No. That would imply that the #4s# orbital is not filled, but it is. Nickel has two #4s# electrons, and eight #3d# electrons.
As it turns out, as nickel "fills" its orbitals with electrons, the energy gap between the #4s# and #3d# is small enough (here, it is #"4.71 eV"# [Appendix B.9], a bit over 1/3 of the #"H"# atom ionization energy).
So, the last two electrons that would "fill" nickel's orbitals are favored to be in the #4s# instead to avoid repulsions (since the #3d# orbitals, with higher #l#, are more compact and more repulsive).
This is apparently on the border, as copper has a valence configuration of #3d^10 4s^1#, and zinc has #3d^10 4s^2#, and the energy gap increases quite a bit when we get there:
For copper, the #3d# orbitals are borderline in energy, so that we can fill the #3d# before filling the #4s#.
For zinc, the #3d# orbitals are so low in energy (dropped by about #"3.93 eV"# from copper) that they are filled first, and then the #4s# are filled afterwards.
