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

1 Medium

1 of 85Question permalink

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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.

2 Medium

2 of 85Question permalink

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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.

3 Medium

3 of 85Question permalink

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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.

4 Medium

4 of 85Question permalink

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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.

5 Medium

5 of 85Question permalink

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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.

6 Medium

6 of 85Question permalink

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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.

7 Medium

7 of 85Question permalink

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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.

8 Medium

8 of 85Question permalink

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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.

9 Medium

9 of 85Question permalink

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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 $\Delta G_f^\circ$ .

10 Medium

10 of 85Question permalink

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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.

11 Medium

11 of 85Question permalink

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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.

12 Medium

12 of 85Question permalink

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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.

13 Medium

13 of 85Question permalink

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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.

14 Medium

14 of 85Question 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.

15 Medium

15 of 85Question permalink

Analyzing +4 | -1

Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

View answer and explanation

Correct option: C

Metal sulphides are less stable than the corresponding oxides

Option C (Metal sulphides are less stable than the corresponding oxides) is correct because the higher stability of oxides compared to sulfides makes roasting favorable, as it converts the sulfide to a more stable oxide.

Solution approach

Correct Answer + Core Reason: Metal sulphides are less stable than the corresponding oxides. This makes roasting favorable.

Step-by-Step Solution:

Step 1: Understand the context of roasting sulphide ores.\nRoasting is the process of heating a sulphide ore in the presence of air to convert it into its oxide. This is often done before carbon reduction.

Step 2: Consider the thermodynamic stability of sulphides and oxides.\nMetal oxides are generally more thermodynamically stable than metal sulphides. This is why roasting is preferred.

Step 3: Analyze the given options.
- Option A: Metal sulphides being thermodynamically more stable than $\mathrm{C}{\mathrm{S}_2}$ is a valid point, but not the primary reason.
- Option B: $\mathrm{C}{\mathrm{O}_2}$ being thermodynamically more stable than $\mathrm{C}{\mathrm{S}_2}$ is also a valid point, but not the primary reason.
- Option D: $\mathrm{C}{\mathrm{O}_2}$ being more volatile than $\mathrm{C}{\mathrm{S}_2}$ is a contributing factor, but not the main reason.
- The primary reason is the higher stability of metal oxides compared to metal sulphides.

Why other options are wrong:
- Option A: While relevant, it's not the core reason for roasting.
- Option B: While relevant, it's not the core reason for roasting.
- Option D: While relevant, it's not the core reason for roasting.

Key Concepts:
- Roasting: Converting sulphide ore to oxide by heating in air.
- Thermodynamic Stability: Oxides are generally more stable than sulphides.

Memory Aids:
- Oxides are "Overall" more stable than sulphides.

Quick Tips:
- Remember that roasting converts sulphides to more stable oxides before carbon reduction.

16 Medium

16 of 85Question permalink

Understanding +4 | -1

Other forms of iron can be produced from:

View answer and explanation

Correct option: C

Pig iron

Option C (Pig iron) is correct because pig iron is the raw product of smelting iron ore in a blast furnace, and other forms of iron are derived from it through further processing.

Solution approach

Correct Answer + Core Reason: Pig iron. Other forms of iron are produced from pig iron.

Step-by-Step Solution:
Step 1: Understand the different forms of iron.
Step 2: Pig iron is the crude iron obtained from a blast furnace. Cast iron, wrought iron, and steel are all produced from pig iron through various refining processes.

Why other options are wrong:
- Option A: Cast iron is a type of iron produced from pig iron, not the other way around.
- Option B: Wrought iron is a type of iron produced from pig iron, not the other way around.
- Option D: Steel is an alloy of iron produced from pig iron, not the other way around.

Key Concepts:
- Different forms of iron
- Iron production

Memory Aids:
- Pig iron is the starting material for other forms of iron.

Quick Tips:
- Remember the different types of iron and their production methods.

17 Medium

17 of 85Question permalink

Remembering +4 | -1

Blister copper is

View answer and explanation

Correct option: D

Cu having 1% impurity

Option D is correct because Blister copper is copper containing about 1% impurities, mainly dissolved gases.

Solution approach

Correct Answer + Core Reason: Cu having 1% impurity. Blister copper contains about 1% impurities.

Step-by-Step Solution:

Step 1: Recall the composition of blister copper.

Step 2: Blister copper is copper containing about 1% impurities, mainly dissolved gases like $SO_2$ .

Step 3: Compare the given options with the known composition of blister copper.

Why other options are wrong:
- Option A: Impure Cu is a general term, but blister copper has a specific impurity level.
- Option B: Cu alloy is a mixture of copper with other metals, not the same as blister copper.
- Option C: Pure Cu is refined copper, not blister copper.

