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

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

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Answer 1 · 2017-09-27T17:41:01

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

See how the $3 d$ $3d$ orbitals are lower in energy than the $4 s$ $4s$ 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 $(n - 1) d$ $(n-1)d$ and $n s$ $ns$ orbital potential energies criss-cross rather unpredictably:

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

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

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