GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

61 of 85Question permalink

Remembering +4 | -1

The maximum temperature that can be achieved in blast furnace is:

View answer and explanation

Correct option: D

Upto 2200 K

Option D is correct because the maximum temperature that can be achieved in a blast furnace is up to 2200 K.

Solution approach

Correct Answer + Core Reason: Upto 2200 K. The maximum temperature that can be achieved in a blast furnace is approximately 2200 K.

Step-by-Step Solution:

Step 1: Blast Furnace Temperature: The temperature inside a blast furnace varies, but the maximum temperature is crucial for the reduction of iron ore.

Step 2: Temperature Range: The maximum temperature typically reaches around 2200 K (approximately 1900 °C).

Step 3: Conclusion: The maximum temperature is up to 2200 K.

Why other options are wrong:
- Option A: 1900 K is too low for the maximum temperature in a blast furnace.
- Option B: 5000 K is much higher than the actual temperature in a blast furnace.
- Option C: 1200 K is too low for the maximum temperature in a blast furnace.

Key Concepts:
- Blast Furnace Operation
- Temperature in Blast Furnace

Memory Aids:
- Remember the approximate temperature range in a blast furnace.

Quick Tips:
- Know the typical operating conditions of a blast furnace.

62 Medium

62 of 85Question permalink

Understanding +4 | -1

Which one of the following is the most correct statement?

View answer and explanation

Correct option: B

Brass is a substitutional alloy, while steel is an interstitial alloy

Option B (Brass is a substitutional alloy, while steel is an interstitial alloy) is correct because brass has atoms of similar size replacing each other, while steel has smaller atoms fitting into the spaces between iron atoms.

Solution approach

Correct Answer + Core Reason: Brass is substitutional, steel is interstitial.

Step-by-Step Solution:

Step 1: Define substitutional and interstitial alloys:
- Substitutional alloy: Atoms of one metal are replaced by atoms of another metal of similar size in the crystal lattice.
- Interstitial alloy: Smaller atoms fit into the spaces (interstices) between the larger atoms in the crystal lattice.

Step 2: Consider the composition of brass and steel:
- Brass: An alloy of copper (Cu) and zinc (Zn). Cu and Zn have similar atomic sizes, so it's a substitutional alloy.
- Steel: An alloy of iron (Fe) and carbon (C). Carbon atoms are much smaller than iron atoms, so they fit into the interstices, making it an interstitial alloy.

Why other options are wrong:
- Option A: Brass is an interstitial alloy, while steel is a substitutional alloy - Incorrect.
- Option C: Brass and steel are both substitutional alloy - Incorrect.
- Option D: Brass and steel are both interstitial alloy - Incorrect.

Key Concepts:
- Alloys
- Substitutional alloys
- Interstitial alloys

Memory Aids:
- Steel: Small carbon in iron's spaces

Quick Tips:
- Consider the relative sizes of the atoms in the alloy to determine if it's substitutional or interstitial.

63 Medium

63 of 85Question permalink

Understanding +4 | -1

After partial roasting, the sulphide of copper is reduced by:

View answer and explanation

Correct option: D

Self reduction

Option D (Self reduction) is correct because, after partial roasting, some of the copper sulfide is converted to copper oxide or copper sulfate, which then reacts with the remaining copper sulfide to form copper metal.

Solution approach

Correct Answer + Core Reason: Self reduction. Copper sulfide is reduced by copper oxide or copper sulfate formed during partial roasting.

Step-by-Step Solution:

Step 1: Understand the process of partial roasting.\nPartial roasting involves heating the sulfide ore in air, but not to completion. This converts some of the sulfide ore to oxide and/or sulfate.

Step 2: Consider the reactions that occur.\nAfter partial roasting, the following reactions can occur:
$2\mathrm{CuS} + 3{\mathrm{O}_2} \to 2\mathrm{CuO} + 2\mathrm{S}{\mathrm{O}_2}$
$\mathrm{CuS} + 2{\mathrm{O}_2} \to \mathrm{CuS}{\mathrm{O}_4}$

Step 3: Understand self-reduction.\nSelf-reduction occurs when the oxide or sulfate formed reacts with the remaining sulfide to produce the metal.
$2\mathrm{CuO} + \mathrm{Cu}_2\mathrm{S} \to 4\mathrm{Cu} + \mathrm{S}{\mathrm{O}_2}$
$\mathrm{CuS}{\mathrm{O}_4} + \mathrm{Cu}_2\mathrm{S} \to 3\mathrm{Cu} + 2\mathrm{S}{\mathrm{O}_2}$

Why other options are wrong:
- Option A: Cyanide process is used for gold and silver extraction, not copper.
- Option B: Electrolysis is used for refining copper, not for the initial reduction of the sulfide ore.
- Option C: Reduction with carbon is not the primary method for reducing copper sulfide after partial roasting. Self-reduction is the key process.

Key Concepts:
- Partial Roasting: Converting some sulfide to oxide/sulfate.
- Self-Reduction: Reaction between oxide/sulfate and remaining sulfide to form the metal.

Memory Aids:
- Self-reduction: "The ore reduces itself".

Quick Tips:
- Remember that copper extraction often involves self-reduction after partial roasting.

