If I am to write the quantum numbers for the highest energy electron in Cr, or any of the transition elements in its respective row, would n=3, l=2?

I understand for the main group elements, the highest energy electron in that row it would be n=4, l= 0 or l=1 depending if its from column 1-2, or 13-18. I am confused when it comes to naming the quantum numbers for the highest energy electron of an element from the transition group.

1 Answer
Truong-Son N.
Oct 1, 2017

No, it would be of the #4s# orbital, with #n = 4# and #l = 0#.

And that should make sense, since you should have been told that the #4s# electrons leave first upon ionizing a first-row transition metal. That only makes sense if the #4s# was higher in energy.

Any first-row transition metal has access to #4s# and #3d# valence atomic orbitals. The #4s# orbitals are higher in energy than the #3d# when they are filled and lower in energy when they are empty.

And so, for the neutral element #"Cr"#, which has electron configuration

#[Ar] 3d^5 4s^1#,

it has one spin-up #4s# electron as its highest energy electron.

A reasonable explanation for why the #3d# subshell is half-filled is given here. (Note that half-filled subshells don't really mean anything to the atom.)

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