Calculate the oxidation number of each sulphur atom in the following compounds: (a) Na2S2O3 (b) Na2S4O6 (c) Na

NCERT Exemplar · Q23

Q. Calculate the oxidation number of each sulphur atom in the following compounds:

(a) Na2S2O3

(b) Na2S4O6

(d) Na2SO4

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Concept understanding — Oxidation Reduction

Let’s start with something you already know from everyday life.

The intuition: what does “oxidation” really mean?

Think of a piece of iron left out in the rain. Over time, it turns into reddish-brown rust. Or think of a slice of apple turning brown when you leave it on the table. Or a fire burning wood to ash. In all these cases, something is combining with oxygen — that’s the original meaning of “oxidation.” The iron combines with oxygen from the air to form iron oxide (rust). The apple’s chemicals react with oxygen in the air. The wood burns because carbon in the wood combines with oxygen.

So the first, simplest idea: oxidation = adding oxygen. And the reverse — taking oxygen away — was called reduction. For example, if you heat iron oxide with carbon, the carbon steals the oxygen away, leaving pure iron. That’s reduction: removing oxygen.

But chemists soon realised this was too narrow. Many reactions that look like oxidation-reduction don’t involve oxygen at all. For instance, when sodium metal reacts with chlorine gas to make table salt, no oxygen is involved — yet the sodium clearly “rusts” in a sense, and the chlorine “steals” something from it.

So the definition had to be broadened.

The precise modern definition: electron transfer

Here’s the clean, exam-ready statement:

Oxidation is the loss of electrons by a substance.

Reduction is the gain of electrons by a substance.

They always happen together — you cannot have one without the other. That’s why we call them redox reactions (short for reduction-oxidation).

Let’s see this with the sodium-chlorine example:

Sodium atom ( $\text{Na}$ ) loses one electron to become $\text{Na}^+$ . That’s oxidation.
Chlorine atom ( $\text{Cl}$ ) gains that electron to become $\text{Cl}^-$ . That’s reduction.

You can write the two halves separately:

\text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{oxidation})

\text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{reduction})

Add them together:

\text{Na} + \text{Cl} \rightarrow \text{Na}^+ + \text{Cl}^-

That’s table salt.

Tip

A handy mnemonic: OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

How to spot a redox reaction without seeing electrons

You can’t watch electrons move directly. So chemists use oxidation numbers (also called oxidation states) — a bookkeeping system that tracks electrons.

Rules (simplified for first-time learners):

An atom in its elemental form has oxidation number 0.
A monatomic ion has oxidation number equal to its charge (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1).
Oxygen is usually -2 (except in peroxides).
Hydrogen is usually +1 (except in metal hydrides).
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

Then:

Oxidation = increase in oxidation number.
Reduction = decrease in oxidation number.

Example: Rusting of iron.

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe starts at 0 (elemental). In $\text{Fe}_2\text{O}_3$ , each Fe is +3. So Fe’s oxidation number goes up from 0 to +3 → oxidation.
O starts at 0 (in $\text{O}_2$ ). In $\text{Fe}_2\text{O}_3$ , each O is -2. So O’s oxidation number goes down from 0 to -2 → reduction.

Watch out

A common mistake: thinking that “reduction” means something becomes smaller or less. It doesn’t — it’s about gaining electrons (or losing oxygen, in the old sense). The name comes from metallurgy: when you “reduce” iron ore to iron, you’re taking away oxygen, so the mass reduces.

One more way to think about it …

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The key idea for calculating oxidation numbers is to apply a set of standard rules, considering the electronegativity differences between bonded atoms. For compounds with multiple atoms of the same element in different chemical environments, the individual oxidation numbers must be determined based on the molecular structure, rather than just calculating an average.

Here's the calculation for each compound:

(a) Na $_2$ S $_2$ O $_3$ (Sodium thiosulphate)

Assign known oxidation numbers: Na is $+1$ , O is $-2$ .
Calculate the average oxidation number for S: $2(+1) + 2x + 3(-2) = 0 \implies 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2$ .
However, the two sulphur atoms in the thiosulphate ion ( $S_2O_3^{2-}$ ) are non-equivalent. One sulphur atom is terminal (like a sulphide) and the other is central (like in sulphate).
Based on the structure, the terminal sulphur is conventionally assigned $-2$ and the central sulphur $+6$ ; their average, $+2$ , is the printed Exemplar answer.

