Ask A Scientist Chemistry Archive

name         Joe M.
status       educator
age          50s

Question -   In filling electron orbitals in multi-electron systems,
the 4s, for example, is filled before the 3d.  However, in losing
electrons in oxidation, the transition elements lose the 4s before the 3d
(or so I have gleaned from many textbooks!).  We are never given a reason.
The 3d has less negative energy, so it should be lost first if we consider
only energy.  Is it that the 4s is at times further from the nucleus?
I have always been confused about this and would appreciate a clear answer
in order to explain to me students.
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Hi, the 3d does not actually have less negative energy than the 4s. It was
once believed that this was the case, but many experimental measurements
and theoretical calculations have shown that this is not the case. If
spin is not included in a calculation of atomic energy levels you do
indeed always get 3d less than 4s, but when you put electrons into
the states according to the Pauli Exclusion  Principle, the order
inverts.

The usual explanation is that the core electrons
shield the nucleus partially, so that the aspect of a valence orbital
which becomes most important is how well it can penetrate the core
electron density. The 3d electrons penetrate more efficiently than
4s electrons would (they have a smaller principal quantum number after all)
and so the occupied 4s subshell gets pushed up to a less negative energy
than the 3d subshell.
The reason that 4s generally has 2 electrons as you go through the
first period of transition metals is
because of the special stability achieved when all of an atom's subshells
are either filled (like 4s1) or half-filled (like 4s2).
This stability of half-filled and filled subshells does not always occur but
it is clearly the case that it exists and is a useful rule of thumb.

The stabilization is principally due to two effects. The first is  known as
"spin-orbit coupling." Spin-orbit coupling refers to the fact
that electrons have magnetic properties intrinsically (spin) and
extrinsically  (each orbital has a different magnetic moment) and so
the spin magnetic moment interacts with the orbital magnetic moment.
The second effect is "exchange" and has to do with the fact that when
two Fermions (like electrons) exchange places, the wave function of the
atom must change sign. For reasons which are very  hard to explain without
writing down the Schrodinger equation and solving it using Hartree-Fock
theory, this effect tends to lower the energy proportional to the
number of identical spin states.

I am sorry that this was not a simple answer but I hope that it helped.

Prof. Topper
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