To solve this problem, we need to analyze the coordination complexes $[Co(NH_3)_6]^{3+}$ and $[CoF_6]^{3-}. \newline$Statement I: Both complexes are octahedral. Let's verify this:$\newline$• $[Co(NH_3)_6]^{3+}$ is an octahedral complex because it has six ligands (NH$_3$) around the central Co ion.$\newline$• $[CoF_6]^{3-}$ is also octahedral with six fluoride (F) ligands around the Co ion.$\newline$Therefore, Statement I is true.$\newline \newline$Statement II: Let's analyze the magnetic behavior of each complex:$\newline$• $[Co(NH_3)_6]^{3+}$ involves Co in the +3 oxidation state, which has an electronic configuration of $d^6. \newline$• NH$_3$ is a strong field ligand, causing pairing of electrons in the $t_{2g}$ orbitals.$\newline$• This results in no unpaired electrons, making $[Co(NH_3)_6]^{3+}$ diamagnetic.$\newline \newline$• $[CoF_6]^{3-}$ also has Co in the +3 oxidation state with a $d^6$ configuration.$\newline$• F is a weak field ligand, which does not cause pairing of electrons.$\newline$• This results in four unpaired electrons, making $[CoF_6]^{3-}$ paramagnetic.$\newline \newline$Therefore, Statement II is true.$\newline \newline$Since both statements are true, the correct answer is Option 3.