Key Concepts:
- Metallurgy of copper
- Blister copper

Memory Aids:
- Blister copper has blisters due to escaping gases.

Quick Tips:
- Remember the approximate impurity level in blister copper.

18 Medium

18 of 85Question permalink

Understanding +4 | -1

Which one of the following statements is false?

View answer and explanation

Correct option: B

The ore is freed from almost all nonmetallic impurities.

Option B is correct because roasting removes moisture and converts sulfide ores to oxides, but does not remove almost all nonmetallic impurities.

Solution approach

Correct Answer + Core Reason: The ore is freed from almost all nonmetallic impurities. Roasting primarily removes volatile impurities and converts sulfides to oxides, but doesn't eliminate all nonmetallic impurities.

Step-by-Step Solution:

Step 1: Understand the process of roasting.

Step 2: Roasting involves heating an ore in the presence of air to remove volatile impurities and convert the ore to its oxide form.

Step 3: Consider the extent of impurity removal. Roasting removes moisture, sulfur, and other volatile substances, but it does not completely eliminate all nonmetallic impurities.

Step 4: Conclude that the statement claiming almost complete removal of nonmetallic impurities is false.

Why other options are wrong:
- Option A: During roasting, moisture is indeed removed from the ore. This is a true statement.
- Option C: Calcination is typically carried out in the absence or limited supply of air, not with a blast of air. This is a true statement.
- Option D: Concentrated zinc blende (ZnS) is subjected to calcination (or roasting) during its extraction by pyrometallurgy to convert it to ZnO. This is a true statement.

Key Concepts:
- Roasting: Heating an ore in the presence of air.
- Calcination: Heating an ore in the absence or limited supply of air.
- Pyrometallurgy: Extraction of metals using heat.

Memory Aids:
- Roasting: Air is involved (think of roasting food). Calcination: Absence of air.

Quick Tips:
- Pay attention to the degree of impurity removal mentioned in the statements. Processes like roasting and calcination remove some, but not all, impurities.

19 Medium

19 of 85Question permalink

Remembering +4 | -1

In blast furnace, iron oxide is reduced by

View answer and explanation

Correct option: B

Carbon monoxide

Option B (Carbon monoxide) is correct because in a blast furnace, iron oxide is primarily reduced by carbon monoxide (CO) formed from the partial combustion of coke.

Solution approach

Correct Answer + Core Reason: Carbon monoxide. In blast furnace, iron oxide is reduced mainly by carbon monoxide.

Step-by-Step Solution:

Step 1: Describe the blast furnace process.

Step 2: Coke (carbon) is burned in the blast furnace with a hot blast of air to produce carbon monoxide (CO).

Step 3: The carbon monoxide then reduces the iron oxide (Fe2O3) to iron (Fe).

Step 4: The main reducing agent is carbon monoxide, not carbon directly.

Why other options are wrong:
- Option A: Hot blast of air is incorrect. The hot blast of air provides oxygen for the combustion of coke.
- Option C: Carbon is incorrect. Carbon reacts with oxygen to form carbon monoxide, which is the actual reducing agent.
- Option D: Silica is incorrect. Silica acts as a flux to remove impurities.

Key Concepts:
- Blast furnace process.
- Reduction of iron oxide.
- Role of carbon monoxide.

Memory Aids:
- CO is the key reducing agent in the blast furnace.

Quick Tips:
- Remember the chemical reactions occurring in the blast furnace.

20 Medium

20 of 85Question permalink

Analyzing +4 | -1

Upon heating with

\mathrm{Cu_2 S}

, the reagent(s) that do not give copper metal is:

View answer and explanation

Correct option: A

\mathrm{CuFeS_2}

Option A (

\mathrm{CuFeS_2}

) is correct because

\mathrm{CuFeS_2}

when heated with

\mathrm{Cu_2S}

will not produce copper metal; it forms a mixture of copper and iron sulfides.

Solution approach

Correct Answer + Core Reason: $\mathrm{CuFeS_2}$ . $\mathrm{CuFeS_2}$ does not yield copper metal upon heating with $\mathrm{Cu_2S}$ .

Step-by-Step Solution:

Step 1: Identify the reaction type: This is a metallurgical process question, specifically related to the extraction of copper.

