Question

What is the oxidation number of S in H₂S₂O₈ ?

• A. +5
• B. +4
• C. +7
• D. +6

Answer 1

The Correct Option is D

We are asked to find the oxidation number of sulfur (S) in the compound H₂S₂O₈ (Peroxodisulfuric acid).

Step-by-Step Solution: 

Step 1: Let the oxidation number of sulfur (S) be x.

Step 2: Oxidation number of hydrogen (H) is +1.

Step 3: Oxidation number of oxygen (O) is usually -2, but in peroxo linkages, it is -1.

Note: H₂S₂O₈ (peroxodisulfuric acid) contains a peroxo linkage (–O–O–), having two oxygen atoms with an oxidation number of -1.

Structure can be represented as: HO–SO₂–O–O–SO₂–OH

• 2 hydrogen atoms: each (+1) ⇒ total = +2
• 2 sulfur atoms: each = x ⇒ total = 2x
• 6 normal oxygen atoms (not in peroxo linkage): each (-2) ⇒ total = -12
• 2 peroxo oxygen atoms: each (-1) ⇒ total = -2

Step 4: The total charge on H₂S₂O₈ is zero, thus:

\[ (+2) + (2x) + (-12) + (-2) = 0 \]

Simplifying, we get:

\[ 2x - 12 = 0 \]

\[ 2x = +12 \]

\[ x = +6 \]

Conclusion:

The oxidation number of sulfur (S) in H₂S₂O₈ is +6.

Correct Answer: Option (D): +6

Answer 2

Let the oxidation number of Sulfur (S) be \( x \). We know the standard oxidation numbers for Hydrogen (H) is +1 and for Oxygen (O) is typically -2.

If we assume the standard oxidation number of -2 for all oxygen atoms in \( \text{H}_2\text{S}_2\text{O}_8 \), the sum of oxidation numbers would be: $$ 2 \times (\text{Oxidation number of H}) + 2 \times (\text{Oxidation number of S}) + 8 \times (\text{Oxidation number of O}) = 0 $$ $$ 2 \times (+1) + 2 \times (x) + 8 \times (-2) = 0 $$ $$ 2 + 2x - 16 = 0 $$ $$ 2x - 14 = 0 $$ $$ 2x = +14 $$ $$ x = +7 $$ (This calculation is shown in the handwritten part of the image).

However, Sulfur belongs to Group 16, and its maximum possible oxidation number is +6 (equal to its number of valence electrons). An oxidation number of +7 is not possible for Sulfur. This indicates that not all oxygen atoms have an oxidation state of -2. Specifically, it suggests the presence of a peroxide linkage (-O-O-).

The structure of peroxydisulfuric acid (\( \text{H}_2\text{S}_2\text{O}_8 \), also known as Marshall's acid) confirms this. It contains one peroxide bond:

\( \text{HO}_3\text{S} - \text{O} - \text{O} - \text{SO}_3\text{H} \)

In this structure:

• There are 2 Hydrogen atoms, each with an oxidation state of +1.
• There are 2 Sulfur atoms, each with an oxidation state of \( x \).
• There are 6 Oxygen atoms bonded only to Sulfur (or Sulfur and Hydrogen), each with an oxidation state of -2.
• There are 2 Oxygen atoms in the peroxide linkage (-O-O-), each with an oxidation state of -1.

Now, we set up the equation summing the actual oxidation numbers: $$ 2 \times (+1) + 2 \times (x) + 6 \times (-2) + 2 \times (-1) = 0 $$ $$ 2 + 2x - 12 - 2 = 0 $$ $$ 2x - 12 = 0 $$ $$ 2x = +12 $$ $$ x = +6 $$

Therefore, the oxidation number of Sulfur (S) in \( \text{H}_2\text{S}_2\text{O}_8 \) is +6.

Comparing this value with the given options, the correct option is (D).

Final Answer: The final answer is \( \boxed{+6} \)