JEE Redox Reactions Questions

Given 500 mL of 1.2 M KI mixed with 500 mL of 0.2 M $$KMnO_4$$ in basic medium, the liberated iodine is titrated with 0.1 M $$Na_2S_2O_3$$.

We begin by calculating the moles of the reactants. Moles of KI = $$0.5 \times 1.2 = 0.6$$ mol and moles of $$KMnO_4$$ = $$0.5 \times 0.2 = 0.1$$ mol.

In basic medium, $$MnO_4^-$$ oxidizes $$I^-$$ to $$I_2$$ according to the reaction: $$2MnO_4^- + 6I^- + 4H_2O \to 2MnO_2 + 3I_2 + 8OH^-$$. Here Mn goes from +7 to +4 (gaining 3e each, total 6e for 2 Mn atoms) and I goes from -1 to 0 (losing 1e each, total 6e for 6 I atoms), so electrons are balanced.

From stoichiometry, 2 mol $$MnO_4^-$$ requires 6 mol $$I^-$$. Since we have 0.1 mol $$KMnO_4$$, it requires $$0.1 \times 3 = 0.3$$ mol $$I^-$$, whereas 0.6 mol $$I^-$$ is available, making $$KMnO_4$$ the limiting reagent.

Because 2 mol $$MnO_4^-$$ produces 3 mol $$I_2$$, 0.1 mol $$MnO_4^-$$ yields $$0.1 \times \frac{3}{2} = 0.15$$ mol $$I_2$$.

The liberated iodine is then titrated with thiosulfate according to $$I_2 + 2Na_2S_2O_3 \to 2NaI + Na_2S_4O_6$$, so the moles of $$Na_2S_2O_3$$ needed are $$2 \times 0.15 = 0.30$$ mol, which at 0.1 M corresponds to a volume of $$\frac{0.30}{0.1} = 3$$ L.

The answer is 3.

The moles of permanganate present are

$$n(MnO_4^-)=M\times V=0.2\times0.5=0.1\text{ mol}.$$

The moles of iodide present are

$$n(I^-)=1.5\times0.5=0.75\text{ mol}.$$

In basic medium, the relevant reaction is

$$2MnO_4^-+6I^-+4H_2O\rightarrow2MnO_2+3I_2+8OH^-.$$

From the stoichiometry,

$$2\text{ mol }MnO_4^- \rightarrow 3\text{ mol }I_2.$$

Hence, the moles of iodine liberated are

$$n(I_2)=\frac{3}{2}\times0.1=0.15\text{ mol}.$$

The liberated iodine is titrated with sodium thiosulphate according to

$$I_2+2S_2O_3^{2-}\rightarrow2I^-+S_4O_6^{2-}.$$

Thus,

$$1\text{ mol }I_2\rightarrow2\text{ mol }S_2O_3^{2-}.$$

Therefore, the moles of thiosulphate required are

$$n(S_2O_3^{2-})=2\times0.15=0.3\text{ mol}.$$

Since ($$300\text{ mL}=0.3\text{ L}$$),

$$x=\frac{n}{V}=\frac{0.3}{0.3}=1\text{ M}.$$

Hence, the value of (x) is 1.

We analyze both statements about $$KMnO_4$$ titrations.

Statement I: In oxalic acid vs $$KMnO_4$$ titration, heating to 60 degrees C is needed initially, but not in FAS vs $$KMnO_4$$ titration.

The reaction of oxalic acid with $$KMnO_4$$ is slow at room temperature. Initial heating to about 60 degrees C is required to start the reaction. Once some $$Mn^{2+}$$ is formed, it acts as an autocatalyst and the reaction proceeds without further heating. For FAS (Ferrous Ammonium Sulphate) vs $$KMnO_4$$, the reaction is fast at room temperature, so no heating is needed. Statement I is TRUE.

Statement II: In oxalic acid vs $$KMnO_4$$, $$MnSO_4$$ formed initially acts as catalyst. In FAS vs $$KMnO_4$$, heating oxidizes $$Fe^{2+}$$ to $$Fe^{3+}$$ by atmospheric oxygen, introducing error.

The first part is correct: $$Mn^{2+}$$ ions (from $$MnSO_4$$) act as an autocatalyst for the permanganate-oxalate reaction. The second part is also correct: heating the FAS solution would cause aerial oxidation of $$Fe^{2+}$$ to $$Fe^{3+}$$, which would consume some $$Fe^{2+}$$ before it reacts with $$KMnO_4$$, leading to errors. Statement II is TRUE.

Both statements are true. The answer is Option B) Both Statement I and Statement II are true.

In acidic medium, the purple permanganate ion $$MnO_4^-$$ present in $$KMnO_4$$ is reduced to the colourless $$Mn^{2+}$$ ion.

Step 1 - Oxidation state of Mn in the reactant:
In $$MnO_4^-$$ let the oxidation state of Mn be $$x$$. Using $$O$$ in state $$-2$$:
$$x + 4(-2) = -1 \;\;\Rightarrow\;\; x - 8 = -1 \;\;\Rightarrow\;\; x = +7$$

Step 2 - Oxidation state of Mn in the product:
In $$Mn^{2+}$$ the oxidation state is $$+2$$.

Step 3 - Difference $$X$$ between the two oxidation states:
$$X = 7 - 2 = 5$$

Step 4 - Brown-red precipitate in the acetate test:
On adding neutral $$FeCl_3$$ to an acetate solution, deep-red $$Fe(CH_3COO)_3$$ is first formed. On boiling it hydrolyses to a brown-red basic ferric acetate, commonly written as $$Fe(OH)(CH_3COO)_2$$. In both species iron is in the $$+3$$ oxidation state, i.e. present as $$Fe^{3+}$$.

Electronic configuration of $$Fe$$: $$[Ar]\,3d^6\,4s^2$$.
For $$Fe^{3+}$$ three electrons are removed (first the two 4s, then one 3d):
$$Fe^{3+} : [Ar]\,3d^5$$

Step 5 - Number of $$d$$ electrons $$Y$$ in $$Fe^{3+}$$:
$$Y = 5$$

Step 6 - Required sum:
$$X + Y = 5 + 5 = 10$$

Therefore, the value of $$X + Y$$ is $$10$$.

Balance the half-reaction: $$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$$.

In acidic medium, $$Cr_2O_7^{2-}$$ is reduced to $$Cr^{3+}$$, each Cr going from oxidation state +6 to +3. Since there are two Cr atoms on the left, the product on the right must be 2$$Cr^{3+}$$, so A = $$Cr^{3+}$$.

To balance oxygen, note the 7 oxygen atoms in $$Cr_2O_7^{2-}$$; these require 7$$H_2O$$ on the right, giving Z = 7.

Those 7$$H_2O$$ molecules contribute 14 hydrogen atoms on the right, so we add 14$$H^+$$ on the left to balance hydrogen, yielding X = 14.

Next, balance charge by adding electrons. The left side has charge $$(-2) + 14(+1) + Y(-1) = 12 - Y$$, while the right side has charge $$2(+3) + 0 = +6$$. Setting these equal gives $$ 12 - Y = 6 \implies Y = 6 $$.

Thus the balanced equation is $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$, and checking charges shows both sides sum to +6.

X = 14, Y = 6, Z = 7, A = $$Cr^{3+}$$.

The correct answer is Option D: 14, 6, 7 and $$Cr^{3+}$$.

Higher reduction potential means stronger oxidizing agent.

$$BrO_4^-: E° = 1.74$$ V (highest), $$IO_4^-: E° = 1.65$$ V, $$ClO_4^-: E° = 1.19$$ V (lowest).

Oxidising power order: $$BrO_4^- > IO_4^- > ClO_4^-$$.

The answer is Option (2): $$\boxed{BrO_4^- > IO_4^- > ClO_4^-}$$.

We need to identify which product is NOT formed when lead sulphide (PbS) reacts with dilute nitric acid ($$HNO_3$$).

First, the balanced chemical equation for this reaction is:

$$ 3PbS + 8HNO_3(\text{dilute}) \rightarrow 3Pb(NO_3)_2 + 3S + 2NO\uparrow + 4H_2O $$

Next, we examine why these products form.

In this reaction:

- $$PbS$$ is oxidized: $$S^{2-}$$ in PbS is oxidized to elemental sulphur $$S^0$$ (oxidation state changes from $$-2$$ to $$0$$).

- $$HNO_3$$ acts as the oxidizing agent: $$N^{+5}$$ in $$HNO_3$$ is reduced to $$N^{+2}$$ in $$NO$$ (nitric oxide).

- $$Pb^{2+}$$ combines with $$NO_3^-$$ to form lead nitrate $$Pb(NO_3)_2$$.

Now, we identify which products are formed and which are not.

The products of this reaction are:

- Lead nitrate ($$Pb(NO_3)_2$$) — formed

- Sulphur ($$S$$) — formed

- Nitric oxide ($$NO$$) — formed

- Water ($$H_2O$$) — formed

Nitrous oxide ($$N_2O$$) is NOT a product of this reaction. Nitrous oxide would require nitrogen to be reduced to the $$+1$$ oxidation state, but with dilute $$HNO_3$$ and PbS, the reduction product is nitric oxide ($$NO$$, where nitrogen is in the $$+2$$ state).

The correct answer is Option (2): Nitrous oxide.

Iron(III) catalyses the reaction between iodide and persulphate ions. Identify the correct mechanism.

The overall reaction is $$2I^- + S_2O_8^{2-} \rightarrow I_2 + 2SO_4^{2-}$$.

In the catalytic cycle of Fe³⁺/Fe²⁺, Fe³⁺ oxidizes I⁻: $$2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$$. Fe²⁺ is then re-oxidized by persulphate (Fe²⁺ reduces persulphate): $$2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}$$.

So: A (Fe³⁺ oxidises iodide) and D (Fe²⁺ reduces persulphate) are correct.

The correct answer is Option (4): A and D only.

We need to determine the product when $$MnO_4^-$$ oxidises $$I^-$$ in alkaline medium.

Understand the oxidising power of permanganate in alkaline medium.

In alkaline medium, permanganate ion ($$MnO_4^-$$) is a strong oxidising agent. It is reduced from Mn(+7) to Mn(+6) (as $$MnO_4^{2-}$$) or to MnO$$_2$$ depending on conditions.

Determine the oxidation product of iodide.

In alkaline medium, $$MnO_4^-$$ being a strong enough oxidiser can take iodide ($$I^-$$, oxidation state -1) all the way up to iodate ($$IO_3^-$$, oxidation state +5). The reaction is:

$$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$$

Verify the oxidation states.

- Iodine goes from -1 in $$I^-$$ to +5 in $$IO_3^-$$ (loss of 6 electrons)

- Manganese goes from +7 in $$MnO_4^-$$ to +4 in $$MnO_2$$ (gain of 3 electrons each, 6 total for 2 atoms)

Electrons are balanced.

The correct answer is Option (4): $$IO_3^-$$.

We apply the oxidation number method to balance the redox reaction between lead sulfide and ozone.

Step 1: Oxidation half‐reaction
Sulfur in $$PbS$$ changes its oxidation state from $$-2$$ to $$+6$$ in $$SO_4^{2-}$$. In acidic medium the half‐reaction is written as:
$$S^{2-} + 4\,H_2O \;\longrightarrow\; SO_4^{2-} + 8\,H^+ + 8\,e^-$$ $$-(1)$$

Step 2: Reduction half‐reaction
Ozone is reduced to oxygen, with each oxygen atom remaining at oxidation state $$0$$. The half‐reaction in acidic medium is:
$$O_3 + 2\,H^+ + 2\,e^- \;\longrightarrow\; O_2 + H_2O$$ $$-(2)$$

Step 3: Equalize electrons
The oxidation half‐reaction (1) involves 8 electrons, while the reduction half‐reaction (2) involves 2 electrons. To balance electrons, multiply equation (2) by 4:
$$4\,O_3 + 8\,H^+ + 8\,e^- \;\longrightarrow\; 4\,O_2 + 4\,H_2O$$ $$-(3)$$

Step 4: Add half‐reactions
Adding equations (1) and (3) and cancelling common species ($$8\,e^-$$, $$8\,H^+$$ and $$4\,H_2O$$) gives:
$$S^{2-} + 4\,O_3 \;\longrightarrow\; SO_4^{2-} + 4\,O_2$$ $$-(4)$$

Step 5: Include lead ion
Since the sulfide ion comes from $$PbS$$ and the sulfate ion forms $$PbSO_4$$, we combine with $$Pb^{2+}$$ to get the overall equation:
$$PbS + 4\,O_3 \;\longrightarrow\; PbSO_4 + 4\,O_2$$ $$-(5)$$

From the balanced equation (5), 1 mole of $$PbS$$ reacts with $$X=4$$ moles of $$O_3$$ to produce $$Y=4$$ moles of $$O_2$$. Therefore,

$$X + Y = 4 + 4 = 8$$

The reaction: 2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 10CO₂ + 8H₂O

Mole ratio: KMnO₄ : H₂C₂O₄ = 2 : 5

Moles of oxalic acid = 0.020 × 2 = 0.04 mol

Moles of KMnO₄ = (2/5) × 0.04 = 0.016 mol

Molarity of KMnO₄ = 0.016/0.002 = 8 M

The answer is 8.

Species that can undergo disproportionation have an element in an intermediate oxidation state:

$$H_2O_2$$: O is in -1 state (can go to 0 and -2) — YES

$$ClO_3^-$$: Cl is in +5 — YES (can disproportionate)

$$P_4$$: P is in 0 state — YES (goes to -3 and +5 in alkali)

$$Cl_2$$: Cl is in 0 state — YES (goes to -1 and +1)

$$Ag$$: metallic, 0 state — NO (already in lowest practical state)

$$Cu^{+1}$$: Can go to Cu(0) and Cu(+2) — YES

$$F_2$$: F is in 0 state, but F only goes to -1 (no positive states) — NO

$$NO_2$$: N is in +4 — YES (goes to +5 and +3)

$$K^+$$: Cannot disproportionate — NO

Total = 6 (H$$_2$$O$$_2$$, ClO$$_3^-$$, P$$_4$$, Cl$$_2$$, Cu$$^{+1}$$, NO$$_2$$).

Therefore, the answer is $$\boxed{6}$$.

We need to find the number of moles of $$H^+$$ ions required by 1 mole of $$MnO_4^-$$ to oxidise oxalate ion to $$CO_2$$.

First, we write the half-reactions for the permanganate-oxalate reaction in acidic medium. The reduction half is $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$, in which manganese goes from +7 to +2 by gaining five electrons, and the oxidation half for oxalate is $$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$$, as each oxalate ion loses two electrons while carbon is oxidised from +3 to +4.

Next, balancing the electrons requires multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5, which yields $$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$ and $$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$$.

Combining these two balanced half-reactions gives the overall ionic equation $$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$$.

Since 2 moles of $$MnO_4^-$$ consume 16 moles of $$H^+$$ in this balanced equation, one mole requires $$\frac{16}{2} = 8$$ moles of $$H^+$$.

The correct answer is 8.

$$K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \xrightarrow{\Delta} 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$$

Rule: In any neutral molecule, the sum of oxidation states equals zero. Let the oxidation state of chromium in $$CrO_2Cl_2$$ be $$x$$. Then
$$x + 2\times(\text{ox. state of O}) + 2\times(\text{ox. state of Cl}) = 0\quad-(1)$$

We know: Oxygen has oxidation state $$-2$$ and chlorine has oxidation state $$-1$$. Substituting these into $$(1)$$ gives:
$$x + 2(-2) + 2(-1) = 0$$
$$x = 6$$

Therefore, the oxidation state of chromium in the product $$CrO_2Cl_2$$ is +6.

Potassium ferricyanide is $$K_3[Fe(CN)_6]$$, in which iron is in the $$+3$$ oxidation state. When iodide ion $$I^-$$ is added, a redox equilibrium is set up in which iron(III) is reduced to iron(II) and iodide is oxidised to iodine. The iron(II) product is potassium ferrocyanide $$K_4[Fe(CN)_6]$$, designated as the complex P.

Step 1 - Write the half-reactions
Reduction: $$[Fe(CN)_6]^{3-} + e^- \rightarrow [Fe(CN)_6]^{4-}$$
Oxidation: $$I^- \rightarrow \dfrac{1}{2}I_2 + e^-$$

Step 2 - Equalise electrons and add the half-reactions
Each half-reaction involves one electron, so they can be added directly:

$$[Fe(CN)_6]^{3-} + I^- \rightarrow [Fe(CN)_6]^{4-} + \dfrac{1}{2}I_2$$

Step 3 - Remove the fractional coefficient
Multiply the entire equation by 2:

$$2\,[Fe(CN)_6]^{3-} + 2\,I^- \rightarrow 2\,[Fe(CN)_6]^{4-} + I_2$$

Replacing the complex ions with their potassium salts gives:

$$2\,K_3[Fe(CN)_6] + 2\,KI \rightarrow 2\,K_4[Fe(CN)_6]\;(\text{= P}) + I_2$$

Step 4 - Determine the stoichiometry for potassium iodide
From the balanced equation, 1 mol of $$KI$$ is required to form 1 mol of $$K_4[Fe(CN)_6]$$. Therefore, to obtain 2 mol of $$K_4[Fe(CN)_6]$$ (complex P) we need:

$$2\;\text{mol P} \times \dfrac{1\;\text{mol }KI}{1\;\text{mol P}} = 2\;\text{mol }KI$$

Hence, the number of moles of potassium iodide required is 2.

Final answer: 2

In permanganate (MnO₄⁻), manganese has an oxidation state of +7.

If the change in oxidation state is 3, then Mn goes from +7 to +4, forming MnO₂.

The behavior of permanganate depends on the medium:

- In acidic medium: Mn goes from +7 to +2 (change of 5), forming Mn²⁺

- In neutral/weakly alkaline medium: Mn goes from +7 to +4 (change of 3), forming MnO₂

- In strongly alkaline medium: Mn goes from +7 to +6 (change of 1), forming MnO₄²⁻

Since the change in oxidation state is 3 (from +7 to +4), the reaction occurs in aqueous neutral medium.

The correct answer is Aqueous neutral.

The water-gas shift reaction is:

$$CO + H_2O \to CO_2 + H_2$$

In this reaction:

• Carbon monoxide (CO) is oxidized to carbon dioxide (CO₂) — carbon's oxidation state changes from +2 to +4
• Water (H₂O) is reduced — hydrogen is reduced from +1 to 0 (H₂)

This matches option 1: Carbon monoxide is oxidized to carbon dioxide.

Statement I says that in redox titration, the indicators used are sensitive to change in pH of the solution. This is incorrect. Redox indicators respond to changes in the oxidation potential (electrode potential) of the solution, not to pH changes. Examples include diphenylamine and potassium permanganate, which change colour based on their oxidation state.

Statement II says that in acid-base titration, the indicators used are sensitive to change in oxidation potential. This is also incorrect. Acid-base indicators such as phenolphthalein and methyl orange are sensitive to changes in the pH of the solution, not to oxidation potential.

Both statements have the properties of the indicators swapped — redox indicators track oxidation potential while acid-base indicators track pH. Since both Statement I and Statement II are incorrect, the answer is Option B.

Statement I: Ellingham diagrams can be constructed for formation of oxides, sulphides and halides of metals.

This is correct. While Ellingham diagrams are most commonly used for oxides, they can also be drawn for sulphides and halides.

Statement II: It consists of plots of $$\Delta H°$$ vs T for formation of oxides of elements.

This is incorrect. Ellingham diagrams plot $$\Delta G°$$ (Gibbs free energy change) vs T, not $$\Delta H°$$ (enthalpy change).

Statement I is correct but Statement II is incorrect.

This matches option 3.

In acidic medium, $$\mathrm{KMnO_4}$$ is reduced to $$\mathrm{Mn^{2+}}$$, while $$\mathrm{H_2S}$$ is oxidised to elemental sulphur. Write the two half-reactions and balance them for mass and charge.

Reduction half-reaction
$$\mathrm{MnO_4^- + 8\,H^+ + 5\,e^- \;\rightarrow\; Mn^{2+} + 4\,H_2O}$$ $$-(1)$$

Oxidation half-reaction
Sulphur changes from $$-2$$ in $$\mathrm{H_2S}$$ to $$0$$ in $$\mathrm{S}$$, so the increase is $$2$$ electrons per molecule.
$$\mathrm{H_2S \;\rightarrow\; S + 2\,H^+ + 2\,e^-}$$ $$-(2)$$

To make the electrons equal, take LCM of $$5$$ and $$2,$$ i.e. $$10$$ electrons.
Multiply $$(1)$$ by $$2$$ and $$(2)$$ by $$5$$:

$$\mathrm{2\,MnO_4^- + 16\,H^+ + 10\,e^- \;\rightarrow\; 2\,Mn^{2+} + 8\,H_2O}$$ $$-(3)$$
$$\mathrm{5\,H_2S \;\rightarrow\; 5\,S + 10\,H^+ + 10\,e^-}$$ $$-(4)$$

Add $$(3)$$ and $$(4)$$; the $$10\,e^-$$ cancel:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 16\,H^+ \;\rightarrow\; 5\,S + 10\,H^+ + 2\,Mn^{2+} + 8\,H_2O}$$

Subtract $$10\,H^+$$ from both sides to obtain the net balanced redox equation in acidic medium:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 6\,H^+ \;\rightarrow\; 5\,S + 2\,Mn^{2+} + 8\,H_2O}$$ $$-(5)$$

The problem states that exactly $$5$$ moles of $$\mathrm{H_2S}$$ react. Comparing with $$(5)$$:

Number of moles of $$\mathrm{H_2O}$$ produced, $$x = 8$$.

Total electrons transferred per $$(5)$$ are $$10$$; hence for the given scale the number of moles of electrons involved, $$y = 10$$.

Therefore $$x + y = 8 + 10 = 18$$.

Final Answer: 18

Calcination is the process of heating an ore in the absence of air (or limited supply of air) to remove volatile substances like water, CO₂, etc.

Option 1: CaCO₃ → CaO + CO₂: This is calcination (removal of CO₂). ✓

Option 2: Fe₂O₃·xH₂O → Fe₂O₃ + xH₂O: This is calcination (removal of water). ✓

Option 3: 2PbS + 3O₂ → 2PbO + 2SO₂: This is roasting (heating a sulphide ore in the presence of oxygen/air). ✗ NOT calcination.

Option 4: CaCO₃·MgCO₃ → CaO + MgO + 2CO₂: This is calcination. ✓

The answer is option 3, which is roasting, not calcination.

During the preparation of ferrous ammonium sulphate (Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O), prolonged heating is avoided.

This is because Fe²⁺ ions are easily oxidized to Fe³⁺ ions when heated in the presence of air and water. Prolonged heating would cause oxidation of ferrous (Fe²⁺) to ferric (Fe³⁺) ions, which would contaminate the product.

The correct answer is Option 4: Prevent oxidation.

We need to identify which halide ions can be oxidised by oxygen in acidic medium.

The reduction potential of O$$_2$$ in acidic medium:

$$O_2 + 4H^+ + 4e^- \to 2H_2O$$, $$E° = +1.23$$ V

For oxidation of X$$^-$$ to X$$_2$$:

- $$2Cl^- \to Cl_2 + 2e^-$$, $$E° = -1.36$$ V (very difficult to oxidise)

- $$2Br^- \to Br_2 + 2e^-$$, $$E° = -1.07$$ V

- $$2I^- \to I_2 + 2e^-$$, $$E° = -0.54$$ V (easiest to oxidise)

For a reaction to be spontaneous: $$E°_{cell} = E°_{cathode} - E°_{anode} > 0$$

For I$$^-$$: $$E°_{cell} = 1.23 - 0.54 = 0.69$$ V > 0 $$\checkmark$$ (spontaneous)

For Br$$^-$$: $$E°_{cell} = 1.23 - 1.07 = 0.16$$ V > 0, but this is marginal and in practice O$$_2$$ does not readily oxidise Br$$^-$$ in dilute acidic solutions.

For Cl$$^-$$: $$E°_{cell} = 1.23 - 1.36 = -0.13$$ V < 0 $$\times$$ (not spontaneous)

In practice, only iodide is readily oxidised by O$$_2$$ in acidic medium. This is why KI solutions turn yellow/brown on standing in air.

The correct answer is Option C: I$$^-$$ only.

Oxidation number (O.N.) is obtained by assigning the bonding electrons of every hetero-atomic bond to the more electronegative partner and adding the formal charge. Whenever an atom is bonded only to atoms of the same element, that bond contributes $$0$$ to its O.N.

Case 1: $$\mathbf{Br_3O_8}$$
Let the oxidation number of the three bromine atoms be $$x_1,x_2,x_3$$ (they need not be identical). Oxygen (except in peroxides and superoxides) has $$-2$$. Total charge is zero, so
$$x_1+x_2+x_3+8(-2)=0\;\;\Longrightarrow\;\;x_1+x_2+x_3=+16$$ Whatever individual values the three bromine atoms possess, none of them can be $$0$$ because their sum is $$+16$$. Number of atoms in O.N. $$0 = 0$$.

Case 2: $$\mathbf{F_2O}$$
Fluorine is the most electronegative element, so $$\text{O.N.(F)}=-1$$. Let O.N.(O) = $$x$$. $$x+2(-1)=0\;\;\Longrightarrow\;\;x=+2$$ No atom has O.N. $$0$$ here. Number of atoms in O.N. $$0 = 0$$.

Case 3: $$\mathbf{H_2S_4O_6}$$ (tetrathionic acid)
This molecule contains a chain $$HO_3S\!-\!S\!-\!S\!-\!SO_3H$$. • The two terminal sulphur atoms are each bonded to three oxygens and one sulphur. Such an environment is the same as in the dithionate ion, whose sulphur is known to be $$+5$$. Assign O.N.(terminal S) = $$+5$$.
• Let the O.N. of each internal sulphur be $$y$$.
For the whole neutral acid (with $$2$$ hydrogens at $$+1$$): $$2(+1)+2(+5)+2y+6(-2)=0$$ $$2+10+2y-12=0\;\;\Longrightarrow\;\;2y=0\;\;\Longrightarrow\;\;y=0$$ Thus both internal sulphur atoms are in the zero oxidation state. Number of atoms in O.N. $$0 = 2$$.

