Does an element with an electron configuration of #["Noble gas"]ns^2(n-1)d^8# have #10# valence electrons?

1 Answer
Oct 9, 2017

It wouldn't make sense to say there are #10# valence electrons; if you think about it, have you ever seen a #+10# oxidation state occur in any metal? The highest I've seen is #+9# on iridium and #+8# on osmium. Otherwise, they usually peak out at #+7# or below.


Examples of elements with this (expected) electron configuration would be:

  • #"Ni"#
  • #"Pd"# (nope, it's #s^0d^10#!)
  • #"Pt"# (nope, it's #s^1d^9#!)

The #(n-1)d# orbitals this late in the transition metal series are significantly lower in energy than the corresponding #ns# orbitals. As a result, it is more favorable that through ionizations, the electron(s) that come out first are from the #ns# orbitals.

In other words, for late transition metals, the #(n-1)d# orbitals are sufficiently core-like, and the #ns# electrons are typically valence. This is especially true for #"Ni"#, which is the only example of an #ns^2(n-1)d^8# configuration.

See here for data used below.

Graphed from Data in Appendix B.9

Graphed from Data in Appendix B.9

Graphed from Data in Appendix B.9