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?

1 Answer
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 #(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):

Graphed from Data in 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.