JEE Redox Reactions Questions

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.

The preparation of potassium permanganate ($$KMnO_4$$) from $$MnO_2$$ involves two steps. First, $$MnO_2$$ is fused with $$KOH$$ in the presence of an oxidizing agent like $$KNO_3$$ (or atmospheric oxygen):

$$2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$$

In this step, $$Mn$$ is oxidized from +4 to +6 oxidation state, forming potassium manganate ($$K_2MnO_4$$), which is a green compound.

Then, $$K_2MnO_4$$ is oxidized to $$KMnO_4$$ (Mn goes from +6 to +7) by electrolytic oxidation or treatment with chlorine or ozone.

The product of the first step is $$K_2MnO_4$$.

The correct answer is Option D: $$K_2MnO_4$$.

We need to identify the species that does NOT undergo a disproportionation reaction among the given chlorine oxyanions.

A disproportionation reaction is one in which the same element is simultaneously oxidised and reduced. For a species to undergo disproportionation, the element must be in an intermediate oxidation state so that it can go both up and down in oxidation state.

- $$ClO^-$$ (hypochlorite): Cl is in +1 oxidation state

- $$ClO_2^-$$ (chlorite): Cl is in +3 oxidation state

- $$ClO_3^-$$ (chlorate): Cl is in +5 oxidation state

- $$ClO_4^-$$ (perchlorate): Cl is in +7 oxidation state

For disproportionation to occur, the element must be able to both increase and decrease its oxidation state. The range of chlorine oxidation states is from -1 (in $$Cl^-$$) to +7 (in $$ClO_4^-$$).

- $$ClO^-$$ (+1): Can go up (to +3, +5, +7) and down (to 0, -1). Can disproportionate. Example: $$3ClO^- \rightarrow ClO_3^- + 2Cl^-$$

- $$ClO_2^-$$ (+3): Can go up and down. Can disproportionate.

- $$ClO_3^-$$ (+5): Can go up and down. Can disproportionate.

- $$ClO_4^-$$ (+7): This is the highest oxidation state of chlorine. It cannot be oxidised further. Since it can only be reduced (not oxidised), it cannot undergo disproportionation.

The correct answer is Option (4): $$ClO_4^-$$.

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 $$10$$.

We need to find the volume of the sealed container of $$CO_2$$.

We start by writing the chemical reactions involved.

$$CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O$$

Next, the excess $$Ca(OH)_2$$ is neutralized by hydrochloric acid:

$$Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$$

Then we calculate the moles of excess $$Ca(OH)_2$$ from the titration data.

Moles of HCl = 0.1 M × 0.040 L = 0.004 mol

Moles of excess $$Ca(OH)_2$$ = $$\frac{0.004}{2} = 0.002$$ mol

The problem states that all of the $$CO_2$$ reacted with exactly half of the initial amount of $$Ca(OH)_2$$.

If the initial $$Ca(OH)_2$$ is N mol, then the reacted portion is N/2 mol and the excess is also N/2 mol.

Therefore, $$N/2 = 0.002$$ mol, so $$N = 0.004$$ mol.

It follows that the moles of $$CO_2$$ are equal to the moles of $$Ca(OH)_2$$ that reacted, which is N/2 = 0.002 mol.

Finally, applying the ideal gas law at STP (273 K, 1 atm):

$$V = n \times 22400 \text{ cm}^3/\text{mol} = 0.002 \times 22400 = 44.8 \text{ cm}^3$$

Rounding to the nearest integer gives $$x \approx 45$$ cm³.

Therefore, the volume of the sealed container of $$CO_2$$ is 45 cm³.

We need to identify which species cannot function as an oxidising agent.

Key Concept: An oxidising agent accepts electrons (gets reduced). For this, the element must have the ability to decrease its oxidation state.

$$N^{3-}$$: Nitrogen is in its lowest possible oxidation state (-3). Since it cannot be reduced further (there is no lower oxidation state for nitrogen), it cannot act as an oxidising agent. It can only act as a reducing agent (losing electrons to go to higher oxidation states).

$$SO_4^{2-}$$: Sulphur is in +6 state. It can be reduced (e.g., to $$SO_2$$, $$S$$, or $$H_2S$$). It can act as an oxidising agent, especially in concentrated $$H_2SO_4$$.

$$BrO_3^-$$: Bromine is in +5 state. It can be reduced to $$Br^-$$ (0 or -1). It is a known oxidising agent.

$$MnO_4^-$$: Manganese is in +7 state. It is one of the strongest oxidising agents, capable of being reduced to $$Mn^{2+}$$, $$MnO_2$$, or $$MnO_4^{2-}$$.

The correct answer is Option 1: $$N^{3-}$$.

We need to find the pair where both central atoms exhibit $$sp^2$$ hybridization.

$$BF_3$$: B has 3 bond pairs, 0 lone pairs. Hybridization = $$sp^2$$. ✓

$$NO_2^-$$: N has 2 bond pairs + 1 lone pair = 3 electron groups. Hybridization = $$sp^2$$. ✓

$$H_2O$$: O has 2 bond pairs + 2 lone pairs = 4 electron groups. Hybridization = $$sp^3$$. ✗

$$NO_2$$: N has 2 bond pairs + 1 unpaired electron = 3 regions. Hybridization = $$sp^2$$. ✓

$$NH_2^-$$: N has 2 bond pairs + 2 lone pairs = 4 electron groups. Hybridization = $$sp^3$$. ✗

The pair where both are $$sp^2$$ is $$BF_3$$ and $$NO_2^-$$.

The correct answer is Option 2: $$BF_3$$ and $$NO_2^-$$.

We need to determine the mathematical representation of an antibonding sigma molecular orbital ($$\sigma^*$$).

Recall the LCAO (Linear Combination of Atomic Orbitals) method

When two atomic orbitals $$\psi_A$$ and $$\psi_B$$ combine, they form two molecular orbitals:

- Bonding MO ($$\sigma$$): formed by constructive (in-phase) combination

- Antibonding MO ($$\sigma^*$$): formed by destructive (out-of-phase) combination

Write the mathematical expressions

The bonding molecular orbital is formed by the addition of the two wave functions:

$$\sigma = \psi_A + \psi_B$$

(with a normalisation constant, which is omitted here for simplicity)

The antibonding molecular orbital is formed by the subtraction of the two wave functions:

$$\sigma^* = \psi_A - \psi_B$$

Understand the physical meaning

In the bonding MO, the wave functions reinforce each other in the internuclear region, leading to increased electron density between the nuclei and a stable bond. In the antibonding MO ($$\sigma^*$$), the wave functions cancel in the internuclear region, creating a node (region of zero electron density) between the nuclei. This results in a destabilising effect that weakens bonding.

The correct answer is Option (2): $$\psi_A - \psi_B$$.

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 evaluate two statements about disproportionation reactions.

A disproportionation reaction is one where the same element is simultaneously oxidized and reduced — it goes to both a higher and a lower oxidation state.

Statement I asserts that $$S_8$$ solid undergoes a disproportionation reaction under alkaline conditions to form $$S^{2-}$$ and $$S_2O_3^{2-}$$. In elemental sulfur $$S_8$$, sulfur has oxidation state 0. In $$S^{2-}$$, its oxidation state is $$-2$$ (reduction), and in $$S_2O_3^{2-}$$ (thiosulfate), the average oxidation state of sulfur is $$+2$$ (oxidation). Since sulfur in state 0 simultaneously goes to $$-2$$ and $$+2$$, this is indeed a disproportionation reaction. The reaction with NaOH can be written as:

$$S_8 + 12\,\text{NaOH} \rightarrow 4\,\text{Na}_2\text{S} + 2\,\text{Na}_2\text{S}_2\text{O}_3 + 6\,\text{H}_2\text{O}$$

Thus, Statement I is correct.

Statement II claims that $$ClO_4^-$$ can undergo disproportionation under acidic conditions. In $$ClO_4^-$$ (perchlorate), chlorine is in the $$+7$$ oxidation state, which is the highest possible for chlorine. Disproportionation requires simultaneous oxidation and reduction, but chlorine at $$+7$$ cannot be further oxidized. Therefore, $$ClO_4^-$$ cannot undergo disproportionation, making Statement II incorrect.

The correct answer is Option A: Statement I is correct but Statement II is incorrect.