64 Medium

64 of 85Question permalink

Understanding +4 | -1

Forth floatation method is successful in separating impurities from ores because

View answer and explanation

Correct option: D

The pure ore is not easily wetted by water as by pine oil, cresylic acid, etc

Option D is correct because in froth floatation, the ore is preferentially wetted by oil and not by water, allowing it to be separated from impurities.

Solution approach

Correct Answer + Core Reason: The pure ore is not easily wetted by water as by pine oil, cresylic acid, etc. This is the principle behind froth floatation.

Step-by-Step Solution:

Step 1: Understand the principle of froth floatation.

Step 2: In froth floatation, the ore particles are preferentially wetted by oil (like pine oil) while the gangue particles are wetted by water.

Step 3: This difference in wettability allows the ore particles to be carried to the surface with the froth, separating them from the gangue.

Why other options are wrong:
- Option A: The pure ore is not soluble in water; it's wetted by oil.
- Option B: The ore's density relative to water is not the primary factor; wettability is.
- Option C: The impurities are wetted by water, not dissolved in it.

Key Concepts:
- Froth floatation
- Wettability of ore and gangue particles

Memory Aids:
- Oil wets ore, water wets waste.

Quick Tips:
- Focus on the surface properties of the ore and gangue in froth floatation.

65 Medium

65 of 85Question permalink

Analyzing +4 | -1

For the formation of

{Cr}_2 {O}_3

and

{Al}_2 {O}_3

, values of

\Delta_f G

are

540\, \mathrm{k\,J\,mol}^{-1}

and

-827 \,\mathrm{kJ\,mol}^{-1}

respectively. What will be the reaction of

{Cr}_2 {O}_3

by

{Al}

?

View answer and explanation

Correct option: A

Reduction of

{Cr}_2 {O}_3

, by

{Al}

will take place

Option A is correct because the negative difference in Gibbs free energy indicates that aluminium can reduce chromium(III) oxide.

Solution approach

Correct Answer + Core Reason: Reduction of ${Cr}_2 {O}_3$ , by ${Al}$ will take place. Since the Gibbs free energy of formation of $Al_2O_3$ is more negative than that of $Cr_2O_3$ , aluminium can reduce $Cr_2O_3$ .

Step-by-Step Solution:

Step 1: Write the reactions and their Gibbs free energy changes:
$Cr_2O_3 \rightarrow 2Cr + \frac{3}{2}O_2$ , $\Delta G = +540 \mathrm{kJ/mol}$
$2Al + \frac{3}{2}O_2 \rightarrow Al_2O_3$ , $\Delta G = -827 \mathrm{kJ/mol}$

Step 2: Write the overall reaction:
$Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3$

Step 3: Calculate the Gibbs free energy change for the overall reaction:
$\Delta G_{reaction} = \Delta G_{f}(Al_2O_3) - \Delta G_{f}(Cr_2O_3) = -827 - 540 = -1367 \mathrm{kJ/mol}$

Step 4: Interpret the result: Since the Gibbs free energy change for the overall reaction is negative, the reaction is spontaneous, meaning aluminium can reduce $Cr_2O_3$ .

Why other options are wrong:
- Option B: Oxidation of $Cr_2O_3$ by Al will not take place because the Gibbs free energy change is negative, indicating reduction.
- Option C: The reaction is feasible because the Gibbs free energy change is negative.
- Option D: The reaction is feasible because the Gibbs free energy change is negative.

Key Concepts:
- Gibbs free energy
- Reduction potential
- Thermodynamics

Memory Aids:
- A more negative Gibbs free energy of formation indicates a more stable oxide.
- Aluminium is a strong reducing agent.

Quick Tips:
- Compare the Gibbs free energies of formation to determine the feasibility of a redox reaction.

66 Medium

66 of 85Question permalink

Analyzing +4 | -1

Assertion : Extraction of ir on metal from iron oxide ore is carried out by heating with coke.

Reason : The reaction,

{Fe}_2 {O}_3(s) \longrightarrow {Fe}(s)+3 / 2 {O}_2(s)

is a spontaneous process.

View answer and explanation

Correct option: C

If the Assertion is correct but Reason is incorrect.

Option C is correct because the extraction of iron involves heating with coke, which is true, but the given reaction is not spontaneous.

Solution approach

Correct Answer + Core Reason: If the Assertion is correct but Reason is incorrect. The extraction of iron from its oxide ore is indeed carried out by heating with coke, but the given reaction is not spontaneous under standard conditions.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The assertion states that iron is extracted from iron oxide ore by heating with coke. This is a correct statement. Coke (carbon) acts as a reducing agent.

Step 2: Analyze the Reason: The reason states that the reaction $Fe_2O_3(s) \longrightarrow Fe(s) + 3/2 O_2(g)$ is a spontaneous process. This is incorrect. The formation of iron from its oxide requires a reducing agent because the reverse reaction (oxidation of iron) is thermodynamically favorable under standard conditions.

Step 3: Determine the relationship between Assertion and Reason: The assertion is correct, but the reason is incorrect. Therefore, the reason is not the correct explanation for the assertion.

Why other options are wrong:
- Option A: Incorrect because the reason is not correct.
- Option B: Incorrect because the reason is not correct.
- Option D: Incorrect because the assertion is correct.

