🎯 The Core Chemical Rule: The electronic configurations of transition metal ions are determined by removing electrons first from the $4s$ subshell and then from the $3d$ subshell. We count unpaired electrons using Hund's Rule across the five $3d$ orbitals.

✅ Step-by-Step Cation Analysis:
• In the +3 Oxidation State: Cobalt ($\text{Co}^{3+}$) has a $3d^6$ configuration (4 unpaired electrons), and Manganese ($\text{Mn}^{3+}$) has a $3d^4$ configuration (4 unpaired electrons). Thus, the number of unpaired electrons is perfectly equal.
• In the +4 Oxidation State: Cobalt ($\text{Co}^{4+}$) shifts to a $3d^5$ configuration (5 unpaired electrons), whereas Manganese ($\text{Mn}^{4+}$) shifts to a $3d^3$ configuration (3 unpaired electrons). Since $\text{X}^{4+}$ has more unpaired electrons than $\text{Y}^{4+}$, Element X is Cobalt (Co) and Element Y is Manganese (Mn).

⚙️ Industrial Uses:
• Cobalt (X): Extensively utilized in building modern lithium-ion and dry cell batteries.
• Manganese (Y): Formulated with iron as ferromanganese to manufacture ultra-tough railway tracks.

🎯 The Core Chemical Rule: Ground-state transition atoms share identical numbers of unpaired electrons when one has a less-than-half-filled subshell and the other is more-than-half-filled. This points directly to the pairings of $\text{Ti}$ / $\text{Ni}$ (2 unpaired) or $\text{V}$ / $\text{Co}$ (3 unpaired).

✅ Identifying the Elements:
• The Electron Decrease: Moving from $\text{X}^{2+}$ to $\text{X}^{3+}$ decreases the count of unpaired electrons. This only happens in less-than-half-filled subshells where losing an electron lowers the count (e.g., $\text{V}^{2+}$ [$3d^3$, 3 unpaired] to $\text{V}^{3+}$ [$3d^2$, 2 unpaired]). Therefore, Element X is an earlier transition metal ($\text{Ti}$ or $\text{V}$).
• Deducing Element Y: Because Element Y shares the same ground-state unpaired electron count but does not match this reduction behavior, it represents the later transition metal counterpart ($\text{Ni}$ or $\text{Co}$).

📊 Periodic Trends Evaluation:
• Atomic Mass (Option c): Atomic mass regularly increases across the series from left to right. Since Y sits further right in the series than X, the atomic mass of Y is larger than X.

🎯 The Core Chemical Rule: Industrial alloy design combines different structural types. Interstitial components add tensile strength and rigidity, while substitutional components modify specific chemical and physical traits like elasticity and corrosion resistance.

✅ Structural Anatomy of Vanadium Steel:
• The Interstitial Steel Base: Small carbon atoms fit tightly into the interstitial spaces of the iron crystal structure, blocking metal layers from sliding and reinforcing structural hardness.
• The Substitutional Vanadium Modification: Added vanadium atoms swap into original iron lattice positions because vanadium and iron have closely matching atomic radii and shared lattice geometries, yielding high structural elasticity.

💡 Application Insight: This combination makes vanadium steel ideal for car springs, as it withstands massive mechanical stress and deformation without breaking.

🎯 The Core Chemical Rule: For a compound to reduce $\text{Mn}^{7+}$ down to $\text{Mn}^{2+}$, it must contain a transition element that readily oxidizes by losing electrons to achieve a highly stable, half-filled ($3d^5$) or completely filled ($3d^{10}$) subshell.

✅ Why Iron(II) Sulfate Succeeds:
• The Reactivity of Iron: In $\text{FeSO}_4$, iron exists as the $\text{Fe}^{2+}$ ion ($3d^6$). It smoothly oxidizes to $\text{Fe}^{3+}$ ($3d^5$) by sacrificing one electron. This resulting $3d^5$ system is exceptionally stable because its five \(3d\) orbitals are uniformly half-filled.
• Thermodynamic Driving Force: Because $\text{Fe}^{2+}$ is highly driven to transform into stable $\text{Fe}^{3+}$, it functions as an excellent reducing agent capable of driving the reduction of $\text{Mn}^{7+}$.

❌ Why Other Options Fail:
• $\text{ScCl}_3$ & $\text{Ti(NO}_3)_4$: Exist as $\text{Sc}^{3+}$ ($3d^0$) and $\text{Ti}^{4+}$ ($3d^0$). They are in their highest, fully empty oxidation states and cannot undergo further oxidation.
• $\text{ZnCl}_2$: Features $\text{Zn}^{2+}$ ($3d^{10}$). Its completely full d-subshell resists losing more electrons.

