Q.What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

Concept: Oxidation Reduction — The oxidation number is the charge an atom would have if all bonds were ionic. For polyatomic species, the sum of oxidation numbers equals the overall charge. For neutral molecules, the sum is zero.

1. KI₃ — Iodine in KI₃ exists as $\text{I}_3^-$ (triiodide ion). The average follows from $3x = -1 \Rightarrow x = -\frac{1}{3}$. Structurally, $\text{I}_3^-$ is an $\text{I}_2$ molecule bound to an $\text{I}^-$ ion, so two iodine atoms are in the 0 state and one is $-1$ — the fractional value is just their average.
   ✓Final answer

   The oxidation number of I in KI₃ is $-\frac{1}{3}$ (average); the real states are 0, 0 and $-1$ (an I₂ unit bound to I⁻).

2. H₂S₄O₆ — Tetrathionic acid. Let $x$ be the oxidation number of each S. H is $+1$, O is $-2$. Total: $2(+1) + 4x + 6(-2) = 0 \Rightarrow 2 + 4x - 12 = 0 \Rightarrow 4x = 10 \Rightarrow x = +2.5$. The two central S atoms are bonded to each other (oxidation number $0$ each), and each terminal S is $+5$ (like in sulphite), giving an average of $+2.5$.
   ✓Final answer

   The oxidation number of S in H₂S₄O₆ is $+2.5$ (average); terminal S are $+5$, central S are $0$.

3. Fe₃O₄ — Magnetite. O is $-2$. Let $x$ be the oxidation number of Fe. $3x + 4(-2) = 0 \Rightarrow 3x - 8 = 0 \Rightarrow x = +\frac{8}{3}$. This is an average; Fe₃O₄ is actually FeO·Fe₂O₃, so one Fe is $+2$ and two Fe are $+3$.
   ✓Final answer

   The oxidation number of Fe in Fe₃O₄ is $+\frac{8}{3}$ (average); one Fe is $+2$, two Fe are $+3$.

4. CH₃CH₂OH — Ethanol. H is $+1$, O is $-2$. Let $x$ be the oxidation number of each C. For C₂H₆O: $2x + 6(+1) + (-2) = 0 \Rightarrow 2x + 6 - 2 = 0 \Rightarrow 2x = -4 \Rightarrow x = -2$. The two C atoms are not equivalent: the CH₃ carbon is $-3$, the CH₂OH carbon is $-1$, average $-2$.
   ✓Final answer

   The oxidation number of C in CH₃CH₂OH is $-2$ (average); CH₃ carbon is $-3$, CH₂OH carbon is $-1$.

5. CH₃COOH — Acetic acid. H is $+1$, O is $-2$. For C₂H₄O₂: $2x + 4(+1) + 2(-2) = 0 \Rightarrow 2x + 4 - 4 = 0 \Rightarrow x = 0$. The CH₃ carbon is $-3$, the COOH carbon is $+3$, average $0$.
   ✓Final answer

   The oxidation number of C in CH₃COOH is $0$ (average); CH₃ carbon is $-3$, COOH carbon is $+3$.

The Core Idea: Oxidation Numbers Are a Bookkeeping Tool

Oxidation numbers (or oxidation states) are not real charges — they are a formalism we use to track electron distribution in compounds. The rules are designed so that:

• More electronegative atoms "take" the electrons in a bond.
• The sum of oxidation numbers equals the overall charge on the molecule or ion.
• Common elements (Group 1 = +1, Group 2 = +2, oxygen usually −2, hydrogen usually +1) serve as anchors.

When a molecule has two or more identical atoms in different environments (like the two carbons in ethanol or the four sulphurs in tetrathionic acid), the average oxidation number is often not meaningful. The real insight comes from looking at the structure — that's where the "rationalisation" the question asks for lives.

Step-by-Step Solutions

1. (a) KI₃ — Potassium Triiodide

Assign known oxidation numbers:

• K is in Group 1: always +1.
• The overall molecule is neutral, so the sum of oxidation numbers = 0.

Let the oxidation number of each I atom be $x$ (assuming they are all equivalent for now).

Then: $+1 + 3x = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$.

Rationalisation using structure:

KI₃ actually exists as $\ce{K+ I3^-}$ — and the linear triiodide ion is best read as an $\ce{I2}$ molecule coordinated to an $\ce{I^-}$ ion ($\ce{I2 <- I^-}$). Since all three atoms are the same element, bonds between them contribute nothing to the oxidation numbers; what matters is where the ionic charge sits:

• The two atoms of the $\ce{I2}$ unit: oxidation number = 0 (exactly as in free iodine).
• The iodine that brought the negative charge: oxidation number = –1 (exactly as in an iodide ion).

