Q.Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7 2– and NO3 –. Suggest structure of these compounds. Count for the fallacy.

Concept: Oxidation Number Calculation — The key is to distinguish between formal oxidation number (based on electronegativity rules) and actual bonding structure, which may show peroxide linkages or multiple bonds.

Step 1: H₂SO₅ (peroxymonosulphuric acid)

Normal rules: H = +1, O = –2 (except peroxides). Let S = $x$.

If all O were –2: $2(+1) + x + 5(-2) = 0 \Rightarrow x = +8$ — impossible (S max is +6).

Structure shows one peroxide bond (–O–O–), so two O atoms have –1 each, three O have –2.

Equation: $2(+1) + x + 3(-2) + 2(-1) = 0 \Rightarrow x = +6$.

Fallacy: Naive calculation gives +8; actual oxidation number of S is +6.

Step 2: Cr₂O₇²⁻ (dichromate ion)

Let Cr = $x$. O is –2 (no peroxide).

$2x + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6$. …

1. Oxidation Number Calculation: The Core Idea

Oxidation numbers are a bookkeeping tool. We assign each atom in a compound a charge assuming all bonds are ionic — the more electronegative atom “takes” both electrons from a bond. For a neutral molecule, the sum of oxidation numbers is zero; for an ion, it equals the ion’s charge.

The trick is that oxygen almost always has an oxidation number of –2, except in peroxides (where it is –1) or with fluorine. Similarly, hydrogen is usually +1, except in metal hydrides (–1). But when a compound has a special structure — like a peroxo (–O–O–) linkage — the simple rule fails if we don’t account for it.

Let’s tackle each compound.

2. H₂SO₅ (Peroxomonosulphuric Acid, Caro’s Acid)

Step 1: Naive calculation (the fallacy)

If we assume all oxygens are –2 and hydrogen is +1, then for H₂SO₅:

Let oxidation number of S be $x$.

But sulphur’s maximum oxidation number is +6 (it has 6 valence electrons). So +8 is impossible — this tells us our assumption is wrong.

Step 2: The actual structure

H₂SO₅ contains a peroxo bond (–O–O–). In a peroxide, each oxygen in the –O–O– unit has oxidation number –1, not –2. The structure is:

HO–SO₂–O–O–H

That is: one sulphur atom bonded to two terminal oxygens (each –2), one hydroxyl oxygen (–2, but bonded to H), and a peroxo group (–O–O–) where each oxygen is –1.

Count the oxygens:

• Two terminal oxygens (double-bonded to S): each –2 → total –4
• One hydroxyl oxygen (single-bonded to S and H): –2
• Two peroxo oxygens (in –O–O–): each –1 → total –2

Step 3: Correct calculation

Let S oxidation number be $x$.

Hydrogens: two H, each +1 → +2

Oxygens: three at –2 (terminal + hydroxyl) → –6; two at –1 (peroxo) → –2

Total from oxygens: –8

Equation: $+2 + x + (-8) = 0 \quad \Rightarrow \quad x - 6 = 0 \quad \Rightarrow \quad x = +6$

Fallacy count: The fallacy is assuming all oxygens are –2. The correct structure shows two oxygens are in a peroxo bond (–1 each).

3. Cr₂O₇²⁻ (Dichromate Ion)

Step 1: Naive calculation

Oxygen is –2. Let Cr oxidation number be $x$.

This gives +6, which is plausible (Cr max is +6). No fallacy here — the calculation is correct.

Step 2: Structure confirmation …