The covalency and oxidation state respectively of boron in $\left[\mathrm{BF}_{4}\right]^{-}$, are :

In the tetrafluoroborate anion ($\left[\mathrm{BF}_{4}\right]^{-}$), boron is bonded to four fluorine atoms through covalent bonds. Therefore, the covalency of boron in this ion is 4. <br/><br/> The oxidation state of boron can be calculated by considering the charges on the atoms involved in the anion. Fluorine has an oxidation state of -1, and there are four fluorine atoms in the anion, which contributes a total of -4. Since the overall charge on the tetrafluoroborate anion is -1, the oxidation state of boron must be +3 in order to balance the charges. <br/><br/> So, the covalency and oxidation state of boron in $\left[\mathrm{BF}_{4}\right]^{-}$ are 4 and 3, respectively.