GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Hard

Remembering +4 | -1

Mac Arthur and Forest cyanide process is used in the extraction of:

View answer and explanation

Correct option: B

Ag and Au

Option B (Ag and Au) is correct because the Mac Arthur and Forest cyanide process is specifically designed for the extraction of silver and gold from their ores.

Solution approach

Correct Answer + Core Reason: Ag and Au. The cyanide process is used for leaching silver and gold ores.

Step-by-Step Solution:

Step 1: The Mac Arthur and Forest cyanide process involves leaching the ore with a dilute solution of sodium cyanide or potassium cyanide.

Step 2: This process selectively dissolves silver and gold, forming soluble cyanide complexes:
${\mathrm{4Au}} + {8\mathrm{CN}^ - } + {\mathrm{O}_2} + {2{\mathrm{H}_2}\mathrm{O}} \to 4{[{\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}}}]^ - } + 4\mathrm{OH}^ -$
${\mathrm{4Ag}} + {8\mathrm{CN}^ - } + {\mathrm{O}_2} + {2{\mathrm{H}_2}\mathrm{O}} \to 4{[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}}]^ - } + 4\mathrm{OH}^ -$

Step 3: The gold and silver are then recovered from the solution by reduction, typically using zinc dust.

Why other options are wrong:
- Option A: Cu (Copper) is extracted using different methods, such as froth flotation and electrolytic refining.
- Option C: Fe (Iron) is extracted using reduction in a blast furnace.
- Option D: Cr (Chromium) is extracted from chromite ore through a series of chemical processes, not the cyanide process.

Key Concepts:
- Extraction of metals
- Leaching
- Complex formation

Memory Aids:
- Cyanide process is specifically for gold and silver.

Quick Tips:
- Remember the specific extraction methods for common metals.

2 Medium

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Remembering +4 | -1

Which form of iron has lowest percentage of carbon?

View answer and explanation

Correct option: B

Wrought iron

Option B is correct because Wrought iron has the lowest percentage of carbon (0.02-0.08%) among the given options.

Solution approach

Correct Answer + Core Reason: Wrought iron. Wrought iron has the lowest percentage of carbon.

Step-by-Step Solution:

Step 1: Recall the carbon content of different forms of iron.

Step 2: Compare the carbon content of the given options:
- Cast iron: 2-4% carbon
- Wrought iron: 0.02-0.08% carbon
- Steel: 0.25-2% carbon

Step 3: Wrought iron has the lowest carbon percentage.

Why other options are wrong:
- Option A: Cast iron has a higher carbon percentage than wrought iron.
- Option C: Steel has a higher carbon percentage than wrought iron.
- Option D: The percentages are not the same; wrought iron has the lowest.

Key Concepts:
- Forms of iron
- Carbon content

Memory Aids:
- Wrought iron is tough and has very little carbon.

Quick Tips:
- Remember the carbon content ranges for different forms of iron.

3 Hard

3 of 186Question permalink

Analyzing +4 | -1

Assertion (A): Wolframite impurities are separated from cassiterite by electromagnetic separation.

Reason (R): Cassiterite being magnetic is attracted by the magnet and forms a separate heap.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C is correct because while the assertion is true, the reason is false. Wolframite is magnetic and is separated from non-magnetic cassiterite, not the other way around.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. Wolframite is magnetic and is separated from non-magnetic cassiterite.

Step-by-Step Solution:

Step 1: Analyze the assertion: Wolframite impurities are separated from cassiterite by electromagnetic separation.
- This is true. Electromagnetic separation is used to separate magnetic and non-magnetic substances.

Step 2: Analyze the reason: Cassiterite being magnetic is attracted by the magnet and forms a separate heap.
- This is false. Wolframite is magnetic, while cassiterite is non-magnetic.

Step 3: Determine the correctness of both statements.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Electromagnetic separation
- Magnetic properties of minerals

Memory Aids:
- Wolframite is magnetic.
- Cassiterite is non-magnetic.

Quick Tips:
- Remember the magnetic properties of common minerals.

4 Medium

4 of 186Question permalink

Remembering +4 | -1

Which one is an ore of sodium?

View answer and explanation

Correct option: D

Soda ash

Option D (Soda ash) is the correct answer because Soda ash (Na2CO3) is a common ore of sodium.

Solution approach

Correct Answer + Core Reason: Soda ash. Soda ash ( $\mathrm{Na}_2\mathrm{CO}_3$ ) is an ore of sodium.

Step-by-Step Solution:
Step 1: Identify the chemical formulas of the given options.
Step 2: Recall that Soda ash ( $\mathrm{Na}_2\mathrm{CO}_3$ ) is a common ore of sodium.

Why other options are wrong:
- Option A: Sylvine (KCl) is an ore of potassium, not sodium.
- Option B: Siderite ( $\mathrm{FeCO}_3$ ) is an ore of iron, not sodium.
- Option C: Spodumene ( $\mathrm{LiAlSi}_2\mathrm{O}_6$ ) is an ore of lithium, not sodium.

Key Concepts:
- Ores of sodium, potassium, lithium, and iron.

Memory Aids:
- Soda ash contains sodium.

Quick Tips:
- Remember the common ores of important metals.

5 Medium

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Understanding +4 | -1

Aluminothermic process is used for the extraction of metals, whose oxides are:

View answer and explanation

Correct option: B

Not easily reduced by carbon

Option B (Not easily reduced by carbon) is correct because the aluminothermic process uses aluminum to reduce oxides that are more stable and difficult to reduce with carbon.

