Question

Why is scandium outer electron configuration `4s^2 3d^1` and not just `3d^3`?

Answer 1

Because God wanted it that way...?

Explanation:

Qualitatively, we can assume that the `4s` orbital should be slightly stabilized with respect to the `3d` orbitals in that the `s` orbitals have some small probability of occurrence near the scandium nucleus. There are some very few examples of reduced, i.e. `"hypovalent"` scandium metal complexes, and their electronic structure is not very well explained.

Answer 2

See below:

Explanation:

The 3d sub level is just below the 4s sub level for the 1st transition series so you might correctly think that these should fill first to give `sf(3d^3)`.

However there is more than one force at work here. Electrons in the same sub shell which have their spins parallel are stabilised by a factor called "Exchange Energy". This is a quantum mechanical effect for which there is no classical analogue.

Another counter factor at work is the coulombic repulsion experienced by electrons in the same sub shell. In this case the coulombic repulsion outweighs the extra stability afforded by the exchange energy which tends to "push" the electrons up into the larger 4s orbital where repulsion is less.

This means that the 4s are the outer electrons which define the atomic radius of the atom and are the first electrons to be lost when ions form.