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Given the reactions: (i) H2O2 + NH2OH -> H2O + A (ii) Na2O2 + Fe(II) salt (fused) -> B + Na+ (iii) Ce(SO4)2 + H2O2 -> C + O2 + H+ Let x = oxidation state of N in A, y = oxidation state of Fe in B, z = oxidation state of Ce in C. Find (x + y + z) / (atomicity of compound A).

1. 2
2. 3
3. 4
4. 5

Correct answer: 4

Solution

Reaction (i): H2O2 + NH2OH -> H2O + NO. Check: H2O2 is oxidant, NH2OH (N at -1) -> NO (N at +2). Balance: H2O2 + NH2OH -> 2H2O + NO. Atoms: Left H=4,N=1,O=3; Right H=4,N=1,O=3. Balanced. x = +2. Reaction (ii): Na2O2 oxidises Fe2+ to Fe3+; B = Fe2O3; y = +3. Reaction (iii): Ce4+ (oxidant) is reduced by H2O2 (reductant, O goes from -1 to 0 as O2); C = Ce2(SO4)3; z = +3. x+y+z = 2+3+3 = 8. Atomicity of NO = 2. Answer = 8/2 = 4.

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