Solution approach

Correct Answer + Core Reason: Not easily reduced by carbon. Aluminothermic process is used for metals whose oxides are difficult to reduce with carbon.

Step-by-Step Solution:

Step 1: Understand the aluminothermic process.\nThe aluminothermic process involves using aluminum as a reducing agent to extract metals from their oxides. This method is used for metals whose oxides are very stable and cannot be easily reduced by carbon.

Step 2: Consider the reactivity of aluminum.\nAluminum is a strong reducing agent and can reduce oxides of metals that are less reactive than aluminum.

Step 3: Analyze the given options.
- If the oxides were fusible, it wouldn't necessarily require aluminothermic process.
- If the oxides were easily reduced by hydrogen, it wouldn't require aluminothermic process.
- If the oxides were strongly basic, it wouldn't directly relate to the need for aluminothermic process.
- The key reason is that the oxides are not easily reduced by carbon, requiring a stronger reducing agent like aluminum.

Why other options are wrong:
- Option A: Fusibility of the oxide is not the primary reason for using the aluminothermic process.
- Option C: Reducibility by hydrogen is not the primary reason for using the aluminothermic process.
- Option D: Basic nature of the oxide is not the primary reason for using the aluminothermic process.

Key Concepts:
- Aluminothermic Process: Reduction of metal oxides using aluminum.
- Reducing Agent: Aluminum is a strong reducing agent.

Memory Aids:
- Aluminothermic: "Aluminum" reduces "difficult" oxides.

Quick Tips:
- The aluminothermic process is used for extracting metals from very stable oxides that cannot be reduced by carbon.

6 Easy

6 of 186Question permalink

Understanding +4 | -1

In the extraction of copper, the metal formed in the Bessemer’s converter is due to the reaction:

View answer and explanation

Correct option: D

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 D (

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

) is correct because this reaction represents the self-reduction of cuprous oxide by cuprous sulfide in the Bessemer converter to produce copper.

Solution approach

Correct Answer + Core Reason: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$ is the reaction in Bessemer's converter.

Step-by-Step Solution:

Step 1: Recall the reactions in the extraction of copper.

Step 2: In the Bessemer converter, $Cu_2O$ reacts with $Cu_2S$ to produce copper and $SO_2$ .

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

Why other options are wrong:
- Option A: $Cu_2S \rightarrow 2Cu + S$ is not the primary reaction in the Bessemer converter.
- Option B: $2Cu_2O \rightarrow 4Cu + O_2$ is not the primary reaction in the Bessemer converter.
- Option C: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$ is the roasting process, not the final step in the Bessemer converter.

Key Concepts:
- Copper extraction
- Bessemer converter
- Self-reduction

Memory Aids:
- Remember the balanced equation for the final step.

Quick Tips:
- Understand the sequence of reactions in copper extraction.

7 Medium

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Remembering +4 | -1

Silver containing lead as impurity is purified by

View answer and explanation

Correct option: B

Cupellation

Option B (Cupellation) is correct because cupellation is a process used to purify silver containing lead as an impurity.

Solution approach

Correct Answer + Core Reason: Cupellation. Cupellation is used to purify silver containing lead impurity.

Step-by-Step Solution:

Step 1: Define Cupellation: Cupellation is a refining process in metallurgy, where metallic ores or concentrates are treated under very high temperatures and have controlled operations to separate noble metals, like silver or gold, from base metals like lead, copper, zinc, arsenic, antimony, or tellurium.

Step 2: In cupellation, the lead is oxidized to lead oxide (PbO), which is either volatilized or absorbed into the cupel, leaving the silver behind.

Why other options are wrong:
- Option A: Poling is used to refine copper and tin, not silver.
- Option C: Lavigation is a process of separating valuable minerals from gangue by using gravity and water, not for refining silver.
- Option D: Distillation is used for metals with low boiling points like zinc and mercury.

Key Concepts:
- Cupellation
- Refining of Silver

Memory Aids:
- Cupellation: Silver + Lead -> Pure Silver (Lead is oxidized and removed)

Quick Tips:
- Remember that cupellation is specifically used for silver refining when lead is present as an impurity.

8 Hard

8 of 186Question permalink

Analyzing +4 | -1

Statement-I: Leaching of gold with cyanide ion in absence of

air/{O_2}

leads to cyano complex of

Au\left( {III} \right).

Statement-II: Zinc is oxidized during the displacement reaction carried out for gold extraction.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: D

Statement-I is incorrect but statement-II is correct

Statement-I is incorrect because leaching of gold with cyanide in the presence of air/O2 leads to the formation of the

Au(I)

complex,

[Au(CN)_2]^-

. Statement-II is correct because zinc is oxidized to

Zn^{2+}

during the displacement reaction.

Solution approach

Correct Answer + Core Reason: Statement-I is incorrect but statement-II is correct. Leaching of gold forms $Au(I)$ complex, and zinc is oxidized in the displacement reaction.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Leaching of gold with cyanide ion in the *presence* of air/O2 leads to the formation of the cyano complex of $Au(I)$ , not $Au(III)$ . The reaction is: $4Au + 8CN^- + O_2 + 2H_2O \rightarrow 4[Au(CN)_2]^- + 4OH^-$ . Therefore, Statement-I is incorrect.

Step 2: Analyze Statement-II: During the displacement reaction, zinc reduces the gold complex, and in the process, zinc itself is oxidized. The reaction is: $2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$ . Zinc goes from oxidation state 0 to +2, so it is oxidized. Therefore, Statement-II is correct.

Why other options are wrong:
- Option A: Statement-I is incorrect.
- Option B: Statement-I is incorrect.
- Option C: Statement-II is correct.

