The Oxidation State Trap: Finding the Oxidation Number of S in H2S2O8
If you blindly apply the standard mathematical formula to find the oxidation state of Sulphur in Marshall's Acid (H2S2O8), you are going to get the wrong answer. This is a classic "trap" question that appears repeatedly in JEE Main, JEE Advanced, and NEET exams.
Let's dive into why the mathematical formula fails here, and how understanding the chemical structure saves you from negative marking.
Video Tutorial: Breaking Down the Structure
Watch Abhishek Sengar sir from CHEMCA demonstrate the mathematical error and reveal the correct structural solution!
The Mathematical Trap
Let's try finding the oxidation number (x) of Sulphur using the basic rules where H = +1 and O = -2:
2(+1) + 2x + 8(-2) = 0
2 + 2x - 16 = 0
2x = 14
x = +7 (WRONG!)
Why is +7 impossible? Sulphur belongs to Group 16 of the periodic table. It only has 6 valence electrons. Therefore, the maximum possible oxidation state of Sulphur is +6. It cannot lose 7 electrons!
The Solution: The Peroxo Linkage
When your calculated oxidation number exceeds the group's maximum limit, it indicates a structural exception—most commonly, a Peroxide (O-O) bond. In a peroxide bond, the oxidation state of oxygen is -1, not -2.
If we look at the structure of H2S2O8 (Peroxodisulfuric acid):
- There are two central Sulphur atoms.
- They are connected by an oxygen-oxygen single bond in the middle (-O-O-). This is the peroxo linkage.
- Because these two oxygens are bonded to each other, they don't draw electrons from each other. They only draw 1 electron from the adjacent Sulphur atoms. Thus, their oxidation state is -1.
- The remaining 6 oxygen atoms are normal oxides with an oxidation state of -2.
(2 H at +1) + (2 S at x) + (6 normal O at -2) + (2 peroxo O at -1) = 0
2(+1) + 2x + 6(-2) + 2(-1) = 0
2 + 2x - 12 - 2 = 0
2x = 12 ⟶ x = +6 (Correct!)
Practice Questions for JEE & NEET
If you understood Marshall's acid, try solving these two famous sister exceptions. Remember the golden rule: check the maximum oxidation state!
Question 1: Calculate the correct oxidation number of Chromium (Cr) in CrO5 (Chromium pentoxide).
Step-by-step Solution:
- The Trap: Math formula gives x + 5(-2) = 0 ⟶ x = +10. (Impossible, Cr is in group 6, max is +6).
- The Structure: CrO5 has a famous "Butterfly structure". It contains two peroxo linkages (4 peroxo oxygens total).
Fig 1: Chromium Pentoxide Butterfly Structure (Green = Peroxo O's)
- Recalculate: 4 oxygens are at -1 (peroxide), and 1 oxygen is double-bonded at -2 (oxide).
- x + 4(-1) + 1(-2) = 0
- x - 4 - 2 = 0
- Answer: Cr = +6
Question 2: What is the oxidation number of Sulphur in Caro's Acid (H2SO5)?
Step-by-step Solution:
- The Trap: Math formula gives 2(+1) + x + 5(-2) = 0 ⟶ x = +8. (Impossible, S max is +6).
- The Structure: Peroxomonosulfuric acid (H2SO5) contains one peroxo linkage (-O-O-H).
Fig 2: Caro's Acid Structure (Green = Peroxo O's)
- Recalculate: Out of 5 oxygens, 3 are normal oxides (-2) and 2 are peroxides (-1).
- 2(+1) + x + 3(-2) + 2(-1) = 0
- 2 + x - 6 - 2 = 0
- x - 6 = 0
- Answer: S = +6
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