Key Concepts:
- Extraction of iron
- Reducing agents
- Spontaneity of reactions
- Thermodynamics

Memory Aids:
- Coke (carbon) is used as a reducing agent in iron extraction.
- The decomposition of iron oxide into iron and oxygen is not spontaneous.

Quick Tips:
- Remember that metal extraction often involves reduction processes using reducing agents.

67 Medium

67 of 85Question permalink

Remembering +4 | -1

The percentage of carbon is same in:

View answer and explanation

Correct option: A

Cast iron and pig iron

Option A (Cast iron and pig iron) is correct because both cast iron and pig iron typically contain a high percentage of carbon, ranging from 3-5%.

Solution approach

Correct Answer + Core Reason: Cast iron and pig iron have similar carbon percentages.

Step-by-Step Solution:

Step 1: Recall the carbon content in different types of iron.

Step 2: Pig iron has a high carbon content (3-5%). Cast iron also has a high carbon content (2-4%). Steel has a lower carbon content (0.25-2%). Wrought iron has a very low carbon content (less than 0.1%).

Why other options are wrong:
- Option B: Cast iron has a higher carbon content than steel.
- Option C: Pig iron has a higher carbon content than steel.
- Option D: Pig iron has a much higher carbon content than wrought iron.

Key Concepts:
- Types of iron
- Carbon content in iron

Memory Aids:
- Pig and Cast iron are "carbon-rich".

Quick Tips:
- Memorize the carbon content ranges for different types of iron.

68 Medium

68 of 85Question permalink

Remembering +4 | -1

The most abundant ore of iron is:

View answer and explanation

Correct option: A

Haematite

Option A is correct because Haematite (

Fe_2O_3

) is the most abundant ore of iron.

Solution approach

Correct Answer + Core Reason: Haematite. Haematite ( $Fe_2O_3$ ) is the most abundant ore of iron.

Step-by-Step Solution:

Step 1: Recall the common ores of iron.

Step 2: Identify the most abundant ore of iron.

Step 3: Analyze each option:
- Haematite ( $Fe_2O_3$ ) is the most abundant ore of iron.
- Limonite ( $Fe_2O_3 \cdot nH_2O$ ) is an ore of iron.
- Magnetite ( $Fe_3O_4$ ) is an ore of iron.
- Siderite ( $FeCO_3$ ) is an ore of iron.

Why other options are wrong:
- Option B: Limonite is an ore of iron, but not the most abundant.
- Option C: Magnetite is an ore of iron, but not the most abundant.
- Option D: Siderite is an ore of iron, but not the most abundant.

Key Concepts:
- Ores of metals
- Iron ores

Memory Aids:
- Haematite is reddish-brown and contains iron oxide.

Quick Tips:
- Memorize the common ore formulas.

69 Medium

69 of 85Question permalink

Remembering +4 | -1

Silver obtained by argentiferrous lead is purified by:

View answer and explanation

Correct option: C

Cupellation

Option C (Cupellation) is correct because cupellation is a process where base metals are oxidized and removed, leaving behind the purified silver.

Solution approach

Correct Answer + Core Reason: Cupellation. Cupellation is used to purify silver obtained from argentiferous lead.

Step-by-Step Solution:

Step 1: Recall the cupellation process: Cupellation involves heating the impure silver in a cupel (a shallow, porous dish) in the presence of air. Base metals oxidize and are absorbed into the cupel, leaving behind pure silver.

Step 2: Evaluate the other options: Distillation, froth floatation, and reacting with KCN are not suitable for this purification.

Step 3: Conclusion: Cupellation is the correct answer.

Why other options are wrong:
- Option A (Distillation): Distillation is used for separating liquids based on boiling points, not for purifying silver in this context.
- Option B (Froth floatation): Froth floatation is used for separating sulfide ores from gangue, not for purifying silver.
- **Option D (Reacting with $\mathrm{KCN}$ ): Reacting with KCN is used in the cyanide process for gold extraction, not for silver purification after lead extraction.

Key Concepts:
- Silver purification
- Cupellation
- Metallurgical processes

Memory Aids:
- Cupellation = Silver Purification

Quick Tips:**
- Know the specific applications of different metallurgical processes.

70 Medium

70 of 85Question permalink

Analyzing +4 | -1

Select the incorrect match.

View answer and explanation

Correct option: D

Both (1) and (3)

Option D (Both (1) and (3)) is correct because highly electropositive metals are extracted by electrolytic reduction, not smelting, and less electropositive metals can be extracted by chemical reduction, not necessarily electrolytic extraction.

Solution approach

Correct Answer + Core Reason: Both (1) and (3). Highly electropositive metals are extracted by electrolytic reduction, and less electropositive metals can be extracted by chemical reduction.

Step-by-Step Solution:

Step 1: Analyze each match.

Step 2: Highly electropositive metals (like Na, K, Al) have a strong tendency to lose electrons and form positive ions. They are extracted by electrolytic reduction because chemical reduction is not feasible.

Step 3: Noble metals (like Au, Ag) are extracted by the cyanide process, which involves complex formation and displacement.

Step 4: Less electropositive metals (like Fe, Cu) can be extracted by chemical reduction methods like smelting, using reducing agents such as carbon or CO. Electrolytic extraction is not always necessary.