Key Concepts:
- Leaching of gold
- Displacement reaction
- Oxidation and reduction

Memory Aids:
- Gold forms a $Au(I)$ complex during leaching with cyanide.
- Zinc is oxidized during the displacement reaction.

Quick Tips:
- Remember the oxidation states of metals in their complexes.

9 Easy

9 of 186Question permalink

Understanding +4 | -1

Leaching is a process of:

View answer and explanation

Correct option: B

Concentration

Option B (Concentration) is correct because leaching is a method used to selectively dissolve the desired ore, thus concentrating it.

Solution approach

Correct Answer + Core Reason: Concentration. Leaching selectively dissolves the desired ore, leaving impurities behind.

Step-by-Step Solution:

Step 1: Define leaching. Leaching is a process where a solid material is dissolved in a liquid solvent to extract the desired component.

Step 2: Explain its application in metallurgy. In metallurgy, leaching is used to selectively dissolve the ore, leaving behind the gangue (impurities).

Step 3: Examples of leaching agents. Common leaching agents include acids, bases, and cyanide solutions.

Why other options are wrong:
- Option A: Reduction. Reduction is a chemical process where a substance gains electrons. While reduction can occur during metal extraction, leaching itself is primarily a separation technique, not a reduction process.
- Option C: Refining. Refining is the process of purifying a metal after it has been extracted from its ore. Leaching occurs before refining.
- Option D: Oxidation. Oxidation is a chemical process where a substance loses electrons. While oxidation can be involved in some leaching processes, the primary purpose of leaching is concentration, not oxidation.

Key Concepts:
- Leaching
- Concentration
- Solvent Extraction

Memory Aids:
- Leaching: \"Leech\" - to extract something by dissolving it.

Quick Tips:
- Remember that leaching is a method for selectively dissolving the desired ore, thus concentrating it.

10 Easy

10 of 186Question permalink

Understanding +4 | -1

In the extraction of copper metal is formed in the Bessemer converter due to reaction

View answer and explanation

Correct option: A

C{u_2}S + 2C{u_2}O \to 6Cu + S{O_2}

Option A (

Cu_2S + 2Cu_2O \to 6Cu + SO_2

) is correct because this reaction represents the self-reduction of copper oxides and sulfides in the Bessemer converter.

Solution approach

Correct Answer + Core Reason: $Cu_2S + 2Cu_2O \to 6Cu + SO_2$ . This reaction represents the final step in copper extraction in the Bessemer converter.

Step-by-Step Solution:

Step 1: In the Bessemer converter, the matte (mainly $Cu_2S$ and $FeS$ ) is treated with silica lining and hot air.

Step 2: The iron sulfide is oxidized to iron oxide, which then reacts with silica to form slag.

Step 3: The remaining copper sulfide is partially oxidized to copper oxide, which then reacts with the remaining copper sulfide to form copper metal and sulfur dioxide.

Step 4: The reaction is: $Cu_2S + 2Cu_2O \rightarrow 6Cu + SO_2$

Why other options are wrong:
- Option B: $Cu_2S \to 2Cu + S$ is not the primary reaction in the Bessemer converter.
- Option C: $Fe + Cu_2O \to FeO + 2Cu$ is not the main reaction occurring in the Bessemer converter.
- Option D: $2Cu_2O \to 4Cu + O_2$ is not the primary reaction in the Bessemer converter.

Key Concepts:
- Metallurgy of Copper
- Bessemerization
- Self-reduction

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

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

11 Medium

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Remembering +4 | -1

Cyanide process is used to extraction of

View answer and explanation

Correct option: A

Ag

Option A (Ag) is correct because the cyanide process is used to extract gold and silver from their ores by forming soluble cyanide complexes.

Solution approach

Correct Answer + Core Reason: Cyanide process is used for Silver (Ag) extraction.

Step-by-Step Solution:

Step 1: Recall the cyanide process.

Step 2: The cyanide process involves leaching silver ore with a cyanide solution to form a soluble silver cyanide complex, $[Ag(CN)_2]^−$ .

Why other options are wrong:
- Option B: Nickel (Ni) is extracted using the Mond process.
- Option C: Platinum (Pt) is extracted using different methods, often involving aqua regia.
- Option D: Zinc (Zn) is extracted using electrolytic methods or reduction with carbon.

Key Concepts:
- Extraction of metals
- Cyanide process for Silver and Gold

Memory Aids:
- Cyanide is used to extract "shiny" metals like silver and gold.

Quick Tips:
- Remember that the cyanide process is specific to gold and silver extraction.

12 Easy

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Remembering +4 | -1

Vapour phase refining of nickel is carried out by using

View answer and explanation

Correct option: D

\mathrm{CO}

Option D (CO) is correct because carbon monoxide (CO) is used in the Mond process for vapor phase refining of nickel.

Solution approach

Correct Answer + Core Reason: CO. Carbon monoxide is used in the Mond process for refining nickel.

Step-by-Step Solution:

Step 1: Recall the Mond process: The Mond process involves reacting impure nickel with carbon monoxide (CO) to form volatile nickel tetracarbonyl ( $Ni(CO)_4$ ).

Step 2: The nickel tetracarbonyl is then heated to decompose it, yielding pure nickel and carbon monoxide, which can be recycled.

Step 3: The reaction is: $Ni + 4CO \rightleftharpoons Ni(CO)_4$

Why other options are wrong:
- Option A: $I_2$ is used in the Van Arkel method for refining Zr and Ti, not Ni.
- Option B: $Cl_2$ is not typically used in vapor phase refining of Ni.
- Option C: $HCl$ is not typically used in vapor phase refining of Ni.

