Please explain the oxidation state of S in S4O6^2- Related: MCQs Question Bank (Competition Preparation) - Redox Reaction, Class 11, Chemistry

Question

Please explain the oxidation state of S in S4O6^2- Related:  MCQs Question Bank (Competition Preparation) - Redox Reaction, Class 11, Chemistry

Answer

Answer

The oxidation state +2.5 is just the average oxidation state for the S atom. According to the structure, the symmetry suggests a −1 on each bridging sulfur ( blue ) (just like the bridging O atoms in a peroxide), and a +6 ( red ) on each central sulfur (like in sulfate). Indeed, +6−12=2.5

Answer

Oxidation State of S in S4O6^2-

The oxidation state of an element is the charge it would have if all the electrons in its bonds belonged to the more electronegative atom. The oxidation state can be positive, negative, or zero. In S4O6^2-, the oxidation state of sulfur (S) needs to be determined.

Step 1: Determine the overall charge of the ion

The overall charge of the ion is 2-.

Step 2: Determine the oxidation state of oxygen (O)

Oxygen has an oxidation state of -2 in most compounds, except in peroxides and superoxides where it has an oxidation state of -1 and -1/2 respectively. In S4O6^2-, there are six oxygen atoms with an overall charge of -12.

-12 = 6x(-2)

Step 3: Determine the oxidation state of sulfur (S)

Let the oxidation state of sulfur in S4O6^2- be x.

The total charge contributed by sulfur is:

4x + 6(-2) = -2

4x - 12 = -2

4x = 10

x = 10/4

x = 2.5

Therefore, the oxidation state of sulfur in S4O6^2- is +2.5.

Conclusion

The oxidation state of sulfur in S4O6^2- is +2.5.

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