Identify which reactions are disproportionation reactions.

A disproportionation reaction is one where the same element in a single oxidation state is simultaneously oxidised and reduced to two different oxidation states.

Reaction (1): $$Cu^+ \rightarrow Cu^{2+} + Cu$$

$$Cu^+$$ (+1) is oxidised to $$Cu^{2+}$$ (+2) and reduced to $$Cu$$ (0). The same element (Cu) in the same oxidation state (+1) undergoes both oxidation and reduction. This IS a disproportionation reaction.

Reaction (2): $$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

Mn in $$MnO_4^{2-}$$ is in +6 state. It goes to +7 in $$MnO_4^-$$ (oxidation) and +4 in $$MnO_2$$ (reduction). Same element, same initial state, both oxidised and reduced. This IS a disproportionation reaction.

Reaction (3): $$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$$

Mn goes from +7 (in $$KMnO_4$$) to +6 (in $$K_2MnO_4$$) and +4 (in $$MnO_2$$). Both are reductions. The oxygen goes from -2 to 0 (oxidation). Here Mn is only reduced (to two different states), while O is oxidised. This is NOT disproportionation of a single element in one state.

Reaction (4): $$2MnO_4^- + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^+$$

Mn starts in two different oxidation states (+7 and +2) and goes to one state (+4). This is a comproportionation (reverse of disproportionation), NOT a disproportionation reaction.

The correct answer is Option A: 1, 2.

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 $$Cr^{3+}$$.

Ethylene ($$C_2H_4$$) has the structure: $$H_2C=CH_2$$.

Counting sigma bonds:

- C-C bond: 1 sigma bond (part of the double bond)

- C-H bonds: 4 sigma bonds (2 on each carbon)

Total sigma bonds = 5.

Counting pi bonds:

- C=C double bond contains 1 pi bond

Total pi bonds = 1.

The correct answer is Option (4): 5 and 1.

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 evaluate two statements about hybridisation of certain molecules and complex ions.

Analysis of Statement I: "PF$$_5$$ and BrF$$_5$$ both exhibit sp$$^3$$d hybridisation."

PF$$_5$$: P has 5 valence electrons and forms 5 bonds with F atoms. There are 5 bond pairs and 0 lone pairs, giving sp$$^3$$d hybridisation and a trigonal bipyramidal geometry. ✓

BrF$$_5$$: Br has 7 valence electrons. With 5 F atoms, there are 5 bond pairs and 1 lone pair = 6 electron pairs total. This requires sp$$^3$$d$$^2$$ hybridisation (not sp$$^3$$d), giving a square pyramidal geometry.

Since BrF$$_5$$ is sp$$^3$$d$$^2$$, not sp$$^3$$d, Statement I is FALSE.

Analysis of Statement II: "Both SF$$_6$$ and [Co(NH$$_3$$)$$_6$$]$$^{3+}$$ exhibit sp$$^3$$d$$^2$$ hybridisation."

SF$$_6$$: S has 6 valence electrons and forms 6 bonds with F atoms. There are 6 bond pairs and 0 lone pairs. This gives sp$$^3$$d$$^2$$ hybridisation and an octahedral geometry. ✓

[Co(NH$$_3$$)$$_6$$]$$^{3+}$$: Co$$^{3+}$$ has the configuration [Ar]3d$$^6$$. With NH$$_3$$ being a strong field ligand, the d-electrons pair up to occupy only 3 of the 5 d-orbitals, leaving 2 inner d-orbitals available for bonding. The hybridisation is d$$^2$$sp$$^3$$ (inner orbital complex), not sp$$^3$$d$$^2$$ (which would use outer d-orbitals).

Since [Co(NH$$_3$$)$$_6$$]$$^{3+}$$ uses d$$^2$$sp$$^3$$ (inner orbital) hybridisation, not sp$$^3$$d$$^2$$, Statement II is FALSE.

The correct answer is Option (4): Both Statement I and Statement II are false.

1. Complete Octet (8 Electrons)

These central atoms follow the standard rule by sharing electrons until they reach a stable count of 8.

• $$\text{CO}_2$$ (Carbon): Carbon has 4 valence electrons and forms two double bonds with Oxygen. $$4 + 4 = 8$$ electrons.
• $$\text{CH}_4$$ (Carbon): Carbon forms 4 single bonds with Hydrogen. $$4 \times 2 = 8$$ electrons.
• $$\text{SiF}_4$$ (Silicon): Like Carbon, Silicon is in Group 14 and forms 4 single bonds. $$4 \times 2 = 8$$ electrons.
• $$\text{C}_2\text{H}_6$$ (Carbons): Each Carbon forms 3 bonds with Hydrogen and 1 bond with the other Carbon. $$4 \times 2 = 8$$ electrons.
• $$\text{CHCl}_3$$ & $$\text{CBr}_4$$ (Carbon): Carbon forms 4 single bonds with halogens. $$4 \times 2 = 8$$ electrons.

2. Exceptions: Incomplete Octet (< 8 Electrons)

These central atoms are stable even though they lack a full shell of 8 electrons.

• $$\text{BF}_3$$ (Boron): Boron is in Group 13. It shares its 3 valence electrons to form 3 single bonds.
  • Total: 6 electrons.
• $$\text{BeF}_2$$ (Beryllium): Beryllium is in Group 2. It shares its 2 valence electrons to form 2 single bonds.
  • Total: 4 electrons.

3. Exceptions: Expanded Octet (> 8 Electrons)

Elements in Period 3 or below have empty $$d$$-orbitals that allow them to hold more than 8 electrons.

• $$\text{H}_2\text{SO}_4$$ (Sulfur): Sulfur forms 6 bonds (two double bonds with Oxygen, two single bonds with -OH).
  • Total: 12 electrons.
• $$\text{PCl}_5$$ (Phosphorus): Phosphorus forms 5 single bonds with Chlorine.
  • Total: 10 electrons.
• $$\text{ClO}_2$$ (Chlorine): Chlorine is bonded to two oxygens and retains lone pairs/unpaired electrons. In many Lewis representations, it exceeds 8 to reduce formal charge.

4. Exceptions: Odd-Electron Molecules

• $$\text{NO}_2$$ (Nitrogen): Nitrogen has 5 valence electrons. Because 5 is an odd number, it is impossible to pair every electron.
  • Total: Nitrogen ends up with 7 valence electrons (one unpaired electron). This makes it a free radical.

We need to find the number of molecules with bond order 2 from: $$C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$$.

Using Molecular Orbital Theory, the bond order is calculated as $$\text{Bond Order} = \frac{N_b - N_a}{2}$$ where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of antibonding electrons.

Since $$C_2$$ (12 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4$$, the bond order is $$(8-4)/2 = 2$$. ✓

Now $$O_2$$ (16 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$$, so the bond order is $$(10-6)/2 = 2$$. ✓

For $$Be_2$$ (8 electrons), the bond order is $$(4-4)/2 = 0$$. ✗

For $$Li_2$$ (6 electrons), the bond order is $$(4-2)/2 = 1$$. ✗

For $$Ne_2$$ (20 electrons), the bond order is $$(10-10)/2 = 0$$. ✗

For $$N_2$$ (14 electrons), the bond order is $$(10-4)/2 = 3$$. ✗

For $$He_2$$ (4 electrons), the bond order is $$(2-2)/2 = 0$$. ✗

Therefore the molecules with bond order 2 are $$C_2$$ and $$O_2$$.

This gives the answer $$\boxed{2}$$.

We need to find the total number of species from the given list that have exactly one unpaired electron. We use Molecular Orbital Theory (MOT) to determine the electronic configurations.

Recall: The molecular orbital filling order for diatomic molecules is:

For $$Z \leq 7$$ (like N, C): $$\sigma_{1s}, \sigma^*_{1s}, \sigma_{2s}, \sigma^*_{2s}, \pi_{2p_x} = \pi_{2p_y}, \sigma_{2p_z}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$

For $$Z > 7$$ (like O): $$\sigma_{1s}, \sigma^*_{1s}, \sigma_{2s}, \sigma^*_{2s}, \sigma_{2p_z}, \pi_{2p_x} = \pi_{2p_y}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$

1. $$N_2$$ (14 electrons):

Configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p_z})^2$$

All electrons are paired. Unpaired electrons = 0. Not selected.

