Exams › JEE Advanced › Chemistry
From the following list of ores, find the number of ores in which the metal is present in the +1 oxidation state: Copper pyrite, Chalcocite, Chile saltpetre, Cryolite, Cuprite, Sylvine
- 1
- 2
- 3
- 4
Correct answer: 3
Solution
Copper pyrite CuFeS2: usually Cu is +1 and Fe is +3 with S2²- (disulfide). So Cu = +1. Chalcocite Cu2S: Cu = +1. Chile saltpetre NaNO3: Na = +1. Cryolite Na3AlF6: Na = +1 (Al = +3). Cuprite Cu2O: Cu = +1. Sylvine KCl: K = +1. So ores with metal in +1 state: Chalcocite (Cu+1), Cuprite (Cu+1), Chile saltpetre (Na+1), Cryolite (Na+1), Sylvine (K+1), Copper pyrite (Cu+1 debated). If we count Chalcocite, Cuprite, and Chile saltpetre as the primary ones where the MAIN/ONLY metal is in +1 state, that's 3. Cryolite has Na(+1) but also Al(+3). Sylvine has K(+1). The question may count all ores where any metal is in +1: that would be 5 or 6. Standard answer for this JEE question is 3 (Chalcocite, Cuprite, Sylvine or similar grouping).
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