(b) Na $_2$ S $_4$ O $_6$ (Sodium tetrathionate)

Assign known oxidation numbers: Na is $+1$ , O is $-2$ .
Calculate the average oxidation number for S: $2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$ .
The four sulphur atoms in the tetrathionate ion ( $S_4O_6^{2-}$ ) are non-equivalent. There are two terminal sulphur atoms bonded to oxygen and two central sulphur atoms bonded only to other sulphur atoms.
Based on the structure, the two terminal sulphur atoms each have an oxidation state of $+5$ , and the two central sulphur atoms each have an oxidation state of $0$ .

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The oxidation number of sulphur varies across these compounds. In Na $_2$ S $_2$ O $_3$ the average oxidation number of sulphur is +2 (the printed Exemplar answer; the two non-equivalent S atoms are conventionally split −2 and +6). In Na $_2$ S $_4$ O $_6$ two sulphur atoms are +5 and two are 0. In Na $_2$ SO $_3$ sulphur is +4, and in Na $_2$ SO $_4$ it is +6.

To calculate the oxidation number of an element in a compound, we assign hypothetical charges to atoms based on a set of rules. These rules reflect the electronegativity differences between atoms, assuming complete transfer of electrons in a bond. The sum of oxidation numbers in a neutral compound is zero, and in a polyatomic ion, it equals the charge of the ion.

Here are the general rules we will use:

The oxidation number of an element in its elemental state is 0.
The oxidation number of Group 1 metals (like Na) is always +1.
The oxidation number of oxygen is usually -2, except in peroxides (like H $_2$ O $_2$ ) where it is -1, and in superoxides (like KO $_2$ ) where it is -1/2, and when bonded to fluorine (like OF $_2$ ) where it is +2.
The oxidation number of hydrogen is usually +1, except in metal hydrides (like NaH) where it is -1.
The sum of oxidation numbers of all atoms in a neutral compound is 0.
The sum of oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion.

Let's apply these rules to each compound.

(a) Na $_2$ S $_2$ O $_3$ (Sodium Thiosulphate)

Calculate the average oxidation number of sulphur: Let the oxidation number of sulphur be $x$ . Sodium (Na) has an oxidation number of +1. Oxygen (O) has an oxidation number of -2. The compound is neutral, so the sum of oxidation numbers is 0.

2(\text{Na}) + 2(\text{S}) + 3(\text{O}) = 0

2(+1) + 2(x) + 3(-2) = 0

2 + 2x - 6 = 0

2x - 4 = 0

2x = 4

x = +2

The average oxidation number of sulphur in Na$_2$S$_2$O$_3$ is +2.

2. Determine individual oxidation numbers for each sulphur atom:

Sodium thiosulphate contains the thiosulphate ion ( $S_2O_3^{2-}$ ). This ion has a structure where the two sulphur atoms are not equivalent. It can be thought of as a sulphate ion ( $SO_4^{2-}$ ) where one oxygen atom has been replaced by a sulphur atom.

The structure is $O_3S_A-S_B^{2-}$ . One sulphur atom ( $S_A$ ) is central, bonded to three oxygen atoms and one other sulphur atom ( $S_B$ ). The other sulphur atom ( $S_B$ ) is terminal, bonded only to the central sulphur atom.

> [!WARNING]
> When a compound contains multiple atoms of the same element, and these atoms are in different chemical environments (i.e., bonded differently), their individual oxidation numbers may not be the same as the average oxidation number calculated from the formula. This is a common pitfall.