Step 2: Consider the reactions of each option with $\mathrm{Cu_2S}$ :
- A. $\mathrm{CuFeS_2}$ : This is chalcopyrite. Heating it with $\mathrm{Cu_2S}$ does not directly yield copper metal.
- B. $\mathrm{CuO}$ : $\mathrm{CuO + Cu_2S \rightarrow 3Cu + SO_2}$
- C. $\mathrm{Cu_2O}$ : $\mathrm{Cu_2O + Cu_2S \rightarrow 6Cu + SO_2}$
- D. $\mathrm{CuSO_4}$ : $\mathrm{2CuSO_4 + Cu_2S \rightarrow 4Cu + 3SO_2}$

Step 3: Based on the reactions, $\mathrm{CuO}$ , $\mathrm{Cu_2O}$ , and $\mathrm{CuSO_4}$ can produce copper metal when heated with $\mathrm{Cu_2S}$ , while $\mathrm{CuFeS_2}$ does not.

Step 4: Conclusion: $\mathrm{CuFeS_2}$ is the correct answer.

Why other options are wrong:
- **Option B ( $\mathrm{CuO}$ ):** $\mathrm{CuO}$ reacts with $\mathrm{Cu_2S}$ to produce copper metal.
- **Option C ( $\mathrm{Cu_2O}$ ):** $\mathrm{Cu_2O}$ reacts with $\mathrm{Cu_2S}$ to produce copper metal.
- **Option D ( $\mathrm{CuSO_4}$ ):** $\mathrm{CuSO_4}$ reacts with $\mathrm{Cu_2S}$ to produce copper metal.

Key Concepts:
- Extraction of metals
- Copper metallurgy
- Reactions of metal oxides and sulfides

Memory Aids:
- Remember the self-reduction reactions of copper oxides and sulfates with copper sulfide.

Quick Tips:
- Know the common ores of copper and their reactions in the extraction process.

21 Medium

21 of 85Question permalink

Analyzing +4 | -1

Identify the statement which is not correct for copper sulphate:

View answer and explanation

Correct option: B

It reacts with

{KCl}

to give

{Cu}_2 {Cl}_2

Option B (It reacts with

{KCl}

to give

{Cu}_2 {Cl}_2

) is correct because copper sulfate does not react with KCl to form Cu2Cl2.

Solution approach

Correct Answer + Core Reason: It reacts with ${KCl}$ to give ${Cu}_2 {Cl}_2$ . Copper sulfate does not react with KCl to form Cu2Cl2.

Step-by-Step Solution:

Step 1: Consider the chemical properties of copper sulfate (CuSO4).

Step 2: Evaluate the reactions of CuSO4 with the given reagents:
- NaOH and glucose: CuSO4 reacts with NaOH and glucose to give Cu2O (reduction of Cu2+ to Cu+).
- KCl: CuSO4 does not react with KCl to form Cu2Cl2. The reaction is not thermodynamically favorable.
- Strong heating: CuSO4 decomposes on strong heating to give CuO, SO2, and O2.
- KI: CuSO4 reacts with KI to liberate I2 (oxidation of I- to I2).

Step 3: Conclude that the statement claiming CuSO4 reacts with KCl to give Cu2Cl2 is incorrect.

Why other options are wrong:
- Option A: It reacts with $\mathrm{NaOH}$ and glucose to give $\mathrm{Cu}_2 \mathrm{O}$ is a correct statement. CuSO4 is reduced to Cu2O in this reaction.
- Option C: It gives $\mathrm{CuO}$ on strong heating is a correct statement. CuSO4 decomposes to CuO at high temperatures.
- Option D: It reacts with $\mathrm{KI}$ to give $\mathrm{I}_2$ is a correct statement. CuSO4 oxidizes I- to I2.

Key Concepts:
- Chemical properties of copper sulfate.
- Redox reactions.
- Thermal decomposition of copper sulfate.

Memory Aids:
- Remember the common reactions of copper sulfate with different reagents.

Quick Tips:
- Consider the redox potentials of the species involved in the reactions.

22 Medium

22 of 85Question permalink

Analyzing +4 | -1

Which of the following becomes paramagnetic on heating at

850 \mathrm{~K}

?

View answer and explanation

Correct option: D

(2) & (3) both

Option D ((2) & (3) both) is correct because both Fe3O4 and ZnFe2O4 become paramagnetic on heating due to changes in their magnetic ordering.

Solution approach

Correct Answer + Core Reason: (2) & (3) both. Both $Fe_3O_4$ and $ZnFe_2O_4$ become paramagnetic on heating at 850 K due to disruption of their magnetic ordering.

Step-by-Step Solution:

Step 1: $\mathrm{MnO}$ : MnO is antiferromagnetic at low temperatures and becomes paramagnetic above its Néel temperature. However, the question specifies 850 K, and MnO typically transitions to paramagnetic at a much lower temperature. So, MnO is already paramagnetic at 850K.

