Question

Given `(n-1)d^3 ns^2` and `(n-1)d^5 ns^2`, of the 2 electron configuration which one would exhibit higher oxidation state explain?

Answer 1

Well, oxidation states are hypothetical charges assuming full transfer of a certain number of valence electrons. If the element is capable of losing more valence electrons (when the conditions are right), they can lose that many.

The `(n-1)d^3 ns^2` and `(n-1)d^5 ns^2` transition metals are early enough in the series that their `(n-1)d` orbitals are similar enough in energy to their `ns` orbitals, making them accessible.

That allows all those electrons to be valence if need be. To show how close in energy they are, here are the first-row transition metal energies (Appendix B.9):

For perspective, the first ionization energy of `"H"` is `"13.61 eV"` and the first ionization energy of `"N"` is `"14.53 eV"`.

Here are representative actual elements with these configurations...

`(n-1)d^color(blue)(3) ns^color(blue)(2)`: `" V"`, `"Nb"`, `"Ta"` (`Z = 23, 41, 73`, respectively)

`(n-1)d^color(blue)(5) ns^color(blue)(2)`: `" Cr"`, `"Tc"`, `"Re"` (`Z = 25, 43, 75`, respectively)

• The first set do indeed form a maximum oxidation state of `color(blue)(+5)`, using their `(n-1)d` and `ns` electrons.
• The second set do indeed form a maximum oxidation state of `color(blue)(+7)`, using their `(n-1)d` and `ns` electrons.

And all of their possible oxidation states can be seen here.