Check: $0 + 0 + (-1) = -1$ — exactly the charge on $\ce{I3^-}$ — and with $\ce{K+}$ the compound is neutral. Correct.

Final for (a): The average is $-\frac{1}{3}$, but the actual oxidation numbers are: two I at 0 (the I₂ unit) and one I at –1.

2. (b) H₂S₄O₆ — Tetrathionic Acid

Assign known oxidation numbers:

• H: +1 each (2 H → +2 total)
• O: −2 each (6 O → −12 total)
• Let the average oxidation number of each S be $x$.

Sum: $2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$.

Again, a fractional average. The structure of tetrathionic acid ($\ce{H2S4O6}$) is a chain of four sulphurs:

    O       O
    ||      ||
HO - S - S - S - S - OH
    ||      ||
    O       O

Each terminal sulphur is bonded to three oxygens (two via S=O double bonds and one via a single S–OH bond) and to one neighbouring sulphur. Each central sulphur is bonded only to two other sulphurs, with no oxygens attached.

Assign using structure:

• Each terminal S: two S=O double bonds contribute $2 \times (+2) = +4$ (oxygen is more electronegative, so each double bond gives S a $+2$ contribution); one S–OH single bond contributes $+1$; the S–S single bond to the neighbouring sulphur is shared equally between identical atoms, contributing $0$. Total for terminal S: $+4 + 1 + 0 = +5$.
• Each central S: bonded only to two other sulphurs (one terminal, one central), both shared equally → $0$.

Check: total from S = $2(+5) + 2(0) = +10$. Adding H ($+2$) and O ($-12$): $+10 + 2 - 12 = 0$, which correctly balances the neutral molecule $\ce{H2S4O6}$.

Final for (b): Average = +2.5; actual: terminal S = +5, central S = 0.

3. (c) Fe₃O₄ — Magnetite

Assign known:

• O: −2 each (4 O → −8 total)
• Let average Fe oxidation number be $x$.

$3x - 8 = 0 \implies 3x = 8 \implies x = +\frac{8}{3}$.

Fractional again. Fe₃O₄ is a mixed-valence compound: it contains both Fe²⁺ and Fe³⁺ ions. The formula can be written as $\ce{FeO·Fe2O3}$ — one FeO (Fe²⁺) and one Fe₂O₃ (two Fe³⁺).

So: one Fe at +2, two Fe at +3. Average: $\frac{2 + 3 + 3}{3} = \frac{8}{3}$.

Final for (c): Average = $+\frac{8}{3}$; actual: one Fe²⁺ (+2) and two Fe³⁺ (+3).

4. (d) CH₃CH₂OH — Ethanol

Assign known:

• H: +1 each (6 H → +6 total)
• O: −2 (1 O → −2)
• Let average C oxidation number be $x$.

$2x + 6 - 2 = 0 \implies 2x + 4 = 0 \implies x = -2$.

Average is −2, but the two carbons are very different.

Structure:

H   H
|   |
H - C - C - O - H
|   |
H   H

• The CH₃ carbon (left): bonded to 3 H (each H is less electronegative, so each C–H bond gives C a −1 contribution → −3 from H) and 1 C (shared equally → 0). So CH₃ carbon = −3.
• The CH₂OH carbon (right): bonded to 2 H (−2), 1 C (0), and 1 O (O is more electronegative, so the C–O bond gives C a +1 contribution). So CH₂OH carbon = −2 + 0 + 1 = −1.

Check: $-3 + (-1) = -4$ total from C, plus $+6$ from H and $-2$ from O gives 0. Correct.

Final for (d): Average = −2; actual: CH₃ carbon = −3, CH₂OH carbon = −1.

5. (e) CH₃COOH — Acetic Acid

Assign known:

• H: +1 each (4 H → +4 total)
• O: −2 each (2 O → −4 total)
• Let average C oxidation number be $x$.

$2x + 4 - 4 = 0 \implies 2x = 0 \implies x = 0$.

Average is 0, but the two carbons are in vastly different oxidation states.

Structure:

H   O
|   ||
H - C - C - O - H
|
H

• The CH₃ carbon (left): bonded to 3 H (−3) and 1 C (0) → −3.
• The COOH carbon (right): bonded to 1 C (0), 1 O via double bond (each bond in a double bond counts separately: two bonds to O, each gives +1 → +2 from the double-bonded O), and 1 O via single bond (+1 from the O in the –OH group). So total: $0 + 2 + 1 = +3$.

Check: $-3 + 3 = 0$ from C, plus $+4$ from H and $-4$ from O gives 0. Correct.

Final for (e): Average = 0; actual: CH₃ carbon = −3, COOH carbon = +3.