🎯 The Core Chemical Rule: To determine the true chemical identity of an unknown cation, calculate its atomic number ($Z$) by adding the positive charge value back to the electron count of its core configuration.

✅ Deciphering the Cations:
• Element X: Given $\text{X}^{2+} = [\text{Ar}] 3d^6$, we add 2 electrons back to the $4s$ subshell to find the neutral atom configuration: $[\text{Ar}] 4s^2 3d^6$. This corresponds to Iron (Fe, $Z=26$).
• Element Y: Given $\text{Y}^{4+} = [\text{Ar}] 3d^6$, we add 4 total electrons back (2 to $4s$ and 2 to $3d$) to find the neutral atom configuration: $[\text{Ar}] 4s^2 3d^8$. This corresponds to Nickel (Ni, $Z=28$).

🔍 Row Evaluation:
• Iron (X) is structural and used to build durable reinforced concrete structures.
• Nickel (Y) possesses exactly 5 stable isotopes (unlike cobalt, which features 12 radioactive isotopes). This matches Row (a).

🎯 The Core Chemical Rule: To distinguish between two salts, adding dilute hydrochloric acid ($\text{HCl}$) must yield contrasting visual observations. This usually means a reaction occurs with only one salt, or one forms a unique insoluble precipitate while the other stays completely dissolved.

✅ Why Option (c) is Correct:
• With Sodium Carbonate ($\text{Na}_2\text{CO}_3$): The salt dissolves smoothly with effervescence from escaping $\text{CO}_2$ gas, producing a completely clear, fully soluble sodium chloride ($\text{NaCl}$) solution.
• With Lead(II) Bicarbonate ($\text{Pb(HCO}_3)_2$): It similarly generates $\text{CO}_2$ gas bubbles, but the liberated lead ions ($\text{Pb}^{2+}$) instantly bond with the chloride ions ($\text{Cl}^-$) from the test acid. This triggers the sudden precipitation of a dense white precipitate of lead(II) chloride ($\text{PbCl}_2 \downarrow$) because $\text{Pb}^{2+}$ belongs to Cation Analytical Group I.

❌ Why Other Options Fail:
• Option (a): Both salts are soluble sodium compounds that react identically to produce a clear solution alongside $\text{CO}_2$ gas.
• Option (b): $\text{HCl}$ cannot displace the stable sulfate anion from sodium sulfate, nor does it react with sodium chloride. Absolutely no reaction takes place.
• Option (d): Both barium carbonate and magnesium bicarbonate dissolve cleanly in $\text{HCl}$, yielding identical $\text{CO}_2$ gas bubbles and uniform, transparent liquid phases.

🎯 The Core Chemical Rule: An unknown salt is identified by treating its solid form with a strong acid to detect volatile acid radicals (anions), and treating its aqueous solution to detect basic radicals (cations) that produce characteristic insoluble precipitates.

✅ Analyzing Anion and Cation Tests for Salt (X):
• 1. Anion Determination (Acid Radical): When hot concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) is added to solid calcium bromide ($\text{CaBr}_2$), hydrogen bromide gas ($\text{HBr}$) is initially evolved. The hot concentrated acid then partially oxidizes $\text{HBr}$, releasing a mixture of gases consisting of reddish-brown bromine vapors ($\text{Br}_2$) and sulfur dioxide gas ($\text{SO}_2$).
• 2. Cation Determination (Basic Radical): When dilute sulfuric acid is added to the aqueous solution of the salt, calcium cations ($\text{Ca}^{2+}$) instantly combine with sulfate anions ($\text{SO}_4^{2-}$). This triggers the formation of a distinct white precipitate of calcium sulfate ($\text{CaSO}_4 \downarrow$), which is insoluble in dilute acids.

❌ Why Other Options Fail:
• $\text{NaI}$: While the iodide anion produces a gaseous mixture ($\text{I}_2$ and $\text{SO}_2$), the sodium cation ($\text{Na}^+$) forms highly soluble salts. It will never create a precipitate with dilute sulfuric acid.
• $\text{CaCO}_3$: Calcium carbonate is an insoluble rock-solid salt in water. It cannot be used to prepare a clear initial salt solution, and it dissolves instantly with vigorous effervescence when exposed to dilute acids.
• $\text{Pb(SO}_3)_2$: The sulfite group ($\text{SO}_3^{2-}$) reacts readily with weak dilute acids and does not require hot concentrated conditions. It evolves only a single gas ($\text{SO}_2$) instead of a gaseous mixture.