Key Concepts:
- Mond Process
- Vapor Phase Refining
- Nickel Tetracarbonyl ( $Ni(CO)_4$ )

Memory Aids:
- Mond: Ni + CO -> $Ni(CO)_4$ -> Pure Ni

Quick Tips:
- Remember the reactants and products in the Mond process.
- Understand the purpose of forming a volatile compound in vapor phase refining.

13 Medium

13 of 186Question permalink

Understanding +4 | -1

In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with:

View answer and explanation

Correct option: B

Sulphur dioxide

Option B (Sulphur dioxide) is correct because in the final step, Cu2O is reduced to Cu by SO2 during self-reduction of copper sulphide ore.

Solution approach

Correct Answer + Core Reason: Sulphur dioxide. $Cu_2O$ is reduced to $Cu$ by $SO_2$ during self-reduction.

Step-by-Step Solution:

Step 1: Copper is extracted from its sulfide ore through a series of steps including roasting and reduction.

Step 2: Partial roasting converts some of the copper sulfide (Cu2S) to copper oxide (Cu2O) and copper sulfate (CuSO4).

Step 3: The remaining copper sulfide then reacts with the copper oxide in a self-reduction process:
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Step 4: In this process, $Cu_2O$ is reduced to $Cu$ and $Cu_2S$ is oxidized to $SO_2$ .

Why other options are wrong:
- Option A: Copper (I) sulfide is a reactant, not the reducing agent.
- Option C: Iron (II) sulfide is involved in earlier stages of the process, but not in the final reduction of cuprous oxide.
- Option D: Carbon monoxide is not typically used in the final reduction step of copper extraction from sulfide ores.

Key Concepts:
- Extraction of copper from sulfide ores
- Roasting
- Self-reduction

Memory Aids:
- Self-reduction: $Cu_2O + Cu_2S \rightarrow Cu + SO_2$

Quick Tips:
- Remember the self-reduction step in copper extraction.

14 Medium

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Remembering +4 | -1

Which of the following metal is correctly matched with its ore?
Metal

\;\;\;\;

Ore

View answer and explanation

Correct option: A

Zinc

\;\;\;\;

Calamine

Option A (Zinc - Calamine) is correct because Calamine (

ZnCO_3

) is a well-known ore of Zinc.

Solution approach

Correct Answer + Core Reason: Zinc - Calamine. Calamine ( $ZnCO_3$ ) is an ore of Zinc.

Step-by-Step Solution:

Step 1: Consider Zinc and Calamine: Calamine is zinc carbonate ( $ZnCO_3$ ), a common ore of zinc.

Step 2: Consider Silver and Ilmenite: Ilmenite ( $FeTiO_3$ ) is an ore of iron and titanium, not silver.

Step 3: Consider Magnesium and Cassiterite: Cassiterite ( $SnO_2$ ) is an ore of tin, not magnesium.

Step 4: Consider Tin and Azurite: Azurite ( $Cu_3(CO_3)_2(OH)_2$ ) is an ore of copper, not tin.

Why other options are wrong:
- Option B: Ilmenite is an ore of iron and titanium, not silver.
- Option C: Cassiterite is an ore of tin, not magnesium.
- Option D: Azurite is an ore of copper, not tin.

Key Concepts:
- Metal Ores
- Calamine ( $ZnCO_3$ )

Memory Aids:
- Calamine: Zinc Carbonate

Quick Tips:
- Memorize common metal ores and their corresponding metals.

15 Medium

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Understanding +4 | -1

The salt which is least likely to be found in minerals is:

View answer and explanation

Correct option: B

Acetate

Option B (Acetate) is correct because acetate salts are organic salts and less likely to be found in minerals compared to inorganic salts like sulfates, chlorides, and sulfides.

Solution approach

Correct Answer + Core Reason: Acetate. Acetate salts are organic and less common in minerals compared to inorganic salts.

Step-by-Step Solution:

Step 1: Consider the common types of salts found in minerals.

Step 2: Minerals are primarily composed of inorganic compounds.

Step 3: Evaluate the given options: Sulphates, Chlorides, and Sulphides are common inorganic salts found in minerals.

Step 4: Acetates are organic salts and are less likely to be found in mineral deposits.

Step 5: Conclude that Acetate is the least likely to be found in minerals.

Why other options are wrong:
- Option A: Sulphate salts are commonly found in minerals like gypsum (CaSO4·2H2O) and barite (BaSO4).
- Option C: Chloride salts are commonly found in minerals like halite (NaCl) and sylvite (KCl).
- Option D: Sulphide salts are commonly found in minerals like pyrite (FeS2) and galena (PbS).

Key Concepts:
- Mineral composition: Primarily inorganic compounds.
- Common types of salts: Sulphates, Chlorides, Sulphides.
- Organic vs. Inorganic compounds.

Memory Aids:
- Think of common minerals and their composition (e.g., rock salt is NaCl).

Quick Tips:
- Consider the origin and formation of minerals, which typically involve inorganic processes.

16 Medium

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Understanding +4 | -1

Considering Ellingham diagram, which of the following metals can be used to reduce alumina?

View answer and explanation

Correct option: C

\mathrm{Mg}

Option C (Mg) is correct because, according to the Ellingham diagram, magnesium has a lower Gibbs free energy of oxide formation than aluminum, making it a stronger reducing agent.

Solution approach

Correct Answer + Core Reason: $\mathrm{Mg}$ . Magnesium has a lower Gibbs free energy of oxide formation than aluminum.

