The number of d-electrons in Ni (Atomic number = 28) is equal to that type of

The atomic number of Nickel (Ni) = 28

Thus, the electronic configuration of Ni is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$ or$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{8}4{\mathrm{s}}^2$

 and the number of d electrons in Ni is 8.

Now,

Option A: For Ni²⁺, number of electrons = 26 and the configuration is- $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^8$

Thus, the number of s and p electrons in Ni²⁺ is 18.

Option B: The electronic configuration of Argon (Ar) is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^6$. Thus, the number of p- electrons in Ar is 12.

Option C: For Ni²⁺, number of electrons = 26 and the configuration is- $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^8$. Thus, there are 8 d- electrons in Ni²⁺.

Option D: As atomic number of Nitrogen is 7, that means it has 7 electrons.

Hence, the number of d-electrons in Ni (Atomic number = 28) is equal to that type of d-electrons in ${\mathrm{Ni}}^{2+}$.