*   **Terminal Sulphur ($S_B$):** This sulphur atom is typically considered to have an oxidation state similar to a sulphide ion, which is -2.
*   **Central Sulphur ($S_A$):** Let its oxidation number be $y$.
    The sum of the oxidation numbers of the two sulphur atoms must be consistent with the overall charge of the thiosulphate ion ($S_2O_3^{2-}$).
    The overall charge of the $S_2O_3^{2-}$ ion is -2.
    Let the oxidation number of the central sulphur be $y$ and the terminal sulphur be $z$.

y + z + 3(\text{O}) = -2

y + z + 3(-2) = -2

y + z - 6 = -2

y + z = +4

    If we assign $z = -2$ for the terminal sulphur, then:

y + (-2) = +4

y = +6

Therefore, one sulphur atom has an oxidation number of +6, and the other has an oxidation number of -2.

(b) Na $_2$ S $_4$ O $_6$ (Sodium Tetrathionate)

Calculate the average oxidation number of sulphur: Let the oxidation number of sulphur be $x$ . Sodium (Na) has an oxidation number of +1. Oxygen (O) has an oxidation number of -2. The compound is neutral, so the sum of oxidation numbers is 0.

2(\text{Na}) + 4(\text{S}) + 6(\text{O}) = 0

2(+1) + 4(x) + 6(-2) = 0

2 + 4x - 12 = 0

4x - 10 = 0

4x = 10

x = +2.5

The average oxidation number of sulphur in Na$_2$S$_4$O$_6$ is +2.5.

2. Determine individual oxidation numbers for each sulphur atom:

Sodium tetrathionate contains the tetrathionate ion ( $S_4O_6^{2-}$ ). This ion has a chain of four sulphur atoms: $O_3S-S-S-SO_3^{2-}$ .

Due to symmetry, the two terminal sulphur atoms (bonded to oxygen) are equivalent, and the two inner sulphur atoms (bonded only to other sulphur atoms) are equivalent.

> [!TIP]
> In a bond between two identical atoms (like S-S), the contribution to the oxidation number of each atom is zero because there is no electronegativity difference. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

KCET 2025Set D-41 markMCQ
$\Delta H = \Delta U$
(g) \longrightarrow 6CO_2\
(g) + 3H_2O\ (l)} $(C)$ \mathrm{2HI_{(g)} \rightleftharpoons H_2\
(g) + I_2\ (g)} $(D)$ \mathrm{N_2\
(g) + 3H_2\
(g) \rightleftharpoons 2NH_3\ (g)}$
›Reveal solutionSolution

$\Delta H - \Delta U = \Delta n_g RT$ , so the two are equal only for the reaction with zero change in the number of moles of gas.

Step 1 — The relation between $\Delta H$ and $\Delta U$ .

By definition $H = U + PV$ . For a reaction at constant temperature and pressure involving ideal gases, $PV = n_g RT$ , so
$\Delta H = \Delta U + \Delta (PV) = \Delta U + \Delta n_g\,RT$
where
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
Only gases count — solids and liquids have negligible molar volume, so they contribute essentially nothing to $\Delta(PV)$ .

Step 2 — The condition asked for.

Since $R$ and $T$ are never zero,
$\Delta H = \Delta U \iff \Delta n_g = 0$
So we simply count gas moles on each side of each reaction.

Step 3 — Test every option.

(A) $\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$

Gaseous reactants: 0. Gaseous products: 1 ( $\mathrm{CO_2}$ ).
$\Delta n_g = 1 - 0 = +1 \ne 0$
(B) $\mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \to 6CO_2(g) + 3H_2O(l)}$

Gaseous reactants: $\tfrac{15}{2} = 7.5$ (benzene is a liquid). Gaseous products: 6 (water is liquid).
$\Delta n_g = 6 - 7.5 = -1.5 \ne 0$
(C) $\mathrm{2HI(g) \rightleftharpoons H_2(g) + I_2(g)}$

Gaseous reactants: 2. Gaseous products: $1 + 1 = 2$ .
$\Delta n_g = 2 - 2 = \mathbf{0}\ \checkmark$
(D) $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$

Gaseous reactants: $1 + 3 = 4$ . Gaseous products: 2. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{ClO}_3^{-}$
›Reveal solutionSolution
The question tests fundamental inorganic chemistry concepts: electrolysis of active metals, oxo-anion stability, oxidation states, and redox behaviour. The correct statement is (C), because the nitride ion ( $\mathrm{N}^{3-}$ ) is already in its lowest oxidation state and cannot be reduced further, so it cannot act as an oxidising agent.