Step-by-Step Solution:

Step 1: Understand the Ellingham diagram.\nThe Ellingham diagram plots the Gibbs free energy of formation ( $\Delta G_f^\circ$ ) of metal oxides as a function of temperature. A metal can reduce the oxide of another metal if its $\Delta G_f^\circ$ curve lies below that of the other metal's oxide.

Step 2: Consider the reduction of alumina ( $\mathrm{Al}_2\mathrm{O}_3$ ).\nWe need to find a metal whose oxide has a more negative $\Delta G_f^\circ$ than alumina.

Step 3: Analyze the given options.
- Copper (Cu) and Iron (Fe) generally have higher $\Delta G_f^\circ$ values for their oxides compared to aluminum oxide.
- Zinc (Zn) has a $\Delta G_f^\circ$ value closer to aluminum, but Magnesium (Mg) has a significantly lower (more negative) $\Delta G_f^\circ$ value.
- Magnesium (Mg) can reduce alumina because its oxide formation is thermodynamically more favorable.

Why other options are wrong:
- Option A: Copper (Cu) is a weaker reducing agent than aluminum and cannot reduce alumina.
- Option B: Iron (Fe) is a weaker reducing agent than aluminum and cannot reduce alumina.
- Option D: Zinc (Zn) is a weaker reducing agent than aluminum and cannot reduce alumina.

Key Concepts:
- Ellingham Diagram: Gibbs free energy of formation of metal oxides.
- Reducing Agent: A metal that can reduce the oxide of another metal.

Memory Aids:
- Lower on the Ellingham diagram = Stronger reducing agent.

Quick Tips:
- The metal with a more negative $\Delta G_f^\circ$ for its oxide can reduce the metal oxide with a less negative $\Delta G_f^\circ$ .

17 Medium

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Understanding +4 | -1

In the metallurgy of iron, when limestone is added to the blast furnace, the calcium ions end up in

View answer and explanation

Correct option: A

Calcium silicate (slag)

Option A (Calcium silicate (slag)) is correct because limestone (CaCO3) decomposes to CaO, which then reacts with silica (SiO2) to form calcium silicate slag.

Solution approach

Correct Answer + Core Reason: Calcium silicate (slag). Limestone decomposes to CaO which reacts with silica to form slag.

Step-by-Step Solution:

Step 1: Limestone (CaCO3) is added to the blast furnace as a flux.

Step 2: At high temperatures, CaCO3 decomposes to CaO and CO2: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Step 3: CaO then reacts with silica (SiO2), an impurity in the ore, to form calcium silicate (CaSiO3), which is the slag: $CaO(s) + SiO_2(s) \rightarrow CaSiO_3(l)$

Step 4: The slag is molten and floats on top of the molten iron, allowing it to be easily separated.

Why other options are wrong:
- Option B: Gangue refers to the original impurities in the ore, not the product of the reaction with limestone.
- Option C: Metallic calcium is not formed in this process. CaO reacts to form calcium silicate.
- Option D: Calcium carbonate is the starting material (limestone), not the final product containing calcium ions.

Key Concepts:
- Blast furnace in iron metallurgy
- Role of limestone as a flux
- Slag formation

Memory Aids:
- Limestone (CaCO3) -> CaO -> Slag (CaSiO3)

Quick Tips:
- Remember that the purpose of limestone is to remove silica impurities as slag.

18 Easy

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Understanding +4 | -1

Metallurgy is the process of:

View answer and explanation

Correct option: C

Extracting the metal from the ore

Option C (Extracting the metal from the ore) is correct because metallurgy is the science and technology of extracting metals from their ores.

Solution approach

Correct Answer + Core Reason: Extracting the metal from the ore. Metallurgy is the process of extracting metals.

Step-by-Step Solution:
Step 1: Define Metallurgy.
Step 2: Metallurgy is the process of extracting metals from their ores and refining them for use.

Why other options are wrong:
- Option A: Concentrating the ore is a part of metallurgy, but not the entire process.
- Option B: Roasting the ore is a part of metallurgy, but not the entire process.
- Option D: Adding carbon to the ore in a blast furnace is a specific step in iron extraction, not the definition of metallurgy itself.

Key Concepts:
- Metallurgy
- Metal extraction

Memory Aids:
- Metallurgy involves extracting metals.

Quick Tips:
- Understand the different processes involved in metallurgy.

19 Medium

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Understanding +4 | -1

Alloy formation gives rise to:

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because alloy formation generally leads to decreased corrosion, increased hardness, and decreased conductivity.

Solution approach

Correct Answer + Core Reason: All are correct. Alloy formation generally decreases corrosion, increases hardness, and decreases conductivity.

Step-by-Step Solution:

Step 1: Understand the concept of alloy formation.

Step 2: Alloy formation involves mixing two or more metals (or a metal and a non-metal) to create a new material with enhanced properties.

Step 3: Consider the effects of alloy formation on different properties:
- Corrosion: Alloys often have improved corrosion resistance compared to pure metals.
- Hardness: Alloys are generally harder than their constituent metals.
- Conductivity: The presence of different metals in an alloy disrupts the regular arrangement of atoms, leading to decreased electrical conductivity.

Step 4: Conclude that all the given statements are correct.

Why other options are wrong:
- Option A: Decrease in corrosion is a correct statement, but not the only correct one.
- Option B: Increase in hardness is a correct statement, but not the only correct one.
- Option C: Decrease in conductivity is a correct statement, but not the only correct one.

Key Concepts:
- Alloy formation: Mixing metals to enhance properties.
- Effects of alloying: Corrosion resistance, hardness, conductivity.

