Magnetic moment $$2.84$$  $$BM$$  is given by
$$\left( {{\text{At}}{\text{.}}\,{\text{no}}{\text{.}}\,Ni = 28,Ti = 22,} \right.$$     $$\left. {Cr = 24,Co = 27} \right)$$

Solution :
Magnetic moment, $$\mu = \sqrt {n\left( {n + 2} \right)} BM$$     where,
$$n=$$ number of unpaired electrons
$$\eqalign{ & \mu = 2.84\left( {{\text{given}}} \right) \cr & \therefore 2.84 = \sqrt {n\left( {n + 2} \right)} B.M \cr & {\left( {2.84} \right)^2} = n\left( {n + 2} \right) \cr & 8 = {n^2} + 2n \cr & {n^2} + 2n - 8 = 0 \cr & {n^2} + 4n - 2n - 8 = 0 \cr & n\left( {n + 4} \right) - 2\left( {n + 4} \right) = 0 \cr & n = 2 \cr} $$
$$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$     ( two unpaired electrons )
$$T{i^{3 + }} = \left[ {Ar} \right]3{d^1}4{s^0}$$     ( one unpaired electrons )
$$C{r^{3 + }} = \left[ {Ar} \right]3{d^3},$$     ( three unpaired electrons )
$$C{o^{2 + }} = \left[ {Ar} \right],3{d^7},4{s^0}$$     ( three unpaired electrons )
So, only $$N{i^{2 + }}$$ has 2 unpaired electrons.