Solution

In square planar geometry, $$Ni^{2+}$$ orbitals are $$\boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed { }$$ The empty orbital is for hybridization. There are no unpaired electrons here. Magnetic Moment: $$\sqrt {n(n+2)} \ BM$$ where $$'n'$$ is number of unpaired electrons. Here, magnetic moment $$= \sqrt {0(0+1)}=0$$ In octahedral geometry, $$Ni^{2+}$$ obitals are $$\boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow} \boxed {\uparrow}$$ Here, it has $$2$$ unpaired electrons. $$Magnetic \ Moment = \sqrt {2(2+2)} = \sqrt {2(4)} = 2 \sqrt 2 = 2.828 = 2.83$$