Memory Aids:
- Alloys are generally stronger and more resistant than pure metals.

Quick Tips:
- Remember the general trends in properties upon alloy formation.

20 Easy

20 of 186Question permalink

Remembering +4 | -1

Which one is malachite from the following?

View answer and explanation

Correct option: D

CuC{O_3}.\,Cu{(OH)_2}

Option D (

CuCO_3.\,Cu(OH)_2

) is correct because it represents the chemical formula of malachite, a copper carbonate hydroxide mineral.

Solution approach

Correct Answer + Core Reason: $CuCO_3.\,Cu(OH)_2$ is the chemical formula for malachite.

Step-by-Step Solution:

Step 1: Recall the definition of malachite.

Step 2: Malachite is a copper carbonate hydroxide mineral.

Step 3: Identify the chemical formula that corresponds to copper carbonate hydroxide.

Step 4: The chemical formula for malachite is $CuCO_3.\,Cu(OH)_2$ .

Why other options are wrong:
- Option A: $CuFeS_2$ is the chemical formula for chalcopyrite, a copper iron sulfide mineral.
- Option B: $Cu(OH)_2$ is the chemical formula for copper(II) hydroxide, but malachite also contains carbonate.
- Option C: $Fe_3O_4$ is the chemical formula for magnetite, an iron oxide mineral.

Key Concepts:
- Mineral identification
- Chemical formulas of minerals

Memory Aids:
- None applicable

Quick Tips:
- Memorize common mineral formulas.

21 Easy

21 of 186Question permalink

Remembering +4 | -1

Which of the following is correct?

View answer and explanation

Correct option: C

Calomel is used medically as purgative

Option C is correct because calomel (Hg2Cl2) is used medically as a purgative (laxative).

Solution approach

Correct Answer + Core Reason: Calomel is used medically as purgative. Calomel (Hg2Cl2) has been used as a purgative.

Step-by-Step Solution:

Step 1: Recall the properties and uses of calomel.

Step 2: Calomel (Hg2Cl2) is mercurous chloride, not mercuric chloride.
- It has been used medically as a purgative.
- It is not freely soluble in water.
- It is not widely used as an antiseptic.

Why other options are wrong:
- Option A: Calomel is mercurous chloride (Hg2Cl2), not mercuric chloride (HgCl2).
- Option B: Calomel is not widely used as an antiseptic.
- Option D: Calomel is not freely soluble in water.

Key Concepts:
- Properties and uses of calomel

Memory Aids:
- Calomel: Remember its use as a purgative.

Quick Tips:
- Know the different forms of mercury chlorides and their uses.

22 Hard

22 of 186Question permalink

Understanding +4 | -1

Which of the following statements are incorrect?

I. Zinc can be extracted by self-reduction.

II. A depressant prevents certain type of particles to come to the froth.

III. Copper matte contains

{ZnS}

and

{Cu}_2{S}

.

IV. The solidified copper obtained from reverberatory furnace has blistered appearance due to evolution of

{SO}_2

during the extraction.

View answer and explanation

Correct option: C

I and III

Option C (I and III) is correct because Zinc cannot be extracted by self-reduction, and copper matte contains Cu2S and FeS, not ZnS.

Solution approach

Correct Answer + Core Reason: I and III. Zinc cannot be extracted by self-reduction and copper matte contains $Cu_2S$ and $FeS$ .

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: I. Zinc cannot be extracted by self-reduction because the reduction potential of ZnO to Zn is too negative. It requires a stronger reducing agent like carbon.

Step 3: II. A depressant prevents certain types of particles from coming to the froth in froth flotation. This statement is correct.

Step 4: III. Copper matte contains primarily $Cu_2S$ and $FeS$ , not $ZnS$ .

Step 5: IV. The solidified copper obtained from the reverberatory furnace has a blistered appearance due to the evolution of $SO_2$ during the extraction. This statement is correct.

Why other options are wrong:
- Option A: Statement II is correct.
- Option B: Statement II is correct.
- Option D: Statement IV is correct.

Key Concepts:
- Self-reduction
- Froth flotation
- Copper matte composition

Memory Aids:
- Zinc needs carbon reduction, not self-reduction.

Quick Tips:
- Remember the composition of copper matte and the conditions for self-reduction.

23 Medium

23 of 186Question permalink

Understanding +4 | -1

Which of the following statements regarding the metallurgy of magnesium using electrolytic method is not correct?

View answer and explanation

Correct option: D

Molten magnesium is heavier than the electrolyte

Option D (Molten magnesium is heavier than the electrolyte) is correct because molten magnesium is lighter than the electrolyte (MgCl2 + NaCl + NaF).

Solution approach

Correct Answer + Core Reason: Molten magnesium is heavier than the electrolyte. Molten magnesium is actually lighter than the electrolyte used in the process.

Step-by-Step Solution:

Step 1: Understand the electrolytic method for magnesium extraction.

Step 2: The electrolyte typically consists of magnesium chloride (MgCl2) mixed with other salts like NaCl and NaF to lower the melting point.

Step 3: Consider the density of molten magnesium relative to the electrolyte. Molten magnesium is less dense than the molten electrolyte mixture.

Step 4: Therefore, molten magnesium floats on top of the electrolyte and can be easily collected.

Step 5: Conclude that the statement claiming molten magnesium is heavier than the electrolyte is incorrect.