2. $$O_2$$ (16 electrons):

Configuration: ...$$(\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$$

The two electrons in $$\pi^*_{2p}$$ occupy one each of $$\pi^*_{2p_x}$$ and $$\pi^*_{2p_y}$$ by Hund's rule. Unpaired electrons = 2. Not selected.

3. $$C_2^-$$ (13 electrons):

$$C_2$$ has 12 electrons; adding 1 gives 13. Configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p_z})^1$$

Unpaired electrons = 1. Selected.

4. $$O_2^-$$ (17 electrons):

$$O_2$$ has 16 electrons; adding 1 gives 17. Configuration: ...$$(\pi^*_{2p})^3$$

Three electrons in two degenerate $$\pi^*$$ orbitals: 2 in one, 1 in the other. Unpaired electrons = 1. Selected.

5. $$O_2^{2-}$$ (18 electrons):

$$O_2$$ has 16 electrons; adding 2 gives 18. Configuration: ...$$(\pi^*_{2p})^4$$

Both $$\pi^*$$ orbitals are fully filled. Unpaired electrons = 0. Not selected.

6. $$H_2^+$$ (1 electron):

Configuration: $$(\sigma_{1s})^1$$

One electron in the bonding orbital. Unpaired electrons = 1. Selected.

7. $$CN^-$$ (14 electrons):

C has 6 electrons, N has 7, plus 1 for the negative charge = 14 electrons. This is isoelectronic with $$N_2$$.

Configuration is the same as $$N_2$$: all electrons paired. Unpaired electrons = 0. Not selected.

8. $$He_2^+$$ (3 electrons):

Two He atoms contribute 4 electrons; removing 1 gives 3. Configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^1$$

Unpaired electrons = 1. Selected.

Summary: Species with exactly 1 unpaired electron: $$C_2^-$$, $$O_2^-$$, $$H_2^+$$, $$He_2^+$$.

The total count is $$\boxed{4}$$.

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,

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

In $$NO_2^-$$ (nitrite ion), nitrogen is the central atom.

Valence electrons: N has 5, each O has 6, plus 1 for the negative charge. Total = 5 + 6(2) + 1 = 18 valence electrons.

Lewis structure of $$NO_2^-$$:

N is bonded to 2 oxygen atoms with one double bond and one single bond (or resonance structures with 1.5 bond order each). Nitrogen has one lone pair.

Valence electrons around nitrogen:

- 2 bonding pairs (from double bond) + 1 bonding pair (from single bond) = 3 bond pairs = 6 electrons from bonds

- 1 lone pair = 2 electrons

Total = 8 electrons around nitrogen.

The answer is 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 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 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.

We need to count the species with the central atom using sp$$^2$$ hybrid orbitals from the given list.

sp$$^2$$ hybridisation occurs when the central atom has 3 electron domains (regions of electron density), giving a trigonal planar arrangement (or bent if lone pairs are present).

Analyze each species:

(i) $$NH_3$$: N has 3 bond pairs + 1 lone pair = 4 electron domains → sp$$^3$$. ✗

(ii) $$SO_2$$: S has 2 bond pairs + 1 lone pair = 3 electron domains → sp$$^2$$. ✓

(iii) $$SiO_2$$: Each Si is bonded to 4 O atoms in a tetrahedral arrangement → sp$$^3$$. ✗

(iv) $$BeCl_2$$: Be has 2 bond pairs + 0 lone pairs = 2 electron domains → sp. ✗

(v) $$C_2H_4$$: Each C has 3 electron domains (2 C-H + 1 C=C) → sp$$^2$$. ✓

(vi) $$C_2H_2$$: Each C has 2 electron domains (1 C-H + 1 C≡C) → sp. ✗

(vii) $$BCl_3$$: B has 3 bond pairs + 0 lone pairs = 3 electron domains → sp$$^2$$. ✓

(viii) $$HCHO$$: C has 3 electron domains (2 C-H + 1 C=O) → sp$$^2$$. ✓

(ix) $$C_6H_6$$: Each C has 3 electron domains → sp$$^2$$. ✓

(x) $$BF_3$$: B has 3 bond pairs + 0 lone pairs → sp$$^2$$. ✓

(xi) $$C_2H_4Cl_2$$: Each C has 4 electron domains → sp$$^3$$. ✗

Count: SO$$_2$$, C$$_2$$H$$_4$$, BCl$$_3$$, HCHO, C$$_6$$H$$_6$$, BF$$_3$$ = 6 species.

The answer is 6.

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 $$\boxed{3}$$.

Determine what "25 volume" hydrogen peroxide means in terms of mass of $$H_2O_2$$ per litre.

"25 volume" $$H_2O_2$$ means 1 L of this solution releases 25 L of $$O_2$$ at STP upon complete decomposition.

The decomposition reaction is $$2H_2O_2 \rightarrow 2H_2O + O_2$$.

Since 1 mole of $$H_2O_2$$ produces $$\frac{1}{2}$$ mole of $$O_2$$, which corresponds to 11.2 L at STP, the volume strength is $$11.2 \times M$$ where M is the molarity of $$H_2O_2$$. Equating this to 25 gives $$25 = 11.2 \times M$$, so $$M = \frac{25}{11.2} = 2.232$$ mol/L.

The molar mass of $$H_2O_2$$ is 34 g/mol, so the mass per litre is $$2.232 \times 34 \approx 75.9$$ g, which is approximately 75 g.

Comparing with the given options, 250 g/L (Option A) is too high; 75 g/L (Option B) matches our calculation $$\checkmark$$; 25 g per 100 mL = 250 g/L (Option C) is too high; and 25 g/L (Option D) is too low. Thus the correct answer is Option B.

We have the reaction:

$$2Au(CN)_2^-(aq) + Zn(s) \rightarrow 2Au(s) + Zn(CN)_4^{2-}(aq)$$

In $$Au(CN)_2^-$$, Au is in the +1 oxidation state, while in the product Au(s), Au is in the 0 oxidation state — so Au is reduced. Meanwhile, Zn goes from 0 (in Zn) to +2 (in $$Zn(CN)_4^{2-}$$), so Zn is oxidized. Since both oxidation and reduction occur, this is a redox reaction.

Now, Zn (a more reactive metal) displaces Au (a less reactive metal) from its complex ion, so this is also a displacement reaction.

This is not a decomposition reaction because no compound is breaking down into simpler substances on its own. It is also not a combination reaction because no two substances are combining to form a single product.

Hence, the correct answer is Option A and B only.

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.

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.

Assertion (A): An aqueous solution of KOH used in volumetric analysis should be standardized or its concentration should be checked before use.

Reason (R): KOH solution absorbs atmospheric $$CO_2$$ on standing.

The reason statement is correct because potassium hydroxide reacts readily with carbon dioxide present in air.

$$2KOH+CO_2\rightarrow K_2CO_3+H_2O$$

As a result, part of the KOH is converted into potassium carbonate, causing the concentration of the KOH solution to change with time.

The assertion is also correct. KOH is not a primary standard because it absorbs both moisture and $$CO_2$$ from the atmosphere. Therefore, the concentration calculated during preparation does not remain constant for long periods.

Before using KOH solution in volumetric analysis, its exact concentration must be determined by standardization against a suitable primary standard.

Thus, the concentration of KOH solution is checked before use because it absorbs atmospheric $$CO_2$$ and undergoes the reaction

$$2KOH+CO_2\rightarrow K_2CO_3+H_2O$$

Therefore, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation of Assertion (A).

$$H_2O_2$$ acts as a reducing agent when it is oxidized, i.e., when the oxygen in $$H_2O_2$$ (oxidation state $$-1$$) is converted to a higher oxidation state.

Examining the reactions:

Option 1: $$2NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_2$$

Here, oxygen in $$H_2O_2$$ (oxidation state $$-1$$) is converted to $$O_2$$ (oxidation state $$0$$). The oxygen is oxidized, so $$H_2O_2$$ acts as a reducing agent.

In the other options, $$H_2O_2$$ acts as an oxidizing agent (oxygen goes from $$-1$$ to $$-2$$).

We need to identify the starting material for the convenient laboratory preparation of deuterated hydrogen peroxide (D$$_2$$O$$_2$$).

