b
Magnetic moment is given by \(\mu=\sqrt{n(n+2)}\)

Here, \(n=\) number of unpaired electrons

\(\Rightarrow 2.83=\sqrt{n(n+2)}\)

\(\Rightarrow(2.83)^{2}=n(n+2)\)

\(0=8.00+n^{2}+2 n\)

\(n^{2}+ 2n-8=0\)

\((n+4)(n-2)=0\)

\(n=2\)      \(n=-\,4\)

\(-\,4\) not be consider

Hence, \(Ni^{2+}\) possesses a magnetic moment of \(2.83\,\mathrm{B} . \mathrm{M}\)