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

1 Answer
Truong-Son N.
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

Related questions
See all questions in s,p,d,f Orbitals
Impact of this question
8593 views around the world
You can reuse this answer
Creative Commons License