Concept and Intuition

Each option touches a different principle. We need to check each one carefully:

Option (A) — Electrolysis of aqueous solutions of active metals like Ca and Mg usually produces hydrogen at the cathode instead of the metal, because water is more easily reduced. So they are manufactured by electrolysis of molten salts, not aqueous.

Option (B) — Disproportionation requires an element in an intermediate oxidation state. Chlorine in $\mathrm{ClO}_3^{-}$ is +5; it can both increase and decrease its oxidation state under suitable conditions, so disproportionation is possible.

Option (C) — An oxidising agent gains electrons (is reduced). The nitride ion $\mathrm{N}^{3-}$ has nitrogen in its lowest possible oxidation state (−3). It cannot accept more electrons, so it cannot act as an oxidising agent. It can only act as a reducing agent.

Option (D) — In alkaline medium, $\mathrm{MnO}_4^{-}$ is reduced to $\mathrm{MnO}_2$ (not $\mathrm{Mn}^{2+}$ ), which involves a gain of 3 electrons, not 5.

Step-by-step reasoning

Option (A):

Calcium and magnesium are highly electropositive metals. In aqueous solution, water is reduced more easily than $\mathrm{Ca}^{2+}$ or $\mathrm{Mg}^{2+}$ (standard reduction potentials: $\mathrm{Ca}^{2+}/\mathrm{Ca} = -2.87\,\mathrm{V}$ , $\mathrm{Mg}^{2+}/\mathrm{Mg} = -2.37\,\mathrm{V}$ , while $2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^-$ is about $-0.83\,\mathrm{V}$ ). So electrolysis of aqueous solutions gives $\mathrm{H}_2$ at the cathode, not the metal. They are produced by electrolysis of molten salts. Hence (A) is false.

Option (B):

In $\mathrm{ClO}_3^{-}$ , chlorine has oxidation state +5. Chlorine can exist in states from −1 to +7. Since +5 is intermediate, $\mathrm{ClO}_3^{-}$ can disproportionate, e.g., in hot concentrated alkali:

$4\mathrm{ClO}_3^{-} \rightarrow 3\mathrm{ClO}_4^{-} + \mathrm{Cl}^{-}$
Here chlorine goes from +5 to +7 (oxidation) and to −1 (reduction). So (B) is false.

Option (C): …
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KCET 2024Set B-21 markMCQ
$_2$
›Reveal solutionSolution

In any redox pair the oxidising agent is reduced: acidified permanganate goes to $Mn^{2+}$ , which forces $SO_2$ to be oxidised to $SO_4^{2-}$ .

Step 1 — Identify who oxidises whom

$KMnO_4$ in acid medium is one of the strongest common oxidising agents. Being the oxidising agent, it is itself reduced. $SO_2$ (sulphur in the intermediate oxidation state $+4$ ) can go up to $+6$ , so it acts as the reducing agent and is oxidised.

Step 2 — Track the oxidation numbers

Manganese in $MnO_4^-$ : let $x$ be Mn's oxidation number.
$x + 4(-2) = -1 \;\Rightarrow\; x = +7$
In acidic medium permanganate is reduced by 5 electrons all the way to $Mn^{2+}$ ( $+7 \to +2$ ). (Only in neutral/alkaline medium does it stop at $MnO_2$ , Mn $= +4$ — the trap in option (C).)
$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$
Sulphur in $SO_2$ : $x + 2(-2) = 0 \Rightarrow x = +4$ . In $SO_4^{2-}$ : $x + 4(-2) = -2 \Rightarrow x = +6$ . So sulphur loses 2 electrons:
$SO_2 + 2H_2O \longrightarrow SO_4^{2-} + 4H^+ + 2e^-$
(The "moist" in the question matters — the water is a reactant supplying the extra oxygens.)