Case 4: $$\mathbf{H_2S_5O_6}$$ (pentathionic acid)
Structure: $$HO_3S\!-\!S\!-\!S\!-\!S\!-\!SO_3H$$ (a five-sulphur chain). • Terminal sulphur atoms again have O.N. $$+5$$.
• Let the three internal sulphur atoms have O.N.s $$y_1,y_2,y_3$$.
Overall charge $$0$$ gives $$2(+1)+2(+5)+(y_1+y_2+y_3)+6(-2)=0$$ $$2+10+(y_1+y_2+y_3)-12=0\;\;\Longrightarrow\;\;y_1+y_2+y_3=0$$ In a pure $$S\!-\!S$$ chain an individual sulphur is normally either $$0$$ or $$-1$$. A convenient integral solution is $$y_1=0,\;y_2=0,\;y_3=0$$ - fully consistent with the bonding scheme (each of those three S atoms is attached only to other sulphurs). Hence all three internal sulphur atoms are in O.N. $$0$$. Number of atoms in O.N. $$0 = 3$$.

Case 5: $$\mathbf{C_3O_2}$$ (carbon sub-oxide, $$O=C=C=C=O$$)
Let the oxidation numbers of the three carbons be $$x_1,x_2,x_3$$ (left to right). Each oxygen is $$-2$$, so total for both oxygens is $$-4$$: $$x_1+x_2+x_3-4=0\;\;\Longrightarrow\;\;x_1+x_2+x_3=+4$$ The two terminal carbonyl carbons (each double-bonded to an oxygen and singly to another carbon) are in O.N. $$+2$$. Therefore $$x_1=+2,\;x_3=+2 \;\;\Rightarrow\;\;x_2=0$$ for the central carbon. Number of atoms in O.N. $$0 = 1$$.

Total count of atoms in oxidation state 0:
$$0\;(Br_3O_8) + 0\;(F_2O) + 2\;(H_2S_4O_6) + 3\;(H_2S_5O_6) + 1\;(C_3O_2) = 6$$

Hence, the required sum is 6.

In acidic medium, KMnO₄ reacts with KI:

$$2KMnO_4 + 10KI + 8H_2SO_4 \to 2MnSO_4 + 5I_2 + 6K_2SO_4 + 8H_2O$$

Manganese changes from +7 (in KMnO₄) to +2 (in MnSO₄).

Total change in oxidation state = 7 - 2 = 5.

This matches the answer key value of $$\mathbf{5}$$.

KMnO$$_4$$ is titrated with ferrous ammonium sulphate in the presence of dilute H$$_2$$SO$$_4$$. We need to find the number of water molecules produced for 2 molecules of KMnO$$_4$$.

The half-reactions in acidic medium are:

Reduction: $$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Oxidation: $$\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$$

Multiplying oxidation half-reaction by 5 and adding:

$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$

$$2\text{MnO}_4^- + 16\text{H}^+ + 10\text{Fe}^{2+} \to 2\text{Mn}^{2+} + 10\text{Fe}^{3+} + 8\text{H}_2\text{O}$$

The number of water molecules produced for 2 molecules of KMnO$$_4$$ is $$\mathbf{8}$$.

We are given that $$Fe_{0.93}O$$ is heated in the presence of oxygen and converts to $$Fe_2O_3$$. We need to find the number of correct statements from A, B, C, D.

We begin by determining the composition of $$Fe_{0.93}O$$. In $$Fe_{0.93}O$$, one mole of the compound contains 0.93 moles of iron and 1 mole of oxygen. The iron ions are a mixture of $$Fe^{2+}$$ and $$Fe^{3+}$$. Let $$x$$ = moles of $$Fe^{2+}$$ and $$y$$ = moles of $$Fe^{3+}$$. From the total iron: $$x + y = 0.93 \quad \cdots (i)$$
From charge balance (total positive charge = total negative charge from $$O^{2-}$$): $$2x + 3y = 2 \quad \cdots (ii)$$

Next, substituting $$x = 0.93 - y$$ from equation (i) into equation (ii) gives: $$2(0.93 - y) + 3y = 2 \implies 1.86 - 2y + 3y = 2 \implies y = 0.14$$
Then $$x = 0.93 - 0.14 = 0.79$$
This shows that there are 0.79 moles of $$Fe^{2+}$$ and 0.14 moles of $$Fe^{3+}$$ per mole of $$Fe_{0.93}O$$.

Now we verify each statement.

Statement A: The equivalent weight of $$Fe_{0.93}O$$ is $$\frac{\text{Molecular weight}}{0.79}$$. When $$Fe_{0.93}O$$ is oxidized to $$Fe_2O_3$$, the 0.79 moles of $$Fe^{2+}$$ are oxidized to $$Fe^{3+}$$, each losing one electron. The total change in oxidation state per mole of $$Fe_{0.93}O$$ is 0.79. Therefore, the n-factor is 0.79, and the equivalent weight = $$\frac{M}{0.79}$$. Statement A is CORRECT.

Statement B: The moles of $$Fe^{2+}$$ and $$Fe^{3+}$$ are 0.79 and 0.14 respectively. This is exactly what we calculated earlier. Statement B is CORRECT.

Statement C: $$Fe_{0.93}O$$ is metal deficient with a lattice comprising cubic close-packed arrangement of $$O^{2-}$$ ions. $$Fe_{0.93}O$$ is a non-stoichiometric compound derived from FeO (which has the NaCl structure with FCC arrangement of $$O^{2-}$$ ions). It is metal-deficient because some $$Fe^{2+}$$ sites are vacant, with charge balance maintained by some $$Fe^{2+}$$ converting to $$Fe^{3+}$$. The $$O^{2-}$$ ions form a cubic close-packed (CCP/FCC) lattice. Statement C is CORRECT.

Statement D: The percentage composition of $$Fe^{2+}$$ and $$Fe^{3+}$$ is 85% and 15% respectively. $$\%Fe^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.95\% \approx 85\%$$
$$\%Fe^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05\% \approx 15\%$$ Statement D is CORRECT.

All 4 statements are correct. The answer is 4.

We need to analyze each statement about $$Fe_{0.93}O$$.

In $$Fe_{0.93}O$$, let the number of $$Fe^{2+}$$ ions be $$x$$ and $$Fe^{3+}$$ ions be $$y$$ per formula unit.

$$x + y = 0.93$$ ... (i)

For charge neutrality with one $$O^{2-}$$: $$2x + 3y = 2$$ ... (ii)

From (i): $$x = 0.93 - y$$. Substituting in (ii):

$$2(0.93 - y) + 3y = 2$$

$$1.86 - 2y + 3y = 2$$

$$y = 0.14$$, so $$x = 0.79$$

Statement A: When $$Fe_{0.93}O$$ is oxidized to $$Fe_2O_3$$, each $$Fe^{2+}$$ loses 1 electron. The n-factor (number of electrons lost per formula unit) = number of $$Fe^{2+}$$ ions = 0.79. So equivalent weight = Molecular weight / 0.79. This is correct.

Statement B: From our calculation, moles of $$Fe^{2+} = 0.79$$ and $$Fe^{3+} = 0.14$$ per mole of $$Fe_{0.93}O$$. This is correct.

Statement C: $$Fe_{0.93}O$$ has a structure similar to FeO (rock salt type) where $$O^{2-}$$ ions form a cubic close packed (FCC) arrangement. It is a metal-deficient defect (some Fe sites are vacant or have Fe³⁺ instead of Fe²⁺). This is correct.

Statement D: % of $$Fe^{2+} = \frac{0.79}{0.93} \times 100 = 84.95\% \approx 85\%$$

% of $$Fe^{3+} = \frac{0.14}{0.93} \times 100 = 15.05\% \approx 15\%$$

This is correct.

All four statements are correct.

The answer is 4.

We need to find the sum of oxidation states of bromine in bromic acid and perbromic acid.

For bromic acid ($$HBrO_3$$), let the oxidation state of Br be $$x$$:

$$+1 + x + 3(-2) = 0$$, so $$x = +5$$.

For perbromic acid ($$HBrO_4$$), let the oxidation state of Br be $$y$$:

$$+1 + y + 4(-2) = 0$$, so $$y = +7$$.

Hence, the sum of oxidation states = $$5 + 7 = 12$$. So, the answer is $$12$$.

We need to find the number of electrons involved in the reduction of permanganate (MnO$$_4^-$$) to manganese dioxide (MnO$$_2$$). In MnO$$_4^-$$, let Mn have oxidation state $$x$$. Then $$x + 4(-2) = -1$$, so $$x = +7$$, and in MnO$$_2$$, $$x + 2(-2) = 0$$, so $$x = +4$$.

The change in oxidation state from +7 to +4 is a decrease of 3, so 3 electrons are involved.

The balanced half-reaction is $$ MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O $$. Verification: charge on left = $$-1 + 4 - 3 = 0$$, charge on right = 0. Oxygen atoms: left = 4, right = 2 + 2 = 4. Hydrogen atoms: left = 4, right = 4.

The number of electrons is 3.

Hydrogen peroxide (H₂O₂) tends to decompose into water and oxygen over time:

$$2H_2O_2 \rightarrow 2H_2O + O_2$$

To prevent this decomposition, a stabilizer is added that can inhibit the catalytic decomposition.

Analysis of options:

Option A: Urea

Urea is a well-known stabilizer for H₂O₂. It forms a stable adduct (urea hydrogen peroxide, UHP) with H₂O₂ and acts as a negative catalyst, slowing down the decomposition. It is commonly used in pharmaceutical and industrial applications to stabilize H₂O₂ solutions.

Option B: Formaldehyde

Formaldehyde is a reducing agent and can react with H₂O₂ rather than stabilize it.

Option C: Formic acid

Formic acid can react with H₂O₂ to form performic acid and is not used as a stabilizer.

Option D: Ethanol

Ethanol is an organic solvent that does not serve as an effective stabilizer for H₂O₂.

Therefore, the correct answer is Option A (Urea).

Molar mass of $$Cu(NO_3)_2 = 63 + 2(14) + 6(16) = 187 \text{ g mol}^{-1}$$.

Moles of $$Cu(NO_3)_2$$ taken $$n_{Cu^{2+}} = \frac{3.74 \text{ g}}{187 \text{ g mol}^{-1}} = 0.020 \text{ mol}$$.

With excess $$KI$$ the reaction is $$2\,Cu^{2+} + 4\,I^- \;\longrightarrow\; 2\,CuI(s) + I_2(aq)$$.

From the stoichiometry, $$2$$ mol of $$Cu^{2+}$$ give $$1$$ mol of $$I_2$$. Hence moles of $$I_2$$ formed $$n_{I_2} = \frac{0.020}{2} = 0.010 \text{ mol}$$.

The brown colour of the filtrate is due to this $$I_2$$. Passing $$H_2S$$ through it: $$I_2 + H_2S \;\longrightarrow\; 2\,HI + S(s)$$.

Here, $$1$$ mol of $$I_2$$ produces $$1$$ mol of elemental sulfur $$S$$. Therefore, moles of precipitated sulfur $$n_S = 0.010 \text{ mol}$$.

Mass of sulfur obtained $$m_S = n_S \times M_S = 0.010 \times 32 = 0.32 \text{ g}$$.

Hence, the amount of precipitate X is 0.32 g.

We need to determine the product when dihydrogen ($$H_2$$) reacts with copper(II) oxide ($$CuO$$). Hydrogen gas is a reducing agent; when it reacts with metal oxides, it reduces the metal oxide, so this redox reaction involves $$H_2$$ acting as a reducing agent and $$CuO$$ getting reduced.

$$CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(l)$$

Copper in $$CuO$$ has an oxidation state of +2, hydrogen reduces $$Cu^{2+}$$ to $$Cu^0$$ (metallic copper) and is itself oxidized from 0 to +1 in $$H_2O$$. This reaction is used as a standard test for the reducing property of hydrogen gas. When $$H_2$$ is passed over heated $$CuO$$ (black), the black powder turns reddish-brown as metallic copper ($$Cu$$) is formed.

Other possible products such as $$Cu(OH)_2$$ would require an aqueous base environment rather than a direct reaction of $$H_2$$ with $$CuO$$; $$Cu_2O$$ would indicate partial reduction, but $$H_2$$ completely reduces $$CuO$$ to metallic $$Cu$$; and copper hydride ($$CuH_2$$) is not formed under these conditions.

Therefore, the correct answer is Option B: $$Cu(s)$$.

We need to determine the product when $$MnO_4^-$$ oxidises thiosulphate ($$S_2O_3^{2-}$$) in neutral or alkaline solution.

In neutral or alkaline medium, permanganate is reduced to $$MnO_2$$:

$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \quad \cdots (i)$$

Mn changes from $$+7$$ to $$+4$$, gaining 3 electrons.

In $$S_2O_3^{2-}$$, the average oxidation state of S is:

$$2x + 3(-2) = -2 \implies x = +2$$

The average oxidation state of sulphur is $$+2$$.

With a strong oxidizing agent like $$MnO_4^-$$ in alkaline medium, thiosulphate undergoes complete oxidation to sulphate ($$SO_4^{2-}$$), where S is in the $$+6$$ state:

$$S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+ + 8e^-$$

Converting to basic medium by adding $$10OH^-$$ to both sides:

$$S_2O_3^{2-} + 5H_2O + 10OH^- \rightarrow 2SO_4^{2-} + 10H_2O + 8e^-$$

Simplifying:

$$S_2O_3^{2-} + 10OH^- \rightarrow 2SO_4^{2-} + 5H_2O + 8e^- \quad \cdots (ii)$$

Each sulphur atom changes from $$+2$$ to $$+6$$, losing 4 electrons per atom (8 electrons total for 2 atoms).

Multiply equation (i) by 8 and equation (ii) by 3 to equalize electrons (24 electrons):

$$8MnO_4^- + 16H_2O + 24e^- \rightarrow 8MnO_2 + 32OH^-$$

$$3S_2O_3^{2-} + 30OH^- \rightarrow 6SO_4^{2-} + 15H_2O + 24e^-$$

Adding both half-reactions:

$$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$$

Conclusion:

In neutral or alkaline solution, $$MnO_4^-$$ completely oxidises thiosulphate to $$SO_4^{2-}$$.

Hence, the correct answer is Option D.

We are told that in neutral or faintly alkaline medium, $$KMnO_4$$ oxidises thiosulphate to sulphate. We need to find the overall change in oxidation state of manganese.

In $$KMnO_4$$, the oxidation state of Mn is $$+7$$. Now, in a neutral or faintly alkaline medium, $$KMnO_4$$ is reduced to $$MnO_2$$ (manganese dioxide), where Mn has an oxidation state of $$+4$$.

We can verify this: in strongly acidic medium, $$MnO_4^-$$ is reduced to $$Mn^{2+}$$ (change of 5). In strongly alkaline medium, it is reduced to $$MnO_4^{2-}$$ (change of 1). But in neutral or faintly alkaline medium, the product is $$MnO_2$$.

The change in oxidation state of manganese is:

$$\Delta = +7 - (+4) = 3$$

Hence, the overall change in oxidation state of manganese is 3.

Hence, the correct answer is Option D: 3.

We need to find the products when hydrogen peroxide reacts with acidified potassium permanganate ($$KMnO_4$$).

In acidic medium, $$KMnO_4$$ acts as a strong oxidising agent and $$H_2O_2$$ acts as a reducing agent (it gets oxidised).

Since $$MnO_4^-$$ is reduced in acidic medium, the reduction half-reaction is $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$. Meanwhile, $$H_2O_2$$ is oxidised according to $$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$$.

Substituting and balancing electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5 gives the balanced half-reactions: $$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$ and $$5H_2O_2 \rightarrow 5O_2 + 10H^+ + 10e^-$$.

Combining these half-reactions yields the overall reaction $$2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$$.

Therefore, the products are $$Mn^{2+}$$, $$H_2O$$, and $$O_2$$. In acidic medium, $$KMnO_4$$ is reduced to $$Mn^{2+}$$ (not $$Mn^{4+}$$). $$Mn^{4+}$$ as $$MnO_2$$ is formed only in neutral or weakly basic medium.

Therefore, the correct answer is Option D: $$Mn^{2+}$$, $$H_2O$$, $$O_2$$ only.

We need to find the difference in oxidation states of the two ions formed when Manganese(VI) disproportionates in acidic solution. Manganese in the +6 oxidation state exists as the manganate ion: $$MnO_4^{2-}$$ (where Mn is +6).

In disproportionation, the same species is simultaneously oxidized and reduced. In acidic solution, $$MnO_4^{2-}$$ disproportionates as:

$$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$

In $$MnO_4^-$$ (permanganate ion): Let the oxidation state of Mn be $$y$$. Then $$y + 4(-2) = -1$$, giving $$y = +7$$. In $$MnO_2$$ (manganese dioxide): Let the oxidation state of Mn be $$z$$. Then $$z + 2(-2) = 0$$, giving $$z = +4$$.

Difference in oxidation states = $$+7 - (+4) = 3$$. The answer is 3.

We need to find the volume of KMnO$$_4$$ solution left in the burette after titration with Mohr's salt.

KMnO$$_4$$ solution concentration = 0.01 M, Mohr's salt (FeSO$$_4$$(NH$$_4$$)$$_2$$SO$$_4$$ · 6H$$_2$$O) solution is 20.0 mL of 0.05 M, and the burette capacity is 50 mL with an initial reading of 0.

In acidic medium: $$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$ So 1 mole of KMnO$$_4$$ reacts with 5 moles of Fe$$^{2+}$$.

$$\text{mmol of Fe}^{2+} = 0.05 \times 20.0 = 1.0 \text{ mmol}, \quad \text{mmol of KMnO}_4 = \frac{1.0}{5} = 0.2 \text{ mmol}$$

$$V = \frac{\text{mmol}}{M} = \frac{0.2}{0.01} = 20 \text{ mL}$$

Volume left = Total volume − Volume used = 50 − 20 = 30 mL. Hence, the answer is 30 mL.

We are given 20 mL of 0.02 M hypo (sodium thiosulphate, $$Na_2S_2O_3$$) solution, 10 mL of copper sulphate ($$CuSO_4$$) solution, and excess KI, and we need to find the molarity of $$Cu^{2+}$$.

Since $$2Cu^{2+} + 4I^- \rightarrow 2CuI \downarrow + I_2$$ in the presence of excess iodide ions, the liberated $$I_2$$ is then titrated with sodium thiosulphate according to $$I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$$.

Substituting the values for the titration gives $$\text{Moles of } Na_2S_2O_3 = 0.02 \times \frac{20}{1000} = 4 \times 10^{-4} \text{ mol}$$, and because 1 mol of $$I_2$$ reacts with 2 mol of $$S_2O_3^{2-}$$, the moles of $$I_2$$ are $$\text{Moles of } I_2 = \frac{4 \times 10^{-4}}{2} = 2 \times 10^{-4} \text{ mol}$$.

Since 2 mol of $$Cu^{2+}$$ produce 1 mol of $$I_2$$, it follows that $$\text{Moles of } Cu^{2+} = 2 \times (2 \times 10^{-4}) = 4 \times 10^{-4} \text{ mol}$$.

Finally, dividing by the 10 mL (0.01 L) of $$Cu^{2+}$$ solution gives $$\text{Molarity} = \frac{\text{Moles}}{\text{Volume in litres}} = \frac{4 \times 10^{-4}}{10/1000} = \frac{4 \times 10^{-4}}{0.01} = 4 \times 10^{-2} \text{ M}$$. Therefore, the molarity of $$Cu^{2+}$$ is 4 $$\times 10^{-2}$$ M.

We need to find the percentage of $$MnO_2$$ in a 2.0 g sample, given that the liberated $$Cl_2$$ reacts with KI and the liberated $$I_2$$ requires 60.0 mL of 0.1 M $$Na_2S_2O_3$$.

When $$MnO_2$$ reacts with HCl according to $$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$$ one mole of $$MnO_2$$ produces one mole of $$Cl_2$$. The generated $$Cl_2$$ then oxidizes KI in the reaction $$Cl_2 + 2KI \rightarrow 2KCl + I_2$$ liberating one mole of $$I_2$$ per mole of $$Cl_2$$. The liberated $$I_2$$ is titrated with $$Na_2S_2O_3$$ according to $$I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI$$ where each mole of $$I_2$$ consumes two moles of $$Na_2S_2O_3$$.

The moles of $$Na_2S_2O_3$$ used in the titration are given by $$\text{Moles of } Na_2S_2O_3 = M \times V = 0.1 \times \frac{60}{1000} = 6 \times 10^{-3} \text{ mol}$$. Therefore, $$\text{Moles of } I_2 = \frac{\text{Moles of } Na_2S_2O_3}{2} = \frac{6 \times 10^{-3}}{2} = 3 \times 10^{-3} \text{ mol}$$. Since one mole of $$Cl_2$$ yields one mole of $$I_2$$, the moles of $$Cl_2$$ are also $$3 \times 10^{-3}$$ mol, which means the sample contained $$3 \times 10^{-3}$$ mol of $$MnO_2$$.

Using the molar mass of $$MnO_2$$ (55 + 2(16) = 87 g/mol), the mass of $$MnO_2$$ in the sample is $$3 \times 10^{-3} \times 87 = 0.261 \text{ g}$$. Hence, the percentage of $$MnO_2$$ in the 2.0 g sample is $$\frac{0.261}{2.0} \times 100 = 13.05\%$$, which rounds to 13%.

The correct answer is 13%.

We are told that $$KIO_4$$ (potassium periodate) is a stronger oxidizing agent that reacts with hydrogen peroxide ($$H_2O_2$$), causing $$H_2O_2$$ to get oxidized with the evolution of $$O_2$$. We need to find the oxidation number of iodine in the product after this reaction.

In $$KIO_4$$, potassium has an oxidation state of +1 and each oxygen has -2. So the oxidation state of iodine is: $$+1 + x + 4(-2) = 0$$, which gives $$x = +7$$. So iodine starts with an oxidation state of +7.

The reaction is:

$$KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2$$

Here, $$H_2O_2$$ is oxidized (oxygen goes from -1 in $$H_2O_2$$ to 0 in $$O_2$$), while $$KIO_4$$ is reduced. The iodine in $$KIO_4$$ (oxidation state +7) is reduced to iodine in $$KIO_3$$ (potassium iodate).

In $$KIO_3$$: $$+1 + x + 3(-2) = 0$$, which gives $$x = +5$$.

So the oxidation number of iodine changes from +7 in $$KIO_4$$ to +5 in $$KIO_3$$.

Hence, the correct answer is 5.

The balanced chemical equation for the reaction between potassium permanganate and oxalic acid in acidic medium is:

$$\mathrm{2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O}$$

Oxalic acid acts as the reducing agent and is oxidised to $$\mathrm{CO_2}$$.

To calculate the change in oxidation number of carbon, determine its oxidation state in both reactant and product.

In oxalic acid $$\mathrm{(H_2C_2O_4)}$$:

Let oxidation state of carbon be $$\mathrm{x}$$.

$$\mathrm{2(+1) + 2x + 4(-2) = 0}$$

$$\mathrm{2 + 2x - 8 = 0}$$

$$\mathrm{2x = 6}$$

$$\mathrm{x = +3}$$

Thus, oxidation state of carbon in oxalic acid is:

$$\mathrm{+3}$$

In carbon dioxide $$\mathrm{(CO_2)}$$:

Let oxidation state of carbon be $$\mathrm{y}$$.

$$\mathrm{y + 2(-2) = 0}$$

$$\mathrm{y - 4 = 0}$$

$$\mathrm{y = +4}$$

Thus, oxidation state of carbon in $$\mathrm{CO_2}$$ is:

$$\mathrm{+4}$$

Therefore, change in oxidation number of carbon:

$$\mathrm{= (+4) - (+3)}$$

$$\mathrm{= 1}$$

$$\boxed{\mathrm{1}}$$

We need to find the difference in oxidation state of chromium in chromate (CrO$$_4^{2-}$$) and dichromate (Cr$$_2$$O$$_7^{2-}$$) salts. Let the oxidation state of Cr in chromate be $$x$$.

$$ x + 4(-2) = -2 $$

$$ x - 8 = -2 $$

$$ x = +6 $$

Let the oxidation state of Cr in dichromate be $$y$$.

$$ 2y + 7(-2) = -2 $$

$$ 2y - 14 = -2 $$

$$ 2y = 12 $$

$$ y = +6 $$

The difference in oxidation states:

$$ x - y = +6 - (+6) = 0 $$

Therefore, the difference in oxidation state of chromium in chromate and dichromate salts is 0.

We need to find the oxidation state of manganese in the product formed when potassium permanganate reacts with hydrogen peroxide in basic medium.

In basic medium, $$KMnO_4$$ reacts with $$H_2O_2$$ as follows: $$2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 2KOH + 3O_2 + 2H_2O$$.

Since the permanganate ion is reduced in this reaction, the manganese-containing product is manganese dioxide, $$MnO_2$$, which appears as a brown precipitate.

Let the oxidation state of Mn in $$MnO_2$$ be $$x$$. Substituting into the charge balance gives $$x + 2(-2) = 0$$.

From the above equation, we have $$x - 4 = 0$$, which leads to $$x = +4$$.

Therefore, the oxidation state of manganese changes from +7 in $$KMnO_4$$ to +4 in $$MnO_2$$.

Hence, the answer is 4.

We need to evaluate both statements about $$H_2O_2$$ and the hydrogen economy.

Consider Statement I: $$H_2O_2$$ can act as both an oxidising and a reducing agent in basic medium. In $$H_2O_2$$, the oxidation state of oxygen is $$-1$$, which lies between $$0$$ (in $$O_2$$) and $$-2$$ (in $$H_2O$$ or $$OH^-$$). Because of this intermediate oxidation state, $$H_2O_2$$ can either gain electrons (be reduced, acting as an oxidising agent) or lose electrons (be oxidised, acting as a reducing agent). When acting as an oxidising agent in basic medium, oxygen goes from $$-1$$ to $$-2$$, for example: $$H_2O_2 + 2e^- \rightarrow 2OH^-$$. When acting as a reducing agent in basic medium, oxygen goes from $$-1$$ to $$0$$, for example: $$H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$$. This is seen when $$H_2O_2$$ decolourises acidified $$KMnO_4$$ (reducing agent) or when it oxidises $$I^-$$ to $$I_2$$ (oxidising agent). Statement I is true.

Consider Statement II: In the hydrogen economy, energy is transmitted in the form of dihydrogen. The hydrogen economy refers to a system in which hydrogen gas ($$H_2$$, dihydrogen) is used as the primary energy carrier. Energy from various sources (solar, nuclear, etc.) is used to produce $$H_2$$ (for example, by electrolysis of water), and this $$H_2$$ is then stored, transported, and used in fuel cells or combustion to release energy where needed. Thus, dihydrogen serves as the medium through which energy is transmitted and distributed. Statement II is true.