Why other options are wrong:
- Option A: Match 1 is incorrect.
- Option B: Match 3 is incorrect.
- Option C: Match 1 is incorrect.

Key Concepts:
- Electrolytic reduction
- Smelting
- Cyanide process
- Electropositivity

Memory Aids:
- Highly electropositive = Electrolytic
- Noble metals = Cyanide

Quick Tips:
- Remember the extraction methods for different types of metals based on their electropositivity.

71 Medium

71 of 85Question permalink

Understanding +4 | -1

In the electrolytic refining of zinc

View answer and explanation

Correct option: D

Acidified zinc sulphate is the electrolyte.

Option D (Acidified zinc sulphate is the electrolyte) is correct because in electrolytic refining of zinc, an acidified zinc sulfate solution is used as the electrolyte.

Solution approach

Correct Answer + Core Reason: Acidified zinc sulphate is the electrolyte. Acidified $ZnSO_4$ solution is used as the electrolyte in the electrolytic refining of zinc.

Step-by-Step Solution:

Step 1: In electrolytic refining, an impure metal is used as the anode and a thin strip of pure metal as the cathode.

Step 2: The electrolyte is a solution containing ions of the metal being refined.

Step 3: For zinc refining, the electrolyte is an acidified solution of zinc sulfate ( $ZnSO_4$ ).

Step 4: At the anode, zinc dissolves into the solution: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$

Step 5: At the cathode, zinc ions are reduced to pure zinc: $Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)$

Why other options are wrong:
- Option A: The anode is made of impure zinc, not graphite.
- Option B: The impure metal is at the anode, not the cathode.
- Option C: Metal ions are reduced at the cathode, not the anode.

Key Concepts:
- Electrolytic refining
- Anode and cathode reactions
- Electrolyte

Memory Aids:
- Anode: Oxidation, Cathode: Reduction

Quick Tips:
- Remember that the electrolyte must contain ions of the metal being refined.

72 Medium

72 of 85Question permalink

Analyzing +4 | -1

Which of the following properties does not belong to the complex formed in Mond's process.

View answer and explanation

Correct option: C

It is square planar

Option C (It is square planar) is correct because the complex formed in Mond's process,

\mathrm{[Ni(CO)_4]}

, is tetrahedral, not square planar.

Solution approach

Correct Answer + Core Reason: It is square planar. The complex formed in Mond's process, $\mathrm{[Ni(CO)_4]}$ , is tetrahedral.

Step-by-Step Solution:

Step 1: Identify the complex formed in Mond's process.\nThe complex formed in Mond's process is tetracarbonylnickel(0), $\mathrm{[Ni(CO)_4]}$ .

Step 2: Determine the properties of $\mathrm{[Ni(CO)_4]}$ .
- It is volatile at around $43^\circ C$ , close to $100^\circ C$ .
- It is diamagnetic because Ni is in 0 oxidation state and CO is a strong field ligand, leading to pairing of electrons.
- It follows the 18-electron rule: Ni (10) + 4 CO (2 x 4) = 18 electrons.
- It has a tetrahedral geometry.

Step 3: Analyze the given options.
- Option A: It is diamagnetic - True.
- Option B: It follows the 18 electron rule - True.
- Option D: Volatile at $100^{\circ} \mathrm{C}$ - Approximately true.
- Option C: It is square planar - False. It is tetrahedral.

Why other options are wrong:
- Option A: The complex is diamagnetic.
- Option B: The complex follows the 18-electron rule.
- Option D: The complex is volatile at around $43^\circ C$ , which can be considered close to $100^\circ C$ .

Key Concepts:
- Mond's Process: A process for purifying nickel.
- Coordination Complexes: Compounds formed by the coordination of ligands to a central metal atom.
- 18-Electron Rule: A rule that predicts the stability of coordination complexes.

Memory Aids:
- $\mathrm{[Ni(CO)_4]}$ : "Nickel" and "Carbonyl" are "Tetrahedral".

Quick Tips:
- Remember the geometry of $\mathrm{[Ni(CO)_4]}$ is tetrahedral.

73 Medium

73 of 85Question permalink

Remembering +4 | -1

Which metal is purified by Pattinson’s process?

View answer and explanation

Correct option: A

Ag

Option A (Ag) is correct because Pattinson's process is used to purify silver (Ag) from lead (Pb).

Solution approach

Correct Answer + Core Reason: Ag. Pattinson's process is used to purify silver.

Step-by-Step Solution:

Step 1: Define Pattinson's Process: Pattinson's process is a method for removing lead from silver. It involves partial freezing of a molten mixture of lead and silver. As the mixture cools, lead crystallizes out first, leaving a silver-enriched liquid.

Step 2: The process is repeated multiple times to further enrich the silver content.

Why other options are wrong:
- Option B: Au is typically purified by electrolytic refining or the Miller process.
- Option C: Fe is refined using processes like smelting and various steelmaking methods.
- Option D: Sb is refined by processes like distillation or electrolytic refining.

Key Concepts:
- Pattinson's Process
- Silver Refining

Memory Aids:
- Pattinson: Silver Purification from Lead

Quick Tips:
- Remember that Pattinson's process relies on the difference in freezing points between lead and silver.

