**Match Column I (Compound) with Column II (Oxidation state of underlined element):** <br> **Column — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Match Column I (Compound) with Column II (Oxidation state of underlined element):** <br> **Column I:** <br> (A) $\underline{Fe}_3O_4$ <br> (B) $\underline{Pb}_3O_4$ <br> (C) $K\underline{O}_2$ <br> (D) $Na_2\underline{S}_4O_6$ <br> **Column II:** <br> (p) +2.5 <br> (q) +8/3 <br> (r) -0.5 <br> (s) +8/3
Answer: A
💡 Solution & Explanation
**Logic:** <br> (A) Magnetite ($FeO.Fe_2O_3$): Average O.S. = $(2+2 \times 3)/3 = 8/3$. <br> (B) Red Lead ($2PbO.PbO_2$): Average O.S. = $(2 \times 2 + 4)/3 = 8/3$. <br> (C) Potassium Superoxide ($K^+[O_2]^- $): $O$ O.S. = $-1/2 = -0.5$. <br> (D) Sodium Tetrathionate: $2(+1) + 4x + 6(-2) = 0 \Rightarrow 4x = 10 \Rightarrow x = +2.5$.
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