The covalency and oxidation state respectively of boron in [BF4]-, are:

🧠 Covalency vs Oxidation State

Two different concepts:

• Covalency = total number of bonds the central atom forms (whether normal covalent or coordinate).
• Oxidation state = formal charge from electronegativity-based electron assignment.

For $[\mathrm{BF_4}]^-$: boron is bonded to 4 fluorines (3 normal B–F + 1 coordinate from $\mathrm{F^-}$). So covalency = 4. OS calculation: $x + 4(-1) = -1 \Rightarrow x = +3$.

🗺️ Both Numbers

Covalency. Boron makes 4 bonds in $[\mathrm{BF_4}]^-$. Three are formed in $\mathrm{BF_3}$; the fourth comes from a fluoride donating its lone pair into boron's empty p-orbital (Lewis acid–base adduct). Total bonds at B = 4.

Oxidation state. F is more electronegative than B, so each F takes both bond electrons → F gets OS $-1$. Then $x + 4(-1) = -1 \Rightarrow x = +3$. So OS = +3.

So pair = (4, +3) → option (4).

⚡ The Lewis-Acid Adduct Pattern

$\mathrm{BF_3}$ + $\mathrm{F^-}$ → $[\mathrm{BF_4}]^-$ is a textbook coordinate bond reaction: covalency rises by 1 (3 → 4) but oxidation state stays at +3. Same pattern for $\mathrm{BF_3 + NH_3 \to F_3B \leftarrow NH_3}$: covalency 4, OS +3.

⚠️ Don't Conflate the Two Numbers

Covalency tracks bond count; OS tracks formal charge. Adding a coordinate bond changes covalency but not OS. Many students see four bonds and write OS +4 (option 3) — a common slip on exam day.

$\boxed{\text{Answer: (4) 4 and 3}}$