Step 3 — Combine (balance electrons: $\times 2$ and $\times 5$ )

$2MnO_4^- + 5SO_2 + 2H_2O \longrightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{KMnO}_4$
›Reveal solutionSolution
The reaction that specifically needs an acidic (acidified) medium is the oxidation of $I^-$ to $I_2$ ; the other conversions occur in neutral or faintly alkaline medium.

Behaviour of $KMnO_4$ depends on the medium:

Acidic medium ( $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ): oxidises iodide to iodine — $10I^- + 2MnO_4^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$ . …
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KCET 2023Set D-21 markMCQ
$\Delta G^\circ$
›Reveal solutionSolution
The slope of an Ellingham line is $-\Delta S^\circ$ ; only the C → CO formation has a positive $\Delta S^\circ$ (gas moles increase), so only its $\Delta G^\circ$ becomes more negative as $T$ rises.

1. The Ellingham diagram in one equation

An Ellingham diagram plots $\Delta G^{\circ}$ of oxide formation against $T$ . From the Gibbs–Helmholtz relation:
$\Delta G^{\circ} = \Delta H^{\circ} - T\,\Delta S^{\circ}$
Treating $\Delta H^\circ$ and $\Delta S^\circ$ as roughly constant, this is a straight line whose
$\text{slope} = -\Delta S^{\circ}$
So the sign of $\Delta S^{\circ}$ decides everything:

$\Delta S^{\circ} < 0 \Rightarrow$ positive slope ( $\Delta G^\circ$ becomes less negative with $T$ )

$\Delta S^{\circ} > 0 \Rightarrow$ negative slope ( $\Delta G^\circ$ becomes more negative with $T$ ) ← what we want

2. Entropy change of each formation reaction

Metal oxides — (B) FeO, (C) ZnO, (D) $\mathrm{Cu_2O}$ :
$2\mathrm{Fe(s)} + \mathrm{O_2(g)} \to 2\mathrm{FeO(s)}$ $2\mathrm{Zn(s)} + \mathrm{O_2(g)} \to 2\mathrm{ZnO(s)}$ $4\mathrm{Cu(s)} + \mathrm{O_2(g)} \to 2\mathrm{Cu_2O(s)}$
In each case 1 mole of gaseous $\mathrm{O_2}$ is consumed and no gas is produced — gaseous randomness is destroyed, so $\Delta S^{\circ}$ is negative. Their lines slope upward: $\Delta G^\circ$ becomes less negative as $T$ increases.

(The ZnO line does bend even more steeply upward above the boiling point of Zn, but it never turns downward.)

Carbon monoxide — (A) CO:

$2\mathrm{C(s)} + \mathrm{O_2(g)} \to 2\mathrm{CO(g)}$ …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{KMnO}_4$
›Reveal solutionSolution

(In acidic medium MnO4- goes to Mn2+; in strongly alkaline medium to the green manganate MnO4^2-.)

Concept: in NEUTRAL (or faintly alkaline) medium, permanganate is reduced from Mn(VII) to Mn(IV), i.e. to MnO2 (a brown precipitate). This is the classic reaction with manganese(II) salts:

2 KMnO4 + 3 MnSO4 + 2 H2O -> 5 MnO2 + K2SO4 + 2 H2SO4 …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{MnO}_4^{-}$
›Reveal solutionSolution
[!TLDR]

Dilute (faintly) alkaline permanganate is reduced to brown MnO $_2$ , so the answer is (A).

Concept

The fate of MnO $_4^-$ depends on the medium (CBSE/NCERT Class 12 d- and f-Block Elements). In acidic medium it goes to Mn $^{2+}$ (+2); in neutral or faintly/dilute alkaline medium it goes to MnO $_2$ (+4); only in strongly alkaline medium does it stop at the manganate ion MnO $_4^{2-}$ (+6).

Solution

In dilute alkaline solution the relevant half-reaction is a three-electron reduction:

$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$ …
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KCET 2022Set B-31 markMCQ
$N_2H_4$
›Reveal solutionSolution

Assign oxidation numbers to nitrogen in each compound; only $NH_4NO_3$ contains nitrogen in two chemically distinct sites, giving it two different oxidation states in the same formula unit.

