JEE Main 2023 · 13 April · evening · Single correct · +4 / −1
The covalency and oxidation state respectively of boron in [BF₄]⁻, are :
- (a) 3 and 4
- (b) 4 and 3
- (c) 4 and 4
- (d) 3 and 5
Answer and solution
Answer (b) 4 and 3
Solution
Chemistry ma’am · 3 steps
Covalency counts the bonds boron makes; oxidation state comes from balancing the charges. Keep them separate.
- Covalency. Boron forms three normal B–F bonds. The extra F⁻ then gives a lone pair into boron's empty p orbital, a coordinate bond. That makes 4 bonds in all, so the covalency is 4.
- Oxidation state. Each F is -1 and the whole ion carries -1. Let boron be x.
x + 4(-1) = -1 ⇒ x = +3
- So. Boron has covalency 4 and oxidation state +3, which is option (b).
The idea to keep. Covalency is the number of bonds an atom forms, including coordinate bonds. Oxidation state is pure charge bookkeeping. The two need not match.
Answer from the NTA final answer key. Solution by PrepWiser.