Solution

$$Co^{+2}$$ contains $$3d^{7}$$ electrons in the outermost orbit. As it is showing the square planar geometry, the electronic configuration will be $$d_{xz}^{2}$$, $$d_{yz}^{2}$$, $$d_{z^{2}}^{2}$$, $$d_{xy}^{1}$$. Hence the number of unpaired electrons ($$n$$) will be $${1}$$. By substituting this value in the formula of magnetic moment $$\mu =\sqrt { n\left( n+2 \right) } $$. $$\mu $$ will be $$1.73\ B.M.$$