🎯 The Core Chemical Rule: Barium chloride ($\text{BaCl}_2$) is used as a main reagent to precipitate specific anions. The key to distinguishing between the resulting barium precipitates lies in their solubility; some dissolve cleanly in dilute hydrochloric acid ($\text{HCl}$), while others remain completely insoluble.

✅ Analyzing Precipitates and Solubility:
• With Phosphate ($\text{PO}_4^{3-}$): Barium chloride reacts with phosphate solutions to form a white precipitate of barium phosphate [$\text{Ba}_3(\text{PO}_4)_2 \downarrow$]. This precipitate dissolves smoothly in dilute hydrochloric acid.
• With Carbonate ($\text{CO}_3^{2-}$): Barium chloride reacts with soluble carbonate solutions to form a white precipitate of barium carbonate [$\text{BaCO}_3 \downarrow$]. This precipitate also dissolves readily in dilute acids with accompanying carbon dioxide ($\text{CO}_2$) gas evolution.

❌ Why Other Options Fail:
• Sulfate ($\text{SO}_4^{2-}$): Barium chloride reacts with sulfate groups to yield a white precipitate of barium sulfate [$\text{BaSO}_4 \downarrow$]. However, this precipitate is strictly insoluble in dilute hydrochloric acid. This immediately eliminates options (a), (b), and (d).
• Nitrate ($\text{NO}_3^-$): All nitrate salts are highly soluble in water, so no precipitate forms when mixed with barium chloride solution.

🎯 The Core Chemical Rule: An unknown cation is identified by observing its precipitation behavior with specific reagents. Cations are classified into analytical groups based on the solubility rules of their chlorides, sulfates, acetates, or other salts.

✅ Testing the Cation with the Three Reagents:
• With Hydrochloric Acid (HCl): It forms a **white precipitate**. This indicates that the cation belongs to Analytical Group I, which precipitates as insoluble chlorides. This directly matches the lead(II) cation, forming lead(II) chloride (PbCl₂ ↓).
• With Sodium Sulfate (Na₂SO₄): It yields a **white precipitate** due to the combination of the cation with sulfate ions, forming lead(II) sulfate (PbSO₄ ↓), a well-known insoluble white solid.
• With Acetic Acid (CH₃COOH): **No precipitate** forms because lead(II) acetate [Pb(CH₃COO)₂] is highly soluble in water, leaving the solution perfectly clear, which fully aligns with the diagram.

❌ Why Other Options Fail:
• Cu²⁺ & Fe²⁺: Both copper(II) chloride ($\text{CuCl}_2$) and iron(II) chloride ($\text{FeCl}_2$) are highly soluble in water and will not form a precipitate with hydrochloric acid.
• Ca²⁺: Calcium chloride ($\text{CaCl}_2$) is extremely soluble in water, meaning no precipitate occurs upon adding hydrochloric acid.

🎯 The Core Chemical Rule: To separate a mixture of anions smoothly via fractional precipitation, each added reagent must selectively precipitate exactly **one single remaining anion** at a time without interfering with the others.

✅ Analyzing the Separation Step-by-Step:
• Step 1 — Adding Copper Nitrate [$\text{Cu(NO}_3)_2$]: Copper ions ($\text{Cu}^{2+}$) selectively react with the sulphide ions ($\text{S}^{2-}$) to precipitate **black copper(II) sulphide** ($\text{CuS} \downarrow$). Meanwhile, copper chloride and copper sulfate remain fully soluble. After filtering out the $\text{CuS}$ precipitate, the solution contains only sulphate and chloride.
• Step 2 — Adding Calcium Hydroxlide [$\text{Ca(OH)}_2$]: Calcium ions ($\text{Ca}^{2+}$) selectively form a **white precipitate of calcium sulphate** ($\text{CaSO}_4 \downarrow$). On the other hand, calcium chloride ($\text{CaCl}_2$) is extremely soluble. Filtering out this precipitate leaves only chloride ions behind in the mixture.
• Step 3 — Adding Silver Nitrate [$\text{AgNO}_3$]: Finally, silver ions ($\text{Ag}^+$) precipitate the last remaining anion, chloride ($\text{Cl}^-$), as a **white precipitate of silver chloride** ($\text{AgCl} \downarrow$).