Since both statements are true, the correct answer is Option (2): Both statement I and statement II are true.

Statement I says sodium hydride (NaH) can be used as an oxidising agent. In NaH, hydrogen is in the $$-1$$ oxidation state. To act as an oxidising agent, hydrogen would need to gain electrons (become more negative), but it is already at its lowest oxidation state of $$-1$$. Instead, NaH acts as a strong reducing agent, as $$H^-$$ readily donates electrons to become $$H_2$$ (oxidation state 0). Therefore, Statement I is false.

Statement II says the lone pair of electrons on nitrogen in pyridine makes it basic. Pyridine has a nitrogen atom with a lone pair in an $$sp^2$$ hybrid orbital that lies in the plane of the ring and is not part of the aromatic $$\pi$$ system. This lone pair is available to accept a proton ($$H^+$$), making pyridine a base. Therefore, Statement II is true.

Since Statement I is false and Statement II is true, the correct answer is option (3).

In basic medium, $$\text{H}_2\text{O}_2$$ can act as both an oxidising agent (getting reduced to $$\text{OH}^-$$) and a reducing agent (getting oxidised to $$\text{O}_2$$). The relevant half-reactions in basic medium are:

As an oxidising agent: $$\text{H}_2\text{O}_2 + 2e^- \rightarrow 2\text{OH}^-$$

As a reducing agent: $$\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^-$$ (in basic medium, $$\text{H}^+$$ ions combine with $$\text{OH}^-$$, so this becomes $$\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^-$$).

(A) $$\text{Mn}^{2+} \rightarrow \text{Mn}^{4+}$$: Manganese is oxidised from +2 to +4. Here $$\text{H}_2\text{O}_2$$ acts as the oxidising agent. In basic medium, $$\text{Mn}^{2+}$$ reacts with $$\text{H}_2\text{O}_2$$ and $$\text{OH}^-$$ to form $$\text{MnO}_2$$ (which contains $$\text{Mn}^{4+}$$). The half-reaction for Mn is: $$\text{Mn}^{2+} \rightarrow \text{MnO}_2 + 2e^-$$. The $$\text{H}_2\text{O}_2$$ provides the oxidising power by accepting these electrons. This reaction is feasible in basic medium.

(B) $$\text{I}_2 \rightarrow \text{I}^-$$: Iodine is reduced from 0 to $$-1$$. For this reduction, something must supply electrons to $$\text{I}_2$$. Here $$\text{H}_2\text{O}_2$$ acts as the reducing agent, getting oxidised to $$\text{O}_2$$. The half-reactions are: $$\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$$ and $$\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^-$$. In basic medium, the reducing power of $$\text{H}_2\text{O}_2$$ is enhanced by $$\text{OH}^-$$, making this reaction feasible.

(C) $$\text{PbS} \rightarrow \text{PbSO}_4$$: Here sulfur must be oxidised from $$-2$$ in $$\text{PbS}$$ to $$+6$$ in $$\text{PbSO}_4$$. This is an 8-electron change per sulfur atom, requiring an extremely strong oxidising agent. The standard electrode potential of $$\text{H}_2\text{O}_2$$ as an oxidiser is not sufficient to drive such a large change, even in basic medium. Concentrated $$\text{HNO}_3$$ or similar powerful oxidisers are needed for this transformation. Therefore, this reaction is not feasible with $$\text{H}_2\text{O}_2$$ in basic medium.

Since reactions (A) and (B) are feasible but (C) is not, the correct answer is Option D: (A), (B) only.

First, let us recall what photochemical smog is. It is the brown hazy mixture that is produced when primary pollutants such as $$\text{NO}$$ and $$\text{NO}_2$$, emitted from vehicular exhaust, react in the presence of strong sunlight with hydrocarbons present in air. The major secondary products that are finally present in this smog are ozone $$\left(\text{O}_3\right)$$, nitric oxide $$\left(\text{NO}\right)$$, acrolein $$\left(\text{CH}_2\!\!=\!\text{CH}-\text{CHO}\right)$$, formaldehyde $$\left(\text{HCHO}\right)$$ and peroxyacetyl nitrate $$\left(\text{PAN},\ \text{CH}_3\text{COOONO}_2\right)$$.

Now, all these substances—especially ozone and PAN—are very powerful oxidising agents. An oxidising agent has the tendency to remove electrons from, that is, oxidise, the substances it comes in contact with. Natural as well as synthetic rubber contains unsaturated $$\text{C=C}$$ bonds in its long-chain molecules. Oxidising agents attack these double bonds, break the chains and weaken the material. This chemical attack is popularly called “cracking of rubber”.

So we have:

$$\text{O}_3 + \text{Rubber} \;\longrightarrow\; \text{Cracked\ rubber\ fragments}$$

Because photochemical smog supplies the oxidants—$$\text{O}_3,\ \text{PAN},\ \text{NO}$$ etc.—it definitely causes cracking of rubber that is exposed to the atmosphere. Thus the statement in the Assertion (A) is correct.

Reason (R) says that the presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidising. We have just seen that each of these species is indeed an oxidant (ozone and PAN are very strong; acrolein and formaldehyde are milder but still oxidising). Therefore the Reason (R) is also correct.

Furthermore, we notice that the reason directly supplies the chemical cause behind the cracking mentioned in the assertion: because the smog is oxidising, it attacks the $$\text{C=C}$$ bonds of rubber and cracks it. Therefore (R) is the true and sufficient explanation of (A).

Hence, the correct answer is Option 3.

In mildly alkaline medium, the permanganate ion ($$MnO_4^-$$) acts as an oxidizing agent. The thiosulphate ion ($$S_2O_3^{2-}$$) is oxidized to sulphate ion ($$SO_4^{2-}$$) in alkaline conditions.

In the thiosulphate ion $$S_2O_3^{2-}$$, the average oxidation state of sulphur is +2. In mildly alkaline medium, permanganate is a strong enough oxidizer to convert thiosulphate completely to sulphate.

In the sulphate ion $$SO_4^{2-}$$, let the oxidation state of sulphur be $$x$$. Then $$x + 4(-2) = -2$$, giving $$x = +6$$.

Therefore, the product 'A' is $$SO_4^{2-}$$ and the oxidation state of sulphur in 'A' is $$6$$.

We have a titration between acidic potassium permanganate and ferrous sulphate. In acidic medium the relevant half-reactions are first stated:

$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

For the electrons to cancel, we multiply the iron half-reaction by 5 and add:

$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$

Hence, $$1$$ mole of $$\text{KMnO}_4$$ reacts with $$5$$ moles of $$\text{FeSO}_4$$.

The volume of the ferrous sulphate solution used is $$10\ \text{mL}=0.010\ \text{L}$$ and its molarity is $$0.1\ \text{M}$$. By the definition of molarity, $$\text{Molarity} = \dfrac{\text{moles}}{\text{volume in L}}$$, so the moles of $$\text{FeSO}_4$$ taken are

$$n_{\text{FeSO}_4}=0.1\ \text{mol L}^{-1}\times0.010\ \text{L}=0.001\ \text{mol}$$

The stoichiometric ratio tells us that

$$n_{\text{KMnO}_4}=\dfrac{n_{\text{FeSO}_4}}{5}=\dfrac{0.001}{5}=0.0002\ \text{mol}$$

The volume of the potassium permanganate solution is also $$10\ \text{mL}=0.010\ \text{L}$$. Again using $$\text{Molarity} = \dfrac{\text{moles}}{\text{volume}}$$, we get

$$M_{\text{KMnO}_4}=\dfrac{0.0002\ \text{mol}}{0.010\ \text{L}}=0.02\ \text{M}$$

The strength (gram per litre) is obtained from the formula $$\text{Strength} = \text{Molarity}\times\text{Molar mass}$$.

The molar mass of $$\text{KMnO}_4$$ is calculated as $$39\ (\text{for K}) + 55\ (\text{for Mn}) + 4\times16\ (\text{for O}) = 158\ \text{g mol}^{-1}$$.

Substituting, we obtain

$$\text{Strength}=0.02\ \text{mol L}^{-1}\times158\ \text{g mol}^{-1}=3.16\ \text{g L}^{-1}$$

Finally, writing $$3.16\ \text{g L}^{-1}$$ in the form $$\times10^{-2}$$ gives $$3.16\ \text{g L}^{-1}=316\times10^{-2}\ \text{g L}^{-1}$$.

Taking the nearest integer as required, we have $$316$$.

So, the answer is $$316$$.

We are told that 10 mL of a solution containing Fe$$^{2+}$$ ions is titrated with 0.02 M K$$_2$$Cr$$_2$$O$$_7$$ in the presence of dilute H$$_2$$SO$$_4$$, using diphenylamine as the indicator. The titration requires 15 mL of the dichromate solution to reach the end point. Our task is to find the molarity of the Fe$$^{2+}$$ solution and express it in the form $$x \times 10^{-2}\,{\rm M}$$, then determine the integer $$x$$.

First, we recall the balanced redox reaction that occurs in acidic medium between dichromate ions and ferrous ions:

$$\text{Cr}_2\text{O}_7^{2-} + 14\,\text{H}^+ + 6\,\text{Fe}^{2+} \;\longrightarrow\; 2\,\text{Cr}^{3+} + 6\,\text{Fe}^{3+} + 7\,\text{H}_2\text{O}$$

From this equation we see clearly that 1 mole of dichromate reacts with 6 moles of Fe$$^{2+}$$.

Now we proceed step by step, starting with the dichromate solution:

Using the definition of molarity, the number of moles is given by

$$n = M \times V$$

where $$M$$ is molarity in mol L$$^{-1}$$ and $$V$$ is volume in litres.

The dichromate data are:

$$M_{\text{Cr}_2\text{O}_7^{2-}} = 0.02\,{\rm M}, \qquad V_{\text{Cr}_2\text{O}_7^{2-}} = 15\,\text{mL} = 0.015\,\text{L}$$

Substituting, we obtain

$$n_{\text{Cr}_2\text{O}_7^{2-}} = 0.02 \times 0.015 = 0.00030\;\text{mole}$$

According to the stoichiometric ratio 1 : 6, the moles of Fe$$^{2+}$$ oxidised are

$$n_{\text{Fe}^{2+}} = 6 \times n_{\text{Cr}_2\text{O}_7^{2-}} = 6 \times 0.00030 = 0.00180\;\text{mole}$$

The volume of the Fe$$^{2+}$$ solution taken is

$$V_{\text{Fe}^{2+}} = 10\,\text{mL} = 0.010\,\text{L}$$

Again invoking the molarity formula,

$$M_{\text{Fe}^{2+}} = \dfrac{n_{\text{Fe}^{2+}}}{V_{\text{Fe}^{2+}}} = \dfrac{0.00180}{0.010} = 0.18\;\text{M}$$

To express this in the required format, we write

$$0.18\;\text{M} = 18 \times 10^{-2}\;\text{M}$$

Thus, $$x = 18$$.

So, the answer is $$18$$.

In acidic medium, dichromate oxidizes ferrous ions according to the balanced equation: $$Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O$$. This shows that 1 mole of $$Cr_2O_7^{2-}$$ reacts with 6 moles of $$Fe^{2+}$$.

The moles of $$Cr_2O_7^{2-}$$ used = $$20 \times 10^{-3} \times 0.03 = 6 \times 10^{-4}$$ mol. From the stoichiometry, moles of $$Fe^{2+}$$ = $$6 \times 6 \times 10^{-4} = 36 \times 10^{-4}$$ mol.

The molarity of the $$Fe^{2+}$$ solution is $$\frac{36 \times 10^{-4}}{15 \times 10^{-3}} = \frac{36}{15} \times 10^{-1} = 2.4 \times 10^{-1} = 0.24$$ M. Expressed as $$\_\_ \times 10^{-2}$$ M, this is $$24 \times 10^{-2}$$ M. The answer is 24.

In such questions we first recall the definition of a redox (reduction-oxidation) reaction: it is a chemical change in which at least one element undergoes an increase in oxidation number (oxidation) while another element simultaneously shows a decrease in oxidation number (reduction). To decide which of the four given processes satisfies this condition, we calculate the oxidation numbers of every element on both sides of each reaction equation.

Let us begin with Option A, the photochemical conversion of di-oxygen to ozone. The skeletal equation is

$$3\;{\rm O_2}\;\xrightarrow{\text{sunlight}}\;2\;{\rm O_3}$$

In both $$\rm O_2$$ and $$\rm O_3$$ every oxygen atom is in the elemental state, so its oxidation number is $$0$$ on the left as well as on the right. Because no atom exhibits any change in oxidation number, this process is not redox.

Now we analyse Option B, the reaction between $$[{\rm Co(H_2O)_6}]^{3+}{\rm Cl_3^-}$$ and $$\rm AgNO_3$$. The molecular equation is

$$[{\rm Co(H_2O)_6}]{\rm Cl_3} + 3\;{\rm AgNO_3} \;\longrightarrow\;[{\rm Co(H_2O)_6}]{\rm (NO_3)_3} + 3\;{\rm AgCl}\downarrow$$

On both sides cobalt remains in the coordination sphere as $$\rm Co^{3+}$$, silver remains $$\rm Ag^{+}$$ in the ionic reagent and precipitated AgCl, chloride stays $$\rm Cl^-$$, nitrogen is $$+5$$ in each $$\rm NO_3^-$$ ion and oxygen is unchanged. Since no element experiences a change in oxidation number, this exchange of ions is a mere precipitation reaction, not a redox change.

We proceed to Option C, neutralisation of sulphuric acid with sodium hydroxide:

$$\rm H_2SO_4 + 2\,NaOH \;\longrightarrow\; Na_2SO_4 + 2\,H_2O$$

The oxidation numbers are as follows:

For sulphur in $$\rm H_2SO_4$$, applying the rule $$\sum \text{(oxidation numbers)} = \text{charge}$$, we have

$$2(+1) + x + 4(-2) = 0 \;\Longrightarrow\; x = +6$$

In $$\rm Na_2SO_4$$ the calculation

$$2(+1) + x + 4(-2) = 0 \;\Longrightarrow\; x = +6$$

gives the same value $$+6$$ for sulphur. Hydrogen is $$+1$$ in the acid and remains $$+1$$ in water; oxygen is $$-2$$ throughout; sodium is $$+1$$ on both sides. Thus the reaction involves no oxidation-number change and is therefore an acid-base, not a redox, process.

Finally, Option D describes the combination of gaseous dinitrogen with dioxygen at very high temperature (about $$2000\;{\rm K}$$):

$$\rm N_2 + O_2 \;\longrightarrow\; 2\,NO$$

We now assign oxidation numbers step by step. In their elemental forms $$\rm N_2$$ and $$\rm O_2$$ each atom possesses oxidation number $$0$$. For nitric oxide $$\rm NO$$, let the oxidation number of nitrogen be $$x$$; oxygen as usual in a neutral oxide is $$-2$$. Hence

$$x + (-2) = 0 \;\Longrightarrow\; x = +2$$

Therefore, nitrogen changes from $$0$$ to $$+2$$, which is an increase, i.e. oxidation. Concurrently oxygen changes from $$0$$ in $$\rm O_2$$ to $$-2$$ in $$\rm NO$$, a decrease, i.e. reduction. Both oxidation and reduction occur simultaneously in this single reaction, satisfying the criterion for a redox process.

Among the four alternatives, only the formation of nitric oxide in Option D exhibits simultaneous rise and fall of oxidation numbers. The other three processes show no such change.

Hence, the correct answer is Option 4.

First, we recall the fundamental rule for any electrically neutral compound: the algebraic sum of the oxidation numbers of all the atoms present is always equal to zero. We write this rule as $$\text{Sum of all oxidation numbers}=0.$$ We apply this idea one compound at a time.

We start with potassium oxide, $$\mathrm{K_2O}.$$ In ordinary oxides, oxygen always carries the oxidation number $$-2.$$ Let the oxidation number of potassium be $$x.$$ The formula contains two potassium atoms and one oxygen atom, so we write the sum:

$$2\,x + (-2) = 0.$$

Now we isolate $$x$$ by shifting $$-2$$ to the other side:

$$2\,x = +2.$$

Dividing both sides by $$2,$$ we obtain

$$x = +1.$$

Hence, in $$\mathrm{K_2O}$$ each potassium atom has oxidation number $$+1.$$

Next, we examine potassium peroxide, $$\mathrm{K_2O_2}.$$ In a peroxide ion $$\left(\mathrm{O_2^{2-}}\right),$$ each oxygen possesses oxidation number $$-1.$$ Again let the oxidation number of potassium be $$x.$$ The compound has two potassium atoms and two oxygen atoms, therefore

$$2\,x + 2\,(-1) = 0.$$

Multiplying out the terms we have

$$2\,x - 2 = 0.$$

Adding $$2$$ on both sides gives

$$2\,x = +2.$$

Dividing by $$2$$ yields

$$x = +1.$$

Thus, in $$\mathrm{K_2O_2}$$ also, each potassium atom carries oxidation number $$+1.$$

Finally, we turn to potassium superoxide, $$\mathrm{KO_2}.$$ A superoxide ion $$\left(\mathrm{O_2^{-}}\right)$$ bears an overall charge of $$-1,$$ so the average oxidation number of each oxygen is $$-\dfrac12.$$ Denote the oxidation number of potassium by $$x.$$ The formula contains one potassium atom and two oxygen atoms, so the sum statement becomes

$$x + 2\left(-\dfrac12\right) = 0.$$

Simplifying the product $$2\left(-\dfrac12\right)$$ we get $$-1,$$ hence

$$x - 1 = 0.$$

Adding $$1$$ to both sides gives

$$x = +1.$$

Consequently, in $$\mathrm{KO_2}$$ the oxidation number of potassium is again $$+1.$$

We have now determined the oxidation numbers of potassium in all three compounds:

$$\mathrm{K_2O} : +1,\qquad \mathrm{K_2O_2} : +1,\qquad \mathrm{KO_2} : +1.$$

These values correspond exactly to the pattern given in Option B.

Hence, the correct answer is Option B.

We start by recalling that a substance can behave as a reducing agent only when the central atom present in it can be oxidised to a still higher oxidation state, and it can behave as an oxidising agent only when that atom can be reduced to a lower oxidation state. Therefore, to decide whether a compound can show both behaviours, we must inspect the oxidation state of the central element and see whether states both above and below that value are possible for that element.

Now we examine each option one by one, writing the oxidation state of the central atom in every compound.

For $$H_3PO_4$$ we let the oxidation state of phosphorus be $$x$$. Using the fact that the oxidation state of hydrogen is $$+1$$ and that of oxygen is $$-2$$, we write the charge-balance equation

$$3(+1) + x + 4(-2) = 0.$$

Simplifying term by term, we have

$$3 + x - 8 = 0,$$

so

$$x = +5.$$

Phosphorus belongs to group 15, and its highest attainable oxidation state is $$+5$$ (equal to its group number). Hence it cannot be oxidised any further. Because further oxidation is impossible, $$H_3PO_4$$ cannot act as a reducing agent. However, the oxidation state $$+5$$ can certainly decrease to $$+3, +1,$$ or $$-3,$$ so reduction is possible and the compound may show oxidising behaviour. Therefore $$H_3PO_4$$ cannot exhibit both actions simultaneously; it can at best act only as an oxidising agent.

For $$HNO_2$$ we assign the oxidation state of nitrogen as $$y$$:

$$1(+1) + y + 2(-2) = 0 \;\Rightarrow\; 1 + y - 4 = 0 \;\Rightarrow\; y = +3.$$

Nitrogen can be oxidised to $$+5$$ (in $$HNO_3$$) and reduced to $$-3$$ (in $$NH_3$$), so it can act both as reducing and as oxidising agent.

For $$H_2SO_3$$ we let the oxidation state of sulphur be $$z$$:

$$2(+1) + z + 3(-2) = 0 \;\Rightarrow\; 2 + z - 6 = 0 \;\Rightarrow\; z = +4.$$

Sulphur can move up to $$+6$$ (in $$H_2SO_4$$) or down to $$-2$$ (in $$H_2S$$); thus $$H_2SO_3$$ can show both behaviours too.

For $$H_2O_2$$ the oxidation state of oxygen is $$-1$$ (because $$2(+1) + 2x = 0 \Rightarrow x = -1$$). Oxygen can be oxidised to $$0$$ (in $$O_2$$) or reduced to $$-2$$ (in $$H_2O$$), so hydrogen peroxide also acts both as oxidising and reducing agent.

Combining all these observations, the only compound that lacks the ability to behave both as oxidising and reducing agent is $$H_3PO_4$$, since its central atom phosphorus is already at its maximum oxidation state of $$+5$$ and cannot be oxidised any further.

Hence, the correct answer is Option A.

First, recall the relation that connects the standard Gibbs free energy change $$\Delta G^\circ$$ of a reaction with its enthalpy change $$\Delta H^\circ$$ and entropy change $$\Delta S^\circ$$:

$$\Delta G^\circ = \Delta H^\circ - T\,\Delta S^\circ$$

Here $$T$$ is the absolute temperature. This equation clearly shows that $$\Delta G^\circ$$ varies linearly with temperature when $$\Delta H^\circ$$ and $$\Delta S^\circ$$ are taken as temperature-independent over a limited range. Plotting $$\Delta G^\circ$$ on the y-axis against $$T$$ on the x-axis therefore gives a straight line. The slope of this line is $$-\Delta S^\circ$$ and the y-intercept is $$\Delta H^\circ$$. Such a plot for a series of oxidation reactions of metals

$$\text{Metal} + \dfrac{1}{2}\,\text{O}_2 \;\longrightarrow\; \text{Metal oxide}$$

is called an Ellingham diagram. Because each line corresponds to the formation of a particular metal oxide, the diagram directly depicts how the standard Gibbs free energy of formation of that oxide changes with temperature.

We have, therefore, that an Ellingham diagram conveys the temperature dependence of $$\Delta G_f^\circ$$ for metal oxides. It tells us at which temperature a metal oxide becomes thermodynamically unstable (i.e., when its line rises above that of another reaction, indicating that reduction becomes spontaneous). It does not give information about pH, electrode potential at various pressures, or the rate (kinetics) of the reduction.

Looking at the given options:

A. speaks of pH and potential - this is not what an Ellingham diagram shows.
B. speaks of temperature dependence of the standard Gibbs energies of formation of some metal oxides - this matches exactly.
C. refers to pressure dependence of electrode potentials - not shown by Ellingham diagrams.
D. refers to kinetics - Ellingham diagrams are purely thermodynamic, not kinetic.

Hence, the correct answer is Option 2.

Ellingham Diagram:

1. Stability of Oxides:

   On an Ellingham diagram, a lower position (more negative $$\Delta G^\circ$$ value) indicates a more stable oxide. The metal corresponding to the lower curve has a higher affinity for oxygen.

2. Criteria for Reduction:

   For a metal to reduce another metal's oxide, the free energy of formation of its own oxide must be more negative than that of the oxide being reduced. In graphical terms, a metal can reduce the oxide of any metal whose line lies above its own line on the diagram.

Step-by-Step Analysis of the Graph:

• Below $$1400^\circ\text{C}$$:

  The curve for the formation of $$\text{BO}_2$$ ($$\text{B} + \text{O}_2 \rightarrow \text{BO}_2$$) lies below the curve for $$\text{AO}_2$$ ($$\text{A} + \text{O}_2 \rightarrow \text{AO}_2$$). This means that below $$1400^\circ\text{C}$$, $$\text{BO}_2$$ is more stable than $$\text{AO}_2$$, and B can reduce $$\text{AO}_2$$, but A cannot reduce $$\text{BO}_2$$.




• At the Intersection Point ($$T = 1400^\circ\text{C}$$):

  The two curves intersect. At this specific temperature, the standard Gibbs free energy change ($$\Delta G^\circ$$) for both oxidation reactions is equal, meaning the system is in a state of dynamic equilibrium.




• Above $$1400^\circ\text{C}$$:

  The curve for the formation of $$\text{AO}_2$$ cross over and now lies below the curve of $$\text{BO}_2$$. As a result, $$\text{AO}_2$$ becomes thermodynamically more stable than $$\text{BO}_2$$. Therefore, element A can spontaneously reduce $$\text{BO}_2$$ to element B according to the net reaction:

  $$\text{BO}_2 + \text{A} \rightarrow \text{AO}_2 + \text{B} \quad (\Delta G^\circ < 0)$$

Conclusion:

Element A reduces $$\text{BO}_2$$ only in the temperature region where the line for $$\text{AO}_2$$ is below the line for $$\text{BO}_2$$, which occurs when the temperature is greater than $$1400^\circ\text{C}$$.

Answer: Option B — $$> 1400^\circ\text{C}$$

We are told that a 20.0 mL sample of an impure $$H_2O_2$$ solution contains 0.2 g of the impure reagent and that this entire 0.2 g reacts completely with 0.316 g of $$KMnO_4$$ in acidic medium. Our goal is to find what fraction of that 0.2 g is actually pure $$H_2O_2$$, and then convert that fraction into a percentage.

First, let us calculate the amount (in moles) of $$KMnO_4$$ that took part in the reaction. The molar mass of $$KMnO_4$$ is given as 158.

We have: $$\text{moles of }KMnO_4=\frac{\text{mass}}{\text{molar mass}}=\frac{0.316\;\text{g}}{158\;\text{g mol}^{-1}}$$

Simplifying the fraction: $$\frac{0.316}{158}=0.00200$$

So, $$n(KMnO_4)=0.00200\;\text{mol}$$.

Next, we write the redox reaction between $$KMnO_4$$ and $$H_2O_2$$ in acidic solution:

$$2\,MnO_4^- + 5\,H_2O_2 + 6\,H^+ \;\longrightarrow\; 2\,Mn^{2+} + 5\,O_2 + 8\,H_2O$$

From this balanced equation, the stoichiometric ratio is:

$$2\;\text{mol }KMnO_4 \;\longrightarrow\; 5\;\text{mol }H_2O_2$$

Therefore, $$1\;\text{mol }KMnO_4 \;\longrightarrow\; \frac{5}{2}=2.5\;\text{mol }H_2O_2$$.