74 Medium

74 of 85Question permalink

Understanding +4 | -1

Among the following groups of oxides, the group containing oxides that cannot be reduced by carbon to give the respective metals is

View answer and explanation

Correct option: D

\mathrm{CaO},{\mathrm{K}_2}\mathrm{O}

Option D (

\mathrm{CaO},{\mathrm{K}_2}\mathrm{O}

) is correct because calcium and potassium oxides have very high stability and cannot be reduced by carbon at reasonable temperatures.

Solution approach

Correct Answer + Core Reason: $\mathrm{CaO},{\mathrm{K}_2}\mathrm{O}$ . Calcium and potassium are highly electropositive metals and their oxides are difficult to reduce with carbon.

Step-by-Step Solution:

Step 1: Recall the principles of reduction using carbon.

Step 2: Carbon can reduce oxides of less electropositive metals to their respective metals.

Step 3: Consider the electropositivity of the metals in the given options:
- Cu, Pb, Fe, and Zn are relatively less electropositive and their oxides can be reduced by carbon.
- K and Ca are highly electropositive and their oxides are very stable and difficult to reduce by carbon.

Step 4: Conclude that the group containing CaO and K2O cannot be reduced by carbon.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}_2}\mathrm{O},{\mathrm{K}_2}\mathrm{O}$ is incorrect because Cu2O can be reduced by carbon, but K2O cannot.
- Option B: $\mathrm{PbO},\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ is incorrect because both PbO and Fe3O4 can be reduced by carbon.
- Option C: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3},\mathrm{ZnO}$ is incorrect because both Fe2O3 and ZnO can be reduced by carbon.

Key Concepts:
- Reduction using carbon.
- Electropositivity of metals.
- Stability of oxides.

Memory Aids:
- More electropositive metals have more stable oxides.

Quick Tips:
- Consider the position of the metal in the electrochemical series to determine the feasibility of reduction by carbon.

75 Medium

75 of 85Question permalink

Remembering +4 | -1

Extraction of Ag from commercial lead is possible by:

View answer and explanation

Correct option: A

Parke’s process

Option A (Parke’s process) is correct because Parke's process is specifically designed for the desilverization of lead, where zinc is added to extract silver.

Solution approach

Correct Answer + Core Reason: Parke’s process. Parke's process is used for the extraction of Ag from commercial lead.

Step-by-Step Solution:

Step 1: Recall the Parke's process: Parke's process involves adding zinc to molten lead containing silver. Zinc has a higher affinity for silver than lead does, and it also has a lower melting point and is immiscible with lead. The zinc-silver alloy solidifies first and can be separated.

Step 2: Evaluate the other options: Clarke's process, Pattinson's process, and electrolytic process are used for different purposes.

Step 3: Conclusion: Parke's process is the correct answer.

Why other options are wrong:
- Option B (Clarke’s process): Clarke's process is not related to silver extraction from lead.
- Option C (Pattinson’s process): Pattinson's process is an older method for concentrating lead, not extracting silver.
- Option D (Electrolytic process): Electrolytic process can be used for refining metals, but Parke's process is more specific for Ag extraction from Pb.

Key Concepts:
- Silver extraction
- Parke's process
- Desilverization of lead

Memory Aids:
- Parke's = Silver from Lead

Quick Tips:
- Know the specific applications of different metallurgical processes.

76 Medium

76 of 85Question permalink

Understanding +4 | -1

The final step in the metallurgical extraction of Cu metal from Cu pyrites takes place in a Bessemer converter. The reaction taking place is:

View answer and explanation

Correct option: C

2\mathrm{C}{\mathrm{u}_2}\mathrm{O} + \mathrm{C}{\mathrm{u}_2}\mathrm{S} \to 6\mathrm{Cu} + \mathrm{S}{\mathrm{O}_2}

Option C (

2Cu_2O + Cu_2S \to 6Cu + SO_2

) is correct because this is the final self-reduction step in the Bessemer converter during copper extraction.

Solution approach

Correct Answer + Core Reason: $2\mathrm{C}{\mathrm{u}_2}\mathrm{O} + \mathrm{C}{\mathrm{u}_2}\mathrm{S} \to 6\mathrm{Cu} + \mathrm{S}{\mathrm{O}_2}$ . This is the final reaction in the Bessemer converter for copper extraction.

Step-by-Step Solution:

Step 1: In the Bessemer converter, copper matte ( $Cu_2S$ ) is partially oxidized to $Cu_2O$ .

Step 2: The $Cu_2O$ then reacts with the remaining $Cu_2S$ to form copper metal and sulfur dioxide.

Step 3: The reaction is: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Why other options are wrong:
- Option A: $Cu_2S + O_2 \to 2Cu + SO_2$ is not the final step, but an intermediate step.
- Option B: $4Cu_2O + FeS \to 8Cu + FeSO_4$ is not a relevant reaction in the Bessemer converter.
- Option D: $Cu_2S + 2FeO \to 2CuO + 2Fe + SO_2$ is not a reaction that occurs in the Bessemer converter.

Key Concepts:
- Metallurgy of Copper
- Bessemer Converter
- Self-reduction

Memory Aids:
- Remember the self-reduction of copper oxide and sulfide to form copper.

Quick Tips:
- The Bessemer converter is used for the final reduction of copper.