❌ Why Order (c) Fails: If you add silver nitrate first, the $\text{Ag}^+$ ions would unselectively precipitate **all three anions at the same time** because silver sulphide ($\text{Ag}_2\text{S}$), silver chloride ($\text{AgCl}$), and silver sulphate ($\text{Ag}_2\text{SO}_4$) are all highly insoluble solids, ruining the separation process completely.

🎯 The Core Chemical Rule: A physical or chemical dynamic equilibrium is broken and converted into a complete (one-way) process when conditions change such that the reverse reaction becomes physically impossible. In a solution, this happens by entirely eliminating the solid phase required for dynamic dynamic precipitation.

✅ Analytical Breakdown of Choice (b):
• Initially, the system exists as a **reversible physical equilibrium** inside a saturated solution where the dissolution rate equals the crystallization rate onto the solid phase:
    AgNO₃(s) ⇌ Ag⁺(aq) + NO₃^-(aq)
• When you **add an excess of water** (the solvent), the updated volume drops the concentration below saturation, forcing all the remaining undissolved solid AgNO₃ into solution.
• With zero remaining solid crystals present in the vessel, the dynamic background rate of crystallization drops to absolute zero. This forces the physical process into a **complete, one-way dissolution**.

❌ Why Option (c) is Incorrect: The displacement reaction of zinc with concentrated sulfuric acid evolves sulfur dioxide gas (SO₂). Because SO₂ cannot chemically react backwards with zinc sulfate to remake elemental zinc and acid, the reaction is already **complete from the start**. Closing the container does not change its operational mechanism.

🎯 The Core Chemical Rule: The activation energy and energy profiles of reactants or products are heavily governed by thermal factors. Changes in basal energy fields without shifting the absolute inner mechanism dimensions correspond cleanly to external kinetic variables.

✅ Precise Mathematical Curve Analysis:
• In Diagram (1): Reactant energy = 100 kJ, Product energy = 25 kJ, and Activated complex energy (peak) = 175 kJ. Thus, the activation energy for the forward pathway is $E_a = 175 - 100 = \text{\color{#4da3ff}75 kJ}$.
• In Diagram (2): Reactant energy = 225 kJ, Product energy = 150 kJ, and Activated complex energy (peak) = 300 kJ. Thus, the forward activation energy is $E_a = 300 - 225 = \text{\color{#ffb347}75 kJ}$.
• The mathematical activation energy ($E_a$) and enthalpy change ($\Delta H = 25 - 100 = -75\text{ kJ}$) remain completely identical across both figures. However, all absolute chemical energy values have been elevated symmetrically by exactly 125 kJ because supplying thermal motion raises the base internal kinetic energy of all chemical entities, indicating an **increase in temperature**.

❌ Why Other Options Fail:
• Adding a Catalyst (Option c): A catalyst creates an alternate pathway with a lower barrier, lowering only the peak (activated complex) while leaving the base horizontal energy lines for reactants and products completely unchanged.
• Increasing Surface Area or Concentration (Options b and d): These factors increase the collision frequency per unit time, enhancing the reaction rate, but do not alter the potential energy values along the vertical axis of a reaction profile diagram.

🎯 The Core Chemical Rule: The equilibrium constant ($K_c$) is determined by establishing the balanced chemical equation, constructing an ICE (Initial, Change, Equilibrium) table to track concentration variables, and inserting the final equilibrium values into the law of mass action expression.

✅ Step-by-Step ICE Table Construction:
• Balanced Dissociation Equation: $2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g)$
• Initial Concentrations ($C = n/V$): Since the volume is $1\text{ L}$, initial $[\text{SO}_3] = 0.2\text{ M}$. Initial products are $0\text{ M}$.
• Change Count via Decomposition Degree: $10\%$ of the initial $\text{SO}_3$ decomposes: $0.2 \times 0.10 = 0.02\text{ M}$.
    - Change for $\text{SO}_3 = -0.02\text{ M}$
    - Change for $\text{SO}_2 = +0.02\text{ M}$ (due to $2:2$ molar ratio)
    - Change for $\text{O}_2 = +0.01\text{ M}$ (due to $2:1$ molar ratio)
• Equilibrium Concentrations:
    - $[\text{SO}_3]_{eq} = 0.2 - 0.02 = 0.18\text{ M}$
    - $[\text{SO}_2]_{eq} = 0.02\text{ M}$
    - $[\text{O}_2]_{eq} = 0.01\text{ M}$

📊 Calculating the Value of $K_c$:
The equilibrium mathematical expression is:
    $K_c = \frac{[\text{SO}_2]^2 \cdot [\text{O}_2]}{[\text{SO}_3]^2}$
    $K_c = \frac{(0.02)^2 \cdot 0.01}{(0.18)^2} = \frac{4 \times 10^{-4} \cdot 10^{-2}}{0.0324} = \frac{4 \times 10^{-6}}{3.24 \times 10^{-2}} \≈ \{1.23 \times 10^{-4}}$

🎯 The Core Chemical Rule: For weak basic (alkaline) solutions, the concentration of hydroxide ions ([OH^-]) is determined by multiplying the initial molarity of the base (C_b) by its degree of dissociation (α). Taking the negative logarithm of this concentration yields the pOH value.