Using this ratio, the moles of $$H_2O_2$$ that actually reacted are:

$$n(H_2O_2)=2.5 \times n(KMnO_4)=2.5 \times 0.00200=0.00500\;\text{mol}$$

Now we find the mass corresponding to these moles of pure $$H_2O_2$$. The molar mass of $$H_2O_2$$ is given as 34.

$$\text{mass of pure }H_2O_2 = n \times \text{molar mass} = 0.00500\;\text{mol} \times 34\;\text{g mol}^{-1}=0.170\;\text{g}$$

This 0.170 g of pure $$H_2O_2$$ was present in the original impure sample that weighed 0.200 g. Hence the purity percentage is:

$$\text{Purity }(\%)=\frac{\text{mass of pure }H_2O_2}{\text{mass of impure sample}}\times100 =\frac{0.170}{0.200}\times100=85\%$$

So, the answer is $$85\%$$.

First we deal with the reaction that occurs in basic medium:

$$2Fe^{2+}+H_2O_2 \rightarrow xA+yB$$

In basic medium, the usual redox couples are

$$Fe^{3+}/Fe^{2+}\;,\qquad H_2O_2/\,OH^-$$

We write the two half-reactions and state the rule that “electrons lost in oxidation must equal electrons gained in reduction.”

Oxidation half-reaction (ferrous to ferric):

$$Fe^{2+}\;\longrightarrow\;Fe^{3+}+e^-$$

Reduction half-reaction (peroxide to hydroxide in basic medium):

$$H_2O_2+2e^-\;\longrightarrow\;2OH^-$$

To equalise electrons, we multiply the oxidation half by 2 so that both halves involve two electrons:

$$2Fe^{2+}\;\longrightarrow\;2Fe^{3+}+2e^-$$

Adding the two balanced halves gives

$$2Fe^{2+}+H_2O_2\;\longrightarrow\;2Fe^{3+}+2OH^-$$

Hence the products are

$$A=Fe^{3+},\qquad B=OH^-$$

with stoichiometric coefficients

$$x=2,\qquad y=2$$

Now we balance the second reaction that takes place in acidic medium:

$$2MnO_4^-+6H^++5H_2O_2 \rightarrow x'C+y'D+z'E$$

In acidic solution the relevant redox couples are

$$MnO_4^-/Mn^{2+}\;,\qquad H_2O_2/O_2$$

Reduction half-reaction (permanganate to manganous):

$$MnO_4^-+8H^++5e^-\;\longrightarrow\;Mn^{2+}+4H_2O$$

Oxidation half-reaction (hydrogen peroxide to oxygen):

$$H_2O_2\;\longrightarrow\;O_2+2H^++2e^-$$

To match the electron count we take the least common multiple of 5 and 2, which is 10. So,

multiply the permanganate half by 2:

$$2MnO_4^-+16H^++10e^-\;\longrightarrow\;2Mn^{2+}+8H_2O$$

multiply the peroxide half by 5:

$$5H_2O_2\;\longrightarrow\;5O_2+10H^++10e^-$$

Adding the two adjusted half-reactions and cancelling the 10 electrons yields

$$2MnO_4^-+16H^++5H_2O_2\;\longrightarrow\;2Mn^{2+}+8H_2O+5O_2+10H^+$$

Subtracting $$10H^+$$ from both sides gives the fully balanced overall equation in acidic medium:

$$2MnO_4^-+6H^++5H_2O_2\;\longrightarrow\;2Mn^{2+}+8H_2O+5O_2$$

Thus the products are

$$C=Mn^{2+},\qquad D=H_2O,\qquad E=O_2$$

with stoichiometric coefficients

$$x'=2,\qquad y'=8,\qquad z'=5$$

Finally we add all the required coefficients:

$$x+y+x'+y'+z'=2+2+2+8+5=19$$

So, the answer is $$19$$.

First, recall the definition of volume strength. It is the volume (expressed in millilitres, at 273 K and 1 atm) of $$O_2$$ gas that is liberated from 1 mL of a hydrogen-peroxide solution when the peroxide decomposes completely.

The balanced decomposition reaction is

$$2\,H_2O_2 \;\longrightarrow\; 2\,H_2O + O_2$$

From this equation, 2 moles of $$H_2O_2$$ give 1 mole of $$O_2$$.

We are given a solution that is $$8.9\;{\rm M}$$, meaning

$$8.9\;{\rm mol\;H_2O_2\;per\;L\;of\;solution}.$$

So, in exactly $$1\;{\rm L} = 1000\;{\rm mL}$$ of this solution, the moles of peroxide are

$$n_{\!H_2O_2}=8.9.$$

Using the stoichiometric ratio $$\dfrac{1\;{\rm mol}\;O_2}{2\;{\rm mol}\;H_2O_2},$$ the moles of oxygen that can be produced from these 8.9 moles of peroxide are

$$n_{O_2}= \dfrac{8.9}{2}=4.45\;{\rm mol}.$$

To find the volume occupied by these $$O_2$$ moles at 273 K and 1 atm, we invoke the ideal-gas equation

$$PV = nRT.$$

Substituting $$P = 1\;{\rm atm},\; n = 4.45\;{\rm mol},\; R = 0.0821\;{\rm L\,atm\,K^{-1}\,mol^{-1}},\; T = 273\;{\rm K},$$ we obtain

$$V = \dfrac{nRT}{P} = 4.45 \times 0.0821 \times 273.$$

Now multiply step by step:

$$0.0821 \times 273 = 22.4133,$$ $$4.45 \times 22.4133 \approx 99.7392\;{\rm L}.$$

Thus, $$1\;{\rm L}$$ of the solution liberates approximately $$99.74\;{\rm L}$$ of $$O_2$$ gas at the stated conditions.

Convert this gas volume to millilitres because volume strength is stated per millilitre of solution:

$$99.74\;{\rm L} = 99.74 \times 1000 = 9.974 \times 10^{4}\;{\rm mL}.$$

Since these $$9.974 \times 10^{4}\;{\rm mL}$$ of gas come from $$1000\;{\rm mL}$$ of solution, the volume of gas corresponding to 1 mL of solution is

$$\dfrac{9.974 \times 10^{4}\;{\rm mL}}{1000\;{\rm mL}} = 99.74\;{\rm mL}.$$

Rounded to the nearest whole number, this becomes $$100\;{\rm mL}.$$

Therefore, the solution is called a 100-volume hydrogen-peroxide solution.

Hence, the correct answer is Option A (100).

To determine the volume of $$0.02\text{ M }K_2Cr_2O_7$$ required for the titration, we use the concept of equivalent weights in redox reactions.

In an acidic medium, potassium dichromate acts as an oxidizing agent and undergoes the reduction

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O.$$

Since two chromium atoms are reduced from an oxidation state of $$+6$$ to $$+3$$, one mole of $$Cr_2O_7^{2-}$$ gains a total of 6 electrons. Therefore, the $$n$$-factor of $$K_2Cr_2O_7$$ is $$6$$.

Ferrous oxalate ($$FeC_2O_4$$) acts as the reducing agent. In this process,

$$Fe^{2+} \rightarrow Fe^{3+} + e^-,$$

and

$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-.$$

Hence, each mole of $$FeC_2O_4$$ loses a total of

$$1 + 2 = 3 \text{ electrons},$$

so its $$n$$-factor is $$3$$.

The molar mass of ferrous oxalate is

$$56 + 24 + 64 = 144\text{ g mol}^{-1}.$$

Therefore, the number of moles present in $$0.288\text{ g}$$ is

$$\frac{0.288}{144} = 0.002\text{ mol}.$$

The number of equivalents of ferrous oxalate is

$$0.002 \times 3 = 0.006\text{ equivalents}.$$

At the equivalence point, the equivalents of the oxidizing agent and reducing agent are equal. Thus,

$$0.02 \times V(\text{L}) \times 6 = 0.006.$$

Solving for the volume,

$$V(\text{L}) = \frac{0.006}{0.12} = 0.05\text{ L}.$$

Converting to millilitres,

$$V = 0.05 \times 1000 = 100\text{ mL}.$$

Hence, the required volume of $$0.02\text{ M }K_2Cr_2O_7$$ solution is

$$100\text{ mL}.$$

We have one mole each of $$\mathrm{FeC_2O_4},\; \mathrm{Fe_2(C_2O_4)_3},\; \mathrm{FeSO_4}$$ and $$\mathrm{Fe_2(SO_4)_3}$$. In acidic medium, $$\mathrm{KMnO_4}$$ will oxidise:

1. $$\mathrm{Fe^{2+}\; \to \; Fe^{3+}}$$

2. $$\mathrm{C_2O_4^{2-}\; \to \; 2\,CO_2}$$

The ferric compound $$\mathrm{Fe_2(SO_4)_3}$$ already contains $$\mathrm{Fe^{3+}}$$, so it does not undergo further oxidation.

Oxidation of iron(II)

Each $$\mathrm{Fe^{2+}}$$ loses one electron according to the half-reaction $$\mathrm{Fe^{2+} \to Fe^{3+} + e^{-}}$$.

Number of $$\mathrm{Fe^{2+}}$$ ions present:

$$\mathrm{FeC_2O_4}\; \text{contains one}\; Fe^{2+}$$

$$\mathrm{FeSO_4}\; \text{contains one}\; Fe^{2+}$$

So, total $$\mathrm{Fe^{2+}} = 1 + 1 = 2 \text{ mol}$$

Electrons released:

$$2 \text{ mol Fe}^{2+} \times 1 \dfrac{\text{mol e}^-}{\text{mol Fe}^{2+}} = 2 \text{ mol e}^-$$

Oxidation of oxalate

The half-reaction in acid is

$$\mathrm{C_2O_4^{2-} \to 2\,CO_2 + 2\,e^-}$$

Count of oxalate ions:

$$\mathrm{FeC_2O_4}\; \text{has } 1$$

$$\mathrm{Fe_2(C_2O_4)_3}\; \text{has } 3$$

Total oxalate ions $$= 1 + 3 = 4 \text{ mol}$$

Electrons released:

$$4 \text{ mol C}_2\mathrm{O}_4^{2-} \times 2 \dfrac{\text{mol e}^-}{\text{mol C}_2\mathrm{O}_4^{2-}} = 8 \text{ mol e}^-$$

Total electrons to be accepted

$$2 + 8 = 10 \text{ mol e}^-$$

Reduction of permanganate in acid

The standard half-reaction is

$$\mathrm{MnO_4^- + 8\,H^+ + 5\,e^- \to Mn^{2+} + 4\,H_2O}$$

So one mole of $$\mathrm{KMnO_4}$$ consumes $$5$$ moles of electrons.

Moles of $$\mathrm{KMnO_4}$$ required

Using the simple proportion

$$\dfrac{10 \text{ mol e}^-}{5 \dfrac{\text{mol e}^-}{\text{mol KMnO}_4}} = 2 \text{ mol KMnO}_4$$

Hence, the correct answer is Option B.

We begin by assigning oxidation numbers so that the change in oxidation state, and therefore the electron transfer, can be seen explicitly.

The oxalate ion is $$\mathrm{C_2O_4^{2-}}$$. A general rule states that the sum of oxidation numbers of all atoms in an ion equals the net charge on that ion. Oxygen almost always has oxidation number $$-2$$ in its covalent compounds and ions.

So, if the oxidation number of carbon in oxalate is $$x$$, we write

$$2\,x + 4(-2) = -2.$$

Simplifying, we have

$$2x - 8 = -2,$$

and hence

$$2x = 6.$$

Dividing both sides by $$2$$ gives

$$x = +3.$$

Therefore each carbon atom in oxalate has oxidation number $$+3$$.

Now the product formed on oxidation is carbon dioxide, $$\mathrm{CO_2}$$. In $$\mathrm{CO_2}$$ we again let the oxidation number of carbon be $$y$$. Using the same rule, we write

$$y + 2(-2) = 0,$$

because the molecule is neutral. Hence

$$y - 4 = 0,$$

which gives

$$y = +4.$$

So, in going from oxalate to carbon dioxide the oxidation number of carbon increases from $$+3$$ to $$+4$$. An increase in oxidation number by $$1$$ means that one electron is lost by that carbon atom.

Oxalate contains two carbon atoms. The half-reaction for its oxidation may therefore be written as

$$\mathrm{C_2O_4^{2-} \;\longrightarrow\; 2\,CO_2 \;+\; 2\,e^-}.$$

We see from this balanced oxidation half-reaction that the loss of $$2$$ electrons produces $$2$$ molecules of $$\mathrm{CO_2}$$.

Now we focus on the requirement of the question: the number of electrons corresponding to one molecule of $$\mathrm{CO_2}$$. We have

$$\frac{2\ \text{electrons}}{2\ \text{CO}_2} = 1\ \text{electron per CO}_2.$$

Hence exactly $$1$$ electron is involved in the formation of one molecule of carbon dioxide during the reaction of oxalate with permanganate in acidic medium.

Hence, the correct answer is Option C.

We begin by recalling that the oxidation number of oxygen in hydrogen peroxide $$\mathrm{H_2O_2}$$ is $$-1$$(because the total oxidation number of the neutral molecule must be zero and hydrogen is $$+1$$, so $$2(+1)+2(x)=0\Longrightarrow x=-1$$). The value $$-1$$ lies midway between the usual oxidation numbers of oxygen in $$\mathrm{O_2^{2-}}$$ (which is $$-2$$ as in $$\mathrm{H_2O}$$, $$\mathrm{O^{2-}}$$, etc.) and in molecular oxygen $$\mathrm{O_2}$$ (which is $$0$$), as well as between $$-1$$ and $$+2$$ found in $$\mathrm{OF_2}$$. Because of this intermediate oxidation state, $$\mathrm{H_2O_2}$$ can either:

• lose electrons (be oxidised) and act as a reducing agent, or
• gain electrons (be reduced) and act as an oxidising agent.

Now we verify its dual behaviour in both acidic and basic media by writing the standard redox half-reactions.

In acidic medium

Oxidising action (hydrogen peroxide gets reduced):

$$\mathrm{H_2O_2 + 2H^+ + 2e^- \;\longrightarrow\; 2H_2O} \quad -(1)$$

Reducing action (hydrogen peroxide gets oxidised):

$$\mathrm{H_2O_2 \;\longrightarrow\; O_2 + 2H^+ + 2e^-} \quad -(2)$$

Because both half-reactions are feasible, $$\mathrm{H_2O_2}$$ acts as both an oxidising and a reducing agent in acidic solution.

In basic medium

Oxidising action (hydrogen peroxide gets reduced):

$$\mathrm{H_2O_2 + 2e^- \;\longrightarrow\; 2OH^-} \quad -(3)$$

Reducing action (hydrogen peroxide gets oxidised):

$$\mathrm{H_2O_2 + 2OH^- \;\longrightarrow\; O_2 + 2H_2O + 2e^-} \quad -(4)$$

Again, both processes are allowed, so in alkaline solution also it behaves as both oxidising and reducing agent.

Because hydrogen peroxide exhibits both oxidising and reducing properties in both acidic and basic media, we match this description with the options provided.

Option A states: “Oxidizing and reducing agent in both acidic and basic medium.” This is precisely what we have demonstrated.

Hence, the correct answer is Option A.

First, we recall the definition of a disproportionation reaction. A single species undergoes simultaneous oxidation as well as reduction, producing two different products that contain the same element in two different oxidation states. In symbolic terms:

$$\text{A}^{n+} \;\longrightarrow\; \text{A}^{(n+m)+} \;+\; \text{A}^{(n-k)+}$$

Here one portion of A loses electrons (oxidation, oxidation state increases by $$m$$) while another portion gains electrons (reduction, oxidation state decreases by $$k$$).

We now analyse each option by writing the oxidation state of the key element in reactants and products.

Option A: $$2KMnO_4 \;\longrightarrow\; K_2MnO_4 + MnO_2 + O_2$$

In $$KMnO_4$$, manganese has oxidation state $$+7$$ (because $$K$$ is $$+1$$ and each $$O$$ is $$-2$$: $$+1 + x + 4(-2)=0 \implies x=+7$$). In $$K_2MnO_4$$, manganese is $$+6$$ ($$2(+1)+x+4(-2)=0 \implies x=+6$$). In $$MnO_2$$, manganese is $$+4$$ ($$x+2(-2)=0 \implies x=+4$$).

Both changes $$+7 \to +6$$ and $$+7 \to +4$$ are reductions; there is no simultaneous oxidation. Hence this is not disproportionation.

Option B: $$2MnO_4^- + 10I^- + 16H^+ \;\longrightarrow\; 2Mn^{2+} + 5I_2 + 8H_2O$$

$$Mn$$ goes $$+7 \to +2$$ (reduction) while $$I$$ goes $$-1 \to 0$$ (oxidation). Two different elements change oxidation state, so again it is not disproportionation.

Option C: $$2\,CuBr \;\longrightarrow\; CuBr_2 + Cu$$

Let us calculate the oxidation states step by step.

• In $$CuBr$$ the bromide ion is $$-1$$. Setting the sum to zero: $$x+(-1)=0 \implies x=+1$$. So copper is $$+1$$.

• In $$CuBr_2$$ we have $$x+2(-1)=0 \implies x=+2$$. Thus copper becomes $$+2$$ — this is oxidation because the oxidation state increases from $$+1$$ to $$+2$$ (loss of one electron).

• In metallic $$Cu$$ the oxidation state is $$0$$. The change $$+1 \to 0$$ is reduction because the oxidation state decreases (gain of one electron).

We see the same species $$Cu^{+}$$ has simultaneously undergone oxidation ($$+1 \to +2$$) and reduction ($$+1 \to 0$$). This exactly fits the definition of disproportionation.

Option D: $$2 NaBr + Cl_2 \;\longrightarrow\; 2NaCl + Br_2$$

Here $$Cl$$ goes $$0 \to -1$$ (reduction) and $$Br$$ goes $$-1 \to 0$$ (oxidation). Two different elements again; so it is not disproportionation.

Only Option C satisfies the criterion that one element in a single oxidation state splits into two different oxidation states, evidencing both oxidation and reduction of that element.

Hence, the correct answer is Option C.

The phrase “volume strength” of a hydrogen-peroxide solution tells us how many litres of dioxygen at STP are released by one litre of the solution when $$H_2O_2$$ decomposes. The decomposition reaction is first written:

$$2\,H_2O_2 \;\longrightarrow\; 2\,H_2O \;+\; O_2$$

From the balanced equation we see that $$2$$ moles of $$H_2O_2$$ give $$1$$ mole of $$O_2$$. So, for every mole of hydrogen peroxide the moles of oxygen obtained are

$$\text{Moles of }O_2 \;=\;\dfrac{1}{2}\times(\text{moles of }H_2O_2).$$

We are given a $$1\text{ M}$$ solution, which by definition contains $$1$$ mole of solute in $$1$$ litre of solution. Therefore, in $$1$$ litre of this solution the moles of $$H_2O_2$$ present are

$$n_{H_2O_2}=1\;\text{mol}.$$

Using the mole ratio derived above, the moles of dioxygen liberated from this one litre are

$$n_{O_2}= \dfrac{1}{2}\times 1 = 0.5\;\text{mol}.$$

At STP, one mole of any ideal gas occupies $$22.4\;\text{L}$$. Thus the volume of $$O_2$$ produced is calculated with

$$V_{O_2}=n_{O_2}\times 22.4\;\text{L mol}^{-1} = 0.5 \times 22.4 = 11.2\;\text{L}.$$

Because this volume corresponds to one litre of the hydrogen-peroxide solution, the “volume strength” is simply the number of litres of oxygen per litre of solution, namely $$11.2$$. Hence, the solution is termed an “$$11.2$$-volume” hydrogen peroxide.

So, the answer is $$11.2$$.

First, we consider the statement of the assertion. In the blast-furnace extraction of iron, the industry generally employs the ore haematite. Chemically, haematite is represented as $$\text{Fe}_2\text{O}_3$$ and is abundant, cheap, and easily reducible. Because of these practical advantages, haematite really is the ore chosen for large-scale iron production. Therefore, the assertion “For the extraction of iron, haematite ore is used” is true.

Now we examine the reason. The reason claims “Haematite is a carbonate ore of iron.” Let us recall the basic classification of iron ores. An oxide ore has iron combined with oxygen, while a carbonate ore has iron combined with the $$\text{CO}_3^{2-}$$ group.

Specifically,

$$\text{Haematite} = \text{Fe}_2\text{O}_3 \quad(\text{oxide ore})$$

$$\text{Siderite} = \text{FeCO}_3 \quad(\text{carbonate ore})$$

Clearly, haematite contains only iron and oxygen; there is no carbonate group present. Hence haematite is an oxide ore, not a carbonate ore. Consequently, the reason given in the question is false.

We have found that the assertion is correct while the reason is incorrect. The option that matches this situation is Option A: “Only the assertion is correct.”

Hence, the correct answer is Option A.

We begin by recalling that concentrated $$\mathrm{HNO_3}$$ is a very strong oxidising agent. When elemental iodine $$\mathrm{I_2}$$ is treated with hot, concentrated $$\mathrm{HNO_3}$$, the nitric acid oxidises iodine and is itself reduced. The principal oxidation product of iodine in this medium is iodic acid, whose formula is $$\mathrm{HIO_3}$$. We symbolise this reaction as

$$\mathrm{I_2 + 10\,HNO_3 \longrightarrow 2\,HIO_3 + 10\,NO_2 + 4\,H_2O}$$

In this balanced redox equation, the species labelled $$\mathrm{HIO_3}$$ is our substance $$Y$$. The problem now reduces to finding the oxidation state of iodine in $$\mathrm{HIO_3}$$.

To do this, we apply the fundamental rule for oxidation-state calculations: the algebraic sum of oxidation numbers of all atoms in a neutral molecule must equal zero. Let us denote the unknown oxidation state of iodine by $$x$$. Then, recognising that hydrogen carries an oxidation state of $$+1$$ and oxygen carries $$-2$$, we write

$$ \underbrace{(+1)}_{\text{H}} \;+\; \underbrace{(x)}_{\text{I}} \;+\; 3\!\times\!\underbrace{(-2)}_{\text{O}} \;=\; 0 . $$

Now we perform the step-by-step algebra:

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

$$ +1 + x - 6 = 0 $$

$$ x - 5 = 0 $$

$$ x = +5 $$

Thus, the oxidation state of iodine in $$\mathrm{HIO_3}$$ (species $$Y$$) is $$+5$$.

Hence, the correct answer is Option B.

We first recall what an Ellingham diagram is. It is a plot of the standard Gibbs free-energy change $$\Delta G^\circ$$ for the formation of oxides (or sulphides, halides, etc.) versus temperature $$T$$ for a variety of metals. The basic thermodynamic relation used to draw every straight line on the diagram is the fundamental Gibbs-Helmholtz equation

$$ \Delta G^\circ \;=\; \Delta H^\circ \;-\; T\,\Delta S^\circ , $$

where $$\Delta H^\circ$$ is the standard enthalpy change and $$\Delta S^\circ$$ is the standard entropy change of the oxide-formation reaction

$$ \text{Metal} \;+\; \dfrac{1}{2}\,O_2 \;\longrightarrow\; \text{Metal oxide}. $$

The slope of each line equals $$-\Delta S^\circ$$ and the intercept equals $$\Delta H^\circ$$. Because most oxide-formation reactions involve a decrease in entropy (gas is consumed), $$\Delta S^\circ$$ is negative, so the slope $$-\,\Delta S^\circ$$ is positive. Thus, the lines usually rise with increasing temperature.

Now, the sign of $$\Delta G^\circ$$ decides spontaneity:

$$ \Delta G^\circ < 0 \quad\Longrightarrow\quad \text{reaction is spontaneous}; $$

$$ \Delta G^\circ > 0 \quad\Longrightarrow\quad \text{reaction is non-spontaneous}. $$

In metallurgical practice we use the diagram exactly the other way around. Consider the reduction of a metal oxide by carbon (or by another metal):

$$ \text{Metal oxide} \;+\; C \;\longrightarrow\; \text{Metal} \;+\; CO \quad\text{(or } CO_2\text{)}. $$

We can obtain the overall $$\Delta G^\circ$$ for this reduction by subtracting the $$\Delta G^\circ$$ line of the metal from that of the reducer (carbon forming CO/CO2). If the resulting $$\Delta G^\circ$$ is negative at a certain temperature, the reduction is feasible at that temperature.

Thus, the Ellingham diagram is primarily employed to judge whether an oxide can be thermally reduced by some reducing agent such as carbon, carbon monoxide, hydrogen, or another metal. This process—smelting in a blast furnace, electrolytic cell feed preparation, etc.—is broadly classified as thermal reduction or pyrometallurgical reduction.

Let us now examine the given options with this understanding:

• Electrolysis: The feasibility of electrolytic reduction depends on decomposition potential and over-potentials, not on Ellingham lines.
• Zone refining: This relies on the difference in solubility of impurities in solid and liquid phases; Ellingham diagram is not used here.
• Vapour phase refining: This depends on the volatility of metal halides (e.g., Mond process), again unrelated to Ellingham diagram.
• Thermal reduction: Exactly what the Ellingham diagram was devised for—to see if heat plus a reducing agent will convert the oxide to metal.

Therefore, only the fourth option matches.

Hence, the correct answer is Option D.

We recall the general rule that in almost every compound the oxidation state of oxygen is $$-2$$.

For nitric oxide, $$\mathrm{NO}$$: Let the oxidation state of the single nitrogen atom be $$x$$. We have

$$x + (-2) = 0$$

because the molecule is electrically neutral. Solving gives

$$x = +2$$.

For nitrous oxide, $$\mathrm{N_2O}$$: Let each nitrogen atom have the same oxidation state $$x$$. Now

$$2x + (-2) = 0 \quad\Longrightarrow\quad 2x = +2 \quad\Longrightarrow\quad x = +1.$$

For nitrogen dioxide, $$\mathrm{NO_2}$$: The single nitrogen atom has oxidation state $$x$$, so

$$x + 2(-2) = 0 \quad\Longrightarrow\quad x - 4 = 0 \quad\Longrightarrow\quad x = +4.$$

For dinitrogen trioxide, $$\mathrm{N_2O_3}$$: With each nitrogen at oxidation state $$x$$, we write

$$2x + 3(-2) = 0 \quad\Longrightarrow\quad 2x - 6 = 0 \quad\Longrightarrow\quad 2x = +6 \quad\Longrightarrow\quad x = +3.$$

Now we list the values obtained:

$$\mathrm{N_2O}: +1,\qquad \mathrm{NO}: +2,\qquad \mathrm{N_2O_3}: +3,\qquad \mathrm{NO_2}: +4.$$

Arranging them in increasing order we get

$$+1 < +2 < +3 < +4 \;\;\Longrightarrow\;\; \mathrm{N_2O} < \mathrm{NO} < \mathrm{N_2O_3} < \mathrm{NO_2}.$$

Hence, the correct answer is Option A.

