JEE Main 2022 · 25 July · morning · Single correct · +4 / −1
The interhalogen compound formed from the reaction of bromine with excess of fluorine is a :
- (a) hypohalite
- (b) halate
- (c) perhalate
- (d) halite
Answer and solution
Answer (b) halate
Solution
Chemistry ma’am · 3 steps
Find the oxidation state of bromine in the product, then find the oxyanion that has the same one.
- The product. With excess fluorine, bromine goes as far as it can and forms bromine pentafluoride.
Br₂ + 5F₂ → 2BrF₅
- Oxidation state. Fluorine is always -1, so in BrF₅ bromine is +5. The options name the halogen oxyanions, and each one has a fixed oxidation state for the halogen.
Oxyanion Example Oxidation state of Br Matching fluoride hypohalite BrO⁻ +1 BrF halite BrO₂⁻ +3 BrF₃ halate BrO₃⁻ +5 BrF₅ perhalate BrO₄⁻ +7 BrF₇ (does not exist) - So. Bromine is +5 in BrF₅, the same as in the bromate ion BrO₃⁻. The compound corresponds to a halate, option (b).
The idea to keep. To link an interhalogen to an oxyanion, match the central halogen's oxidation state: +1 hypohalite, +3 halite, +5 halate, +7 perhalate.
Answer from the NTA final answer key. Solution by PrepWiser.