✅ Step-by-Step Mathematical Solution:
• 1. Base Concentration (C_b): Since the total solution volume is exactly 1 L, the base concentration matches its number of moles directly:
    C_b = n / V = 0.25 mol / 1 L = 0.25 M
• 2. Hydroxide Ion Concentration ([OH^-]): According to Ostwald's dilution law equations:
    [OH^-] = α × C_b
    [OH^-] = (2 × 10^-2) × 0.25 = 0.005 M = 5 × 10^-3 M
• 3. Computing the pOH: Apply the negative base-10 logarithm function:
    pOH = -log[OH^-]
    pOH = -log(5 × 10^-3) = 3 - log(5) = 3 - 0.7 = 2.3

📊 Analytical Insight: The direct calculation yields a pOH of 2.3, matching option (d). If the question had requested the pH value instead, it would be calculated as 14 - 2.3 = 11.7 (which corresponds to option a).

🎯 The Core Chemical Rule: Sodium hydroxide ($\text{NaOH}$) is a strong base that is already $100\%$ fully ionized in aqueous solution. Diluting a strong base increases the solution volume, which decreases the molar concentration of ions, while the actual amount of substance (total number of dissolved ions) remains unchanged.

✅ Step-by-Step Dilution Analysis:
• Constant Number of Ions: Because $\text{NaOH}$ is completely broken apart into $\text{Na}^+$ and $\text{OH}^-$ ions initially, adding water cannot cause further ionization. Thus, the number of produced ions remains constant.
• Decreasing Concentration: Increasing the solvent volume spreads the existing ions across a larger space, which **decreases the concentration of hydroxide ions** ($[\text{OH}^-]$).
• Logarithmic Values Shift: $\text{pOH}$ is defined via an inverse logarithmic relationship ($\text{pOH} = -\log[\text{OH}^-]$). When the concentration of hydroxide ions ($[\text{OH}^-]$) drops, the $\text{pOH}$ value increases accordingly. (Consequently, the $\text{pH}$ decreases as the solution becomes less basic and moves closer to neutral $7$).

📊 Conclusion: The statement that perfectly tracks these physical chemistry rules is that the total number of produced ions remains constant while the overall $\text{pOH}$ value increases, satisfying option (d).

🎯 The Core Chemical Rule: According to Le Chatelier's principle, altering the volume or pressure shifts the equilibrium position toward the side that counteracts the change. Crucially, **the equilibrium constant ($K_c$) value is altered exclusively by changing the temperature**.

✅ Step-by-Step Reaction Analysis:
• The Fixed $K_c$ Constraint: The question specifies increasing the yield of products "without affecting the $K_c$ value". This condition immediately eliminates option (a), because changing temperature changes $K_c$.
• Counting Gaseous Moles:
    - Reactants side has $1\text{ mol}$ of gas ($\text{CO}_2$). Pure solid carbon ($\text{C}_{(s)}$) is excluded from gas laws.
    - Products side has $2\text{ mol}$ of gas ($\text{CO}$).
• Effect of Increasing Vessel Volume: By **increasing the volume of the reaction vessel**, the internal system pressure drops. The equilibrium counters this change by shifting forward to the side with the larger number of gaseous moles ($1 \to 2$). This forward shift successfully increases the amount of carbon monoxide ($\text{CO}$) gas while keeping $K_c$ unchanged.

❌ Why Other Options Fail:
• Option (b): Carbon is a pure solid; modifying its mass or surface area does not shift the equilibrium position or change gas ratios.
• Option (c): Decreasing the volume increases pressure, driving the system in the reverse direction toward fewer gaseous moles, which decreases the yield of $\text{CO}$.

🎯 The Core Chemical Rule: In electrochemistry, a displacement reaction is non-spontaneous if the added elemental metal has a lower oxidation potential (less active) than the metal ion already dissolved in the solution. The more active metal easily undergoes oxidation, serving as a **strong reducing agent**.

