JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Fluorine forms one oxoacid.
Reason R : Fluorine has smallest size amongst all halogens and is highly electronegative.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (a) Both A and R are correct and R is the correct explanation of A.
- (b) Both A and R are correct but R is NOT the correct explanation of A.
- (c) A is correct but R is not correct.
- (d) A is not correct but R is correct.
Answer and solution
Answer (a) Both A and R are correct and R is the correct explanation of A.
Solution
Chemistry ma’am · 4 steps
Fluorine stops at one oxoacid, HOF, because it is too small and too electronegative to take more oxygens.
- Check Assertion A. Fluorine forms only one oxoacid, hypofluorous acid HOF. Chlorine, by contrast, forms four: HOCl, HClO₂, HClO₃ and HClO₄. So A is correct.
- Check Reason R. Fluorine has the smallest atomic size among the halogens and is the most electronegative element of all. So R is correct.
- Link them. In the higher oxoacids the halogen holds several oxygens and has positive oxidation states of +3, +5 or +7. Fluorine is so electronegative that it never takes a positive oxidation state. It is also so small, with no d orbitals, that it cannot bond to more than one atom. Even in HOF, fluorine stays at -1.
- So. Both statements are correct, and the small size and high electronegativity in R are exactly why fluorine has only one oxoacid. The answer is (a).
The idea to keep. Fluorine never shows a positive oxidation state and cannot expand its octet. When fluorine behaves unlike the other halogens, look to its small size, high electronegativity and lack of d orbitals.
Answer from the NTA final answer key. Solution by PrepWiser.