The "spin-only" magnetic moment [in units of Bohr magneton, $\left({\mu }_{\mathrm{B}}\right)$] of ${\mathrm{Ni}}^{2+}$ in aqueous solution would be (atomic number Ni = 28)

The expression of spin-only magnetic moment is $L=\sqrt{n(n+2)}{\mu }_B$

The configuration of ${\mathrm{Ni}}^{2+}$ is $(3d{)}^8$. It contains two unpaired electrons. Hence $L=\sqrt{2\left(4\right)}{\mu }_{\mathrm{B}}=2.83{\mu }_{\mathrm{B}}$