We start with the compound $$\mathrm{KO_2}$$. Potassium belongs to the alkali-metal group and in all its stable compounds it carries an oxidation number of $$+1$$. We write this fact explicitly:

$$\text{Oxidation number of K}=+1$$

Let the oxidation number of each oxygen atom in $$\mathrm{KO_2}$$ be $$x$$. There are two oxygen atoms, so the total contribution of oxygen to the overall charge is $$2x$$.

The compound $$\mathrm{KO_2}$$ is electrically neutral, therefore the algebraic sum of the oxidation numbers of all the atoms present must be equal to the net charge, which is zero. Using the general rule

$$\sum (\text{oxidation numbers})=\text{overall charge},$$

we can write

$$+1 + 2x = 0.$$

Now we solve this simple linear equation step by step.

Subtract $$1$$ from both sides:

$$2x = -1.$$

Divide both sides by $$2$$ to isolate $$x$$:

$$x = \frac{-1}{2}.$$

So, each oxygen atom in $$\mathrm{KO_2}$$ has an oxidation state of $$- \dfrac{1}{2}.$$

Next, we classify the type of oxygen species. The three common dioxygen anions are:

1. $$\mathrm{O^{2-}}$$  — oxide ion (oxidation state of each O = $$-2$$)
2. $$\mathrm{O_2^{2-}}$$  — peroxide ion (oxidation state of each O = $$-1$$)
3. $$\mathrm{O_2^{-}}$$  — superoxide ion (oxidation state of each O = $$-\dfrac12$$)

Because we have just calculated an oxidation state of $$-\dfrac12$$ for each oxygen atom, the dioxygen fragment must be the superoxide ion $$\mathrm{O_2^{-}}$$.

Therefore, in $$\mathrm{KO_2}$$ the oxygen species is a superoxide, and the oxidation state of each oxygen atom is $$-\dfrac12$$.

Hence, the correct answer is Option B.

We are asked to find the species that accompany the iron-cyanide complex when hydrogen peroxide reacts in two different media.

First, recall the dual nature of hydrogen peroxide:

• In acidic medium it behaves as an oxidising agent and is reduced to water. • In alkaline medium it behaves as a reducing agent and is oxidised to dioxygen.

Let us tackle the acidic case step by step.

In acid, hydrogen peroxide accepts electrons. We write the two half-reactions.

Oxidation half (for the complex): $$[Fe(CN)6]^{4- \;\longrightarrow\; [Fe(CN)6]^{3-} + e^-}$$

Reduction half (for H$$_2$$O$$_2$$). The standard form is stated first: $$H_{2}O_{2} + 2H^+ + 2e^- \;\longrightarrow\; 2H_{2}O$$

To equate electrons we multiply the iron half-reaction by 2:

$$2[Fe(CN)_6]^{4-} \longrightarrow 2[Fe(CN)_6]^{3-} + 2e^-$$

Adding the two halves gives

$$2[Fe(CN)_6]^{4-} + H_2O_2 + 2H^+ \longrightarrow 2[Fe(CN)_6]^{3-} + 2H_2O$$

We see clearly that the only additional product, apart from the ferricyanide ion, is water. Hence in acidic medium H$$_2$$O is formed.

Now we move to the alkaline case where hydrogen peroxide donates electrons. The standard oxidation half-reaction in base is

$$H_{2}O_{2} + 2OH^- \longrightarrow O_{2} + 2H_{2}O + 2e^-$$

The reduction half for the iron complex is simply the reverse of the earlier oxidation step:

$$[Fe(CN)6]^{3- + e^- \;\longrightarrow\; [Fe(CN)6]^{4-}}$$

Again we multiply this half-reaction by 2 to match electrons:

$$2[Fe(CN)_6]^{3-} + 2e^- \longrightarrow 2[Fe(CN)_6]^{4-}$$

Adding the two halves gives

$$2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^- \longrightarrow 2[Fe(CN)_6]^{4-} + O_2 + 2H_2O$$

Here the extra species produced, besides the ferrocyanide ion, are O$$_2$$ and H$$_2$$O. They appear on the product side together.

Collecting our results:

• Acidic medium → additional product: $$H_{2}O$$ • Alkaline medium → additional products: $$O_{2}$$ and $$H_{2}O$$

The option that lists “H$$_2$$O” for the first reaction and “(H$$_2$$O + O$$_2$$)” for the second is Option D.

Hence, the correct answer is Option D.

To decide which pair shows the metals in their highest possible oxidation state, we must first calculate the oxidation number of the metal in every compound listed.

Recall the rule: The algebraic sum of the oxidation numbers of all atoms in a neutral compound is zero, while in an ion it equals the ionic charge. Oxygen almost always has $$-2$$, chlorine almost always has $$-1$$, and the cyanide ion $$CN^-$$ carries a charge of $$-1$$. We now apply this rule to every species.

Option A  $$Mn_2O_7$$ and $$CrO_2Cl_2$$

For $$Mn_2O_7$$ (a neutral molecule):

Let the oxidation number of each manganese atom be $$x$$.

We have two manganese atoms, so their total contribution is $$2x$$.

There are seven oxygens, each at $$-2$$, giving $$7(-2)=-14$$.

The compound is neutral, so the total is zero:

$$2x + (-14) = 0 \;\Longrightarrow\; 2x = +14 \;\Longrightarrow\; x = +7.$$

Thus manganese is in the $$+7$$ state, the highest oxidation state known for Mn.

For $$CrO_2Cl_2$$ (chromyl chloride, also neutral):

Let chromium have oxidation number $$y$$.

Two oxygens contribute $$2(-2)=-4$$, and two chlorines contribute $$2(-1)=-2$$.

Total charge being zero, we write

$$y + (-4) + (-2) = 0 \;\Longrightarrow\; y = +6.$$

Chromium is therefore $$+6$$. Chromium cannot go beyond $$+6$$ in stable compounds, so this is also its highest oxidation state.

Option B  $$[Fe(CN)_6]^{3-}$$ and $$[Cu(CN)_4]^{2-}$$

For $$[Fe(CN)_6]^{3-}$$:

Let Fe be $$z$$. Six $$CN^-$$ ligands give $$6(-1)=-6$$. The complex ion has an overall charge of $$-3$$, so:

$$z + (-6) = -3 \;\Longrightarrow\; z = +3.$$

Iron can exhibit up to $$+6$$ in ferrate $$FeO_4^{2-}$$, so $$+3$$ is not its highest.

For $$[Cu(CN)_4]^{2-}$$:

Let Cu be $$w$$. Four $$CN^-$$ ligands give $$4(-1)=-4$$. The ion’s charge is $$-2$$:

$$w + (-4) = -2 \;\Longrightarrow\; w = +2.$$

Copper can reach $$+3$$ in compounds like $$KCuO_2$$, so $$+2$$ is not its highest.

Option C  $$[NiCl_4]^{2-}$$ and $$[CoCl_4]^{2-}$$

For $$[NiCl_4]^{2-}$$:

Let Ni be $$a$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-2$$:

$$a + (-4) = -2 \;\Longrightarrow\; a = +2.$$

Nickel can exhibit $$+3$$ and $$+4$$ in other species, so $$+2$$ is not the highest.

For $$[CoCl_4]^{2-}$$:

Let Co be $$b$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-2$$:

$$b + (-4) = -2 \;\Longrightarrow\; b = +2.$$

Cobalt easily attains $$+3$$ in many compounds; $$+2$$ is therefore not its highest.

Option D  $$[FeCl_4]^{-}$$ and $$Co_2O_3$$

For $$[FeCl_4]^{-}$$:

Let Fe be $$c$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-1$$:

$$c + (-4) = -1 \;\Longrightarrow\; c = +3.$$

Again, iron can reach $$+6$$, so $$+3$$ is not the highest.

For $$Co_2O_3$$ (neutral):

Let Co be $$d$$. Three oxygens: $$3(-2)=-6$$. The molecule is neutral:

$$2d + (-6) = 0 \;\Longrightarrow\; 2d = +6 \;\Longrightarrow\; d = +3.$$

But cobalt is known up to $$+3$$ only in its stable compounds, so while $$+3$$ is Co's usual maximum, we have already rejected the pair because iron is not at its maximum.

Summarising, only Option A provides both metals in their respective highest oxidation states: manganese at $$+7$$ and chromium at $$+6$$.

Hence, the correct answer is Option A.

We start by recalling the basic idea of redox behaviour for hydrogen peroxide. In $$H_2O_2$$ the oxidation number of each oxygen atom is $$-1$$. Whenever $$H_2O_2$$ itself is converted to $$O_2$$, the oxidation number of its oxygen rises from $$-1$$ to $$0$$, so $$H_2O_2$$ is getting oxidised; therefore it must be acting as a reducing agent in that case. Conversely, whenever $$H_2O_2$$ changes to $$H_2O$$ or $$OH^-$$, the oxidation number of its oxygen falls from $$-1$$ to $$-2$$; hence $$H_2O_2$$ is being reduced and therefore behaves as an oxidising agent.

Now we examine each option one by one, writing the oxidation numbers step by step.

Option A is

$$HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2$$

• In $$HOCl$$ the chlorine is in the $$+1$$ state.
• It becomes $$Cl^-$$ where chlorine is $$-1$$, so chlorine is reduced.
• Simultaneously $$H_2O_2$$ turns into $$O_2$$, changing oxygen from $$-1$$ to $$0$$, i.e. oxygen is oxidised.
Because $$H_2O_2$$ is oxidised, it must be the reducing agent here. So $$H_2O_2$$ is not an oxidising agent in reaction A.

Option B is

$$I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2$$

• Iodine goes from $$0$$ in $$I_2$$ to $$-1$$ in $$I^-$$, so iodine is reduced.
• Again $$H_2O_2$$ becomes $$O_2$$ (oxygen $$-1 \rightarrow 0$$), meaning $$H_2O_2$$ is oxidised and acts as the reducing agent.
Therefore in reaction B, $$H_2O_2$$ is not an oxidising agent.

Option C is

$$PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$$

Let us assign oxidation numbers carefully:

• In $$PbS$$, sulphur is $$-2$$.
• In $$PbSO_4$$, sulphur is $$+6$$ (because the overall charge is zero and $$Pb$$ is $$+2$$).

So sulphur changes from $$-2$$ to $$+6$$, a rise of $$8$$ units; sulphur is oxidised.

Next we track $$H_2O_2$$:

• Oxygen in $$H_2O_2$$ is $$-1$$.
• Oxygen in the product $$H_2O$$ is $$-2$$.

Thus the oxygen of peroxide is reduced. Since $$H_2O_2$$ itself is reduced while the other species (sulphur in $$PbS$$) is oxidised, $$H_2O_2$$ must be supplying the oxygen and causing oxidation of $$PbS$$. That means $$H_2O_2$$ acts as the oxidising agent in reaction C.

Option D is

$$2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-$$

• Manganese goes from $$+7$$ in $$MnO_4^-$$ to $$+4$$ in $$MnO_2$$, so manganese is reduced.
• Yet again $$H_2O_2$$ produces $$O_2$$, oxygen moving from $$-1$$ to $$0$$, so $$H_2O_2$$ is oxidised and therefore a reducing agent here.

Summarising all four cases:

• A - $$H_2O_2$$ is reducing agent
• B - $$H_2O_2$$ is reducing agent
• C - $$H_2O_2$$ is oxidising agent
• D - $$H_2O_2$$ is reducing agent

Hence only reaction C (which corresponds to Option 3) features hydrogen peroxide acting as an oxidising agent.

Hence, the correct answer is Option C.

We begin with the definition of standard reduction potential. For any redox couple written in the reduction direction, for example

$$X^{n+}+ne^- \rightarrow X$$

the listed value $$E^\circ_{X^{n+}/X}$$ is the reduction potential. If we wish to look at the reverse (oxidation) direction, we use the relation

$$E^\circ_{\text{oxidation}}=-\,E^\circ_{\text{reduction}}.$$

A species behaves as a reducing agent when it itself is oxidised, i.e. when the reverse reaction is favoured. The larger (more positive) the oxidation potential, the greater the reducing power. Equivalently, the smaller (more negative) the reduction potential, the greater the reducing power. So, in practice, we simply compare the given reduction potentials and pick the most negative one.

The data supplied are

$$E^\circ_{Cl_{2}/Cl^-}=+1.36\ \text{V},$$

$$E^\circ_{Cr^{3+}/Cr}=-0.74\ \text{V},$$

$$E^\circ_{Cr_{2}O_{7}^{2-}/Cr^{3+}}=+1.33\ \text{V},$$

$$E^\circ_{MnO_{4}^- /Mn^{2+}}=+1.51\ \text{V}.$$

Now we match every option with its relevant couple and note the corresponding reduction potential:

(A) $$Mn^{2+}$$ comes from the couple $$MnO_{4}^- /Mn^{2+}$$ with $$E^\circ=+1.51\ \text{V}.$$

(B) $$Cr^{3+}$$ is part of the couple $$Cr_{2}O_{7}^{2-}/Cr^{3+}$$ with $$E^\circ=+1.33\ \text{V}.$$

(C) $$Cl^-$$ belongs to the couple $$Cl_{2}/Cl^-$$ with $$E^\circ=+1.36\ \text{V}.$$

(D) $$Cr$$ is linked with the couple $$Cr^{3+}/Cr$$ whose reduction potential is $$E^\circ=-0.74\ \text{V}.$$

Among the four numbers  +1.51 V, +1.36 V, +1.33 V and -0.74 V, the smallest (most negative) value is $$-0.74\ \text{V}.$$

Because a more negative reduction potential means a more positive oxidation potential (by $$E^\circ_{\text{oxidation}}=-E^\circ_{\text{reduction}}$$), the couple with $$E^\circ=-0.74\ \text{V}$$ provides the strongest tendency for the reverse (oxidation) reaction. That reverse reaction is

$$Cr \rightarrow Cr^{3+}+3e^-,$$

showing that metallic chromium, $$Cr$$, donates electrons most readily. Therefore $$Cr$$ is the strongest reducing agent among the given choices.

Hence, the correct answer is Option 4.

First, we recall the basic test for a redox (reduction-oxidation) reaction. A reaction is redox if at least one element shows an increase in its oxidation number (oxidation) while another shows a decrease (reduction). We shall therefore calculate the oxidation numbers of the key atoms on both sides of every given equation.

Option A

$$\text{XeF}_2 + \text{PF}_5 \; \rightarrow \; \text{XeF}^{+}\text{PF}_6^{-}$$

For $$\text{XeF}_2$$ we put $$\text{OS}_\text{F}=-1$$. Writing $$x$$ for the oxidation state of xenon, we have

$$x + 2(-1)=0 \;\;\Longrightarrow\;\; x=+2.$$

For $$\text{PF}_5$$, with $$\text{OS}_\text{F}=-1$$ and $$y$$ for phosphorus,

$$y + 5(-1)=0 \;\;\Longrightarrow\;\; y=+5.$$

Now consider $$\text{XeF}^{+}$$. If the oxidation state of Xe is $$x'$$, then

$$x' + (-1)=+1 \;\;\Longrightarrow\;\; x'=+2.$$

For $$\text{PF}_6^{-}$$ we set $$y'$$ for phosphorus, so

$$y' + 6(-1)=-1 \;\;\Longrightarrow\;\; y'=+5.$$

Thus Xe remains $$+2$$ and P remains $$+5$$. No oxidation state changes occur, so Option A is not redox.

Option B

$$\text{XeF}_6 + \text{H}_2\text{O} \; \rightarrow \; \text{XeOF}_4 + 2\text{HF}$$

In $$\text{XeF}_6$$, with $$x$$ for Xe,

$$x + 6(-1)=0 \;\;\Longrightarrow\;\; x=+6.$$

For $$\text{XeOF}_4$$, again letting $$x'$$ be the oxidation state of Xe, we use $$\text{OS}_\text{O}=-2$$ and obtain

$$x' + (-2) + 4(-1)=0 \;\;\Longrightarrow\;\; x'=+6.$$

The oxidation numbers of H and O in $$\text{H}_2\text{O}$$ and $$\text{HF}$$ remain $$+1$$ (for H) and $$-2$$ (for O) or $$-1$$ (for F) throughout. No element changes its oxidation state, hence Option B is not redox.

Option C

$$\text{XeF}_6 + 2\text{H}_2\text{O} \; \rightarrow \; \text{XeO}_2\text{F}_2 + 4\text{HF}$$

In $$\text{XeF}_6$$ we already found $$\text{OS}_\text{Xe}=+6$$. For $$\text{XeO}_2\text{F}_2$$ (put $$x$$ for Xe),

$$x + 2(-2) + 2(-1)=0 \;\;\Longrightarrow\;\; x=+6.$$

Again, the oxidation states of H, O and F do not change. Therefore Option C is not a redox reaction.

Option D

$$\text{XeF}_4 + \text{O}_2\text{F}_2 \; \rightarrow \; \text{XeF}_6 + \text{O}_2$$

For $$\text{XeF}_4$$ (let $$x$$ be Xe):

$$x + 4(-1)=0 \;\;\Longrightarrow\;\; x=+4.$$

For $$\text{XeF}_6$$ (let $$x'$$ be Xe):

$$x' + 6(-1)=0 \;\;\Longrightarrow\;\; x'=+6.$$

Thus Xe rises from $$+4$$ to $$+6$$, showing oxidation.

Now examine $$\text{O}_2\text{F}_2$$, which contains two oxygens and two fluorines. Putting $$y$$ for the average oxidation state of each O atom and $$\text{OS}_\text{F}=-1$$,

$$2y + 2(-1)=0 \;\;\Longrightarrow\;\; y=+1.$$

In $$\text{O}_2$$ the oxidation state of O is its elemental value $$0$$. Hence oxygen is reduced from $$+1$$ to $$0$$.

Because xenon is oxidised and oxygen is reduced, simultaneous oxidation and reduction occur; therefore Option D is a redox reaction.

Since only Option D satisfies the criterion for a redox change,

Hence, the correct answer is Option D.

We start by recalling what the process called “galvanization” actually means in metallurgy and corrosion science. Galvanization is a protective technique in which a base metal, most commonly iron or steel, is covered with a thin, adherent layer of a second metal that is more reactive toward oxidation, so that it preferentially corrodes and thereby shields the underlying iron or steel from rusting.

To understand why a particular metal is chosen, we remember the basic principle of sacrificial protection. A metal that is more easily oxidized, i.e., that lies above iron in the electrochemical series, will oxidize first. Its oxide layer or the metal itself forms a barrier, and any corrosive environment attacks this sacrificial layer rather than the base metal.

Among the given options, let us examine each metal in turn:

Option A gives copper, written chemically as $$Cu$$. Copper stands below iron in the electrochemical (activity) series, meaning it is less reactive than iron. If copper were coated on iron, the iron would actually corrode faster at any pinholes or scratches, because copper would act as a cathodic region. So copper is not used for galvanization.

Option C supplies lead, $$Pb$$. Lead is again less reactive than iron and is soft and toxic; though lead coatings are sometimes used for acid-resistant purposes, the word “galvanization” does not refer to applying lead.

Option D lists chromium, $$Cr$$. Chromium is indeed used for “chromium plating,” also called “chrome plating,” which gives shiny, corrosion-resistant finishes. However that is a different process, commonly referred to as electroplating or hard chrome, not galvanization in the traditional sense.

Option B presents zinc, $$Zn$$. Zinc lies above iron in the electrochemical series, meaning it is more easily oxidized. When a coating of zinc covers iron or steel, the following advantages arise:

1. The zinc layer forms a physical barrier that keeps oxygen and moisture away from the iron surface.
2. Even if the coating is scratched, zinc continues to corrode preferentially, offering sacrificial, cathodic protection to the exposed iron.

Because of these dual benefits, the industrial term “galvanization” specifically signifies applying a coating of zinc to iron or steel articles, often by hot-dip methods or electro-galvanizing.

After analyzing all the choices, only zinc satisfies the exact definition and practical usage of galvanization.

Hence, the correct answer is Option B.

Because copper is less reactive than zinc, it lacks the chemical strength to displace zinc ions from the $$ZnSO_4$$ solution. Therefore, when a block of copper is placed in $$ZnSO_4$$: 

$$Cu(s) + ZnSO_4(aq) \rightarrow \text{No Reaction}$$

During the extraction of copper from its sulphide ores, the concentrated ore is first roasted to convert most of the $$FeS$$ and a part of $$Cu_2S$$ into their oxides. This produces a molten mass containing mainly $$Cu_2S$$, $$Cu_2O$$ and $$FeO$$, along with some unchanged $$FeS$$.

At this stage the mass is transferred to a smelting furnace. Here we deliberately add powdered silica, which is chemically represented as $$SiO_2$$. First, let us recall a fundamental fact from metallurgy: an acidic oxide such as $$SiO_2$$ will combine readily with a basic oxide to form a fusible salt called slag. The overall goal is to remove unwanted basic oxides in the form of this slag, thereby preventing them from contaminating the impure copper (called matte).

Among the oxides present, $$FeO$$ is strongly basic, whereas $$Cu_2O$$ is only weakly basic and, in addition, is actually useful later because it reacts with $$Cu_2S$$ to yield metallic copper. Therefore our primary target for removal is $$FeO$$.

We thus rely on the following chemical reaction, which expresses exactly how silica removes iron(II) oxide:

$$FeO + SiO_2 \;\rightarrow\; FeSiO_3$$

Here $$FeSiO_3$$ is iron(II) silicate, a fusible, lighter slag that floats over the molten mixture and can be separated off easily. Notice that neither $$FeS$$, $$Cu_2S$$ nor $$Cu_2O$$ reacts with $$SiO_2$$ under the furnace conditions because they are not basic oxides; hence they remain in the melt or, in the case of $$Cu_2O$$, participate in a later self-reduction step.

Therefore, the specific species that silica removes is $$FeO$$.

Hence, the correct answer is Option C.

First, recall the basic principle of the froth-flotation method. This method separates gangue from ore by taking advantage of the preferential wettability of the ore particles by oil and that of the gangue particles by water. Because sulphide ores tend to be hydrophobic, they attach to the air bubbles, rise with the froth, and thus get collected, while the gangue sinks.

Therefore, ores that are sulphides are best suited for this process. Let us list the given options with their chemical nature:

We have Galena: $$PbS$$, which is a sulphide ore.

Now consider Malachite: $$CuCO_3\cdot Cu(OH)_2$$, a basic carbonate of copper, not a sulphide, so it is not effectively concentrated by froth flotation.

Next, Magnetite: $$Fe_3O_4$$, an oxide ore of iron, also not suitable for froth flotation.

Finally, Siderite: $$FeCO_3$$, a carbonate ore of iron, likewise unsuitable for froth flotation.

So, among all the options, only Galena is a sulphide ore and therefore responds efficiently to the froth-flotation technique.

Hence, the correct answer is Option A.

Zinc is a highly reactive metal (above hydrogen in the electrochemical series). When it reacts with dilute $$HNO_3$$, the acid is reduced to Nitrous oxide ($$N_2O$$), commonly known as laughing gas.

$$4Zn + 10HNO_3 (\text{dilute}) \longrightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O \uparrow$$

With concentrated $$HNO_3$$, the oxidizing power of the acid is at its peak. Regardless of the metal's reactivity, concentrated nitric acid is predominantly reduced to Nitrogen dioxide ($$NO_2$$), which is characterized by its pungent-smelling brown fumes.

$$Zn + 4HNO_3 (\text{conc.}) \longrightarrow Zn(NO_3)_2 + 2H_2O + 2NO_2 \uparrow$$

First of all, we are told that the unknown element $$X$$ always exhibits an oxidation state of $$+3$$ in all its compounds.

Whenever we write a stable chemical formula, the algebraic sum of all the charges present in the whole formula must be zero, because a compound as a whole is electrically neutral. That is,

$$\text{(Total positive charge)} + \text{(Total negative charge)} = 0.$$

We shall now test each of the four given formulae one by one, using the above neutrality requirement and the fact that the oxidation state of $$X$$ is $$+3$$.

Option A: $$X_2O_3$$

• Each oxygen atom almost always has the oxidation state $$-2.$$
• There are $$3$$ oxygen atoms, so the total negative charge contributed by oxygen is

$$3 \times (-2) = -6.$$

• Let the oxidation state of each $$X$$ be $$+3$$ (as given). There are $$2$$ atoms of $$X,$$ so their total positive charge is

$$2 \times (+3) = +6.$$

• Adding the two totals gives

$$+6 + (-6) = 0.$$

Since the overall charge is zero, the formula $$X_2O_3$$ is perfectly acceptable for an element having a $$+3$$ state.

Option B: $$X_2(SO_4)_3$$

• The sulphate ion $$SO_4^{2-}$$ possesses a charge of $$-2.$$
• There are $$3$$ sulphate ions, hence the total negative charge from all the sulphate ions equals

$$3 \times (-2) = -6.$$

• Again, each $$X$$ is in the $$+3$$ state, and there are $$2$$ such atoms, so the total positive charge is

$$2 \times (+3) = +6.$$

• Adding the two totals gives

$$+6 + (-6) = 0.$$

The neutrality condition is satisfied, therefore $$X_2(SO_4)_3$$ is also a correct formula for an element with oxidation state $$+3.$$

Option C: $$XPO_4$$

• The phosphate ion $$PO_4^{3-}$$ carries a charge of $$-3.$$
• There is only one phosphate ion present, so the total negative charge is simply $$-3.$$
• The compound contains one atom of $$X,$$ each of which is $$+3,$$ so the total positive charge is $$+3.$$

The sum of the charges is

$$+3 + (-3) = 0,$$

hence the formula $$XPO_4$$ is perfectly compatible with $$X^{3+}.$$

Option D: $$X_2Cl_3$$

• Chlorine (in chloride) has the oxidation state $$-1.$$
• There are $$3$$ chloride ions, so their total negative charge is

$$3 \times (-1) = -3.$$

• Each $$X$$ is still $$+3,$$ and there are $$2$$ such atoms, giving a total positive charge of

$$2 \times (+3) = +6.$$

• Adding the two totals gives

$$+6 + (-3) = +3.$$

This sum is not zero; it equals $$+3,$$ which means the formula $$X_2Cl_3$$ is not electrically neutral. Hence this compound cannot exist if $$X$$ is strictly in the $$+3$$ oxidation state.

