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I have seen two electron configurations for $\ce{_{28}Ni}$ in my chemistry textbook: $$[\ce{Ar}]~3d^8~4s^2$$ $$[\ce{Ar}]~3d^9~4s^1$$ Which of those is ACTUALLY the correct one (in real life, reality)? Why exactly?

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NIST suggests here and here (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, as Wikipedia describes, if you don't specify the term symbol for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.

Now, why one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in this previous Chem.SE question and the several nested questions within.

Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.

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