77 Medium

77 of 85Question permalink

Understanding +4 | -1

In purification of bauxite by hall’s process

View answer and explanation

Correct option: A

Bauxite ore is fused with

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}

Option A (Bauxite ore is fused with

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}

) is correct because in Hall's process, bauxite ore is fused with sodium carbonate to convert alumina to soluble sodium aluminate.

Solution approach

Correct Answer + Core Reason: Bauxite ore is fused with $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ in Hall's process.

Step-by-Step Solution:

Step 1: Hall's process is used for the purification of bauxite ore.

Step 2: In this process, bauxite ore is fused with sodium carbonate ( $Na_2CO_3$ ) to convert alumina ( $Al_2O_3$ ) into soluble sodium aluminate ( $NaAlO_2$ ).

Step 3: The reaction is: $Al_2O_3 + Na_2CO_3 \rightarrow 2NaAlO_2 + CO_2$

Step 4: The sodium aluminate is then leached with water, and the impurities are filtered out.

Why other options are wrong:
- Option B: Bauxite ore is heated with $\mathrm{NaOH}$ solution at $50^\circ C$ - This is part of Baeyer's process, not Hall's.
- Option C: Bauxite ore is heated with $\mathrm{NaHC}{\mathrm{O}_3}$ - Sodium bicarbonate is not used in Hall's process.
- Option D: Bauxite ore is fused with coke and heated at $1800^\circ C$ in a current of nitrogen - This describes Serpeck's process, not Hall's.

Key Concepts:
- Hall's process
- Purification of bauxite
- Sodium carbonate fusion

Memory Aids:
- Hall's: $Na_2CO_3$ fusion

Quick Tips:
- Remember the specific reagents and conditions used in Hall's process.

78 Medium

78 of 85Question permalink

Analyzing +4 | -1

{\text{A}}{{\text{g}}_2}{\text{S}} + \mathop {{\text{KCN}}}\limits_{{\text{Excess}}} \to {\left[ {{\text{Ag}}{{({\text{CN}})}_{\text{x}}}} \right]^{ - {\text{n}}}}

\left[\mathrm{Ag}(\mathrm{CN})_{\mathrm{x}}\right]^{-\mathrm{n}}+\mathrm{Zn} \rightarrow\left[\mathrm{Zn}(\mathrm{CN})_{\mathrm{y}}\right]^{-\mathrm{m}}

Find the sum of

\mathrm{x}

and

\mathrm{y}

.

View answer and explanation

Correct option: A

6

Option A is correct because the reaction forms [Ag(CN)2]^- and [Zn(CN)4]^2-, so x=2 and y=4, and their sum is 6.

Solution approach

Correct Answer + Core Reason: 6. The complex formed with silver is [Ag(CN)2]^- and with zinc is [Zn(CN)4]^2-, so x+y = 2+4 = 6.

Step-by-Step Solution:

Step 1: Write the balanced chemical equations:
- $\mathrm{Ag}_2\mathrm{S} + 4\mathrm{KCN} \rightarrow 2\mathrm{K[Ag(CN)_2]} + \mathrm{K_2S}$
- $2[\mathrm{Ag(CN)_2}]^- + \mathrm{Zn} \rightarrow [\mathrm{Zn(CN)_4}]^{2-} + 2\mathrm{Ag}$

Step 2: Determine the values of x and y:
- From the first reaction, x = 2 (since the complex is [Ag(CN)2]^-).
- From the second reaction, y = 4 (since the complex is [Zn(CN)4]^2-).

Step 3: Calculate the sum of x and y:
- x + y = 2 + 4 = 6

Why other options are wrong:
- Option B: Incorrect identification of the complexes or incorrect summation.
- Option C: Incorrect identification of the complexes or incorrect summation.
- Option D: Incorrect identification of the complexes or incorrect summation.

Key Concepts:
- Complex formation in metallurgy
- Cyanide process for silver extraction

Memory Aids:
- Silver forms a complex with two cyanide ions: [Ag(CN)2]^-
- Zinc forms a complex with four cyanide ions: [Zn(CN)4]^2-

Quick Tips:
- Remember the common coordination numbers of metals in complexes.

79 Medium

79 of 85Question permalink

Understanding +4 | -1

Which is incorrect as the uses of lime stone in industries are concerned?

View answer and explanation

Correct option: B

In the extraction of Sn from its ore

Option B is correct because limestone (CaCO3) is used in cement, iron extraction, and glass manufacture, but not in tin extraction.

Solution approach

Correct Answer + Core Reason: In the extraction of Sn from its ore. Limestone is not used in the extraction of Sn.

Step-by-Step Solution:

Step 1: Identify the uses of limestone in industries.

Step 2: Limestone (CaCO3) is used:
- In the manufacture of cement.
- In the extraction of iron from its ore (as a flux to remove silica).
- In the manufacture of glass.

Step 3: Limestone is not used in the extraction of tin (Sn) from its ore.

Why other options are wrong:
- Option A: Limestone is used in making cement.
- Option C: Limestone is used in the extraction of Fe from its ore.
- Option D: Limestone is used in the manufacture of glass.

Key Concepts:
- Uses of limestone in various industries

Memory Aids:
- Cement, Iron, Glass - Limestone

Quick Tips:
- Remember the common uses of important industrial compounds.

