In the complex ion $[CoF_6]^{3-}$, the central cobalt atom is in the $+3$ oxidation state. The ground state electronic configuration of Co (Z = 27) is $[Ar] 3d^7 4s^2$, so $Co^{3+}$ has a $3d^6$ configuration. Fluoride ($F^-$) is a weak field ligand, so it does not cause forced pairing of the inner $3d$ electrons. The complex uses outer $4d$ orbitals and undergoes $sp^3d^2$ hybridization, forming a high-spin (outer orbital) complex. According to Hund's rule, the six $3d$ electrons are arranged to give the maximum number of unpaired electrons (configuration $t_{2g}^4 e_g^2$ in CFT). Therefore, there are 4 unpaired electrons.