All the other three options satisfy charge neutrality, but Option D does not. Hence, the correct answer is Option D.

To determine the number of electrons involved in the redox reaction, we first need to balance the reaction since it is given unbalanced. The reaction is:

$$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{CO}_2$$

We start by identifying the oxidation states of each element:

• In $$\text{Cr}_2\text{O}_7^{2-}$$, oxygen is $$-2$$. Let the oxidation state of Cr be $$x$$. Then $$2x + 7(-2) = -2$$, so $$2x - 14 = -2$$, giving $$2x = 12$$, and $$x = +6$$. Thus, Cr is $$+6$$.
• In $$\text{Fe}^{2+}$$, iron is $$+2$$.
• In $$\text{C}_2\text{O}_4^{2-}$$, oxygen is $$-2$$. Let the oxidation state of C be $$y$$. Then $$2y + 4(-2) = -2$$, so $$2y - 8 = -2$$, giving $$2y = 6$$, and $$y = +3$$. Thus, each carbon is $$+3$$.
• In $$\text{Cr}^{3+}$$, chromium is $$+3$$.
• In $$\text{Fe}^{3+}$$, iron is $$+3$$.
• In $$\text{CO}_2$$, oxygen is $$-2$$, so carbon is $$+4$$.

Now, observe the changes in oxidation states:

• Chromium is reduced from $$+6$$ to $$+3$$, gaining electrons.
• Iron is oxidized from $$+2$$ to $$+3$$, losing electrons.
• Carbon in oxalate is oxidized from $$+3$$ to $$+4$$, losing electrons.

We split the reaction into half-reactions:

Reduction: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$$

Oxidation: $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$$ and $$\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2$$

Balance the reduction half-reaction in acidic medium:

1. Balance Cr atoms: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$$.
2. Balance oxygen by adding water: $$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.
3. Balance hydrogen by adding H⁺: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.
4. Balance charge: Left side charge is $$-2 + 14(+1) = +12$$, right side is $$2(+3) = +6$$. Add 6 electrons to the left: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$.

So, reduction involves 6 electrons per $$\text{Cr}_2\text{O}_7^{2-}$$ ion.

Balance the oxidation half-reactions:

• For iron: $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$. Each $$\text{Fe}^{2+}$$ loses 1 electron.
• For oxalate: $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$$. Oxidation state change: Two carbons go from $$+3$$ to $$+4$$, so total loss of 2 electrons. Balance atoms and charge: Left has charge $$-2$$, right has charge $$0$$. Add 2 electrons to the right: $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$$. Each $$\text{C}_2\text{O}_4^{2-}$$ loses 2 electrons.

Now, combine the half-reactions. Let the number of $$\text{Cr}_2\text{O}_7^{2-}$$ ions be $$a$$, $$\text{Fe}^{2+}$$ ions be $$b$$, and $$\text{C}_2\text{O}_4^{2-}$$ ions be $$c$$.

Electron balance: Electrons gained = electrons lost. Reduction gains $$6a$$ electrons. Oxidation loses $$b$$ electrons from iron and $$2c$$ electrons from oxalate. So:

$$6a = b + 2c$$

The balanced reaction must include H⁺ and H₂O. Using the reduction half-reaction, for each $$\text{Cr}_2\text{O}_7^{2-}$$, we need 14H⁺ and produce 7H₂O. The full reaction is:

$$a \text{Cr}_2\text{O}_7^{2-} + b \text{Fe}^{2+} + c \text{C}_2\text{O}_4^{2-} + 14a \text{H}^+ \rightarrow 2a \text{Cr}^{3+} + b \text{Fe}^{3+} + 2c \text{CO}_2 + 7a \text{H}_2\text{O}$$

Balance atoms and charge:

• Oxygen: Left has $$7a + 4c$$ atoms (from $$\text{Cr}_2\text{O}_7^{2-}$$ and $$\text{C}_2\text{O}_4^{2-}$$), right has $$7a$$ from H₂O and $$4c$$ from CO₂ (since each CO₂ has 2 oxygen atoms, and $$2c$$ CO₂ has $$4c$$ oxygen atoms). So $$7a + 4c = 7a + 4c$$, balanced.
• Hydrogen: Left has $$14a$$ H atoms from H⁺, right has $$14a$$ H atoms from $$7a$$ H₂O, balanced.
• Charge: Left charge is $$-2a + 2b - 2c + 14a = 12a + 2b - 2c$$. Right charge is $$2a \times 3 + b \times 3 = 6a + 3b$$. Set equal: $$12a + 2b - 2c = 6a + 3b$$, simplifying to $$6a - b - 2c = 0$$, or $$b + 2c = 6a$$, which matches the electron balance equation.

Solve $$b + 2c = 6a$$ for positive integers $$a$$, $$b$$, $$c$$. Since the unbalanced reaction shows one of each reactant, we find minimal solutions:

• If $$a = 1$$, then $$b + 2c = 6$$. Possible pairs: $$(b, c) = (4, 1)$$ or $$(2, 2)$$.

Both are valid balanced equations:

1. For $$(a, b, c) = (1, 4, 1)$$: $$\text{Cr}_2\text{O}_7^{2-} + 4\text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 4\text{Fe}^{3+} + 2\text{CO}_2 + 7\text{H}_2\text{O}$$
2. For $$(a, b, c) = (1, 2, 2)$$: $$\text{Cr}_2\text{O}_7^{2-} + 2\text{Fe}^{2+} + 2\text{C}_2\text{O}_4^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 2\text{Fe}^{3+} + 4\text{CO}_2 + 7\text{H}_2\text{O}$$

In both cases, the total electrons transferred are 6, as reduction gains 6 electrons and oxidation loses 6 electrons:

• Case 1: Iron loses $$4 \times 1 = 4$$ electrons, oxalate loses $$1 \times 2 = 2$$ electrons, total 6.
• Case 2: Iron loses $$2 \times 1 = 2$$ electrons, oxalate loses $$2 \times 2 = 4$$ electrons, total 6.

The number of electrons involved is always 6 per mole of $$\text{Cr}_2\text{O}_7^{2-}$$, regardless of the coefficients chosen for the reducing agents.

Hence, the correct answer is Option A.

First, we need to understand the reaction: $$H_{2}SO_{3}(aq) + Sn^{4+(aq) + H2O(l) -> Sn^{2+}(aq) + HSO4^{-}(aq) + 3H^{+}(aq)}$$. To determine which statement is correct, we must identify the oxidizing and reducing agents by analyzing the changes in oxidation numbers.

Recall the rules for oxidation numbers: Hydrogen (H) is usually +1, Oxygen (O) is usually -2, and the sum of oxidation numbers in a neutral compound is zero, while in an ion, it equals the charge. Let's assign oxidation numbers to each element in the reactants and products.

Starting with the reactants:

• In $$H_{2}SO_{3}$$: Hydrogen (H) is +1 (since it's combined with non-metals). Let the oxidation number of Sulfur (S) be $$x$$. Oxygen (O) is -2. So, $$2 \times (+1) + x + 3 \times (-2) = 0$$. Simplifying: $$2 + x - 6 = 0$$ → $$x - 4 = 0$$ → $$x = +4$$. Thus, S in $$H_{2}SO_{3}$$ has an oxidation number of +4.
• In $$Sn^{4+}$$: The ion has a +4 charge, so Sn has an oxidation number of +4.
• In $$H_{2}O$$: Hydrogen is +1, Oxygen is -2.

Now for the products:

• In $$Sn^{2+}$$: The ion has a +2 charge, so Sn has an oxidation number of +2.
• In $$HSO_{4}^{-}$$: Hydrogen (H) is +1. Oxygen (O) is -2. Let the oxidation number of Sulfur (S) be $$y$$. The ion has a -1 charge. So, $$+1 + y + 4 \times (-2) = -1$$. Simplifying: $$1 + y - 8 = -1$$ → $$y - 7 = -1$$ → $$y = +6$$. Thus, S in $$HSO_{4}^{-}$$ has an oxidation number of +6.
• In $$H^{+}$$: Hydrogen has an oxidation number of +1.

Now, compare the oxidation numbers:

• Sulfur (S): From +4 in $$H_{2}SO_{3}$$ to +6 in $$HSO_{4}^{-}$$. This is an increase of +2, indicating oxidation (loss of electrons).
• Tin (Sn): From +4 in $$Sn^{4+}$$ to +2 in $$Sn^{2+}$$. This is a decrease of -2, indicating reduction (gain of electrons).

Since Sulfur in $$H_{2}SO_{3}$$ is oxidized, $$H_{2}SO_{3}$$ is the reducing agent (it causes reduction by being oxidized). Since Tin in $$Sn^{4+}$$ is reduced, $$Sn^{4+}$$ is the oxidizing agent (it causes oxidation by being reduced).

Now, evaluate the options:

• Option A: "Sn^{4+} is the oxidizing agent because it undergoes oxidation" → Incorrect, because Sn^{4+} undergoes reduction, not oxidation.
• Option B: "Sn^{4+} is the reducing agent because it undergoes oxidation" → Incorrect, because Sn^{4+} undergoes reduction, and it is not the reducing agent.
• Option C: "H2SO3 is the reducing agent because it undergoes reduction" → Incorrect, because H2SO3 undergoes oxidation, not reduction.
• Option D: "H2SO3 is the reducing agent because it undergoes oxidation" → Correct, as H2SO3 undergoes oxidation and is therefore the reducing agent.

Hence, the correct answer is Option D.

We begin by recalling that a species behaves as a reducing agent when it itself gets oxidised, that is, when it loses electrons and the oxidation number of (at least one of) its atoms increases.

In hydrogen peroxide, $$H_2O_2$$, the oxidation number of each oxygen atom is $$-1$$. If $$H_2O_2$$ is oxidised, the oxidation number of oxygen must rise from $$-1$$ to $$0$$ (in $$O_2$$) or to $$-2$$ (in $$OH^-$$ or $$H_2O$$ it actually goes down, so that would represent reduction). Oxidation will therefore be accompanied by the loss of electrons, and in a half-reaction those lost electrons will appear on the right-hand side.

Now we analyse each given half-reaction one by one, carefully watching the position of electrons and the change in oxidation number.

(a) $$H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$$

Here the electrons are written on the left. So $$H_2O_2$$ is gaining electrons. This means that $$H_2O_2$$ is being reduced, acting as an oxidising agent, not as a reducing agent.

(b) $$H_2O_2 \;-\; 2e^- \rightarrow O_2 + 2H^+$$

We can rewrite this more explicitly as $$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$$. Now the electrons are on the right; $$H_2O_2$$ is losing electrons. Its oxygen atoms go from $$-1$$ in $$H_2O_2$$ to $$0$$ in $$O_2$$, an increase in oxidation number. Thus $$H_2O_2$$ is oxidised and therefore acts as a reducing agent.

(c) $$H_2O_2 + 2e^- \rightarrow 2OH^-$$

Electrons again appear on the left; $$H_2O_2$$ is gaining electrons, so it is being reduced. Its oxygen atoms change from $$-1$$ in $$H_2O_2$$ to $$-2$$ in $$OH^-$$ (a decrease in oxidation number). Hence $$H_2O_2$$ functions as an oxidising agent, not as a reducing agent.

(d) $$H_2O_2 + 2OH^- \;-\; 2e^- \rightarrow O_2 + 2H_2O$$

Again, writing the electrons on the right yields $$H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$$. The electrons are produced, so $$H_2O_2$$ is losing electrons. The oxidation number of oxygen in $$H_2O_2$$ increases from $$-1$$ to $$0$$ in $$O_2$$, confirming that $$H_2O_2$$ is oxidised and therefore acts as a reducing agent.

Summarising our findings:

  • Reaction (a): oxidising agent.
  • Reaction (b): reducing agent.
  • Reaction (c): oxidising agent.
  • Reaction (d): reducing agent.

So the reactions in which $$H_2O_2$$ acts as a reducing agent are (b) and (d).

Looking at the options provided, option D corresponds to (b) and (d).

Hence, the correct answer is Option D.

$$H_2O_2$$ acts as a reducing agent only in presence of strong oxidising agents (i.e., $$MnO_4^-$$).

$$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$$

The reaction involves hydrogen sulfide (H₂S) reacting with acidic potassium dichromate (K₂Cr₂O₇) solution. Potassium dichromate in acidic medium acts as a strong oxidizing agent, and hydrogen sulfide is a reducing agent. The typical reaction proceeds as follows.

First, the dichromate ion (Cr₂O₇²⁻) is reduced to chromium(III) ions (Cr³⁺) in acidic medium. The reduction half-reaction is:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

Hydrogen sulfide (H₂S) is oxidized to elemental sulfur (S). The oxidation half-reaction is:

$$\text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2\text{e}^-$$

To balance the electrons, multiply the oxidation half-reaction by 3:

$$3\text{H}_2\text{S} \rightarrow 3\text{S} + 6\text{H}^+ + 6\text{e}^-$$

Now, add this to the reduction half-reaction:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S} + 6\text{H}^+ + 6\text{e}^-$$

Cancel the common terms: 6e⁻ on both sides, and 6H⁺ from the right with 6H⁺ from the left (leaving 8H⁺ on the left):

$$\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S}$$

This is the net ionic equation. To write the complete molecular equation, include the potassium ions from K₂Cr₂O₇ and the sulfate ions from the acid (H₂SO₄). The acid provides H⁺ ions, and each H₂SO₄ provides 2H⁺. Since 8H⁺ are needed, 4 molecules of H₂SO₄ are required, which also provide 4 SO₄²⁻ ions.

The potassium ions (2K⁺ from K₂Cr₂O₇) and chromium ions (2Cr³⁺) will combine with sulfate ions. The 2Cr³⁺ ions require 3 SO₄²⁻ ions to form Cr₂(SO₄)₃, and the remaining SO₄²⁻ ion (from the 4 provided) combines with 2K⁺ to form K₂SO₄.

Thus, the balanced molecular equation is:

$$\text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 + 3\text{H}_2\text{S} \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 7\text{H}_2\text{O} + 3\text{S}$$

The products are potassium sulfate (K₂SO₄), chromium(III) sulfate (Cr₂(SO₄)₃), water (H₂O), and elemental sulfur (S).

Now, examining the options:

A. K₂SO₄ is formed, as shown in the products.

B. Cr₂(SO₄)₃ is formed, as chromium is in the +3 oxidation state.

C. S (elemental sulfur) is formed from the oxidation of H₂S.

D. CrSO₄ would imply chromium(II) sulfate, where chromium is in the +2 oxidation state. However, in the reaction, chromium is reduced to +3, not +2, so CrSO₄ is not formed.

Hence, the correct answer is Option D.

For $$CrO_2Cl_2$$ (Chromyl chloride):

Let the oxidation state of $$Cr$$ be $$x$$. 

$$x + 2(\text{O.S. of } O) + 2(\text{O.S. of } Cl) = 0$$

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

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

For $$[Cr(CN)_6]^{3-}$$: 

Let the oxidation state of $$Cr$$ be $$x$$. 

$$x + 6(\text{O.S. of } CN) = -3$$

$$x + 6(-1) = -3$$

$$x - 6 = -3 \implies x = +3$$

For $$Cr_2O_3$$:

Let the oxidation state of $$Cr$$ be $$x$$. 

$$2x + 3(\text{O.S. of } O) = 0$$

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

$$2x = 6 \implies x = +3$$

For $$[MnO_4]^-$$ (Permanganate ion):

Let the oxidation state of $$Mn$$ be $$y$$. 

$$y + 4(\text{O.S. of } O) = -1$$

$$y + 4(-2) = -1$$

$$y - 8 = -1 \implies y = +7$$

We are told that the empirical formula of the oxide is $$M_{0.98}O$$. This means that, for every one oxide ion $$O^{2-}$$ present, there are only $$0.98$$ atoms of the metal $$M$$ in the solid.

The metal occurs in two oxidation states, $$M^{2+}$$ and $$M^{3+}$$. Let us assume that a fraction $$x$$ (in decimal form) of the total metal atoms is present as $$M^{3+}$$. Consequently, the remaining fraction $$1 - x$$ will be present as $$M^{2+}$$.

So, out of the total $$0.98$$ metal atoms,

$$ \text{Number of }M^{3+}\text{ ions} = 0.98\,x, \qquad \text{Number of }M^{2+}\text{ ions} = 0.98\,(1 - x). $$

Next, we impose the requirement of overall electrical neutrality. The oxide ion carries a charge of $$-2$$. Therefore, the sum of positive charges provided by the metal ions must equal $$+2$$.

First, we compute the positive charge contributed by each type of metal ion.

Charge from $$M^{2+}: \quad 2 \times 0.98\,(1 - x) = 1.96\,(1 - x).$$

Charge from $$M^{3+}: \quad 3 \times 0.98\,x = 2.94\,x.$$

The total positive charge is the sum of these two:

$$ \text{Total positive charge} = 1.96\,(1 - x) + 2.94\,x. $$

We now set this equal to the required $$+2$$ charge from one $$O^{2-}$$:

$$ 1.96\,(1 - x) + 2.94\,x \;=\; 2. $$

Let us expand and simplify step by step.

First, expand the term $$1.96\,(1 - x)$$:

$$ 1.96\,(1 - x) = 1.96 - 1.96\,x. $$

Add the other term $$2.94\,x$$:

$$ 1.96 - 1.96\,x + 2.94\,x \;=\; 2. $$

Combine the $$x$$ terms:

$$ 1.96 + (2.94 - 1.96)\,x = 2. $$

Calculate the coefficient of $$x$$ in parentheses:

$$ 2.94 - 1.96 = 0.98, $$

so the equation becomes

$$ 1.96 + 0.98\,x = 2. $$

Now isolate $$x$$ by subtracting $$1.96$$ from both sides:

$$ 0.98\,x = 2 - 1.96 = 0.04. $$

Finally, divide by $$0.98$$ to solve for $$x$$:

$$ x = \frac{0.04}{0.98}. $$

Performing the division gives

$$ x \approx 0.040816. $$

To find the percentage of metal present as $$M^{3+}$$, we multiply the fraction by 100:

$$ \%\,M^{3+} = 0.040816 \times 100 \approx 4.08\%. $$

Hence, the correct answer is Option D.

To determine which species cannot function as an oxidizing agent, we must recall that an oxidizing agent accepts electrons and gets reduced in a redox reaction. Therefore, a species that cannot be reduced (i.e., cannot gain electrons) cannot act as an oxidizing agent. We will examine each option by checking if it can be reduced.

Starting with Option A: I$$^-$$. The iodine atom in I$$^-$$ has an oxidation state of -1, which is the lowest possible oxidation state for iodine. Since it cannot accept more electrons to be reduced further, I$$^-$$ cannot act as an oxidizing agent. Instead, it tends to lose electrons and act as a reducing agent.

Next, Option B: S(s). Sulfur in its elemental form has an oxidation state of 0. Sulfur can be reduced to lower oxidation states, such as -2 in compounds like H$$_2$$S. For example, in the reaction with iron: $$Fe + S -> FeS$$, sulfur is reduced from 0 to -2. Thus, S(s) can function as an oxidizing agent.

Option C: NO$$_3^-$$ (aq). The nitrogen atom in nitrate ion has an oxidation state of +5. Nitrogen can be reduced to lower oxidation states, such as +4 in NO$$_2$$, +2 in NO, 0 in N$$_2$$, or -3 in NH$$_3$$. For instance, in acidic conditions, nitrate oxidizes copper: $$Cu + 4H+ + 2NO_{3}- -> Cu^{2+ + 2NO2 + 2H2O}$$, where NO$$_3^-$$ is reduced. Hence, NO$$_3^-$$ can act as an oxidizing agent.

Option D: Cr$$_2$$O$$_7^{2-}$$. The chromium atoms in dichromate ion have an oxidation state of +6. Chromium can be reduced to lower oxidation states, such as +3 in Cr$$^{3+}$$. For example, in acidic medium, dichromate oxidizes Fe$$^{2+}$$: $$Cr_{2}O_{7}^{2- + 6Fe^{2+} + 14H+ -> 2Cr^{3+} + 6Fe^{3+} + 7H2O}$$, where Cr$$_2$$O$$_7^{2-}$$ is reduced. Therefore, Cr$$_2$$O$$_7^{2-}$$ can function as an oxidizing agent.

From the analysis, I$$^-$$ (Option A) cannot be reduced and thus cannot act as an oxidizing agent, while the other options can. Hence, the correct answer is Option A.

The given redox reaction is:

$$XNa_2HAsO_3 + YNaBrO_3 + ZHCl \rightarrow NaBr + H_3AsO_4 + NaCl$$

We need to find the values of X, Y, and Z by balancing the reaction using the oxidation number method. First, we determine the oxidation states of the elements involved.

In the reactant Na₂HAsO₃:

• Sodium (Na) has an oxidation state of +1.
• Hydrogen (H) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of arsenic (As) be a. The sum of oxidation states in Na₂HAsO₃ is zero: 2*(+1) + (+1) + a + 3*(-2) = 0 → 2 + 1 + a - 6 = 0 → a - 3 = 0 → a = +3.

In the reactant NaBrO₃:

• Sodium (Na) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of bromine (Br) be b. The sum is: +1 + b + 3*(-2) = 0 → 1 + b - 6 = 0 → b - 5 = 0 → b = +5.

In the product NaBr:

• Sodium (Na) has an oxidation state of +1, so bromine (Br) must be -1.

In the product H₃AsO₄:

• Hydrogen (H) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of arsenic (As) be c. The sum is: 3*(+1) + c + 4*(-2) = 0 → 3 + c - 8 = 0 → c - 5 = 0 → c = +5.

Arsenic changes from +3 to +5, losing 2 electrons per atom. Bromine changes from +5 to -1, gaining 6 electrons per atom (since +5 to -1 is a decrease of 6).

Let X be the number of Na₂HAsO₃ molecules and Y be the number of NaBrO₃ molecules. The total electrons lost by arsenic are 2X, and the total electrons gained by bromine are 6Y. For the reaction to be balanced, electrons lost must equal electrons gained:

$$2X = 6Y \rightarrow X = 3Y$$

So, X must be 3 times Y. The smallest integer solution is Y = 1 and X = 3.

Now, substitute X = 3 and Y = 1 into the reaction:

$$3Na_2HAsO_3 + NaBrO_3 + ZHCl \rightarrow NaBr + H_3AsO_4 + NaCl$$

Balance the atoms. On the reactant side:

• Sodium (Na): From 3Na₂HAsO₃ → 6 Na, from NaBrO₃ → 1 Na, total Na = 7.
• Hydrogen (H): From 3Na₂HAsO₃ → 3 H (one per molecule), from ZHCl → Z H, total H = 3 + Z.
• Arsenic (As): From 3Na₂HAsO₃ → 3 As.
• Bromine (Br): From NaBrO₃ → 1 Br.
• Oxygen (O): From 3Na₂HAsO₃ → 9 O, from NaBrO₃ → 3 O, total O = 12.
• Chlorine (Cl): From ZHCl → Z Cl.

On the product side, as written, we have NaBr, H₃AsO₄, and NaCl. However, with 3 As atoms, we need 3 H₃AsO₄. With 1 Br atom, we need 1 NaBr. Sodium atoms on the left are 7, but NaBr provides 1 Na and NaCl provides 1 Na, totaling 2 Na, which is insufficient. Therefore, we adjust the products to include coefficients:

Products should be: NaBr (for Br), 3H₃AsO₄ (for As), and NaCl for the remaining Na and Cl. Sodium required: total Na from reactants is 7. NaBr provides 1 Na, so the remaining 6 Na must come from NaCl. Thus, we need 6 NaCl.

So, the balanced products are: NaBr + 3H₃AsO₄ + 6NaCl.

Now, the product side has:

• Sodium (Na): NaBr → 1 Na, 6NaCl → 6 Na, total Na = 7.
• Hydrogen (H): 3H₃AsO₄ → 9 H.
• Arsenic (As): 3H₃AsO₄ → 3 As.
• Bromine (Br): NaBr → 1 Br.
• Oxygen (O): 3H₃AsO₄ → 12 O.
• Chlorine (Cl): 6NaCl → 6 Cl.

Set reactant and product atoms equal:

• Na: 7 = 7 → balanced.
• H: 3 + Z = 9 → Z = 6.
• As: 3 = 3 → balanced.
• Br: 1 = 1 → balanced.
• O: 12 = 12 → balanced.
• Cl: Z = 6 → balanced.

Thus, X = 3, Y = 1, Z = 6. The balanced equation is:

$$3Na_2HAsO_3 + NaBrO_3 + 6HCl \rightarrow NaBr + 3H_3AsO_4 + 6NaCl$$

Comparing with the options:

• A. 2, 1, 2
• B. 2, 1, 3
• C. 3, 1, 6
• D. 3, 1, 4

Hence, the correct answer is Option C.

We first identify that the reaction takes place in an acidic medium, so we shall balance it by the ion-electron (half-reaction) method.

We have two half-reactions:

1. Reduction of permanganate:
$$MnO_4^- \rightarrow Mn^{2+}$$

2. Oxidation of oxalate:
$$C_2O_4^{2-} \rightarrow CO_2$$

Balancing the reduction half-reaction.

• Balance manganese atoms: already 1 on each side.
• Balance oxygen atoms by adding water. There are 4 oxygen atoms on the left, so we add 4 $$H_2O$$ to the right:
$$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$

• Balance hydrogen atoms by adding $$H^+$$. We now have 8 hydrogens on the right, so we add 8 $$H^+$$ on the left:
$$MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$

• Balance charge by adding electrons. The total charge on the left is $$(-1)+(+8)=+7$$. The total charge on the right is $$+2$$. To reduce the left side from +7 to +2 we add 5 electrons to the left:
$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Balancing the oxidation half-reaction.

• Oxalate already has two carbon atoms, so we write two molecules of carbon dioxide on the right to keep carbon balanced:
$$C_2O_4^{2-} \rightarrow 2CO_2$$

• Oxygen is now balanced (4 on each side). No hydrogen appears, so no $$H^+$$ or $$H_2O$$ are needed here.