Analysis of each option:

Option A: K$$_2$$S$$_2$$O$$_8$$ (Potassium persulfate)

The hydrolysis of potassium persulfate with heavy water (D$$_2$$O) gives D$$_2$$O$$_2$$:

$$ \text{K}_2\text{S}_2\text{O}_8 + 2\text{D}_2\text{O} \to 2\text{KDSO}_4 + \text{D}_2\text{O}_2 $$

In this method, deuterium comes from D$$_2$$O (heavy water), which is readily available. This is the most convenient laboratory method for preparing D$$_2$$O$$_2$$.

Option B: 2-ethylanthraquinol

This is used in the industrial anthraquinone auto-oxidation process for large-scale H$$_2$$O$$_2$$ production. It requires catalytic hydrogenation using D$$_2$$ gas, which is not practical for routine laboratory preparation of D$$_2$$O$$_2$$.

Option C: BaO$$_2$$ (Barium peroxide)

The Merck process uses BaO$$_2$$ with dilute acid: BaO$$_2$$ + H$$_2$$SO$$_4$$ → BaSO$$_4$$ + H$$_2$$O$$_2$$. For D$$_2$$O$$_2$$, one would need D$$_2$$SO$$_4$$ (deuterated sulfuric acid), which is expensive and not conveniently available. This makes it impractical for D$$_2$$O$$_2$$ preparation.

Option D: BaO (Barium oxide)

Barium oxide is a basic oxide, not a peroxide. It cannot directly produce H$$_2$$O$$_2$$ or D$$_2$$O$$_2$$. The reaction BaO + H$$_2$$O → Ba(OH)$$_2$$ only gives a hydroxide.

The correct answer is Option A: K$$_2$$S$$_2$$O$$_8$$.

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.

We are given the reaction:

$$ClO \cdot + NO_2 \rightarrow X \overset{H_2O}{\rightarrow} Y + Z$$

Step 1: Finding X

The chlorine monoxide radical ($$ClO \cdot$$) reacts with nitrogen dioxide ($$NO_2$$). The chlorine monoxide radical bonds to $$NO_2$$ through its oxygen, forming chlorine nitrate:

$$ClO \cdot + NO_2 \rightarrow ClONO_2$$

So $$X = ClONO_2$$ (chlorine nitrate).

This is an important reaction in atmospheric chemistry. Chlorine nitrate acts as a temporary reservoir for both chlorine and nitrogen oxide species in the stratosphere.

Step 2: Hydrolysis of X to find Y and Z

When chlorine nitrate ($$ClONO_2$$) reacts with water, the $$Cl-O$$ bond and the $$O-NO_2$$ bond break. The hydrolysis reaction is:

$$ClONO_2 + H_2O \rightarrow HOCl + HNO_3$$

So $$Y = HOCl$$ (hypochlorous acid) and $$Z = HNO_3$$ (nitric acid).

Verification:

- $$X = ClONO_2$$ (chlorine nitrate) ✓

- $$Y = HOCl$$ (hypochlorous acid) ✓

- $$Z = HNO_3$$ (nitric acid) ✓

This matches Option C: X = $$ClONO_2$$, Y = $$HOCl$$, Z = $$HNO_3$$.

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.

Assertion A: Sharp change in slope at ~1120°C in the Mg → MgO Ellingham line — True. This occurs near the boiling point of Mg (1091°C).

Reason R: Large entropy change associated with change of state — True. When Mg boils, the entropy increases sharply, causing the slope change in the Ellingham diagram.

Both are true and R correctly explains A.

Since,
$$\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}$$

For the forward reaction:

Given:

• Pre-exponential factor for the forward reaction, $$A_f = 10^{15}\text{ s}^{-1} \implies \log A_f = 15$$.
• From the given graph, at $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ (which corresponds to $$T = 500\text{ K}$$), the value of $$\log k_f = 9$$.

$$9 = 15 - \frac{E_{a,f}}{2.303 R} \times 0.002$$
$$\frac{E_{a,f}}{2.303 R} \times 0.002 = 15 - 9 = 6$$
$$\frac{E_{a,f}}{2.303 R} = \frac{6}{0.002} = 3000\text{ K}$$

The relationship between the equilibrium constant ($K$) and the forward and backward rate constants ($k_f$ and $$k_b$$) is:

$$K = \frac{k_f}{k_b} \implies \log K = \log k_f - \log k_b$$

At $$T = 500\text{ K}$$:

• $$\log k_f = 9$$ (from the graph)
• $$\log K = 6$$ (given)

Substituting these values:

$$6 = 9 - \log k_b$$
$$\log k_b = 3 \quad (\text{at } 500\text{ K})$$

For the backward reaction:

Given:

• Pre-exponential factor for the backward reaction, $$A_b = 10^{11}\text{ s}^{-1} \implies \log A_b = 11$$.
• At $$T = 500\text{ K}$$, $$\frac{1}{T} = 0.002\text{ K}^{-1}$$ and $$\log k_b = 3$$.

Substituting these values:

$$3 = 11 - \frac{E_{a,b}}{2.303 R} \times 0.002$$
$$\frac{E_{a,b}}{2.303 R} \times 0.002 = 11 - 3 = 8$$
$$\frac{E_{a,b}}{2.303 R} = \frac{8}{0.002} = 4000\text{ K}$$

Now, Calculate $$|\log k_b|$$ at $$250\text{ K}$$

At $$T = 250\text{ K}$$, the value of $$\frac{1}{T}$$ is:

$$\frac{1}{T} = \frac{1}{250} = 0.004\text{ K}^{-1}$$

Using the Arrhenius equation for the backward reaction at $$250\text{ K}$$:

$$\log k_b = 11 - 4000 \times 0.004$$
$$\log k_b = 11 - 16 = -5$$

Taking the absolute value:

$$|\log k_b| = |-5| = 5$$

Osmotic pressure is given by $$\pi=\dfrac{n_{\text{total}}RT}{V_{\text{solution}}}$$, where
$$n_{\text{total}}$$ = total number of moles of solute after mixing,
$$R = 62\;{\text{L Torr K}}^{-1}{\text{mol}}^{-1}$$ (given),
$$T = 300\;\text{K}$$ (given),
$$V_{\text{solution}}$$ = final volume of the mixed solution in litres.

Step 1: Moles of urea in the 50 mL, 0.2 molal solution

Molality definition: $$m = \dfrac{n}{\text{mass of solvent (kg)}}$$

Let the mass of solvent in this solution be $$w\; \text{g}$$.
Then $$n = m \times \dfrac{w}{1000} = 0.2 \times \dfrac{w}{1000}=0.0002w$$ mol.

Mass of solute $$= n \times M = 0.0002w \times 60 = 0.012w \;\text{g}$$.

Total mass of this solution $$= w + 0.012w = 1.012w \;\text{g}$$.

Using the density (1.012 g mL$$^{-1}$$):
$$\text{mass} = \text{density} \times \text{volume} = 1.012 \times 50 = 50.6 \;\text{g}$$.

So $$1.012w = 50.6 \;\Rightarrow\; w = 50.0\;\text{g}$$.

Therefore, moles of urea in this portion
$$n_1 = 0.0002 \times 50 = 0.01\;\text{mol}$$.

Step 2: Moles of urea in the 250 mL second solution

Given mass of urea = 0.06 g.
$$n_2 = \dfrac{0.06}{60} = 0.001\;\text{mol}$$.

Step 3: Total moles after mixing

$$n_{\text{total}} = n_1 + n_2 = 0.01 + 0.001 = 0.011\;\text{mol}$$.

Step 4: Total volume of the mixed solution

Assuming $$\Delta_{mix}V = 0$$ (volume is additive):
$$V_{\text{solution}} = 50\;\text{mL} + 250\;\text{mL} = 300\;\text{mL} = 0.300\;\text{L}$$.

Step 5: Osmotic pressure

$$\pi = \dfrac{n_{\text{total}} RT}{V_{\text{solution}}} = \dfrac{0.011 \times 62 \times 300}{0.300}$$

First compute the concentration:
$$\dfrac{0.011}{0.300}=0.0366667\;\text{mol L}^{-1}$$.

Then
$$\pi = 0.0366667 \times 62 \times 300 = 0.0366667 \times 18\,600 \approx 682\;\text{Torr}$$.

Hence, the osmotic pressure of the resulting solution at 300 K is 682 Torr.