✅ Ranking the Elements by Activity (Oxidation Potentials):
• From Reaction 1 (Non-spontaneous): The solid metal Z cannot displace Y²⁺ ions. This implies that Z is less active than Y. Therefore: Y > Z.
• From Reaction 2 (Non-spontaneous): The solid metal Y cannot displace X²⁺ ions. This means that Y is less active than X. Therefore: X > Y.
• Final Activity Series: Combining both observations gives a descending activity trend of: X > Y > Z.

📊 Evaluating the Target Reaction:
• The target equation describes elemental metal X reacting with Z²⁺ ions. Since **X is more active than Z** (positioned higher in the electromotive series), it easily displaces Z, meaning the reaction occurs **spontaneously**.
• During this change, metal X loses electrons and undergoes oxidation ($X \to X^{2+} + 2e^-$). By supplying electrons to reduce Z²⁺, the neutral element **X acts as a reducing agent**.

🎯 The Core Chemical Rule: During the **recharging** process, a lithium-ion battery behaves as an electrolytic cell. The chemical reactions that occurred spontaneously during discharge are reversed by an external power source. Consequently, lithium ions ($\text{Li}^+$) migrate from the positive electrode (acting as the anode during charging) to the negative electrode (acting as the cathode during charging).

✅ Analyzing Electrode Behavior During Charging:
• At the Anode (Positive Electrode): The structural material consists of lithium cobalt oxide ($\text{LiCoO}_2$). When recharging, lithium ions leave the lattice layer and exit into the liquid medium. This loss causes a noticeable **decrease in the mass of this anode**.
• Oxidation Mechanism: To preserve overall charge balance as positive $\text{Li}^+$ ions leave, the transition metal framework experiences oxidation, where cobalt(III) ions ($\text{Co}^{3+}$) are oxidized to cobalt(IV) ions ($\text{Co}^{4+}$) as shown in the charging half-reaction:
    $\text{LiCoO}_2 \rightarrow \text{Li}_{1-x}\text{CoO}_2 + x\text{Li}^+ + xe^-$
• At the Cathode (Negative Electrode): The graphite matrix ($\text{C}_6$) intercalates the arriving lithium ions alongside free circuit electrons to form lithium graphite ($\text{LiC}_6$), increasing its mass. The total concentration of lithium ions inside the electrolyte remains overall constant because the rate of ionic entry matches the rate of insertion.

📊 Conclusion: Option (d) perfectly describes the structural physics of the charging state, showing that the active anode mass decreases while the cobalt core is oxidized from $+3$ to $+4$.

🎯 The Core Chemical Rule: Faraday's laws of electrolysis state that the quantity of electricity passed determines the number of moles of electrons transferred. By comparing the moles of electrons to the moles of products formed at each electrode, we can deduce the accurate stoichiometric ratio and valency of the ions involved.

✅ Step-by-Step Quantitative Analysis:
• 1. Calculate the Moles of Electrons Transferred: Using Faraday's constant ($96500\text{ C/mol }e^-$):
    $\text{Moles of }e^- = \frac{\text{Quantity of electricity}}{\text{Faraday's Constant}} = \frac{9650\text{ C}}{96500\text{ C/mol}} = 0.1\text{ mol of }e^-$
• 2. Determine the Cathode Valence Ratio: The problem states that $0.1\text{ mol}$ of copper is deposited.
    $\text{Ratio} = \frac{\text{Moles of }e^-}{\text{Moles of Cu}} = \frac{0.1\text{ mol}}{0.1\text{ mol}} = 1 : 1$
    This indicates $1\text{ mol}$ of electrons reduces $1\text{ mol}$ of copper ions, confirming the presence of the monovalent copper ion ($\text{Cu}^+$). This narrows our options to rows (a) or (c).
• 3. Determine the Anode Gas Ratio: The problem states that $0.05\text{ mol}$ of gas is released.
    $\text{Ratio} = \frac{\text{Moles of }e^-}{\text{Moles of gas}} = \frac{0.1\text{ mol}}{0.05\text{ mol}} = 2 : 1$
    This means $2\text{ mol}$ of electrons are lost per $1\text{ mol}$ of gas produced. Looking at the options:
    - For chlorine gas ($\text{Cl}_2$): $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ ($\text{Ratio } e^- : \text{gas} = 2 : 1$). This perfectly matches our calculation!
    - For oxygen gas ($\text{O}_2$): $2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-$ ($\text{Ratio } e^- : \text{gas} = 4 : 1$).