Why other options are wrong:
- Option A: Electrolyte is magnesium chloride containing a little of $\mathrm{NaCl}\;and\;\mathrm{NaF}$ is a correct statement. These salts are added to lower the melting point of MgCl2.
- Option B: Air tight iron pot acts as a cathode is a correct statement. The iron pot serves as the cathode in the electrolytic cell.
- Option C: Electrolysis is done in the atmosphere of coal gas is a correct statement. Coal gas provides an inert atmosphere to prevent oxidation of magnesium.

Key Concepts:
- Electrolytic extraction of magnesium.
- Electrolyte composition: MgCl2, NaCl, NaF.
- Density of molten magnesium.

Memory Aids:
- Remember that magnesium is a relatively light metal.

Quick Tips:
- Pay attention to the relative densities of the metal and the electrolyte in electrolytic extraction processes.

24 Hard

24 of 186Question permalink

Analyzing +4 | -1

Which of the following statements about the advantages of roasting of sulphide ore before reduction is not true?

View answer and explanation

Correct option: D

Carbon and hydrogen are suitable reducing agents for metal sulphides

Option D is correct because carbon and hydrogen are generally not suitable reducing agents for metal sulfides directly; roasting converts the sulfide to an oxide, which is then reduced.

Solution approach

Correct Answer + Core Reason: Carbon and hydrogen are suitable reducing agents for metal sulphides. This statement is incorrect because metal sulfides are typically roasted to oxides before reduction with carbon or hydrogen.

Step-by-Step Solution:

Step 1: Understand the context: The question asks about the advantages of roasting sulfide ores before reduction.

Step 2: Analyze option A: $\Delta G_f^\circ$ of the sulfide is greater than $CS_2$ and $H_2S$ . This is generally true, making the formation of $CS_2$ or $H_2S$ less favorable than the metal sulfide itself.

Step 3: Analyze option B: $\Delta G_f^\circ$ is negative for roasting of sulfide ore to oxide. This is true, indicating that the roasting process is thermodynamically favorable.

Step 4: Analyze option C: Roasting of the sulfide to oxide is thermodynamically feasible. This is true, as indicated by the negative $\Delta G_f^\circ$ .

Step 5: Analyze option D: Carbon and hydrogen are suitable reducing agents for metal sulfides. This is generally false. Sulfides are first converted to oxides by roasting, and then the oxides are reduced by carbon or hydrogen.

Step 6: Conclusion: Option D is the incorrect statement.

Why other options are wrong:
- Option A: This statement is generally true. The Gibbs free energy of formation of metal sulfides is typically higher (more positive or less negative) than that of $CS_2$ and $H_2S$ , making the formation of $CS_2$ or $H_2S$ less favorable.
- Option B: This statement is true. Roasting sulfide ores to form oxides is a thermodynamically favorable process, indicated by a negative Gibbs free energy change.
- Option C: This statement is true. Because the Gibbs free energy change for roasting is negative, the process is thermodynamically feasible.
- Option D: This statement is incorrect. Metal sulfides are not directly reduced by carbon or hydrogen. They are first converted to oxides by roasting, and then the oxides are reduced.

Key Concepts:
- Thermodynamics of metallurgical processes
- Gibbs free energy of formation
- Roasting of sulfide ores
- Reduction of metal oxides

Memory Aids:
- Remember that sulfides are usually roasted to oxides before reduction.

Quick Tips:
- When dealing with metallurgical processes, consider the thermodynamic feasibility of each step.

25 Easy

25 of 186Question permalink

Understanding +4 | -1

Which method of purification is represented by the following equations

Ti + 2{I_2}\xrightarrow{{523\,K}}Ti{I_4}\xrightarrow{{1700\,K}}Ti + 2{I_2}

View answer and explanation

Correct option: C

Van Arkle

Option C (Van Arkle) is correct because the given equations represent the Van Arkel method, which involves forming a volatile iodide to purify metals.

Solution approach

Correct Answer + Core Reason: Van Arkle. The given reaction is the Van Arkel method for refining metals like Titanium.

Step-by-Step Solution:

Step 1: The Van Arkel method involves converting the impure metal to a volatile iodide at a lower temperature.

Step 2: The volatile iodide is then decomposed at a higher temperature to yield the pure metal.

Step 3: The reactions are: $Ti + 2I_2 \xrightarrow{{523 K}} TiI_4$ and $TiI_4 \xrightarrow{{1700 K}} Ti + 2I_2$

Why other options are wrong:
- Option A: Cupellation is used for refining precious metals like silver and gold.
- Option B: Poling is used for refining copper and involves stirring molten metal with green wood.
- Option D: Zone refining is based on the principle that impurities are more soluble in the molten phase than in the solid phase.

Key Concepts:
- Refining of metals
- Van Arkel method

Memory Aids:
- Van Arkel uses Iodine to purify metals like Ti and Zr.

Quick Tips:
- Remember the specific refining methods for different metals.

26 Easy

26 of 186Question permalink

Remembering +4 | -1

Composition of azurite mineral is

View answer and explanation

Correct option: C

2\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}

Option C is correct because the chemical formula for azurite is

2\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}

.

Solution approach

Correct Answer + Core Reason: $2\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}$ . This is the correct chemical formula for azurite.

Step-by-Step Solution:

Step 1: Recall the chemical formula of azurite.

Step 2: Azurite is a copper carbonate hydroxide mineral with the formula $2\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}$ .

Why other options are wrong:
- Option A: Incorrect formula.
- Option B: Incorrect formula.
- Option D: Incorrect formula.

Key Concepts:
- Mineral formulas
- Chemical composition of azurite

Memory Aids:
- Azurite: Remember the copper carbonate hydroxide formula.

Quick Tips:
- Memorize the formulas of common minerals.

