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Difficulty: MediumOxidation Numbers and IUPAC Nomenclature of Redox Species

In the reaction between octasulfur, S8S_8, and concentrated trioxonitrate(V) acid, sulfur is oxidized to tetraoxosulfate(VI) acid, H2SO4H_2SO_4. What are the oxidation numbers of sulfur in S8S_8 and H2SO4H_2SO_4 respectively?

  1. 00 and +6+6Answer
  2. B
    −2-2 and +6+6
  3. C
    +8+8 and +6+6
  4. D
    00 and +4+4

Answer

The oxidation numbers of sulfur in S8S_8 and H2SO4H_2SO_4 are 00 and +6+6 respectively.
Free uncombined elements carry an oxidation state of zero, so sulfur in S8S_8 is 00. In tetraoxosulfate(VI) acid (H2SO4H_2SO_4), setting the neutral molecule oxidation sum to zero yields 2(+1)+S+4(−2)=02(+1) + S + 4(-2) = 0, giving S=+6S = +6.

Step-by-Step Solution

1
Determine the oxidation state of sulfur in free elemental form (S8S_8).
Oxidation state of sulfur in S8=0S_8 = 0.
By definition, an element in its free or uncombined state has an oxidation state of zero regardless of its atomicity.
2
Calculate the oxidation state of sulfur in H2SO4H_2SO_4.
Oxidation state of sulfur in H2SO4=+6H_2SO_4 = +6.
Assign +1+1 for each hydrogen atom and −2-2 for each oxygen atom. Solving 2(+1)+S+4(−2)=02(+1) + S + 4(-2) = 0 gives +2+S−8=0+2 + S - 8 = 0, hence S=+6S = +6.

Key Concept

Assigning oxidation states to free elemental forms and central atoms in polyatomic oxoacids
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Topics

Oxidation Numbers and IUPAC Nomenclature of Redox Species

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