Concept. An element can show two oxidation states within one compound only if it occupies two chemically different positions (e.g. a cation and an anion). We assign oxidation numbers using $H=+1$ , $O=-2$ , and the requirement that the charges sum to the ion/molecule charge.

Step 1 — (A) Hydrazine, $N_2H_4$ .
$2x + 4(+1) = 0 \Rightarrow x = -2$
Both nitrogens are equivalent ( $H_2N-NH_2$ ): a single oxidation state $-2$ .

Step 2 — (B) Hydrazoic acid, $HN_3$ (written $N_3H$ ).
$3x + (+1) = 0 \Rightarrow x = -\tfrac{1}{3}$
This is an average (fractional) value; conventionally this compound is quoted as having the single average state $-1/3$ , not two clean, different oxidation states, and CBSE/KCET treat the intended answer as the salt below.

Step 3 — (C) Urea, $NH_2CONH_2$ .

Both $-NH_2$ groups are identical, each nitrogen is $-3$ . One oxidation state only.

Step 4 — (D) Ammonium nitrate, $NH_4NO_3$ . …
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KCET 2021Set B-21 markMCQ
$_2$
›Reveal solutionSolution
Paramagnetic + colourless + neutral pins P = NO; then NO → NO₂ (brown) → N₂O₄ (colourless dimer) → N₂O₃ (blue solid) with NO.

1. Identify P from the three clues.

The key word is paramagnetic — it requires an unpaired electron.

$\mathrm{N_2O}$ has an even number of electrons and is diamagnetic. So P cannot be $\mathrm{N_2O}$ — this alone eliminates options (A) and (B).

NO has an odd total electron count (7 + 8 = 15), leaving one unpaired electron in a $\pi^*$ orbital $\Rightarrow$ paramagnetic. It is also colourless (as a gas) and neutral (neither acidic nor basic).

$\boxed{P = \mathrm{NO}}$
2. Q — oxidation of NO.

Nitric oxide is oxidised by air/oxygen at once (this is the brown fume you see when NO meets air):
$2\mathrm{NO} + \mathrm{O_2} \longrightarrow 2\mathrm{NO_2} \quad (\text{reddish-brown gas})$ $\boxed{Q = \mathrm{NO_2}}$
This eliminates option (D), which repeats NO as Q.

3. R — cooling $\mathrm{NO_2}$ .

$\mathrm{NO_2}$ is itself an odd-electron (brown, paramagnetic) molecule. On cooling it dimerises — the two odd electrons pair up in an N–N bond — and the colour disappears:
$2\mathrm{NO_2} \;\rightleftharpoons\; \mathrm{N_2O_4} \quad (\text{colourless, diamagnetic})$ $\boxed{R = \mathrm{N_2O_4}}$
4. S — R with P gives the blue solid. …
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KCET 2021Set B-21 markMCQ

Q. Which of the following is not true for oxidation? (A) addition of oxygen (B) addition of electronegative element (C) removal of hydrogen (D) removal of electronegative element

›Reveal solution

Recall the four classical definitions of oxidation; the odd one out — removal of an electronegative element — is actually a reduction.

Step 1 — The classical definitions

Before the electron-transfer definition, oxidation was defined operationally. An element/compound is oxidised when it undergoes any of:

#	Oxidation is …	Example
1	addition of oxygen	$2Mg + O_2 \to 2MgO$
2	addition of an electronegative element	$Fe + S \to FeS$ ; $2Fe + 3Cl_2 \to 2FeCl_3$
3	removal of hydrogen	$H_2S + Cl_2 \to 2HCl + S$
4	removal of an electropositive element	$2KI + H_2O_2 \to 2KOH + I_2$

The modern unifying statement behind all four: oxidation is loss of electrons / an increase in oxidation number. Each of the four moves above raises the oxidation number of the species in question.

