# 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?

Jul 18, 2017

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 $\left(n - 1\right) {d}^{3} n {s}^{2}$ and $\left(n - 1\right) {d}^{5} n {s}^{2}$ transition metals are early enough in the series that their $\left(n - 1\right) d$ orbitals are similar enough in energy to their $n s$ 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 $\text{H}$ is $\text{13.61 eV}$ and the first ionization energy of $\text{N}$ is $\text{14.53 eV}$.

Here are representative actual elements with these configurations...

$\left(n - 1\right) {d}^{\textcolor{b l u e}{3}} n {s}^{\textcolor{b l u e}{2}}$: $\text{ V}$, $\text{Nb}$, $\text{Ta}$ ($Z = 23 , 41 , 73$, respectively)

$\left(n - 1\right) {d}^{\textcolor{b l u e}{5}} n {s}^{\textcolor{b l u e}{2}}$: $\text{ Cr}$, $\text{Tc}$, $\text{Re}$ ($Z = 25 , 43 , 75$, respectively)

• The first set do indeed form a maximum oxidation state of $\textcolor{b l u e}{+ 5}$, using their $\left(n - 1\right) d$ and $n s$ electrons.
• The second set do indeed form a maximum oxidation state of $\textcolor{b l u e}{+ 7}$, using their $\left(n - 1\right) d$ and $n s$ electrons.