The “spin-only” magnetic moment [in units of Bohr magneton, (${\mu }_B$)] of ${\text{Ni}}^{2+}$ in aqueous solution would be (Atomic number of $\text{Ni}=28$):

${\text{Ni}}^{2+}$ has $8$ electrons in the $d$ -subshell ($d^8$).In aqueous solution, it forms an octahedral complex $[{\text{Ni(H}}_2{\text{O)}}_6{]}^{2+}$.Configuration: $t_{2g}^6{e}_g^2$.Number of unpaired electrons ($n$) $=2$.Magnetic moment $\mu =\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}\approx 2.83\text{BM}$.