**Question:**What is the oxidation number of S in the compound Na₂SO₄?**Explanation:**To determine the oxidation number of sulfur (S) in sodium sulfate (Na₂SO₄), follow these steps:1. **Assign Known Oxidation Numbers:** - Sodium (Na) is part of Group 1 in the periodic table, so its oxidation number is +1. Since there are two sodium atoms, their total contribution is +2. - Oxygen (O) typically has an oxidation number of -2. With four oxygen atoms, the total contribution is -8.2. **Set Up the Equation:** - The compound is neutral, so the sum of the oxidation numbers must equal zero. - Let the oxidation number of sulfur be \( x \). - Equation: \( 2(+1) + x + 4(-2) = 0 \).3. **Solve the Equation:** - Simplify: \( 2 + x - 8 = 0 \). - Further simplify: \( x - 6 = 0 \). - Solve for \( x \): \( x = +6 \).Thus, the oxidation number of sulfur in Na₂SO₄ is +6. There are no graphs or diagrams in this image.

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What is the oxidation number of S in the compound Na₂SO4?

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Concept explainers

Group 13 Elements

In the periodic table, group 13 elements exist in the first p-block. There are six elements present in group 13, which include boron (B), aluminum (Al), gallium (Ga), indium (In), thallium (Tl), and element 113 (ununtrium, (Uut)). Group 13 elements are also designated as the boron family.

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