80 Medium

80 of 85Question permalink

Analyzing +4 | -1

Assertion (A) : Reduction of

{Cr}_2 {O}_3

with aluminium is possible.

Reason (R): The value of

\Delta G^0

for the formation of

{Cr}_2{O}_3

is

-540 \, \mathrm{kJ \, mol}^{-1}

and that of

{Al}_2 {O}_3

is

-827 \mathrm{~kJ} \mathrm{~mol}^{-1}.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A is correct because the negative Gibbs free energy change for the formation of Al2O3 is more negative than that of Cr2O3, making the reduction of Cr2O3 with aluminum thermodynamically feasible.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). The Gibbs free energy change for the formation of $Al_2O_3$ is more negative than that of $Cr_2O_3$ , indicating that aluminum can reduce chromium oxide.

Step-by-Step Solution:

Step 1: Understand the concept of Gibbs free energy and its relation to the feasibility of a reaction.

Step 2: Recall that a reaction is spontaneous (feasible) if the change in Gibbs free energy ( $\Delta G$ ) is negative.

Step 3: Analyze the given information. The $\Delta G^0$ for the formation of $Cr_2O_3$ is -540 kJ/mol, and for $Al_2O_3$ is -827 kJ/mol.

Step 4: Consider the reduction reaction: $Cr_2O_3 + 2Al \rightarrow Al_2O_3 + 2Cr$ . The Gibbs free energy change for this reaction can be approximated as $\Delta G_{reaction} = \Delta G_{f}(Al_2O_3) - \Delta G_{f}(Cr_2O_3)$ .

Step 5: Calculate the $\Delta G_{reaction}$ : $\Delta G_{reaction} = -827 - (-540) = -287$ kJ/mol.

Step 6: Since $\Delta G_{reaction}$ is negative, the reduction of $Cr_2O_3$ by aluminum is feasible.

Step 7: The reason (R) correctly explains the assertion (A) because the more negative $\Delta G^0$ for $Al_2O_3$ formation drives the reduction of $Cr_2O_3$ .

Why other options are wrong:
- Option B: While both assertion and reason are true, the reason *is* the correct explanation.
- Option C: The assertion is true, so this option is incorrect.
- Option D: Both assertion and reason are true, so this option is incorrect.

Key Concepts:
- Gibbs Free Energy ( $\Delta G$ )
- Spontaneity of Reactions
- Thermochemistry
- Reduction Potential

Memory Aids:
- A negative $\Delta G$ indicates a spontaneous (feasible) reaction.
- More negative $\Delta G_f$ means greater stability of the compound.

Quick Tips:
- When comparing the feasibility of reactions based on Gibbs free energy, consider the difference in $\Delta G_f$ values. A more negative difference favors the reaction.

81 Medium

81 of 85Question permalink

Remembering +4 | -1

Cynaide process is used for the extraction of

View answer and explanation

Correct option: D

Both (a) and (b)

Option D is correct because the cyanide process is used for the extraction of both gold (Au) and silver (Ag) from their ores.

Solution approach

Correct Answer + Core Reason: Both (a) and (b). The cyanide process is used for the extraction of both gold (Au) and silver (Ag).

Step-by-Step Solution:

Step 1: Cyanide Process: This process involves leaching the metal ore with a dilute solution of cyanide ions (CN-) to form a soluble metal cyanide complex.

Step 2: Gold and Silver Extraction: Both gold and silver form cyanide complexes, allowing them to be extracted from their ores using this method.

Step 3: Conclusion: The cyanide process is used for both Au and Ag.

Why other options are wrong:
- Option A: While the cyanide process is used for gold, it is also used for silver.
- Option B: While the cyanide process is used for silver, it is also used for gold.
- Option C: Copper is not typically extracted using the cyanide process.

Key Concepts:
- Cyanide Process
- Extraction of Gold and Silver
- Metal Cyanide Complexes

Memory Aids:
- Remember that the cyanide process is used for noble metals like gold and silver.

Quick Tips:
- Know the metals that are extracted using the cyanide process.

82 Medium

82 of 85Question permalink

Analyzing +4 | -1

Assertion (A): At 673 K,

{CO}

is better reducing agent for

{Fe}_2 {O}_3

than coke.

Reason (R):

\Delta G_{\left({CO}, {CO}_2\right)}

is less than

\Delta G_{({C}, {CO})}

at 673 K.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because at 673K, CO is a better reducing agent than coke, and this is because the change in Gibbs free energy for CO to CO2 is more negative than that of C to CO.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). At 673 K, CO is a better reducing agent for $Fe_2O_3$ than coke because $\Delta G_{(CO, CO_2)}$ is less than $\Delta G_{(C, CO)}$ at 673 K.

Step-by-Step Solution:

Step 1: Analyze the assertion: At 673 K, CO is indeed a better reducing agent for $Fe_2O_3$ than coke. This is related to the thermodynamics of the reactions.

Step 2: Analyze the reason: The Gibbs free energy change ( $\Delta G$ ) indicates the spontaneity of a reaction. A more negative $\Delta G$ means a greater driving force for the reaction to occur.

