Magnetic moment of Fex+ ion is \(\sqrt{35}\) B.M.What is the number of unpaired electrons?

5 6 4 3

5

Magnetic moment is given by, u = \(\sqrt {n(n+2) }\) \(\sqrt {35}\) = \(\sqrt {n(n+2) }\) n = 5 (Unpaired electrons) Fe is present in +3 oxidation state i.e. [Ar]3d5 configuration.