Solution

Electronic configuration of $$Fe$$ is $$[Ar] 3d^6 4s^2$$ Oxidation state of Fe in $$[FeF_6]^{3-}$$ is $$x$$. $$ \Rightarrow x - 6 = -3$$ $$x = +3$$ Electronic configuration of $$Fe^{3+} = [Ar]3d^5$$ No. of unpaired electrons in $$Fe^{3+} = 5$$ Magnetic moment, $$\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}$$ $$=\sqrt{35}=5.92\ BM$$ Hence, option B is correct.