Step 3: Since $\Delta G_{(CO, CO_2)}$ is less (more negative) than $\Delta G_{(C, CO)}$ at 673 K, the reaction of CO to $CO_2$ is thermodynamically more favorable than the reaction of C to CO. This makes CO a better reducing agent.

Step 4: Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason is the correct explanation.
- Option C: The reason is true.
- Option D: Both the assertion and the reason are true.

Key Concepts:
- Reducing agent
- Gibbs free energy
- Ellingham diagrams

Memory Aids:
- More negative $\Delta G$ = Better reducing agent

Quick Tips:
- Understand the relationship between Gibbs free energy and the effectiveness of reducing agents.

83 Medium

83 of 85Question permalink

Remembering +4 | -1

The metal extracted by cyanide process is

View answer and explanation

Correct option: A

Silver

Option A is correct because the metal extracted by the cyanide process is primarily silver, although it is also used for gold.

Solution approach

Correct Answer + Core Reason: Silver. The metal extracted by the cyanide process is silver.

Step-by-Step Solution:

Step 1: Cyanide Process Application: The cyanide process is used to extract metals that form stable cyanide complexes.

Step 2: Silver Extraction: Silver is extracted by leaching its ore with a cyanide solution, forming a soluble silver cyanide complex, which is then reduced to obtain silver.

Step 3: Conclusion: Silver is extracted by the cyanide process.

Why other options are wrong:
- Option B: Copper is not typically extracted using the cyanide process.
- Option C: Iron is extracted using a blast furnace with carbon reduction.
- Option D: Sodium is extracted by electrolysis of its molten chloride.

Key Concepts:
- Cyanide Process
- Silver Extraction
- Metal Cyanide Complexes

Memory Aids:
- Remember that the cyanide process is commonly associated with silver and gold extraction.

Quick Tips:
- Know the metals that are extracted using the cyanide process.

84 Medium

84 of 85Question permalink

Understanding +4 | -1

Which of the following metals cannot be extracted by carbon reduction process?

View answer and explanation

Correct option: B

Al

Option B (Al) is correct because aluminum has a strong affinity for oxygen, making its oxide very stable and difficult to reduce with carbon.

Solution approach

Correct Answer + Core Reason: Al. Aluminum oxide is very stable and difficult to reduce with carbon.

Step-by-Step Solution:

Step 1: Understand the carbon reduction process.\nThe carbon reduction process involves using carbon (coke) to reduce metal oxides to their respective metals at high temperatures.

Step 2: Consider the stability of metal oxides.\nThe feasibility of carbon reduction depends on the relative stability of the metal oxide and carbon monoxide (or carbon dioxide). If the metal oxide is very stable, carbon reduction is not feasible.

Step 3: Analyze the given options.
- Zinc (Zn), Mercury (Hg), and Lead (Pb) oxides can be reduced by carbon.
- Aluminum (Al) oxide is highly stable and requires a stronger reducing agent than carbon, such as electrolysis or the aluminothermic process.

Why other options are wrong:
- Option A: Zinc (Zn) can be extracted by carbon reduction.
- Option C: Mercury (Hg) can be extracted by carbon reduction.
- Option D: Lead (Pb) can be extracted by carbon reduction.

Key Concepts:
- Carbon Reduction: Using carbon to reduce metal oxides.
- Stability of Metal Oxides: Determines the feasibility of carbon reduction.

Memory Aids:
- Aluminum oxide is very stable and requires electrolysis for extraction.

Quick Tips:
- Consider the position of the metal in the reactivity series. Metals high in the series (like Al) are difficult to reduce with carbon.

85 Medium

85 of 85Question permalink

Understanding +4 | -1

\mathrm{TiI}_4

on heating gives

View answer and explanation

Correct option: B

{Ti}+2 {I}_2

Option B (

{Ti}+2 {I}_2

) is correct because

\mathrm{TiI}_4

is unstable at high temperatures and decomposes into titanium and iodine.

Solution approach

Correct Answer + Core Reason: ${Ti}+2 {I}_2$ . $\mathrm{TiI}_4$ decomposes into titanium and iodine upon heating.

Step-by-Step Solution:

Step 1: Understand the reaction.
$\mathrm{TiI}_4$ is a volatile compound. When heated, it decomposes into its constituent elements, titanium (Ti) and iodine (I2).

Step 2: Write the balanced chemical equation.\nThe balanced chemical equation for the decomposition is:
$\mathrm{TiI}_4(s) \to \mathrm{Ti}(s) + 2\mathrm{I}_2(g)$

Step 3: Analyze the given options.\nOption B, ${Ti}+2 {I}_2$ , correctly represents the products of the decomposition.

Why other options are wrong:
- Option A: $T i I_2+I_2$ is incorrect as it does not represent the correct decomposition products.
- Option C: ${TiI}_3+\frac{2}{2} I_2$ is incorrect as it does not represent the correct decomposition products.
- Option D: None of these is incorrect because option B is correct.

Key Concepts:
- Decomposition Reaction: A reaction where a compound breaks down into simpler substances.
- Thermal Stability: The ability of a compound to resist decomposition upon heating.

Memory Aids:
- $\mathrm{TiI}_4$ breaks down into its elements upon heating.

Quick Tips:
- Remember that many metal halides decompose upon heating, especially those with larger, more polarizable anions like iodide.

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