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 $$\mathbf{5}$$.

We need to find the volume of HCl solution required to completely neutralize the NaOH obtained from the reaction of metallic sodium with water.

Find moles of Na.

Molar mass of Na = 23 g/mol

Reaction of Na with water.

Moles of NaOH formed = Moles of Na = 0.03 mol (1:1 ratio)

Neutralization reaction.

Moles of HCl required = Moles of NaOH = 0.03 mol

Find the molarity of HCl solution.

HCl solution contains 73 g/L.

Molar mass of HCl = $$1 + 35.5 = 36.5$$ g/mol

Calculate the required volume.

The volume of HCl required is $$\boxed{15}$$ mL.

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 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.

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.

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.

A disproportionation reaction is one in which the same element in a single oxidation state is simultaneously oxidized and reduced to form products with different oxidation states.

Option A: $$2H_2O_2 \rightarrow 2H_2O + O_2$$

Oxygen in H₂O₂ has oxidation state -1. In H₂O, oxygen is -2 (reduced). In O₂, oxygen is 0 (oxidized). This IS a disproportionation reaction.

Option B: $$2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$$

Nitrogen in NO₂ has oxidation state +4. In HNO₃, nitrogen is +5 (oxidized). In HNO₂, nitrogen is +3 (reduced). This IS a disproportionation reaction.

Option C: $$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$$

Mn in MnO₄⁻ has oxidation state +7. In MnO₂, Mn is +4. This is a simple reduction reaction where Mn goes from +7 to +4 by gaining 3 electrons. There is no simultaneous oxidation and reduction of the same element. This is NOT a disproportionation reaction.

Option D: $$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

Mn in MnO₄²⁻ has oxidation state +6. In MnO₄⁻, Mn is +7 (oxidized). In MnO₂, Mn is +4 (reduced). This IS a disproportionation reaction.

The reaction that is NOT an example of a disproportionation reaction is Option C.

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).

We need to identify which reaction is a disproportionation reaction.

What is a disproportionation reaction?

A disproportionation reaction is one in which the same element in a single oxidation state is simultaneously oxidized and reduced.

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

Mn in $$MnO_4^{2-}$$: oxidation state = +6

Mn in $$MnO_4^-$$: oxidation state = +7 (oxidized)

Mn in $$MnO_2$$: oxidation state = +4 (reduced)

Here, Mn(+6) is both oxidized to Mn(+7) and reduced to Mn(+4). This is a disproportionation reaction.

$$MnO_4^- + 4H^+ + 4e^- \to MnO_2 + 2H_2O$$

This is a half-reaction (reduction only), not a disproportionation.

$$10I^- + 2MnO_4^- + 16H^+ \to 2Mn^{2+} + 8H_2O + 6I_2$$

$$I^-$$ is oxidized to $$I_2$$, and $$Mn^{+7}$$ is reduced to $$Mn^{+2}$$. Different elements are oxidized and reduced — this is a simple redox reaction.

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

$$Mn^{+7}$$ is reduced to $$Mn^{+4}$$, and $$S^{+2}$$ is oxidized to $$S^{+6}$$. Different elements are oxidized and reduced — not a disproportionation.

The correct answer is Option A: $$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$.

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.

High purity (>99.95% dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes)

Statement from NCERT. Therefore, B is the right option.

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 evaluate two statements about hydrogen peroxide.

Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions. In acidic medium, $$H_2O_2$$ acts as an oxidizing agent:

$$H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$$

For example, it oxidizes $$Fe^{2+}$$ to $$Fe^{3+}$$ in acidic solution. In basic medium, $$H_2O_2$$ also acts as an oxidizing agent:

$$H_2O_2 + 2e^- \rightarrow 2OH^-$$

For example, it oxidizes $$Fe^{2+}$$ to $$Fe^{3+}$$ in basic solution as well. Statement I is true.

Density of hydrogen peroxide at $$298 \text{ K}$$ is lower than that of $$D_2O$$. The density of $$H_2O_2$$ at $$298 \text{ K}$$ is approximately $$1.44 \text{ g/cm}^3$$, whereas the density of $$D_2O$$ (heavy water) at $$298 \text{ K}$$ is approximately $$1.104 \text{ g/cm}^3$$. Since $$1.44 > 1.104$$, the density of $$H_2O_2$$ is higher than that of $$D_2O$$, not lower. Statement II is false.

The correct answer is Option C: Statement I is true but Statement II is false.

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 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 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.

The key is to identify the paramagnetic gas formed when $$HClO_3$$ reacts with $$HCl$$ and to know how that gas behaves toward $$O_3$$.

Step 1 : Product of $$HClO_3 + HCl$$
Chloric acid reacts with concentrated hydrochloric acid in the following redox (disproportionation) reaction:
$$2\,HClO_3 + HCl \;\longrightarrow\; 2\,ClO_2 \;+\; Cl_2 \;+\; 2\,H_2O$$

$$ClO_2$$ has an odd number of electrons (it contains one unpaired electron), so it is paramagnetic. Hence the “paramagnetic gas” mentioned is $$ClO_2$$.

Step 2 : Reaction of $$ClO_2$$ with $$O_3$$
When chlorine dioxide is treated with ozone, it dimerises to give dichlorine hexoxide:
$$2\,ClO_2 + O_3 \;\longrightarrow\; Cl_2O_6 + O_2$$

The product formed is $$Cl_2O_6$$.

Among the given options, $$Cl_2O_6$$ corresponds to Option C.

Therefore, the correct answer is:
Option C which is: Cl$$_2$$O$$_6$$

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 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 $$\textbf{24}$$.

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.

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.

To determine the oxidation states of nitrogen in each species, we use the rule that oxygen has an oxidation state of $$-2$$ and the sum of oxidation states equals the overall charge.

In $$\text{NO}$$: $$x + (-2) = 0$$, so $$x = +2$$. In $$\text{NO}_2$$: $$x + 2(-2) = 0$$, so $$x = +4$$. In $$\text{N}_2\text{O}$$: $$2x + (-2) = 0$$, so $$x = +1$$. In $$\text{NO}_3^-$$: $$x + 3(-2) = -1$$, so $$x = +5$$.

Arranging in decreasing order: $$\text{NO}_3^- (+5) > \text{NO}_2 (+4) > \text{NO} (+2) > \text{N}_2\text{O} (+1)$$.

This matches option (1): $$NO_3^- > NO_2 > NO > N_2O$$.

We need to identify which substance does NOT produce CO when reacting with water.

Carbon reacts with steam at high temperatures: $$C + H_2O \rightarrow CO + H_2$$ (water gas reaction). So carbon produces CO.

Methane reacts with steam in the steam reforming reaction: $$CH_4 + H_2O \rightarrow CO + 3H_2$$. So methane produces CO.

Propane similarly undergoes steam reforming: $$C_3H_8 + 3H_2O \rightarrow 3CO + 7H_2$$. So propane produces CO.

$$CO_2$$ reacts with water to form carbonic acid: $$CO_2 + H_2O \rightarrow H_2CO_3$$. This reaction does not produce CO. There is no standard reaction between $$CO_2$$ and water that yields CO.

Therefore, water does not produce CO on reacting with $$CO_2$$, and the correct answer is option (1).

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.

Let us examine each statement about heavy water ($$D_2O$$) one by one.

Statement (A): Heavy water is used as a moderator in nuclear reactors. This is correct — $$D_2O$$ slows down fast neutrons effectively without absorbing them, making it an excellent moderator in reactors like CANDU.

Statement (B): Heavy water is obtained as a by-product in the fertilizer industry. This is correct — heavy water is produced as a by-product during the electrolysis of water in fertilizer plants that manufacture hydrogen for ammonia synthesis.

Statement (C): Heavy water is used for the study of reaction mechanisms. This is correct — deuterium-labeled compounds (using $$D_2O$$) are widely used as tracers to study reaction mechanisms through kinetic isotope effects.

Statement (D): Heavy water has a higher dielectric constant than water. This is incorrect — the dielectric constant of heavy water ($$D_2O$$) is about 78.06 at 25°C, which is slightly lower than that of ordinary water ($$H_2O$$), which is about 78.36 at 25°C. So heavy water does NOT have a higher dielectric constant than water.