• Balance charge by adding electrons. The left side has a charge of $$-2$$. The right side is neutral. To make charges equal we add 2 electrons on the right:
$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$$

Equalising the number of electrons. The reduction half needs 5 electrons, the oxidation half releases 2 electrons. Their least common multiple is 10, so we multiply:

• The reduction half-reaction by 2:
$$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$

• The oxidation half-reaction by 5:
$$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$$

Adding the two half-reactions. Because both sides now contain 10 electrons, they cancel out when we add the equations:

$$2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2$$

Re-ordering the terms gives the balanced overall equation:

$$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$$

Comparing with the general form given in the question,

$$x \ MnO_4^- + y \ C_2O_4^{2-} + zH^+ \rightarrow x \ Mn^{2+} + 2y \ CO_2 + \frac{z}{2} H_2O,$$

we read directly:
$$x = 2,\; y = 5,\; z = 16.$$

Hence, the correct answer is Option A.

To determine the oxidation state of sulfur in the given anions, we recall that the oxidation state is the hypothetical charge on an atom if all bonds were ionic. The sum of oxidation states in a compound must equal the overall charge. Oxygen typically has an oxidation state of -2, except in peroxides, which is not the case here. Let the oxidation state of sulfur be denoted by $$x$$ for each anion.

First, consider the sulfate ion $$SO_4^{2-}$$. However, note that the options include $$SO_3^{2-}$$ (sulfite ion) instead of sulfate. Given the correct answer option and the context, it appears there might be a typo in the question, and we should consider sulfite ion $$SO_3^{2-}$$ instead. We will calculate for $$SO_3^{2-}$$, $$S_2O_4^{2-}$$, and $$S_2O_6^{2-}$$ as per the options.

Start with the sulfite ion $$SO_3^{2-}$$:

• It has one sulfur atom and three oxygen atoms.
• The overall charge is -2.
• So, oxidation state of sulfur + 3 × (oxidation state of oxygen) = overall charge.
• $$x + 3 \times (-2) = -2$$
• $$x - 6 = -2$$
• $$x = -2 + 6$$
• $$x = 4$$
• Thus, the oxidation state of sulfur in $$SO_3^{2-}$$ is +4.

Next, the dithionite ion $$S_2O_4^{2-}$$:

• It has two sulfur atoms and four oxygen atoms.
• The overall charge is -2.
• Assuming both sulfur atoms have the same oxidation state, we have 2 × (oxidation state of sulfur) + 4 × (oxidation state of oxygen) = overall charge.
• $$2x + 4 \times (-2) = -2$$
• $$2x - 8 = -2$$
• $$2x = -2 + 8$$
• $$2x = 6$$
• $$x = 3$$
• Thus, the oxidation state of sulfur in $$S_2O_4^{2-}$$ is +3 per sulfur atom.

Finally, the dithionate ion $$S_2O_6^{2-}$$:

• It has two sulfur atoms and six oxygen atoms.
• The overall charge is -2.
• Assuming both sulfur atoms have the same oxidation state, we have 2 × (oxidation state of sulfur) + 6 × (oxidation state of oxygen) = overall charge.
• $$2x + 6 \times (-2) = -2$$
• $$2x - 12 = -2$$
• $$2x = -2 + 12$$
• $$2x = 10$$
• $$x = 5$$
• Thus, the oxidation state of sulfur in $$S_2O_6^{2-}$$ is +5 per sulfur atom.

We have the oxidation states:

• $$S_2O_4^{2-}$$: +3
• $$SO_3^{2-}$$: +4
• $$S_2O_6^{2-}$$: +5

Arranging these in increasing order of oxidation state:

• +3 (from $$S_2O_4^{2-}$$) is less than +4 (from $$SO_3^{2-}$$), which is less than +5 (from $$S_2O_6^{2-}$$).
• So, the order is $$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$$.

Comparing with the options:

• Option A: $$S_2O_6^{2-} < S_2O_4^{2-} < SO_3^{2-}$$ → +5 < +3 < +4, which is incorrect.
• Option B: $$SO_6^{2-} < S_2O_4^{2-} < S_2O_6^{2-}$$ → Note: $$SO_6^{2-}$$ is not standard and likely a typo; ignoring this as it does not match our anions.
• Option C: $$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$$ → +3 < +4 < +5, which matches our order.
• Option D: $$S_2O_4^{2-} < S_2O_6^{2-} < SO_3^{2-}$$ → +3 < +5 < +4, but +5 is not less than +4, so incorrect.

Hence, the correct answer is Option C.

The reducibility of a metal oxide with carbon is predicted with the Ellingham diagram.

1. Ellingham diagram concept: The line for the reaction $$\text{C (s) + }\tfrac12\text{O}_2(g) \rightarrow \text{CO (g)}$$ (or for $$\text{CO (g) + }\tfrac12\text{O}_2(g) \rightarrow \text{CO}_2(g)$$) represents the ability of carbon/CO to remove oxygen.
• If the $$\Delta G^\circ_{f}$$ line for a metal oxide lies above the carbon line at a given temperature, carbon can reduce that oxide because doing so makes the overall $$\Delta G^\circ$$ negative.
• If the oxide line lies below the carbon line, the oxide is more stable than CO/CO₂; carbon cannot reduce it.

2. Oxides of highly electropositive (alkali and alkaline-earth) metals such as Ca and K lie far below the carbon line throughout the temperature range. Their $$\Delta G^\circ_{f}$$ values are thus more negative than that for the formation of CO or CO₂, making them thermodynamically non-reducible by carbon. These oxides can only be reduced by electrolytic methods.

3. Oxides of less electropositive metals—Cu₂O, SnO₂, PbO, Fe₃O₄, Fe₂O₃, ZnO—intersect or stay above the carbon line at moderate or high temperatures. Hence carbon (as coke or charcoal) is routinely used to reduce these oxides in metallurgy (e.g., blast furnace for Fe, roasting of ZnO, smelting of SnO₂, extraction of Cu from Cu₂O, etc.).

4. Examining the given groups:

Case A: Cu₂O and SnO₂  →  reducible by C.

Case B: CaO and K₂O  →  NOT reducible by C (require electrolysis).

Case C: PbO and Fe₃O₄  →  reducible by C.

Case D: Fe₂O₃ and ZnO  →  reducible by C.

Therefore, the only group of oxides that cannot be reduced by carbon is found in Option B.

Option B which is: CaO, $$\text{K}_2\text{O}$$

The unbalanced ionic equation given in the question is

$$\text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} + H^+ \;\longrightarrow\; \text{Mn}^{2+} + CO_2 + H_2O$$

We have to find the smallest whole-number coefficients $$X,\,Y$$ and $$Z$$ for $$\text{MnO}_4^- , \text{C}_2\text{O}_4^{2-}$$ and $$H^+$$ respectively that balance the above redox reaction in acidic medium.

Step 1 : Write the two half-reactions.
Reduction half-reaction (permanganate):
$$\text{MnO}_4^- \;\rightarrow\; \text{Mn}^{2+}$$

Oxidation half-reaction (oxalate):
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; CO_2$$

Step 2 : Balance each half-reaction separately.

(a) Reduction half-reaction
• Balance Mn: already 1 on each side.
• Balance O by adding $$H_2O$$: $$\text{MnO}_4^- \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O$$
• Balance H by adding $$H^+$$: $$\text{MnO}_4^- + 8\,H^+ \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O$$
• Balance charge by adding electrons: LHS charge $$= -1 + 8 = +7$$, RHS charge $$= +2$$. Add 5 e⁻ to the left:
$$\text{MnO}_4^- + 8\,H^+ + 5\,e^- \;\rightarrow\; \text{Mn}^{2+} + 4\,H_2O \qquad -(1)$$

(b) Oxidation half-reaction
• Balance C: already 2 on LHS, 1 on RHS ⇒ write 2 CO₂ on RHS:
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 2\,CO_2$$
• O already balanced (4 each side).
• No H atoms are involved.
• Balance charge by adding electrons: LHS charge $$= -2$$, RHS charge $$= 0$$, so add 2 e⁻ to RHS:
$$\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 2\,CO_2 + 2\,e^- \qquad -(2)$$

Step 3 : Equalise electrons and add the two half-reactions.
The reduction step involves 5 e⁻; the oxidation step, 2 e⁻. The least common multiple of 5 and 2 is 10, so multiply equation (1) by 2 and equation (2) by 5.

$$\begin{aligned} 2\bigl[\text{MnO}_4^- + 8H^+ + 5e^- &\rightarrow \text{Mn}^{2+} + 4H_2O\bigr] \\[4pt] 5\bigl[\text{C}_2\text{O}_4^{2-} &\rightarrow 2CO_2 + 2e^-\bigr] \end{aligned}$$

After multiplication:

Reduction (×2): $$2\text{MnO}_4^- + 16H^+ + 10e^- \;\rightarrow\; 2\text{Mn}^{2+} + 8H_2O$$

Oxidation (×5): $$5\text{C}_2\text{O}_4^{2-} \;\rightarrow\; 10CO_2 + 10e^-$$

Step 4 : Add and cancel electrons.

$$2\text{MnO}_4^- + 16H^+ + 10e^- + 5\text{C}_2\text{O}_4^{2-} \;\longrightarrow\; 2\text{Mn}^{2+} + 8H_2O + 10CO_2 + 10e^-$$

The 10 e⁻ on both sides cancel exactly, giving the balanced net ionic equation:

$$\boxed{\,2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16H^+ \;\longrightarrow\; 2\text{Mn}^{2+} + 10CO_2 + 8H_2O\,}$$

Step 5 : Read off the required coefficients.
The coefficients of $$\text{MnO}_4^- , \text{C}_2\text{O}_4^{2-}$$ and $$H^+$$ are respectively $$X = 2,\; Y = 5,\; Z = 16$$.

Hence, the correct option is:
Option A which is: $$2,\,5,\,16$$.

Roasting is the step in which a metal sulphide $$MS$$ is first converted to its oxide $$MO$$ by heating it in excess air :

$$2\,MS + 3\,O_2 \rightarrow 2\,MO + 2\,SO_2$$

Only afterwards is the oxide reduced (most commonly by carbon/CO). We do not attempt a direct carbon-reduction of the sulphide

$$MS + C \; \longrightarrow \; M + CS_2 \;(\text{or } CS).$$

The feasibility of any pyrometallurgical reaction is judged from the standard Gibbs energies of formation, which are plotted in an Ellingham diagram. Keeping this in mind, let us examine the four statements.

Case 1: Option B — $$CO_2$$ is thermodynamically more stable than $$CS_2$$

The line for $$CO_2$$ lies far below that for $$CS_2$$ on the Ellingham diagram, i.e. $$\Delta G_f^\circ(CO_2)$$ is much more negative. Carbon therefore has a far greater affinity for oxygen than for sulphur. Hence carbon can reduce an oxide easily (because it is itself oxidised to $$CO_2$$/$$CO$$), but it cannot reduce a sulphide (because the oxidation product $$CS_2$$ is much less stable). This factor is directly responsible for preferring roasting over direct carbon reduction, so Option B is significant.

Case 2: Option C — Metal sulphides are less stable than the corresponding oxides

For most metals the $$MO$$ line also lies below the $$MS$$ line. Consequently the conversion $$MS \rightarrow MO$$ (roasting) is spontaneous, often exothermic, and needs no external reducing agent. Thus Option C is again a valid reason for carrying out roasting.

Case 3: Option D — $$CO_2$$ is more volatile than $$CS_2$$

The gaseous product formed during reduction leaves the reaction mixture, driving the equilibrium to the right. $$CO_2$$ (b.p. -78 °C under 1 atm) is far more volatile than $$CS_2$$ (b.p. 46 °C). Therefore the removal of $$CO_2$$ in oxide reduction is much easier than the removal of $$CS_2$$ in a hypothetical sulphide reduction. This volatility factor also favours the oxide route, so Option D is significant.

Case 4: Option A — Metal sulphides are thermodynamically more stable than $$CS_2$$

This statement compares two sulphur-containing species that are both on the sulphur side of the Ellingham diagram. Whether $$MS$$ happens to be more or less stable than $$CS_2$$ does not decide the success of carbon as a reducing agent; what really matters is the relative stabilities of the oxidation products of carbon (i.e. Option B) and of the ores themselves (i.e. Option C), together with the ease of removing the gaseous product (Option D). Hence the fact quoted in Option A is of no practical significance when deciding to roast the sulphide instead of reducing it directly with carbon.

Therefore, the factor that is not relevant for choosing roasting over direct carbon reduction is:

Option A which is: Metal sulphides are thermodynamically more stable than $$CS_2$$.

During the titration of oxalic acid with $$KMnO_4$$, a strongly acidic medium is needed because $$MnO_4^-$$ gets reduced smoothly only in the presence of excess $$H^+$$. Dilute $$H_2SO_4$$ is normally chosen because its anion ($$SO_4^{2-}$$) does not participate in any side redox reaction.

If, instead, $$HCl$$ is used to supply the required $$H^+$$ ions, chloride ions $$\left(Cl^- \right)$$ act as a reducing agent toward the powerful oxidiser $$MnO_4^-$$. The principal side reaction is

$$2 \, MnO_4^- + 10 \, Cl^- + 16 \, H^+ \rightarrow 2 \, Mn^{2+} + 5 \, Cl_2 + 8 \, H_2O$$

Thus, even before oxalic acid can be oxidised, a part of the permanganate is consumed (reduced from $$+7$$ to $$+2$$ oxidation state). Because the titrant is lost in this parasitic reaction, the observed end-point is reached prematurely, giving an erroneously low value for the oxalic-acid concentration.

Hence the titration becomes unsatisfactory in the presence of $$HCl$$ because $$Cl^-$$ ions reduce permanganate to $$Mn^{2+}$$.

Option C which is: reduces permanganate to $$Mn^{2+}$$

The term “oxidizing behaviour” means that the substance itself gets reduced by gaining electrons while causing some other species to lose electrons (get oxidized).
For $$H_2SO_4$$, the sulphur atom is in the $$+6$$ oxidation state. To act as an oxidizing agent, sulphur must come down to a lower oxidation state (e.g. $$+4, +2, 0, -2$$) during the reaction.

Let us inspect each option by writing the oxidation numbers before and after reaction.

Case A:

$$2HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2H_2O$$

• In $$HI$$, iodine is $$-1$$. In $$I_2$$, iodine is $$0$$. Therefore iodine is oxidized from $$-1$$ to $$0$$ (loss of electrons).
• In $$H_2SO_4$$, sulphur is $$+6$$. In $$SO_2$$, sulphur is $$+4$$. Therefore sulphur is reduced from $$+6$$ to $$+4$$ (gain of electrons).
Because $$H_2SO_4$$ gets reduced while oxidizing $$HI$$, this reaction illustrates the oxidizing behaviour of $$H_2SO_4$$.

Case B:

$$Ca(OH)_2 + H_2SO_4 \longrightarrow CaSO_4 + 2H_2O$$

This is a simple acid-base neutralization. Oxidation states of all elements remain unchanged. No redox process occurs, so oxidizing behaviour is absent.

Case C:

$$NaCl + H_2SO_4 \longrightarrow NaHSO_4 + HCl$$

This reaction is an acid-salt displacement. Chlorine stays at $$-1$$, sulphur stays at $$+6$$. Again, no redox change; $$H_2SO_4$$ is not acting as an oxidizing agent here.

Case D:

$$2PCl_5 + H_2SO_4 \longrightarrow 2POCl_3 + 2HCl + SO_2Cl_2$$

This is mainly a dehydrating/chlorinating reaction. Check sulphur: it is $$+6$$ in both $$H_2SO_4$$ and $$SO_2Cl_2$$, so no reduction of sulphur. Hence no oxidizing action is displayed.

Only in Case A does sulphur in $$H_2SO_4$$ undergo reduction and thereby oxidizes another species. Therefore the reaction in Option A is the one that depicts the oxidizing behaviour of sulphuric acid.

Final Answer: Option A which is: $$2HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2H_2O$$

According to the law of equivalence, 

The milliequivalents of the oxidant  = ($$K_2Cr_2O_7$$, $$n=6$$) 

must equal those of the reductant (Mohr's salt, $$n=1$$). 

Using the formula:  $$(M_1 \times n_1 \times V_1) = (M_2 \times n_2 \times V_2)$$, 

we get $$(x \times 10^{-3} \times 6) \times 200 = (0.6 \times 1) \times 750$$. 

This simplifies to $$1.2x = 450$$, 

which yields $$x = 375$$.

We need to find X + Y where X is the number of electrons involved in the oxidation of $$I^-$$ to $$I_2$$ and Y is for $$S^{2-}$$ to S, by acidified $$K_2Cr_2O_7$$.

The half-reaction for oxidation of $$I^-$$ to $$I_2$$ is $$2I^- \rightarrow I_2 + 2e^-$$, and the reduction half-reaction of dichromate under acidic conditions is $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$. Multiplying the oxidation half-reaction by 3 balances the electrons exchanged, giving $$6I^- \rightarrow 3I_2 + 6e^-$$. Thus, X = 6 electrons.

For the oxidation of $$S^{2-}$$ to S, the half-reaction is $$S^{2-} \rightarrow S + 2e^-$$, while the reduction half-reaction remains the same. To balance the electrons, the oxidation half-reaction is multiplied by 3, resulting in $$3S^{2-} \rightarrow 3S + 6e^-$$. Hence, Y = 6 electrons.

$$X + Y = 6 + 6 = 12$$

The answer is 12.

We need to identify products A, B, and C through a sequence of reactions, then find X + Y + Z.

Reaction 1: Identifying product A

$$NaCl + K_2Cr_2O_7 + H_2SO_4 \rightarrow A + KHSO_4 + NaHSO_4 + H_2O$$

This is the classic chromyl chloride test. When a chloride salt is heated with potassium dichromate and concentrated sulphuric acid, chromyl chloride is formed:

$$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 4NaHSO_4 + 2KHSO_4 + 3H_2O$$

Therefore, A = $$CrO_2Cl_2$$ (chromyl chloride, a deep red oily liquid).

Reaction 2: Identifying product B

$$A + NaOH \rightarrow B + NaCl + H_2O$$

When chromyl chloride reacts with sodium hydroxide:

$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

Therefore, B = $$Na_2CrO_4$$ (sodium chromate, a yellow compound).

Reaction 3: Identifying product C

$$B + H_2SO_4 + H_2O_2 \rightarrow C + Na_2SO_4 + H_2O$$

When sodium chromate reacts with sulphuric acid and hydrogen peroxide, it forms chromium pentoxide ($$CrO_5$$), which is a blue-coloured compound:

$$Na_2CrO_4 + H_2SO_4 + 2H_2O_2 \rightarrow CrO_5 + Na_2SO_4 + 3H_2O$$

Therefore, C = $$CrO_5$$ (also known as perchromic acid).

Structure of $$CrO_5$$:

$$CrO_5$$ has the structure $$CrO(O_2)_2$$, meaning it contains:

- 1 oxide ion ($$O^{2-}$$)

- 2 peroxide units ($$O_2^{2-}$$)

Finding X, Y, and Z:

X = number of $$O_2^{2-}$$ units = 2

Y = total number of oxygen atoms in $$CrO_5$$ = 5 (1 from oxide + 2 + 2 from two peroxide units)

Z = oxidation state of Cr in $$CrO_5$$:

Let the oxidation state of Cr be $$x$$. Each oxide contributes $$-2$$, each peroxide unit contributes $$-2$$:

$$x + (-2) + 2(-2) = 0$$

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

So Z = +6.

Final calculation:

$$X + Y + Z = 2 + 5 + 6 = 13$$

The answer is 13.

Permanganate ion $$\left(MnO_4^- \right)$$ is a strong oxidising agent. The species to which it gets reduced and the product obtained from the reducing agent (here $$I^-$$) depend on the pH of the solution.

Useful facts to remember:
  • In acidic medium: $$MnO_4^- \rightarrow Mn^{2+}$$ and $$I^- \rightarrow I_2$$.
  • In neutral (or weakly basic) medium: $$MnO_4^- \rightarrow MnO_2$$ and $$I^- \rightarrow IO_3^-$$.
  • In strongly alkaline medium: $$MnO_4^- \rightarrow MnO_4^{2-}$$ and $$I^- \rightarrow IO_4^-$$.

Because the question specifically mentions a neutral aqueous medium, we focus on the second line above.

Half-reactions in neutral solution:
Reduction half-reaction for permanganate:
$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$ $$-(1)$$
Oxidation half-reaction for iodide:
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$ $$-(2)$$

To combine $$-(1)$$ and $$-(2)$$, multiply $$-(1)$$ by 2 so that the electrons cancel (total 6 e$$^-$$):
$$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$ $$-(3)$$

Add $$-(2)$$ and $$-(3)$$:

$$2MnO_4^- + I^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$

Electrons cancel. Combine, then simplify water and $$H^+/OH^-$$ to obtain the overall balanced equation in neutral medium:

$$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$$

The iodine species formed is $$IO_3^-$$ (iodate ion).

Hence, in a neutral aqueous medium, the reaction of permanganate ion with iodide ion produces iodate ion.

Option B which is: IO$$_3^-$$

We need to find which oxidation reactions are performed by both $$K_2Cr_2O_7$$ and $$KMnO_4$$ in acidic medium.

A. $$I^- \rightarrow I_2$$:

Both $$K_2Cr_2O_7$$ and $$KMnO_4$$ are strong enough oxidizing agents to oxidize iodide to iodine. YES for both.

B. $$S^{2-} \rightarrow S$$:

Both oxidizing agents can oxidize sulfide ions to elemental sulfur in acidic medium. YES for both.

C. $$Fe^{2+} \rightarrow Fe^{3+}$$:

Both $$K_2Cr_2O_7$$ and $$KMnO_4$$ readily oxidize ferrous ions to ferric ions in acidic medium. This is a standard titration reaction for both. YES for both.

D. $$I^- \rightarrow IO_3^-$$:

Oxidizing $$I^-$$ all the way to $$IO_3^-$$ requires a very strong oxidizing agent under specific conditions. Neither $$K_2Cr_2O_7$$ nor $$KMnO_4$$ in normal acidic conditions oxidizes $$I^-$$ beyond $$I_2$$ to $$IO_3^-$$. NO for both.

E. $$S_2O_3^{2-} \rightarrow SO_4^{2-}$$:

While $$KMnO_4$$ can oxidize thiosulfate to sulfate, $$K_2Cr_2O_7$$ typically does not oxidize thiosulfate cleanly to sulfate. Not reliably both.

The common reactions are A, B, and C. The answer is Option D) A, B and C Only.

Nitrous acid, $$HNO_2$$, is an unstable, weak, monobasic acid in which nitrogen has the oxidation state $$+3$$.

In aqueous medium at room temperature it undergoes disproportionation (simultaneous oxidation and reduction of the same species).
To find the products, we balance two half-reactions for nitrogen:

Oxidation half-reaction (N : $$+3 \rightarrow +5$$)
$$HNO_2 + H_2O \rightarrow NO_3^- + 3\,H^+ + 2\,e^-$$

Reduction half-reaction (N : $$+3 \rightarrow +2$$)
$$HNO_2 + e^- \rightarrow NO + H_2O$$

Multiply the reduction half-reaction by $$2$$ so that the electron numbers match, then add:

$$\bigl(HNO_2 + H_2O \rightarrow NO_3^- + 3\,H^+ + 2\,e^- \bigr)$$
$$+$$
$$\bigl(2\,HNO_2 + 2\,e^- \rightarrow 2\,NO + 2\,H_2O \bigr)$$

After canceling the $$2\,e^-$$ on both sides and combining water molecules, we obtain

$$3\,HNO_2 \rightarrow NO_3^- + 2\,NO + H_2O + 3\,H^+$$

In water each $$H^+$$ immediately forms $$H_3O^+$$, so the observable species in the solution (and gas phase) are

$$H_3O^+,\; NO_3^- \text{ and } NO(g)$$

Thus the disproportionation of in-situ generated $$HNO_2$$ at room temperature gives Option A.

Answer - Option A: $$H_3O^+$$, $$NO_3^-$$ and $$NO$$

We need to balance the redox equation in basic medium: $$2MnO_4^- + bI^- + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^-$$.

Mn in $$MnO_4^-$$ is +7 and in $$MnO_2$$ is +4, so manganese is reduced by gaining three electrons. Iodine in $$I^-$$ is -1 and in $$I_2$$ is 0, so iodine is oxidized by losing one electron.

In basic medium the half-reactions are written as follows. The reduction half-reaction is $$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$ and the oxidation half-reaction is $$2I^- \rightarrow I_2 + 2e^-$$.

To equalize the electrons exchanged, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3, giving: $$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$ and $$6I^- \rightarrow 3I_2 + 6e^-$$.

Adding these half-reactions yields the balanced overall equation: $$2MnO_4^- + 6I^- + 4H_2O \rightarrow 3I_2 + 2MnO_2 + 8OH^-$$.

Verification confirms that manganese, iodine, oxygen, hydrogen, and charge are all balanced.

From the balanced equation: $$z = 8$$

Answer: 8

We need to find compounds with sulphur in +4 oxidation state:

- $$SO_3$$: S is +6

- $$H_2SO_3$$: $$2(+1) + x + 3(-2) = 0$$, so $$x = +4$$. Yes.

- $$SOCl_2$$: $$x + (-2) + 2(-1) = 0$$, so $$x = +4$$. Yes.

- $$SF_4$$: $$x + 4(-1) = 0$$, so $$x = +4$$. Yes.

- $$BaSO_4$$: $$(+2) + x + 4(-2) = 0$$, so $$x = +6$$.

- $$H_2S_2O_7$$: This is pyrosulfuric acid. $$2(+1) + 2x + 7(-2) = 0$$, so $$2x = 12$$, $$x = +6$$.

Compounds with +4 oxidation state of S: $$H_2SO_3$$, $$SOCl_2$$, $$SF_4$$ = 3 compounds.

The answer is $$\boxed{3}$$.

Total 12 mole of $$I$$ in RHS side of the reaction. 

$$2 + x = 12$$

$$x = 12 - 2$$

$$x = 10$$

Hence on the left hand side to balance it x must be 10.

H₂O₂ acts as a reducing agent when it gets oxidized (oxygen goes from -1 to 0 or higher oxidation state).

In H₂O₂, the oxidation state of oxygen is -1.

Option A: PbS + 4H₂O₂ → PbSO₄ + 4H₂O. Here O goes from -1 in H₂O₂ to -2 in H₂O. H₂O₂ is reduced → acts as oxidizing agent.

Option B: 2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH⁻. Here O goes from -1 to -2. H₂O₂ is reduced → acts as oxidizing agent.

Option C: HOCl + H₂O₂ → H₃O⁺ + Cl⁻ + O₂. Here O in H₂O₂ goes from -1 to 0 (in O₂). H₂O₂ is oxidized → acts as reducing agent.