📊 Conclusion: Combining the matched conditions ($\text{Cu}^+$ at the cathode and $\text{Cl}_2$ at the anode), Row (c) represents the fully accurate electrolysis mechanism for this cell.

🎯 The Core Chemical Rule: In structural storage problems, standard activity strings are written as decreasing hierarchies. A salt cannot be safely contained if the vessel possesses a higher oxidation potential than the solute, whereas an electrochemical mismatch where the vessel is less reactive allows successful storage.

✅ Step-by-Step Activity Series Deduction:
• Analyzing Metal X: Storing the ionic salt of X inside a container made of Z is impossible due to a spontaneous redox displacement reaction. This shows that the vessel metal Z is more reactive than the solute metal X (oxidation potential: Z > X).
• Analyzing Metal Y: Storing the ionic salt of Y inside a container made of Z is fully successful because no spontaneous displacement takes place. This confirms that the vessel metal Z is less reactive than the solute metal Y (oxidation potential: Y > Z).
• Final Structural Order Matrix: Combining these potential barriers proves that Y holds the highest oxidation capacity, Z occupies the intermediate space, and X remains the least active component. Expressed as a descending reactivity string, it reads: X < Z < Y.

📊 Conclusion: The mathematical formatting tracking this precise operational reactivity descending sequence aligns perfectly with choice (b).

🎯 The Core Chemical Rule: The electrolytes chosen for a salt bridge must contain ions that do not react or form insoluble precipitates with the ions present in either the anode or cathode half-cells. Precipitating out essential ions disrupts electrical neutrality, halting the electric current.

✅ Step-by-Step Precipitation Analysis:
• 1. Components of a Daniell Cell: A standard Daniell cell utilizes a zinc anode dipped in a zinc sulfate (ZnSO₄) solution, and a copper cathode dipped in a copper(II) sulfate (CuSO₄) solution.
• 2. Incompatibility of Calcium Chloride (CaCl₂): When CaCl₂ is placed in the salt bridge, its calcium cations (Ca²⁺) migrate toward the cathode compartment to balance the negative charge. However, the cathode compartment is filled with sulfate ions (SO₄^2-).
• 3. Chemical Precipitation Reaction: The Ca²⁺ ions immediately combine with SO₄^2- ions to form a dense white precipitate of calcium sulfate (CaSO₄ ↓), which is highly insoluble:
    Ca²⁺(aq) + SO₄^2-(aq) → CaSO₄(s) ↓
• This fast precipitation reaction blocks the pores of the salt bridge and depletes the free ions needed to neutralize the accumulation of charges. As a result, the flow of electrons is blocked, and the cell stops operating after a very short time.

📊 Conclusion: Because calcium forms an insoluble precipitate with the sulfate ions of the cell solutions, option (b) represents the correct outcome.

🎯 The Core Chemical Rule: Iron (Fe) only corrodes when it acts as an **Anode** (i.e., when connected to a less active metal with a lower oxidation potential). The rate of iron corrosion increases as the oxidation potential of the connected metal decreases, creating a larger potential difference.

✅ Analyzing the Electrodes and Corrosion Rates:
• Sacrificial Cathodic Protection: In tubes (1) and (3), metals W (+1.27 V) and Y (+0.74 V) have higher oxidation potentials than Iron (+0.44 V). They act as anodes and corrode instead of Iron, meaning **Iron does not corrode at all in tubes (1) and (3)**.
• Iron Corrosion Condition: In tubes (2) and (4), metals Z (−0.4 V) and X (+0.12 V) have lower oxidation potentials than Iron (+0.44 V). Therefore, Iron acts as the anode and undergoes corrosion.
• Comparing Corrosion Speed:
    - **Tube (4) with metal X:** Potential difference = 0.44 − 0.12 = 0.32 V.
    - **Tube (2) with metal Z:** Potential difference = 0.44 − (−0.4) = 0.84 V.
• Since tube (2) has a larger potential difference, Iron corrodes faster in tube (2) than in tube (4). Thus, the rate of corrosion follows the order: (4) < (2).

📊 Conclusion: The rate of corrosion in tube (4) is less than that in tube (2), making option (a) the correct statement.

CH₃C(CH₃)₃ (neopentane) has only single bonds → saturated aliphatic hydrocarbon.

3 C atoms each. Propane has 1 –CH₂– group; cyclopropane has 3 –CH₂– groups. → (1),(3).

Propyne hydration → propanone (a ketone), which resists oxidation and does not decolorize KMnO₄.

