What is magnetic moment of [FeF₆ ]^3– ?

Fe³⁺ =[Ar]3d⁵ electronic configuration
⇒x−6=−3
   x=+3
Configuration of electronic Fe³⁺=[Ar]3d⁵
the number of unpaired electrons in Fe³⁺ = 5
M.M (Magnetic moment), μ = $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ = $\sqrt{5(5+2)}$
= $\sqrt{35}$ = 5.92BM