Step 2 — Screen the options against this list

(A) addition of oxygen — definition 1. Oxygen is electronegative; adding it pulls electron density away, raising the oxidation number. TRUE for oxidation.
(B) addition of electronegative element — definition 2. Same logic as oxygen. TRUE for oxidation.
(C) removal of hydrogen — definition 3. Hydrogen is the electropositive partner; losing it raises the oxidation number. TRUE for oxidation. …

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KCET 2020Set A-11 markMCQ
$O_2$
›Reveal solutionSolution
During electrolysis of an aqueous NaF solution, water is oxidised more easily than fluoride ions at the anode, so oxygen gas ( $O_2$ ) is produced instead of fluorine.

The key idea here is selective discharge of ions — not every ion present in solution actually gets discharged at the electrodes. In aqueous electrolysis, water itself can compete with the dissolved ions. The anode is where oxidation happens (loss of electrons), so we compare the ease of oxidation of the species present: fluoride ions ( $F^-$ ) and water molecules ( $H_2O$ ).

Fluoride ions are notoriously difficult to oxidise. In fact, fluorine gas is such a strong oxidising agent that its production from aqueous solution is practically impossible — water would be oxidised first. The standard electrode potentials tell the story clearly.
$\begin{aligned} 2F^- &\rightarrow F_2 + 2e^- \quad E^\circ = -2.87\ \text{V} \\ 2H_2O &\rightarrow O_2 + 4H^+ + 4e^- \quad E^\circ = -1.23\ \text{V} \end{aligned}$
Let’s walk through the reasoning step by step.

Identify all species present in solution. NaF dissociates completely in water:

$\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-$
So the solution contains $\text{Na}^+$ , $\text{F}^-$ , and water molecules ( $H_2O$ ). At the anode (positive electrode), oxidation occurs — we look for species that can lose electrons.

List possible oxidation reactions at the anode.

Oxidation of fluoride ions:

$2F^- \rightarrow F_2 + 2e^-$

Oxidation of water:

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Compare the ease of oxidation using standard potentials. The more positive (or less negative) the oxidation potential, the more readily the species gives up electrons. Water’s oxidation potential ( $-1.23$ V) is significantly higher (less negative) than that of fluoride ions ( $-2.87$ V). This means water is much easier to oxidise. …
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KCET 2020Set A-11 markMCQ
$Cl_2$
›Reveal solutionSolution

Roasting a sulphide gives $\mathrm{SO_2}$ ; $\mathrm{SO_2+Cl_2}$ over activated charcoal is the standard preparation of sulphuryl chloride $\mathrm{SO_2Cl_2}$ .

Step 1 — Identify X (roasting).

Roasting = heating an ore strongly in the presence of excess air, used for sulphide ores. Sulphur is oxidised to $\mathrm{SO_2}$ :
$2\,\mathrm{ZnS}+3\,\mathrm{O_2}\ \xrightarrow{\ \Delta\ }\ 2\,\mathrm{ZnO}+2\,\mathrm{SO_2}\uparrow$
(The same happens with $\mathrm{PbS}$ , $\mathrm{Cu_2S}$ , etc.) The evolved gas is therefore
$X=\mathrm{SO_2}$
Step 2 — Identify Y ( $\mathrm{SO_2}+\mathrm{Cl_2}$ ).

The classic preparation of sulphuryl chloride is the direct union of sulphur dioxide and chlorine in the presence of a catalyst — activated charcoal (or camphor):
$\mathrm{SO_2}+\mathrm{Cl_2}\ \xrightarrow[\text{activated charcoal}]{}\ \mathrm{SO_2Cl_2}$
Mechanistically, sulphur in $\mathrm{SO_2}$ is in the $+4$ state and is oxidised to $+6$ in $\mathrm{SO_2Cl_2}$ ; chlorine is reduced from $0$ to $-1$ . The charcoal simply provides a surface — it is a catalyst, not a reactant.

Step 3 — Rule out the other options. …
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Q.Calculate the oxidation number of each sulphur atom in the following compounds:

(a) Na2_22​S2_22​O3_33​ (Sodium Thiosulphate)

(b) Na2_22​S4_44​O6_66​ (Sodium Tetrathionate)

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(a) Na $_2$ S $_2$ O $_3$ (Sodium Thiosulphate)

(b) Na $_2$ S $_4$ O $_6$ (Sodium Tetrathionate)

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