To solve this problem, we need to determine which element cannot form the $MF_6^{3-}$ ion.$\newline$The formation of $MF_6^{3-}$ ions involves the central metal atom forming six bonds with fluorine atoms.$\newline$This requires the central atom to have vacant d-orbitals to accommodate the lone pairs from the fluorine atoms.$\newline$Let's analyze each option:$\newline$• Option 1: Boron (B)$\newline$ - Boron belongs to Group 13 and has the electronic configuration $1s^2 2s^2 2p^1. \newline$ - Boron does not have vacant d-orbitals in its valence shell (n=2) to expand its coordination number to 6.$\newline$ - Therefore, boron cannot form $BF_6^{3-}. \newline$• Option 2: Aluminum (Al)$\newline$ - Aluminum has the electronic configuration $[Ne] 3s^2 3p^1. \newline$ - Aluminum can utilize its 3d orbitals to expand its coordination number, allowing it to form $AlF_6^{3-}. \newline$• Option 3: Gallium (Ga)$\newline$ - Gallium has the electronic configuration $[Ar] 3d^{10} 4s^2 4p^1. \newline$ - Gallium can use its 4d orbitals to form $GaF_6^{3-}. \newline$• Option 4: Indium (In)$\newline$ - Indium has the electronic configuration $[Kr] 4d^{10} 5s^2 5p^1. \newline$ - Indium can utilize its 5d orbitals to form $InF_6^{3-}. \newline$Therefore, the element that is unable to form $MF_6^{3-}$ is Boron (B).$\newline$This corresponds to Option 1.