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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.When $Ni$ forms a $Ni^{2+}$ cation,it loses two electr

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$2$

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(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.When $Ni$ forms a $Ni^{2+}$ cation,it loses two electrons from the $4s$ orbital.Thus,the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.In the $3d^8$ configuration,the electrons are filled in the five $3d$ orbitals according to Hund's rule:- The first five electrons occupy the five orbitals singly.- The next three electrons pair up in the first three orbitals.This leaves two orbitals with single electrons.Therefore,there are $2$ unpaired electrons in the $Ni^{2+}$ cation.

How many unpaired electrons are present in the $Ni^{2+}$ cation (atomic number $= 28$)?

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Given below are the quantum numbers for $4$ electrons.
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