Boron is unable to form $BF_{6}^{3-}$ ion. Explain.

(N/A) Boron belongs to the second period and has an electronic configuration of $1s^2 2s^2 2p^1$.
Due to the absence of $d$-orbitals in its valence shell,boron cannot expand its octet beyond $8$ electrons.
Consequently,the maximum covalency of boron is limited to $4$.
Therefore,it cannot form the $BF_{6}^{3-}$ ion,which would require a covalency of $6$.