Option D: Mn²⁺ + H₂O₂ → Mn⁴⁺ + 2OH⁻. Here O goes from -1 to -2. H₂O₂ is reduced → acts as oxidizing agent.

The correct answer is Option 3: HOCl + H₂O₂ → H₃O⁺ + Cl⁻ + O₂.

We need to determine how many electrons are gained during the reduction of the permanganate anion ($$MnO_4^-$$) in alkaline medium.

The product of reduction depends on the medium:

- In acidic medium: $$MnO_4^- \to Mn^{2+}$$ (gain of 5 electrons)

- In neutral/weakly alkaline medium: $$MnO_4^- \to MnO_2$$ (gain of 3 electrons)

- In strongly alkaline medium: $$MnO_4^- \to MnO_4^{2-}$$ (gain of 1 electron)

For the standard alkaline medium reduction, the product is $$MnO_2$$.

In $$MnO_4^-$$: Let the oxidation state of Mn be $$x$$. Then $$x + 4(-2) = -1$$, giving $$x = +7$$.

In $$MnO_2$$: Let the oxidation state of Mn be $$y$$. Then $$y + 2(-2) = 0$$, giving $$y = +4$$.

Change in oxidation state: $$+7 \to +4$$, a decrease of 3.

$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$

We can verify this is balanced: Mn: 1 on each side; O: 4+2=6 on left, 2+4=6 on right; H: 4 on left, 4 on right; Charge: -1-3=-4 on left, -4 on right.

The permanganate anion gains 3 electrons in alkaline medium.

Hypophosphoric acid has the formula $$H_4P_2O_6$$.

To find the oxidation state of phosphorus, let it be $$x$$:

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

$$ 4 + 2x - 12 = 0 $$

$$ 2x = 8 $$

$$ x = +4 $$

The oxidation state of phosphorus in hypophosphoric acid is +4.

We are given the following statements:

Assertion A: Permanganate titrations are not performed in the presence of hydrochloric acid.

Reason R: Chlorine is formed as a consequence of oxidation of hydrochloric acid.

Permanganate ($$KMnO_4$$) is a strong oxidising agent used in redox titrations. In acidic medium, it is typically used with dilute sulphuric acid ($$H_2SO_4$$) rather than hydrochloric acid ($$HCl$$).

The reason is that $$KMnO_4$$ is a strong enough oxidiser to oxidise $$Cl^-$$ ions from HCl to $$Cl_2$$ gas. The reaction is: $$2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$$.

This is problematic because the $$KMnO_4$$ gets consumed not only by the analyte (the substance being titrated, such as $$Fe^{2+}$$ or oxalate) but also by the $$HCl$$. This leads to an error in the titration — more $$KMnO_4$$ is used than what is needed for the analyte alone, giving incorrect results.

So Assertion A is true — permanganate titrations are indeed not performed with HCl. Reason R is also true — chlorine gas is indeed formed because HCl gets oxidised by the strong oxidising agent $$KMnO_4$$. Moreover, R correctly explains A, because it is precisely the oxidation of HCl to $$Cl_2$$ that causes the problem and is the reason why HCl is avoided.

Hence, the correct answer is Option A: Both A and R are true and R is the correct explanation of A.

The reaction of $$H_2O_2$$ with $$KMnO_4$$ in acidic medium is a well-known redox reaction.

The balanced equation is:

$$2KMnO_4 + 5H_2O_2 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O$$

Analysis of oxidation states:

- In $$KMnO_4$$, manganese is in the $$+7$$ oxidation state.

- In the product $$MnSO_4$$, manganese is in the $$+2$$ oxidation state.

In acidic medium, $$KMnO_4$$ acts as a strong oxidising agent and is reduced from $$Mn^{7+}$$ to $$Mn^{2+}$$. Meanwhile, $$H_2O_2$$ acts as a reducing agent and is oxidised to $$O_2$$.

Note: In neutral or weakly acidic medium, $$KMnO_4$$ is reduced to $$Mn^{4+}$$ (as $$MnO_2$$), but in strongly acidic medium, the reduction goes all the way to $$Mn^{2+}$$.

Therefore, the correct answer is Option A: $$Mn^{2+}$$.

We need to identify the reaction where H$$_2$$O$$_2$$ acts as a reducing agent in basic medium. When H$$_2$$O$$_2$$ acts as a reducing agent, it gets oxidized (oxygen goes from -1 oxidation state in H$$_2$$O$$_2$$ to 0 in O$$_2$$).

Analysis of each option:

Option A: HOCl + H$$_2$$O$$_2$$ $$\rightarrow$$ H$$_3$$O$$^+$$ + Cl$$^-$$ + O$$_2$$

H$$_2$$O$$_2$$ is oxidized to O$$_2$$ (reducing agent), but this reaction produces H$$_3$$O$$^+$$, indicating an acidic medium, not basic.

Option B: PbS + 4H$$_2$$O$$_2$$ $$\rightarrow$$ PbSO$$_4$$ + 4H$$_2$$O

Here, oxygen in H$$_2$$O$$_2$$ (-1) goes to O in H$$_2$$O (-2). So H$$_2$$O$$_2$$ is reduced and acts as an oxidizing agent, not a reducing agent.

Option C: 2MnO$$_4^-$$ + 3H$$_2$$O$$_2$$ $$\rightarrow$$ 2MnO$$_2$$ + 3O$$_2$$ + 2H$$_2$$O + 2OH$$^-$$

H$$_2$$O$$_2$$ is oxidized to O$$_2$$ (oxygen: -1 $$\rightarrow$$ 0), so it acts as a reducing agent. The presence of OH$$^-$$ in the products confirms basic medium. Also, Mn goes from +7 in MnO$$_4^-$$ to +4 in MnO$$_2$$, confirming MnO$$_4^-$$ is reduced.

Option D: Mn$$^{2+}$$ + H$$_2$$O$$_2$$ $$\rightarrow$$ Mn$$^{4+}$$ + 2OH$$^-$$

Here, Mn is oxidized from +2 to +4, and oxygen in H$$_2$$O$$_2$$ (-1) goes to OH$$^-$$ (-2). So H$$_2$$O$$_2$$ acts as an oxidizing agent.

Hence, the correct answer is Option C, where H$$_2$$O$$_2$$ acts as a reducing agent in basic medium.

We need to find the overall change in oxidation number of manganese when $$KMnO_4$$ reacts with oxalic acid in acidic medium.

The balanced reaction of $$KMnO_4$$ with oxalic acid ($$H_2C_2O_4$$) in acidic medium is:

$$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O$$

In $$KMnO_4$$:

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

$$x = +7$$

So, the oxidation state of Mn in $$KMnO_4$$ is +7.

The purple colour of $$KMnO_4$$ disappears, which means $$Mn^{+7}$$ is reduced. In acidic medium, $$KMnO_4$$ is reduced to $$Mn^{2+}$$ (present as $$MnSO_4$$).

So, the oxidation state of Mn in the product is +2.

Change in oxidation number = Initial oxidation state - Final oxidation state

$$= +7 - (+2) = 5$$

The overall change in the oxidation number of manganese is 5 (it decreases from +7 to +2).

Hence, the correct answer is Option A: $$5$$.

We are given that 20 mL of 0.02 M $$K_2Cr_2O_7$$ is used to titrate 10 mL of $$Fe^{2+}$$ solution in acidic medium.

The balanced redox reaction in acidic medium is:

$$K_2Cr_2O_7 + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 2K^+ + 7H_2O$$

From the stoichiometry, 1 mole of $$K_2Cr_2O_7$$ reacts with 6 moles of $$Fe^{2+}$$.

Moles of $$K_2Cr_2O_7$$ used:

$$n_{K_2Cr_2O_7} = M \times V = 0.02 \times \frac{20}{1000} = 4 \times 10^{-4} \text{ mol}$$

Moles of $$Fe^{2+}$$ that react:

$$n_{Fe^{2+}} = 6 \times n_{K_2Cr_2O_7} = 6 \times 4 \times 10^{-4} = 24 \times 10^{-4} \text{ mol}$$

Molarity of $$Fe^{2+}$$ solution:

$$M_{Fe^{2+}} = \frac{n_{Fe^{2+}}}{V} = \frac{24 \times 10^{-4}}{10/1000} = \frac{24 \times 10^{-4}}{10^{-2}} = 24 \times 10^{-2} \text{ M}$$

Therefore, the answer is $$\textbf{24}$$.

First, recall that hydrogen peroxide $$\mathrm{H_2O_2}$$ can behave both as an oxidising agent and as a reducing agent. In a basic medium, when it meets elemental iodine $$\mathrm{I_2}$$, it prefers to act as a reducing agent.

To understand what is formed, we analyse the redox process through the half-reaction method.

For hydrogen peroxide as a reductant, we write the oxidation half-reaction. The well-known relation is:

$$\mathrm{H_2O_2 \;\rightarrow\; O_2 + 2\,H^+ + 2\,e^-}$$

Because we are in basic (alkaline) medium, every $$\mathrm{H^+}$$ must be neutralised by adding an equal number of $$\mathrm{OH^-}$$ ions to both sides. Thus we add $$2\,\mathrm{OH^-}$$:

$$\mathrm{H_2O_2 + 2\,OH^- \;\rightarrow\; O_2 + 2\,H_2O + 2\,e^-}$$

Now we write the reduction half-reaction for iodine. Elemental iodine in its molecular form $$\mathrm{I_2}$$ accepts electrons to become iodide $$\mathrm{I^-}$$. The basic reduction relation is:

$$\mathrm{I_2 + 2\,e^- \;\rightarrow\; 2\,I^-}$$

Both half-reactions involve exactly $$2$$ electrons, so they already balance electrically and no scaling is needed. We now add them term by term:

$$\bigl(\mathrm{H_2O_2 + 2\,OH^- \;\rightarrow\; O_2 + 2\,H_2O + 2\,e^-}\bigr)$$

$$\bigl(\mathrm{I_2 + 2\,e^- \;\rightarrow\; 2\,I^-}\bigr)$$

Adding and cancelling the $$2\,e^-$$ on opposite sides, we obtain the overall ionic equation:

$$\mathrm{H_2O_2 + I_2 + 2\,OH^- \;\rightarrow\; 2\,I^- + 2\,H_2O + O_2}$$

The only iodine-containing species on the product side is $$\mathrm{I^-}$$, i.e. iodide ion.

Hence, the correct answer is Option C.

We start with the general molecular formula of polythionic acid: $$\mathrm{H_2S_xO_6},$$ where $$x = 3, 4, 5.$$

The experimental structures (deduced from salts, X-ray data and stepwise oxidation studies) show a straight chain of sulphur atoms. The two end sulphur atoms are each bonded to three oxygens and one sulphur, forming $$\text{-SO}_3\text{H}$$ groups. The remaining $$x-2$$ sulphur atoms are present only in the S-S chain and are not bonded to oxygen.

Whenever a sulphur atom is bonded to three oxygens in the $$\text{-SO}_3\text{H}$$ environment (exactly like in sulphurous acid fragments), its oxidation number is taken as an unknown $$+a.$$ Each “internal” sulphur that is attached only to neighbouring sulphur atoms carries no hetero-atoms, so its oxidation number is zero.

Hence, in $$\mathrm{H_2S_xO_6}$$ we have:

• 2 hydrogen atoms, each with oxidation number $$+1.$$

• 6 oxygen atoms, each with oxidation number $$-2.$$

• 2 terminal sulphur atoms with oxidation number $$+a$$ (to be found).

• $$(x-2)$$ internal sulphur atoms with oxidation number $$0.$$

Because the molecule is electrically neutral, the algebraic sum of all oxidation numbers must be zero. Writing this balance, we get:

$$2(+1) \;+\; 2(+a) \;+\; (x-2)(0) \;+\; 6(-2) \;=\; 0.$$

Simplifying step by step:

$$2 + 2a + 0 - 12 = 0,$$

$$2a - 10 = 0,$$

$$2a = 10,$$

$$a = +5.$$

Thus each terminal sulphur is in the $$+5$$ oxidation state, while every internal sulphur is in the $$0$$ oxidation state.

Therefore, the possible oxidation states of sulphur found in any polythionic acid $$\mathrm{H_2S_xO_6}$$ are $$0$$ and $$+5$$ only.

Hence, the correct answer is Option C.

First we recall the definition of a disproportionation reaction. In such a redox process the same element present in a single oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower oxidation state. Therefore, a necessary condition for disproportionation is that the element must be in an oxidation state which is intermediate between at least one higher and one lower stable oxidation state. Mathematically, if the element is presently in oxidation state $$x$$ while higher and lower attainable states are $$x_{\text{high}}$$ and $$x_{\text{low}},$$ then

$$x_{\text{low}} < x < x_{\text{high}}$$

must hold. If the element is already in either its maximum or minimum oxidation state, it cannot undergo disproportionation.

Now we examine every species in each option and calculate the oxidation state of the central atom to judge whether the species can disproportionate under suitable conditions.

Option A contains $$\mathrm{ClO_4^-},\; \mathrm{MnO_4^-},\; \mathrm{ClO_2^-}$$ and $$\mathrm{F_2}.$$

For $$\mathrm{ClO_4^-}$$ we have

$$\text{Let the oxidation state of Cl be } x.$$

Since the overall charge is $$-1$$ and each oxygen is $$-2,$$

$$x + 4(-2) = -1 \;\;\Longrightarrow\;\; x = +7.$$

$$+7$$ is the maximum attainable oxidation state of chlorine; no higher state exists. Hence $$\mathrm{ClO_4^-}$$ cannot disproportionate. Therefore the whole set in Option A fails the test.

Option B contains $$\mathrm{MnO_4^{2-}},\; \mathrm{ClO_2},\; \mathrm{Cl_2}$$ and $$\mathrm{Mn^{3+}}.$$ We examine them one by one.

1. For $$\mathrm{MnO_4^{2-}}$$ (manganate ion):

$$x + 4(-2) = -2 \;\;\Longrightarrow\;\; x = +6.$$

Manganese exhibits both higher $$+7$$ (in $$\mathrm{MnO_4^-}$$) and lower $$+4,\,+3,\,+2$$ states. The well-known reaction

$$3\,\mathrm{MnO_4^{2-}} + 2\,\mathrm{H_2O} \;\longrightarrow\; 2\,\mathrm{MnO_4^-} + \mathrm{MnO_2} + 4\,\mathrm{OH^-}$$

shows it disproportionating to $$+7$$ and $$+4$$ states. So $$\mathrm{MnO_4^{2-}}$$ qualifies.

2. For $$\mathrm{ClO_2}$$ (chlorine dioxide):

$$x + 2(-2) = 0 \;\;\Longrightarrow\;\; x = +4.$$

Chlorine has higher $$+5$$ (in $$\mathrm{ClO_3^-}$$) and lower $$+3$$ (in $$\mathrm{ClO_2^-}$$) states. Indeed, in alkaline medium

$$2\,\mathrm{ClO_2} + \mathrm{OH^-} \;\longrightarrow\; \mathrm{ClO_3^-} + \mathrm{ClO_2^-} + \mathrm{H^+}$$

illustrates its disproportionation. Hence $$\mathrm{ClO_2}$$ also fits.

3. For $$\mathrm{Cl_2}$$ (molecular chlorine):

The oxidation state of each chlorine atom is $$0.$$ It can increase to $$+1$$ (in $$\mathrm{ClO^-}$$) and decrease to $$-1$$ (in $$\mathrm{Cl^-}$$), so in basic solution

$$\mathrm{Cl_2} + 2\,\mathrm{OH^-} \;\longrightarrow\; \mathrm{ClO^-} + \mathrm{Cl^-} + \mathrm{H_2O}$$

is a textbook disproportionation. Thus $$\mathrm{Cl_2}$$ meets the criterion.

4. For $$\mathrm{Mn^{3+}}$$:

The oxidation state is $$+3.$$ Manganese can rise to $$+4$$ (in $$\mathrm{MnO_2}$$) and fall to $$+2$$ (in $$\mathrm{Mn^{2+}}$$). The reaction

$$2\,\mathrm{Mn^{3+}} \;\longrightarrow\; \mathrm{Mn^{2+}} + \mathrm{Mn^{4+}}$$

proves its disproportionation capability. So $$\mathrm{Mn^{3+}}$$ also satisfies the condition.

Therefore, every species in Option B can undergo a disproportionation reaction.

Option C includes $$\mathrm{Cr_2O_7^{2-}},\; \mathrm{MnO_4^-},\; \mathrm{ClO_2^-}$$ and $$\mathrm{Cl_2}.$$ For $$\mathrm{Cr_2O_7^{2-}}$$ each chromium is $$+6,$$ which is the maximum common state for chromium; chromium cannot be oxidised further, hence $$\mathrm{Cr_2O_7^{2-}}$$ cannot disproportionate. Thus Option C fails.

Option D again contains $$\mathrm{Cr_2O_7^{2-}},$$ already shown incapable of disproportionation, so this option also fails.

Only Option B features a set in which all the listed species are able to undergo disproportionation.

Hence, the correct answer is Option B.

We need to determine the role of $$H_2O_2$$ in both reactions.

In equation (A): $$HOCl + H_2O_2 \to H_3O^+ + Cl^- + O_2$$

The oxidation state of oxygen in $$H_2O_2$$ is $$-1$$. In the product $$O_2$$, the oxidation state of oxygen is $$0$$. Since the oxidation state increases from $$-1$$ to $$0$$, oxygen is oxidised. A species that gets oxidised acts as a reducing agent. So $$H_2O_2$$ is a reducing agent in reaction (A).

In equation (B): $$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$

Again, the oxidation state of oxygen in $$H_2O_2$$ is $$-1$$. In the product $$O_2$$, the oxidation state of oxygen is $$0$$. The oxidation state increases from $$-1$$ to $$0$$, so oxygen is oxidised. Therefore, $$H_2O_2$$ acts as a reducing agent in reaction (B) as well.

Hence, $$H_2O_2$$ acts as a reducing agent in both equations (A) and (B).

Hence, the correct answer is Option 4.

We have to find in which of the given redox conversions the change in the oxidation state of the central element (or of carbon, in the last case) is exactly five. The required principle is:

Sum of oxidation numbers of all atoms in a species $$= \text{overall charge of that species}.$$

Using this rule, we shall calculate the oxidation number of the relevant atom in both the reactant and the product for every option, then subtract to obtain the magnitude of the change.

Option A : $$Cr_2O_7^{2-} \rightarrow 2\,Cr^{3+}$$

Let the oxidation number of chromium in $$Cr_2O_7^{2-}$$ be $$x$$. There are two Cr atoms and seven oxygen atoms.

Oxygen almost always has $$-2$$ as its oxidation number. Hence

$$2x + 7(-2) = -2.$$

Simplifying, $$2x - 14 = -2 \; \Longrightarrow \; 2x = +12 \; \Longrightarrow \; x = +6.$$

Thus $$Cr$$ changes from $$+6$$ in $$Cr_2O_7^{2-}$$ to $$+3$$ in $$Cr^{3+}$$. Hence

$$\Delta = 6 - 3 = 3.$$

The change is three, not five.

Option B : $$MnO_4^{-} \rightarrow Mn^{2+}$$

Let the oxidation number of manganese in $$MnO_4^{-}$$ be $$x$$. Using the same rule,

$$x + 4(-2) = -1.$$

So $$x - 8 = -1 \; \Longrightarrow \; x = +7.$$

Manganese moves from $$+7$$ in $$MnO_4^-$$ to $$+2$$ in $$Mn^{2+}$$. Therefore

$$\Delta = 7 - 2 = 5.$$

Here the change is exactly five.

Option C : $$CrO_4^{2-} \rightarrow Cr^{3+}$$

Let the oxidation number of chromium in $$CrO_4^{2-}$$ be $$x$$. We write

$$x + 4(-2) = -2.$$

This gives $$x - 8 = -2 \; \Longrightarrow \; x = +6.$$

Going from $$+6$$ to $$+3$$ yields

$$\Delta = 6 - 3 = 3.$$

The change is three, so this option does not fit.

Option D : $$C_2O_4^{2-} \rightarrow 2\,CO_2$$

For the oxalate ion $$C_2O_4^{2-}$$, let the oxidation number of carbon be $$x$$ (both carbons are equivalent).

$$2x + 4(-2) = -2.$$

Thus $$2x - 8 = -2 \; \Longrightarrow \; 2x = +6 \; \Longrightarrow \; x = +3.$$

In $$CO_2$$, oxygen is $$-2$$, so for carbon $$y + 2(-2) = 0 \Longrightarrow y = +4.$$ Therefore carbon changes from $$+3$$ to $$+4$$:

$$\Delta = 4 - 3 = 1.$$

This change is only one.

Comparing all four possibilities, only Option B gives a change of five in oxidation state.

Hence, the correct answer is Option B.

When $$H_2O_2$$ acts as an oxidizing agent, it gets reduced (its oxygen goes from -1 oxidation state to -2). We need to identify the reaction where $$H_2O_2$$ oxidizes another species.

In option (1): $$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$. Here, iodide ($$I^-$$, oxidation state -1) is oxidized to $$I_2$$ (oxidation state 0), and $$H_2O_2$$ (oxygen in -1 state) is reduced to $$H_2O$$ (oxygen in -2 state). So $$H_2O_2$$ acts as an oxidizing agent.

In option (2): $$I_2 + H_2O_2 + 2OH^- \to 2I^- + 2H_2O + O_2$$. Here, $$H_2O_2$$ is oxidized to $$O_2$$ (oxygen goes from -1 to 0), so $$H_2O_2$$ acts as a reducing agent.

In option (3): $$Cl_2 + H_2O_2 \to 2HCl + O_2$$. Here, $$H_2O_2$$ is oxidized to $$O_2$$, so $$H_2O_2$$ acts as a reducing agent.

In option (4): $$KIO_4 + H_2O_2 \to KIO_3 + H_2O + O_2$$. Here too, $$H_2O_2$$ is oxidized to $$O_2$$, acting as a reducing agent.

Therefore, the equation depicting the oxidizing nature of $$H_2O_2$$ is option (1): $$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$.

First, we note that in an acidic medium the permanganate ion $$\text{MnO}_4^-$$ is reduced from the oxidation state $$+7$$ to $$+2$$. According to the change in oxidation number, each mole of $$\text{KMnO}_4$$ therefore gives up $$5$$ electrons, so its n-factor is $$5$$ in such titrations.

The molarity of the potassium permanganate solution is given as $$0.05\ \text{M}$$ and its volume used is $$10.0\ \text{mL}=0.010\ \text{L}$$. We convert this molarity into normality by multiplying by the n-factor:

$$N_1 = M_1 \times n\text{-factor} = 0.05 \times 5 = 0.25\ \text{N}.$$

The number of equivalents of $$\text{KMnO}_4$$ actually used is then

$$\text{Equivalents of }\text{KMnO}_4 = N_1 \times V_1 = 0.25\ \text{N} \times 0.010\ \text{L} = 0.0025\ \text{eq}.$$

During titration these equivalents must exactly equal the equivalents furnished by the oxalic acid dihydrate solution, because in any redox titration the two reactants exchange the same total number of electrons.

Hence for $$10.0\ \text{mL}=0.010\ \text{L}$$ of oxalic acid dihydrate solution we must also have

$$\text{Equivalents of }(\text{COOH})_2\cdot 2\text{H}_2\text{O}=0.0025\ \text{eq}.$$

Oxalic acid (or the oxalate ion that forms in acid medium) is oxidised from carbon oxidation state $$+3$$ to $$+4$$ in carbon dioxide, a change of $$+1$$ per carbon atom and hence $$+2$$ per molecule. Thus each mole of oxalic acid loses $$2$$ electrons, giving it an n-factor of $$2$$ in this reaction.

Its normality therefore is

$$N_2=\frac{\text{equivalents}}{\text{volume in L}}=\frac{0.0025}{0.010}=0.25\ \text{N}.$$

We convert this normality into molarity by dividing by the n-factor:

$$M_2=\frac{N_2}{n\text{-factor}}=\frac{0.25}{2}=0.125\ \text{M}.$$

Now we need the strength, i.e. the mass of oxalic acid dihydrate per litre. The molar mass is calculated as follows:

$$\text{Molar mass of }H_2C_2O_4 = 2(1)+2(12)+4(16)=90\ \text{g mol}^{-1},$$

and with two waters of crystallisation

$$2\text{H}_2\text{O}=2\bigl[2(1)+16\bigr]=36\ \text{g mol}^{-1}.$$

So

$$M_{\text{dihydrate}} = 90+36 = 126\ \text{g mol}^{-1}.$$

Strength (grams per litre) is

$$S = M_2 \times M_{\text{dihydrate}} = 0.125\ \text{mol L}^{-1}\times126\ \text{g mol}^{-1} = 15.75\ \text{g L}^{-1}.$$

To fit the required form $$\text{number}\times10^{-2}\ \text{g L}^{-1}$$ we write

$$15.75\ \text{g L}^{-1}=1575\times10^{-2}\ \text{g L}^{-1}.$$

Rounded to the nearest integer, the number is $$1575$$.

So, the answer is $$1575$$.

In basic medium, $$CrO_4^{2-}$$ oxidises $$S_2O_3^{2-}$$ to $$SO_4^{2-}$$, and $$CrO_4^{2-}$$ itself is reduced to $$Cr(OH)_4^-$$.

First, we determine the change in oxidation states. In $$CrO_4^{2-}$$, chromium is in the +6 oxidation state, and in $$Cr(OH)_4^-$$, chromium is in the +3 oxidation state. So each Cr atom gains 3 electrons.

In $$S_2O_3^{2-}$$, the average oxidation state of sulfur is +2, and in $$SO_4^{2-}$$, sulfur is in the +6 oxidation state. Each sulfur atom loses 4 electrons, and since there are 2 sulfur atoms per $$S_2O_3^{2-}$$, each $$S_2O_3^{2-}$$ ion loses a total of 8 electrons.

Using the equivalence relation (milliequivalents of oxidant = milliequivalents of reductant): $$n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2$$, where $$n_1 = 3$$ (n-factor of $$CrO_4^{2-}$$) and $$n_2 = 8$$ (n-factor of $$S_2O_3^{2-}$$).

Substituting: $$3 \times 0.154 \times V_1 = 8 \times 0.25 \times 40$$

$$0.462 \times V_1 = 80$$

$$V_1 = \frac{80}{0.462} = 173.16 \text{ mL} \approx 173 \text{ mL}$$

The volume of $$0.154$$ M $$CrO_4^{2-}$$ required is $$\mathbf{173}$$ mL.