JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Chemical Bonding and Molecular Structure

In $[{\mathrm{BF}}_4{]}^{-}$, boron forms four covalent bonds with four fluorine atoms, so the covalency is 4. To find the oxidation state: let the oxidation state of B be $x$. Since each F has oxidation state −1 and the overall charge is −1: $x+4(-1)=-1$, giving $x=+3$. Therefore, covalency = 4 and oxidation state = +3.