The covalency and oxidation state respectively of boron in BF$$_4^-$$, are

Boron trifluoride, $$BF_3$$, is a neutral molecule in which boron is bonded to three fluorine atoms. In the tetrafluoroborate ion, $$BF_4^-$$, one extra $$F^-$$ ion donates a lone pair to boron, creating a coordinate (dative) $$\sigma$$-bond. Thus boron is surrounded by four $$F$$ atoms.

Step 1 - Covalency of boron
Covalency is defined as the total number of $$\sigma$$-bonds an atom forms with surrounding atoms. In $$BF_4^-$$ boron forms four B-F $$\sigma$$-bonds (three normal + one coordinate). Therefore the covalency of boron is $$4$$.

Step 2 - Oxidation state of boron
Assign oxidation numbers using the rule that fluorine always has an oxidation state of $$-1$$.

Let the oxidation state of boron be $$x$$. For the ion $$BF_4^-$$:

$$x + 4(-1) = -1$$

$$x - 4 = -1$$

$$x = +3$$

Hence the oxidation state of boron is $$+3$$.

Result
Covalency = $$4$$, Oxidation state = $$+3$$.

Therefore the correct choice is Option D.