The electronic configuration of nickel is

Ni: [Ar] $3d^84s^2$

Here, while writing the configuration, we fill the 3d after 4s.Hence the 28th electron enters the d orbital.


$\ce{Ni+}$: [Ar] $3d^84s^1$

Here we take out the last electron (28th) from 4s. Are there any rules or conventions regarding this or am I conceptually misguided.

  • 2
    $\begingroup$ Yup, the transition metals do some weird things. For Ni, it prefers to keep the 3d shell intact. The point to remember is that real atoms have multiple electrons, so the nice neat picture of orbitals you learn based on hydrogen just don't hold. What is more amazing is that the simple picture holds up pretty well. $\endgroup$
    – Jon Custer
    May 10, 2017 at 16:35

1 Answer 1


When you ionize an element, all the orbitals go down in energy but orbitals with higher angular momentum drop more than those with lower angular momentum. When you start ionizing Ni, the $d$-orbitals drop much more than the $s$-orbitals and so, in the ions, the highest energy orbitals are the $4s$.

Relatedly, I believe (and I just ran a quick calculation to confirm) your electron configuration for $\ce{Ni+}$ is incorrect and should be [Ar]$3d^9$ for the same reason.


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