The correct statement is

The electronic configuration of Boron (atomic number $5$) is $1s^{2} 2s^{2} 2p^{1}$. Its valence shell is the second shell ($n=2$), which contains one $2s$ orbital and three $2p$ orbitals, giving a total of four valence orbitals. Due to the absence of $d$-orbitals in its valence shell, Boron cannot expand its octet and has a maximum covalency of four (e.g., in $[BF_{4}]^{-}$).

Aluminium ($n=3$) has one $3s$, three $3p$, and five $3d$ orbitals in its valence shell, making a total of nine valence orbitals.

Beryllium ($n=2$) has one $2s$ and three $2p$ orbitals, making a total of four valence orbitals.

Magnesium ($n=3$) has vacant $3d$ orbitals in its valence shell, allowing it to expand its covalency beyond four (up to six, as in $[Mg(H_{2}O)_{6}]^{2+}$).

Answer: Boron has a maximum covalency of four.