How come electrons in the d-subshell are considered valence electrons in transition metals?

Question

Similarly, why are f-subshell electrons considered valence electrons in the lanthanides and actinides?

I was under the impression that the outermost electrons, and therefore the electrons with the highest principal quantum number, made up the valence electrons.

Also, if the d-subshell electrons are considered valence electrons in transition metals, how come they're not considered valence electrons for elements in Groups 13-18?

I understand that the highest d-subshell is of higher energy than the highest s-subshell but I fail to see the logic in how it would be considered a valence electron.

Answer 1

The valence electrons are the ones responsible for engaging in chemistry and corresponds to the orbitals which are highest in energy. These electrons are the most easily accessed because they are not buried in the energy well of an atom. For main group elements, the highest principal quantum number p and s orbitals are the highest in energy, as you said, but for the transition metals, the partially filled d orbitals are the highest in energy. Generally, whichever orbitals are being filled last as you go across the periodic table are the ones that should be considered the valence electrons.

Another thing to note, as elements are ionized, the orbitals with higher orbital angular momentum drop in energy faster than orbitals with lower angular momentum. So, in the transition metals, the n d orbitals drop much more than the (n+1)s and p orbitals. Since, transition metals have <10 electrons (the number it would take to fill up all the d orbitals), all of them reside in d orbitals (and none in the s and p) and so these are the highest occupied orbitals and therefore the ones that engage in chemistry.

Answer 2

When you say valence shell, you have to pay careful attention to the word valence.

What does valence mean? Wikipedia says a bit about it's etymology:

"The etymology of the word "valence" traces back to 1425, meaning "extract, preparation", from Latin valentia "strength, capacity", from the earlier valor "worth, value", and the chemical meaning referring to the "combining power of an element" is recorded from 1884, from German "Valenz".

Clearly, valence refers to the combining power of the element.

Which shell decides the combining power of an element? The $(n+1)s$ or $nd$ subshell? To further prove my point, consider these examples:

• Iron has its most prominent oxidation states (valencies) are +2 and +3. The respective electronic configurations are $\ce{[Ar]}3d^6 4s^0$ and $\ce{[Ar]}3d^5 4s^0$. The $\ce{Fe^{3+}}$ is extra stable since it has a half filled $d$ subshell. Does the $s$ orbital play any role here? Absolutely not.
• In acidic medium, the permanganate ion ($\ce{MnO4^-}$) is a powerful oxidizing agent, reducing itself to $\ce{Mn^{2+}}$. Why $\ce{Mn^{2+}}$? Recall that it's electronic configuration is also $\ce{[Ar]}3d^5 4s^0$. Is it because of the $s$ shell? Think carefully.

While you may have heard that valence shell means the outermost shell, it's not always the case, and never the case in transition and inner transition metals.