The only incorrect statement is (D), so the answer is option (C): (D) only.

For any electrically neutral molecule, the sum of the oxidation numbers of all the atoms present is zero. We shall use the usual convention that hydrogen has an oxidation number of $$+1$$ and oxygen has an oxidation number of $$-2$$.

First we consider $$\mathrm{H_4P_2O_7}.$$ Let the oxidation number of each phosphorus atom be $$x.$$ Writing the algebraic sum, we have

$$4(+1) + 2(x) + 7(-2) = 0.$$

This simplifies to $$4 + 2x - 14 = 0,$$ so $$2x - 10 = 0,$$ giving $$2x = 10$$ and finally $$x = +5.$$ Thus, in $$\mathrm{H_4P_2O_7},$$ each phosphorus is in the $$+5$$ oxidation state.

Next we take $$\mathrm{H_4P_2O_5}.$$ Again letting the oxidation number of phosphorus be $$x,$$ we write

$$4(+1) + 2(x) + 5(-2) = 0.$$

Therefore $$4 + 2x - 10 = 0,$$ which gives $$2x - 6 = 0,$$ so $$2x = 6$$ and hence $$x = +3.$$ Consequently, phosphorus is in the $$+3$$ state in $$\mathrm{H_4P_2O_5}.$$

Finally we analyse $$\mathrm{H_4P_2O_6}.$$ Setting the oxidation number of phosphorus equal to $$x,$$ we write

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

Simplifying, $$4 + 2x - 12 = 0,$$ leading to $$2x - 8 = 0,$$ so $$2x = 8$$ and hence $$x = +4.$$ Thus, in $$\mathrm{H_4P_2O_6},$$ phosphorus has the oxidation state $$+4.$$

Collecting our results, the oxidation states of phosphorus in $$\mathrm{H_4P_2O_7},\; \mathrm{H_4P_2O_5}$$ and $$\mathrm{H_4P_2O_6}$$ are $$+5,\; +3$$ and $$+4$$ respectively.

Hence, the correct answer is Option C.

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.

Let us examine each statement about $$H_2O_2$$ (hydrogen peroxide):

(A) $$H_2O_2$$ is used in the treatment of effluents. This is correct. Hydrogen peroxide is widely used as an oxidant for treating industrial and domestic effluents, as it can decompose organic pollutants.

(B) $$H_2O_2$$ acts as both an oxidising and reducing agent. This is correct. In $$H_2O_2$$, oxygen is in the $$-1$$ oxidation state, which is intermediate between $$0$$ (in $$O_2$$) and $$-2$$ (in $$H_2O$$). Therefore, it can be oxidised to $$O_2$$ (acting as a reducing agent) or reduced to $$H_2O$$ (acting as an oxidising agent).

(C) The two hydroxyl groups lie in the same plane. This is incorrect. The structure of $$H_2O_2$$ is non-planar. The two $$O-H$$ bonds are in different planes, with a dihedral angle of about $$111.5°$$ in the gas phase. The molecule has an open-book structure where the two $$H$$ atoms are not in the same plane as the $$O-O$$ bond.

(D) $$H_2O_2$$ is miscible with water. This is correct. Hydrogen peroxide is completely miscible with water in all proportions due to its ability to form hydrogen bonds with water molecules.

Statements (A), (B), and (D) are correct, while (C) is incorrect. The correct answer is option (2).

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): $$2I^- + H_2O_2 + 2H^+ \to I_2 + 2H_2O$$.

We need to identify which form of hydrogen emits low energy $$\beta^-$$ particles.

Among the isotopes of hydrogen: Protium ($$_1^1H$$) has 1 proton and 0 neutrons, Deuterium ($$_1^2H$$) has 1 proton and 1 neutron, and Tritium ($$_1^3H$$) has 1 proton and 2 neutrons.

Protium and Deuterium are stable isotopes and are not radioactive. The proton $$H^+$$ is simply an ion and is also not radioactive.

Tritium ($$_1^3H$$) is the only radioactive isotope of hydrogen. It undergoes beta decay: $$_1^3H \to \, _2^3He + \beta^- + \bar{\nu}_e$$. The maximum energy of the emitted $$\beta^-$$ particles is about 18.6 keV, which is very low compared to beta particles from other radioactive nuclei.

The correct answer is Option (3): Tritium $$_1^3H$$.

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.

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 need to balance the redox equation: $$2MnO_4^- + bC_2O_4^{2-} + cH^+ \rightarrow xMn^{2+} + yCO_2 + zH_2O$$.

In the permanganate ion, Mn is in the +7 oxidation state, and it is reduced to $$Mn^{2+}$$ (oxidation state +2). The change in oxidation state per Mn atom is $$7 - 2 = 5$$. Since there are 2 $$MnO_4^-$$ ions, the total electron gain is $$2 \times 5 = 10$$ electrons.

In the oxalate ion $$C_2O_4^{2-}$$, carbon has an oxidation state of +3. In $$CO_2$$, carbon has an oxidation state of +4. Each carbon atom loses 1 electron, and each oxalate ion has 2 carbon atoms, so each oxalate ion loses 2 electrons.

For electron balance: total electrons lost = total electrons gained. So $$2b = 10$$, giving $$b = 5$$.

Now we can determine the other coefficients. Balancing carbon: $$2 \times 5 = y$$, so $$y = 10$$. Balancing Mn: $$x = 2$$. Balancing oxygen: left side has $$2 \times 4 + 5 \times 4 = 8 + 20 = 28$$ oxygen atoms; right side has $$10 \times 2 + z = 20 + z$$. So $$z = 8$$.

Balancing hydrogen: $$c = 2z = 16$$.

Let us verify the charge balance. Left side: $$2(-1) + 5(-2) + 16(+1) = -2 - 10 + 16 = +4$$. Right side: $$2(+2) = +4$$. The charges balance.

The balanced equation is $$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$$.

Therefore, the value of $$c$$ is $$\mathbf{16}$$.

We need to balance the reaction: $$S_{8}(s) + a \; OH^{-}(aq) \to b \; S^{2-}(aq) + c \; S_2O_3^{2-}(aq) + d \; H_2O(l)$$

This is a disproportionation reaction of sulphur in alkaline medium. Sulphur is simultaneously oxidised and reduced.

In $$S_8$$, the oxidation state of S is 0. In $$S^{2-}$$, it is $$-2$$. In $$S_2O_3^{2-}$$, the average oxidation state of S is $$+2$$.

Balancing sulphur atoms: $$8 = b + 2c$$

Balancing the change in oxidation state (electron transfer): each S going to $$S^{2-}$$ gains 2 electrons, so total electrons gained = $$2b$$. Each S going to $$S_2O_3^{2-}$$ loses 2 electrons, so total electrons lost = $$2 \times 2c = 4c$$. Wait — there are $$2c$$ sulphur atoms each losing 2 electrons, so total electrons lost = $$4c$$.

For electron balance: $$2b = 4c$$, which gives $$b = 2c$$.

Substituting in the sulphur balance: $$8 = 2c + 2c = 4c$$, so $$c = 2$$ and $$b = 4$$.

Now balancing oxygen: on the right side, we have $$c \times 3 = 6$$ oxygens from $$S_2O_3^{2-}$$ and $$d$$ oxygens from $$H_2O$$. On the left side, $$OH^-$$ contributes $$a$$ oxygens. So $$a = 6 + d$$.

Balancing hydrogen: $$a = 2d$$.

From $$a = 6 + d$$ and $$a = 2d$$: $$2d = 6 + d$$, so $$d = 6$$ and $$a = 12$$.

Let us verify the charge balance. Left side: $$-a = -12$$. Right side: $$-2b - 2c = -2(4) - 2(2) = -8 - 4 = -12$$. The charges balance.

The balanced equation is: $$S_8 + 12 \; OH^- \to 4 \; S^{2-} + 2 \; S_2O_3^{2-} + 6 \; H_2O$$

So, the answer is $$12$$.

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.

We have to choose that experimental procedure which delivers dihydrogen whose purity is higher than $$99.95\%$$. In other words, the method must minimise every possible gaseous or vapour impurity that could mix with the $$H_{2}$$ produced.

