or incorrect???

1 Answer
Truong-Son N.
Apr 25, 2018

Well, the so-called #20#th electron is a #3d# electron... not a #4s# electron.

We choose arbitrarily the electron in the #4s# orbital of #"Mn"#, which has #n = 4# and #l = 0#, i.e. it is in the 4th energy level and has zero angular momentum. Thus, it must have #m_l = 0#, since #-l <= m_l <= l#.

However, #"Mn"# has a valence configuration of #3d^5 4s^2#, so the #4s# electron can be either spin up or spin down...

Therefore, #n = 4#, #l = 0#, #m_l = 0#, and #m_s = pm1/2#.

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