27 Easy

27 of 186Question permalink

Remembering +4 | -1

Which of the following pairs of metal is purified by zone refining method ?

View answer and explanation

Correct option: D

Si and Ge

Option D is correct because Silicon (Si) and Germanium (Ge) are semiconductors that are commonly purified using the zone refining method.

Solution approach

Correct Answer + Core Reason: Si and Ge. Zone refining is commonly used to purify semiconductors like Si and Ge.

Step-by-Step Solution:

Step 1: Understand the principle of zone refining.
- Zone refining is a method for purifying materials by melting a small region (zone) and moving it along the material.
- Impurities tend to concentrate in the molten zone.

Step 2: Identify the metals that are commonly purified by zone refining.
- Silicon (Si) and Germanium (Ge) are semiconductors that require high purity for their applications.

Why other options are wrong:
- Option A: Ga and Ni are not typically purified by zone refining.
- Option B: Zr and Ti can be purified by Van Arkel method.
- Option C: Ag and Au are purified by other methods like electrolytic refining.

Key Concepts:
- Zone refining
- Purification of semiconductors

Memory Aids:
- Zone refining: Think of semiconductors (Si, Ge).

Quick Tips:
- Remember the materials that require high purity and the methods used to achieve it.

28 Medium

28 of 186Question permalink

Understanding +4 | -1

Softening of lead means:

View answer and explanation

Correct option: C

Removal of metallic impurities from lead

Option C (Removal of metallic impurities from lead) is correct because softening of lead refers to the removal of impurities like tin, antimony, and arsenic.

Solution approach

Correct Answer + Core Reason: Removal of metallic impurities from lead. Softening of lead refers to removing impurities like Sn, Sb, and As.

Step-by-Step Solution:

Step 1: Define softening of lead. Softening of lead is a process used to remove metallic impurities from lead, making it more malleable and less brittle.

Step 2: Identify common impurities in lead. Common impurities include tin (Sn), antimony (Sb), and arsenic (As).

Step 3: Explain the process of softening. The process typically involves oxidizing the impurities and removing them as slag or by other methods.

Why other options are wrong:
- Option A: Conversion of lead into $\mathrm{PbO}$ . Conversion of lead into PbO is a different process, such as oxidation of lead, and is not related to softening.
- Option B: Conversion of lead into $\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}$ . Conversion of lead into $\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}$ is also a different process and not related to softening.
- Option D: Washing lead with $\mathrm{HN}{\mathrm{O}_3}$ followed by dilute alkali solution. Washing with nitric acid and alkali might be used in some purification processes, but it is not the standard definition of softening lead.

Key Concepts:
- Softening of Lead
- Metallic Impurities
- Purification

Memory Aids:
- Softening: Think of \"softening\" the lead by removing impurities that make it hard or brittle.

Quick Tips:
- Remember that softening of lead specifically refers to the removal of metallic impurities.

29 Hard

29 of 186Question permalink

Remembering +4 | -1

In Mac Arthur forrest method, silver is extracted from the solution of

\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

by the use of

View answer and explanation

Correct option: D

Zn

Option D (Zn) is correct because Zinc is used as a reducing agent to precipitate silver from the complex ion solution in the Mac Arthur Forrest method.

Solution approach

Correct Answer + Core Reason: Zn. Zinc is used as a reducing agent to displace silver from the complex.

Step-by-Step Solution:

Step 1: Understand the Mac Arthur Forrest method. This method involves leaching silver ore with a cyanide solution to form a silver cyanide complex.

Step 2: The silver is then recovered from the solution by adding a more electropositive metal, which acts as a reducing agent.

Step 3: Zinc is more electropositive than silver and can displace it from the complex ion, $\mathrm{Na}[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}]$ . The reaction is:
$\mathrm{2Na}[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}] + \mathrm{Zn} \to \mathrm{Na}_2[{\mathrm{Zn}{{\left( {\mathrm{CN}} \right)}_4}}] + \mathrm{2Ag}$

Step 4: Silver precipitates out as a solid.

Why other options are wrong:
- Option A: Fe (Iron) is not typically used in the Mac Arthur Forrest method for silver extraction.
- Option B: Mg (Magnesium) is a strong reducing agent but not typically used in this specific process.
- Option C: Cu (Copper) is less electropositive than silver and cannot displace it from the cyanide complex.
- Option D: Zn (Zinc) is the correct reducing agent used in this process.

Key Concepts:
- Extraction of metals
- Redox reactions
- Electrochemistry
- Complex formation

Memory Aids:
- Zinc is more reactive than silver, so it can displace it.

Quick Tips:
- Remember the reactivity series to determine which metal can displace another.

30 Medium

30 of 186Question permalink

Remembering +4 | -1

Mond process is used in the extraction of:

View answer and explanation

Correct option: B

Ni

Option B (Ni) is correct because the Mond process is specifically used for the extraction and purification of nickel.

Solution approach

Correct Answer + Core Reason: Mond process is used for Nickel (Ni) extraction.

Step-by-Step Solution:

Step 1: Recall the Mond process.

Step 2: The Mond process involves the formation of volatile nickel carbonyl, $Ni(CO)_4$ , which is then decomposed to yield pure nickel.

Why other options are wrong:
- Option A: Cobalt (Co) is extracted using different methods.
- Option C: Molybdenum (Mo) is extracted using different methods.
- Option D: Zinc (Zn) is extracted using different methods like the electrolytic process.

Key Concepts:
- Extraction of metals
- Mond process for Nickel

Memory Aids:
- Mond = Money = Nickel (association for remembering).

Quick Tips:
- Memorize specific extraction processes for different metals.

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