Why does the #ns# orbital go before the #(n-1)d# orbital when writing transition metal electron configurations?

1 Answer
Sep 27, 2017

It doesn't. The #(n-1)d# orbitals, PARTICULARLY in the last-column transition metals, are consistently lower in energy than the #ns# orbitals... and as such, we should be writing #3d^10 4s^2#, #4d^10 5s^2#, and #5d^10 6s^2#.


See how the #3d# orbitals are lower in energy than the #4s# for the first-row transition metals here:

Data from Appendix B.9 of Inorganic Chemistry by Miessler, Fischer, and Tarr

And you can further see how the Aufbau principle fails for the heavier transition metals, in that the #(n-1)d# and #ns# orbital potential energies criss-cross rather unpredictably:

Data from Appendix B.9 of Inorganic Chemistry by Miessler, Fischer, and Tarr

Data from Appendix B.9 of Inorganic Chemistry by Miessler, Fischer, and Tarr