First we examine the chemical route given in option A. The reaction involved is

$$Zn + 2\,HCl(dil.) \;\longrightarrow\; ZnCl_{2} + H_{2}\uparrow$$

During this laboratory preparation, water vapour from the acid and traces of chloride or even sulphur-containing impurities commonly present in zinc can accompany the evolved hydrogen. These side components bring the purity well below $$99.95\%$$, so option A cannot satisfy the requirement.

Now let us analyse option B, namely the electrolysis of acidified water using platinum (Pt) electrodes. The pertinent half-reactions are

At the cathode: $$2\,H^+ + 2e^- \;\longrightarrow\; H_{2}(g)$$

At the anode: $$2\,H_{2}O \;\longrightarrow\; O_{2}(g) + 4\,H^+ + 4e^-$$

Although platinum is inert, small bubbles of oxygen generated at the anode can diffuse through the electrolyte and contaminate the hydrogen collected at the cathode. Acid mists may also be entrained. The gas produced is therefore not of the exceptionally high purity demanded.

We proceed to option C, the electrolysis of brine (aqueous $$NaCl$$). The half-cell processes are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$2\,Cl^- \;\longrightarrow\; Cl_{2}(g) + 2e^-$$

Chlorine gas is formed simultaneously with hydrogen and can mix with it. Even highly efficient cell designs cannot eliminate slight chlorine carry-over, so the hydrogen obtained from brine electrolysis does not attain $$99.95\%$$ purity.

Finally, we consider option D, the electrolysis of a warm, dilute $$Ba(OH)2$$ solution employing nickel (Ni) electrodes. The reactions are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$4\,OH^- \;\longrightarrow\; O_{2}(g) + 2\,H_{2}O + 4e^-$$

Crucially, barium hydroxide is a strong base whose cation $$Ba^{2+}$$ and anion $$OH^-$$ remain completely in the liquid phase; they are non-volatile and therefore cannot accompany the gaseous products. Nickel electrodes are sturdy yet practically inert under these conditions, so they introduce no foreign gases. The only potential contaminant is oxygen, but the solubility of oxygen in the basic medium is low and the cell can be designed so that the hydrogen outlet is physically isolated from the oxygen-evolution compartment. Consequently the hydrogen collected possesses a purity better than $$99.95\%$$, the standard demanded for applications such as the electronic and metallurgical industries.

Summarising the discussion, options A, B and C each introduce impurities (acid mist, oxygen, chlorine, etc.) that limit hydrogen purity, whereas option D yields dihydrogen of the required exceptionally high purity.

Hence, the correct answer is Option D.

First, we recall the definition of the water-gas shift reaction. In industrial chemistry this name is reserved for the reversible catalytic conversion in which carbon monoxide reacts with steam to give carbon dioxide and hydrogen. Symbolically we write

$$\text{CO}(g) + \text{H}_2\text{O}(g) \;\rightleftharpoons\; \text{CO}_2(g) + \text{H}_2(g).$$

The standard operating temperature for the high-temperature stage of this process is about $$673\;\text{K}$$, and it is always carried out in the presence of a suitable catalyst such as iron oxide promoted with chromium oxide. Thus, any correct option must explicitly contain (i) carbon monoxide on the reactant side, (ii) steam on the reactant side, (iii) carbon dioxide and hydrogen on the product side, and (iv) a temperature near $$673\;\text{K}$$ along with a catalyst mention.

Now we examine the four alternatives one by one.

Option A gives $$\text{CH}_4(g) + \text{H}_2\text{O}(g) \xrightarrow{1270\,\text{K}} \text{CO}(g) + 3\text{H}_2(g).$$ This is the steam-reforming reaction of methane, not the water-gas shift. The reactants and products do not match the required $$\text{CO}/\text{H}_2\text{O}$$ ⇌ $$\text{CO}_2/\text{H}_2$$ pattern.

Option B gives $$2\text{C}(s) + \text{O}_2(g) + 4\text{N}_2(g) \xrightarrow{1273\,\text{K}} 2\text{CO}(g) + 4\text{N}_2(g).$$ This merely represents the combustion of carbon in air to carbon monoxide; it again fails to involve steam and therefore cannot be the water-gas shift reaction.

Option C gives $$\text{C}(s) + \text{H}_2\text{O}(g) \xrightarrow{1270\,\text{K}} \text{CO}(g) + \text{H}_2(g).$$ Although steam is present, the solid carbon reactant shows that this is the water-gas (or carbon-steam) reaction, not the shift reaction. The essential carbon dioxide product is missing, so this option is also ruled out.

Option D gives $$\text{CO}(g) + \text{H}_2\text{O}(g) \xrightarrow{673\,\text{K}}_{\text{catalyst}} \text{CO}_2(g) + \text{H}_2(g).$$ Here every required feature is present: carbon monoxide and steam on the left, carbon dioxide and hydrogen on the right, a temperature around $$673\;\text{K}$$, and an explicit mention of a catalyst. Therefore this reaction exactly matches the definition of the water-gas shift reaction.

All other options fail to satisfy one or more of these essential criteria, so they must be rejected.

Hence, the correct answer is Option D.

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 note the atomic (molar) mass of zinc:

$$M_{\text{Zn}} = 65 \ \text{g mol}^{-1}$$

The mass of zinc given is

$$m_{\text{Zn}} = 5 \ \text{g}$$

Hence the number of moles of zinc present is obtained from the relation

$$n = \dfrac{m}{M}$$

Substituting the values, we get

$$n_{\text{Zn}} = \dfrac{5 \ \text{g}}{65 \ \text{g mol}^{-1}} = \dfrac{5}{65} \ \text{mol} = \dfrac{1}{13} \ \text{mol}$$

Now we consider the two separate reactions.

Reaction (a): Zinc with dilute hydrochloric acid

The balanced chemical equation is

$$\text{Zn} + 2\,\text{HCl} \longrightarrow \text{ZnCl}_2 + \text{H}_2$$

From the equation we see that

1 mol Zn → 1 mol H2

Therefore, the moles of hydrogen evolved in this case will be

$$n_{\text{H}_2,\; \text{from HCl}} = n_{\text{Zn}} = \dfrac{1}{13} \ \text{mol}$$

Reaction (b): Zinc with aqueous sodium hydroxide

The balanced chemical equation is

$$\text{Zn} + 2\,\text{NaOH} \longrightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$$

(The product Na2ZnO2 is called sodium zincate.)

Again, we observe from the equation that

1 mol Zn → 1 mol H2

Thus the moles of hydrogen evolved here are also

$$n_{\text{H}_2,\; \text{from NaOH}} = n_{\text{Zn}} = \dfrac{1}{13} \ \text{mol}$$

Since both reactions produce the same number of moles of hydrogen gas, and equal moles of any gas occupy equal volumes under identical conditions (Avogadro’s law), the volumes of hydrogen obtained from the two experiments are equal.

Therefore,

$$V_{\text{H}_2\;(\text{from HCl})} : V_{\text{H}_2\;(\text{from NaOH})} = 1 : 1$$

Hence, the correct answer is Option B.

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 chlorine reacting with hot and concentrated sodium hydroxide. The balanced equation actually taught in the N.C.E.R.T. text is

$$3\,\text{Cl}_2 + 6\,\text{NaOH}\;(\text{hot, conc.}) \;\longrightarrow\; 5\,\text{NaCl} + \text{NaClO}_3 + 3\,\text{H}_2\text{O}$$

From this single reaction we can immediately identify the two products that contain chlorine in different oxidation states:

$$\text{NaCl}\;(\text{X}) \quad\text{and}\quad \text{NaClO}_3\;(\text{Y}).$$

Compound (X) is $$\text{NaCl}$$. When treated with silver-nitrate solution $$\text{AgNO}_3$$, it forms a curdy white precipitate of $$\text{AgCl}$$, exactly as described in the statement of the question, so our identification of (X) is consistent.

Therefore compound (Y) must be $$\text{NaClO}_3$$. In aqueous solution this salt furnishes the chlorate ion $$\text{ClO}_3^-$$, and it is in this ion that we have to discuss the average bond order of the Cl-O bonds.

First, recall the definition that will be used. For a species that shows resonance, the conventional (Pauling) bond order is defined as

$$\text{Bond order} \;=\; \frac{\text{Total bond multiplicity in all canonical forms}}{\text{Number of equivalent bonds}\;\times\;\text{Number of canonical forms}}.$$

Put colloquially, we count the value (1 for a single, 2 for a double, etc.) of each Cl-O bond in every possible resonance structure, add them all up, and then divide by the number of resonance structures times the three equivalent Cl-O positions. Every Cl-O bond in the real molecule has the same “average” order obtained in this way.

