Q.Answer the following question.
Calculate the oxidation number of underlined atoms.
a. H2S̲O4 b. HN̲O3 c. H3P̲O3 d. K2C̲2O4 e. H2S̲4O6 f. C̲r2O7²⁻ g. NaH2P̲O4
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Start your 14-day free trial to unlock the full solution →a. H2SO4 (S underlined). H = +1 x2 = +2; O = -2 x4 = -8; neutral: (+2) + ON(S) + (-8) = 0, so ON(S) = +6.
b. HNO3 (N underlined). H = +1; O = -2 x3 = -6; neutral: (+1) + ON(N) + (-6) = 0, so ON(N) = +5.
c. H3PO3 (P underlined). H = +1 x3 = +3; O = -2 x3 = -6; neutral: (+3) + ON(P) + (-6) = 0, so ON(P) = +3.
d. K2C2O4 (C underlined, potassium oxalate). K = +1 x2 = +2; O = -2 x4 = -8; neutral: (+2) + 2xON(C) + (-8) = 0, so 2xON(C) = +6, ON(C) = +3 (average, rule 8).
e. H2S4O6 (S underlined, tetrathionic acid). H = +1 x2 = +2; O = -2 x6 = -12; neutral: (+2) + 4xON(S) + (-12) = 0, so 4xON(S) = +10, ON(S) = +2.5 (average, rule 8 -- matching the 'Do you know?' box's tetrathionate example, where the true individual values are two S at 0 and two S at +5). …
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