Chemistry - CHEMICAL BONDING Question with Solution | TestHub
An element may have inequivalent oxidation states within one compound; an average is not automatically the state of every atom. Use O=−2 except O–O peroxide oxygen=−1, H=+1, and assign a homonuclear bond zero contribution. Count species containing at least one element in TWO distinct atom-resolved oxidation states: (i) H₂S₄O₆ with HO₃S–S–S–SO₃H connectivity; (ii) HO–S(=O)₂–SH; (iii) HO–S(=O)₂–O–O–S(=O)₂–OH; (iv) HO–S(=O)₂–O–S(=O)₂–OH; (v) HO–O–N(=O)₂; (vi) H₄P₂O₆, (HO)₂P(=O)–P(=O)(OH)₂; (vii) H₂O₂; (viii) Fe₃O₄, treated as FeO·Fe₂O₃.
Answer:
Solution:
i has terminal S+5 and internal S0; ii has central S+5 and terminal S−1. iii has S+6 at both centres but oxygen−1 and−2, so still qualifies. iv has S+6 and only ordinary oxygen and does not. v has peroxide−1 and ordinary−2 oxygen and qualifies. vi has equivalent P+4, vii only O−1, while viii has Fe+2 and+3. The qualifying set is i, ii, iii, v, viii, giving 5.