Now let us draw the lowest-energy (formal-charge-minimised) resonance structures of $$\text{ClO}_3^-$$. In each of them chlorine is hypervalent (uses a 3d orbital) and expands its octet. The pattern is

Cl has two Cl=O double bonds and one Cl-O− single bond.

Because any one of the three oxygen atoms can be the one that bears the negative charge, we obtain three canonical forms, all of equal weight:

  Structure 1: O=Cl=O with one O−
  Structure 2: O=Cl-O−=O
  Structure 3: O−-Cl=O=O

Let us now tabulate the bond multiplicities.

• In any one canonical form there are exactly two Cl=O double bonds. Each double bond carries a multiplicity of 2.
• In the same canonical form there is exactly one Cl-O single bond. A single bond carries a multiplicity of 1.

Therefore, in a single canonical form the total Cl-O bond multiplicity is

$$2 + 2 + 1 \;=\; 5.$$

There are exactly $$3$$ such canonical forms. Hence the grand total of Cl-O bond multiplicity summed over all canonical forms is

$$5 \times 3 \;=\; 15.$$

The number of equivalent Cl-O bond positions is $$3$$ (because there are three oxygens attached to the same chlorine). The Pauling definition now gives

$$\text{Average bond order} \;=\; \frac{15}{3 \times 3} \;=\; \frac{15}{9} \;=\; \frac{5}{3} \;=\; 1.67.$$\

Thus every Cl-O bond in the chlorate ion possesses an average order of $$\displaystyle 1.67$$, exactly as required.

So, the answer is $$1.67$$.

First, we consider the chromate ion, whose formula is $$\mathrm{CrO_4^{2-}}$$. The ion contains one chromium atom and four oxygen atoms. Every oxygen atom is directly bonded to the single chromium atom. So, counting each chromium-oxygen linkage once, we have

$$$\text{Number of Cr-O bonds in chromate } = 4.$$$

Now we turn to the dichromate ion, whose formula is $$\mathrm{Cr_2O_7^{2-}}$$. We can picture it as two $$\mathrm{CrO_4^{2-}}$$ tetrahedra that share one oxygen atom. This shared (bridging) oxygen links the two chromium atoms together. Let us count the chromium-oxygen bonds one by one.

Each chromium is still surrounded by four oxygen atoms. Three of those oxygens are terminal (they are bonded to only that one chromium), while the fourth oxygen is the single bridging atom that connects to both chromiums. Hence, looking at all the bonds emanating from the two chromiums, we have

$$$\begin{aligned} \text{Terminal Cr-O bonds } &= 2 \text{ chromiums}\times 3 \text{ bonds per chromium}=6,\\[2mm] \text{Bridging Cr-O bonds } &= 2 \text{ bonds} \;(\text{one from each Cr to the same O}).\\ \end{aligned}$$$

Adding these gives

$$$\text{Total Cr-O bonds in dichromate } = 6 + 2 = 8.$$$

Finally, the problem asks for the sum of the chromium-oxygen bonds in both ions. Therefore,

$$$\text{Required sum}= \underbrace{4}_{\text{chromate}} + \underbrace{8}_{\text{dichromate}} = 12.$$$

So, the answer is $$12$$.

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.

First we recall the meaning of “11.2 volume” solution. One volume of the solution (we may conveniently take $$1\ \text{L}$$) liberates 11.2 volumes of oxygen gas at STP, that is $$11.2\ \text{L}$$ of $$\mathrm O_2$$.

Hydrogen peroxide decomposes according to

$$2\mathrm{H_2O_2}\;\longrightarrow\;2\mathrm H_2\mathrm O+\mathrm O_2$$

From the balanced equation we see that

$$2\ \text{mol H}_2\text O_2 \;\longrightarrow\;1\ \text{mol O}_2$$

The molar mass of $$\mathrm{H_2O_2}$$ is obtained from the data given:

$$M(\mathrm{H_2O_2}) = 2(1\,\text{g mol}^{-1}) + 2(16\,\text{g mol}^{-1}) = 34\ \text{g mol}^{-1}$$

Hence

$$2\ \text{mol H}_2\text O_2 = 2 \times 34\ \text{g} = 68\ \text{g}$$

At STP, one mole of any gas occupies $$22.4\ \text{L}$$. Therefore

$$68\ \text{g H}_2\text O_2 \;\longrightarrow\;22.4\ \text{L O}_2$$

Now we set up a simple proportion for the mass of $$\mathrm{H_2O_2}$$ required to give $$11.2\ \text{L O}_2$$:

$$\frac{68\ \text{g}}{22.4\ \text{L}} = \frac{x\ \text{g}}{11.2\ \text{L}}$$

Solving for $$x$$,

$$x = \frac{68 \times 11.2}{22.4}\ \text{g} = 34\ \text{g}$$

Thus, $$34\ \text{g}$$ of $$\mathrm{H_2O_2}$$ are present in the chosen $$1\ \text{L}$$ of solution. Strength in the usual “g per litre” sense is therefore

$$S = 34\ \text{g L}^{-1}$$

To express this as a percentage weight/volume (grams per 100 mL), we divide by 10:

$$\%\,(\text{w/v}) = \frac{34\ \text{g}}{1000\ \text{mL}}\times100\ \text{mL} = 3.4\ \text{g per}\ 100\ \text{mL}$$

Hence the solution is a $$3.4\%$$ w/v solution of hydrogen peroxide.

Hence, the correct answer is Option D.

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.

First, recall what a primary standard is. A primary standard is a substance that possesses a very high purity, is stable in air, has a known and fixed composition, has a reasonably high molar mass (to minimise weighing errors), and reacts completely and stoichiometrically with the reagent being standardised. Such a substance is weighed accurately and then used to determine the exact concentration of a titrant.

We have to select a compound that can accurately standardise an aqueous $$\text{NaOH}$$ solution. Sodium hydroxide absorbs both $$\text{CO}_2$$ and moisture from air, so its concentration changes with time. Hence it cannot be weighed directly to prepare an exactly known solution; instead, it must be titrated against a primary standard acid.

Now, let us examine the given options in the light of the required properties:

$$\textbf{(A) Sodium tetraborate (Borax, } \text{Na}_2\text{B}_4\text{O}_7\cdot10\text{H}_2\text{O}\text{)}$$ Although fairly pure and stable, in basic titrations borax is generally used to standardise strong acids, not strong bases. Moreover, borax itself produces a buffered, weakly basic solution, making end-point detection with phenolphthalein less sharp for direct standardisation of $$\text{NaOH}$$.

$$\textbf{(B) Ferrous ammonium sulphate (Mohr’s salt, } \text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O}\text{)}$$ Mohr’s salt is mainly employed in redox titrations (e.g., with $$\text{KMnO}_4$$ or $$\text{K}_2\text{Cr}_2\text{O}_7$$). It is not an acid compound suitable for a simple acid-base titration with $$\text{NaOH}$$.

**(C)** Oxalic acid (usually the dihydrate, $$\text{(COOH)}_2\cdot2\text{H}_2\text{O}$$) Oxalic acid dihydrate is highly pure, has a fixed and known stoichiometry, is stable on storage, and possesses a reasonably high molar mass of $$126\ \text{g mol}^{-1}$$. It behaves as a diprotic acid, giving a clear stoichiometric reaction with $$\text{NaOH}$$: $$\text{(COOH)}_2 + 2\,\text{NaOH} \;\longrightarrow\; \text{Na}_2\text{C}_2\text{O}_4 + 2\,\text{H}_2\text{O}$$ Because the stoichiometry is exact and the end point with phenolphthalein is sharp, oxalic acid is ideally suited for the standardisation of sodium hydroxide solutions.

$$\textbf{(D) Dilute HCl}$$ Dilute hydrochloric acid is itself a solution of unknown exact concentration and cannot be weighed directly as a solid; hence it cannot serve as a primary standard.

From the above discussion it is evident that only oxalic acid satisfies all the criteria of a primary standard for the titration of $$\text{NaOH}$$.

Hence, the correct answer is Option C.

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.