Ester = isopropyl methanoate → gives 2-propanol + methanamide.

3-methyl-2-butanol has no –CH₂– group and is a secondary alcohol (oxidizable), so it decolorizes KMnO₄.

HBr on 2-methyl-2-butene (Markovnikov) → 2-bromo-2-methylbutane. So X = (b).

Propyl acetate = acetic acid + 1-propanol by esterification.

Y = glycerol; Z = nitroglycerin, used in treatment of heart problems (angina).

Simplest tertiary alcohol = tert-butanol. Dehydrate 2-methyl-1-propanol → 2-methylpropene → catalytic hydration (Markovnikov) → (CH₃)₃C-OH.

Oxidation of –CH₂OH → –COOH (carboxylic acid), which reacts with NaOH and Na₂CO₃.

❸ uses glucose (non-electrolyte); ❹ uses Na₂SO₄ forming insoluble PbSO₄/BaSO₄ → both block ion flow.

🎯 The Core Chemical Rule: Paramagnetic substances possess at least one unpaired electron in their \(d\)-subshell, whereas diamagnetic substances have all their electrons completely paired up (\(3d^0\) or \(3d^{10}\)).

✅ Identifying the Three Consecutive Elements:
• Element X (Group 8): Represents **Nickel (₂₈Ni)**. Its common oxidation state is Ni²⁺ ([Ar] 3d⁸), which has 2 unpaired electrons and is paramagnetic. Group 8 elements never form diamagnetic compounds in their standard ionic states, making all its compounds paramagnetic.
• Element Y (Group IB): Represents **Copper (₂₉Cu)**. In the Cu⁺ state ([Ar] 3d¹⁰), it has no unpaired electrons (diamagnetic), while in the Cu²⁺ state ([Ar] 3d⁹), it has 1 unpaired electron (paramagnetic).
• Element Z (Group IIB): Represents **Zinc (₃₀Zn)**. Its only oxidation state is Zn²⁺ ([Ar] 3d¹⁰), where the \(d\)-subshell is fully filled. With zero unpaired electrons, all its compounds are strictly diamagnetic.

📊 Conclusion: Since Nickel (Ni), Copper (Cu), and Zinc (Zn) are three consecutive elements with atomic numbers 28, 29, and 30 respectively, their corresponding groups are **Group 8, Group IB, and Group IIB**, which perfectly matches Row (b).

n(HCl)=7.3/36.5=0.2 mol → n(CaCO₃)=0.1 mol → mass=10 g. Purity=80%, impurities=20%.

Exp1: acid:base = 2 mmol:1 mmol = 2:1. Exp2: base=5 mmol → acid=10 mmol → 10/40 = 0.250 M.

Ca₃(PO₄)₂: Ksp=(3S)³(2S)²=108S⁵=108×10⁻²⁰=1.08×10⁻¹⁸ ✔. Anion = phosphate.

NaOH neutralizes H⁺ → [H⁺] decreases → pH increases.

n(Cu)=0.12 mol → e⁻=0.24 mol → Q=0.24×96500=23160 C → t=23160/96.5=240 sec.

Y=propanone; reduction → 2-propanol (C₃H₈O); isomer = ethyl methyl ether.

B must be asymmetric → 1-butene (not symmetrical 2-butene). So A=butyl hydrogen sulphate, B=1-butene, C=butane.

Excess Cl₂/FeCl₃ → 2,4,6-trichlorotoluene. 3H₂ hydrogenates ring → methyl cyclohexane.

Acetylene → benzene (cyclic polymerization) → alkylation → sulfonation → neutralization → detergent.

B burns → 2CO₂+2H₂O → C₂H₄ = ethene. C (aromatic, for explosives) means A is long-chain; D is a larger fragment → liquid. So (d).

Identification: A=FeCO₃, C=FeO, D=Fe₂O₃, E=FeSO₄, F=Fe₂(SO₄)₃, B=Fe(OH)₃.

(1) Compounds giving Fe₂O₃ on heating: Iron(III) hydroxide and hydrated iron(III) oxide.

(2) Fe₂O₃ + conc. HCl → Iron(III) chloride (ferric chloride).

(3) Formula of F = Fe₂(SO₄)₃.

• Y = acetic acid (rayon) = CH₃COOH
• X = ethanol = C₂H₅OH
• Z = methyl acetate = CH₃COOCH₃
• L = ethyl methanoate = HCOOC₂H₅

Z and L are isomers, both C₃H₆O₂.