Step 2: $\mathrm{Fe}_3 \mathrm{O}_4$ : $Fe_3O_4$ (magnetite) is ferrimagnetic at room temperature. Upon heating above its Curie temperature (around 858 K), it loses its ferrimagnetism and becomes paramagnetic.

Step 3: $\mathrm{ZnFe}_2 \mathrm{O}_4$ : $ZnFe_2O_4$ is a ferrite that can exhibit different magnetic behaviors depending on the cation distribution. At lower temperatures, it can be ferrimagnetic or antiferromagnetic. Heating to 850 K can disrupt the magnetic ordering, causing it to become paramagnetic.

Why other options are wrong:
- Option A: MnO is already paramagnetic at 850K.
- Option B: Only $Fe_3O_4$ is considered.
- Option C: Only $ZnFe_2O_4$ is considered.

Key Concepts:
- Paramagnetism
- Ferromagnetism
- Ferrimagnetism
- Curie temperature
- Néel temperature

Memory Aids:
- Curie Temperature: Ferro/Ferri -> Para
- Néel Temperature: Anti-ferro -> Para

Quick Tips:
- Know the magnetic properties of common oxides and how temperature affects them.

23 Medium

23 of 85Question permalink

Understanding +4 | -1

The slag obtained during the smelting process in the extraction of copper from copper pyrites is composed mainly of:

View answer and explanation

Correct option: B

\mathrm{FeSi}{\mathrm{O}_3}

Option B (

\mathrm{FeSi}{\mathrm{O}_3}

) is correct because during smelting, iron oxide reacts with silica to form iron silicate slag.

Solution approach

Correct Answer + Core Reason: $\mathrm{FeSi}{\mathrm{O}_3}$ . Iron silicate is the main component of the slag formed during copper extraction from copper pyrites.

Step-by-Step Solution:

Step 1: Describe the smelting process for copper extraction. Smelting involves heating the ore with a reducing agent (e.g., coke) and a flux (e.g., silica).

Step 2: Explain the role of silica. Silica ( $\mathrm{Si}{\mathrm{O}_2}$ ) acts as a flux, reacting with iron oxide (FeO) to form slag.

Step 3: Write the reaction for slag formation. $\mathrm{FeO} + \mathrm{Si}{\mathrm{O}_2} \rightarrow \mathrm{FeSi}{\mathrm{O}_3}$

Step 4: Explain the composition of the slag. The slag is primarily composed of iron silicate ( $\mathrm{FeSi}{\mathrm{O}_3}$ ), which is lighter than the molten copper and can be easily separated.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}_2}\mathrm{S}$ . $\mathrm{C}{\mathrm{u}_2}\mathrm{S}$ is an intermediate product in the extraction of copper, but it is not the main component of the slag.
- Option C: $\mathrm{CuSi}{\mathrm{O}_3}$ . Copper silicate is not the primary component of the slag formed during copper extraction from copper pyrites. Iron silicate is the major component.
- Option D: $\mathrm{Si}{\mathrm{O}_2}$ . While silica is used as a flux, it reacts with iron oxide to form iron silicate slag. Silica itself is not the main component of the slag.

Key Concepts:
- Smelting
- Slag Formation
- Flux
- Copper Extraction

Memory Aids:
- Slag: \"Sl\"ag - \"Si\"lica reacts with \"I\"ron to form \"Sl\"ag.

Quick Tips:
- Remember that slag is formed by the reaction of a flux with impurities in the ore.

24 Medium

24 of 85Question permalink

Application +4 | -1

When the sample of with

\mathrm{Zn}

impurity is to be purified by electrolysis, the appropriate electrodes are:
Cathode Anode

View answer and explanation

Correct option: D

Pure

\mathrm{Cu}

Impure sample

Option D (Pure

\mathrm{Cu}

Impure sample) is correct because in electrolytic refining, the impure metal acts as the anode and the pure metal acts as the cathode.

Solution approach

Correct Answer + Core Reason: Pure $\mathrm{Cu}$ Impure sample. In electrolytic refining, the impure metal is the anode and the pure metal is the cathode.

Step-by-Step Solution:

Step 1: Describe electrolytic refining. Electrolytic refining is a process used to purify metals using electrolysis.

Step 2: Explain the role of the anode. The impure metal (in this case, copper with zinc impurity) is used as the anode. During electrolysis, the metal dissolves into the electrolyte solution.

Step 3: Explain the role of the cathode. A pure copper strip is used as the cathode. Copper ions from the electrolyte are deposited on the cathode, resulting in pure copper.

Step 4: Explain the behavior of impurities. Zinc, being more electropositive than copper, also dissolves into the solution. However, only copper ions are reduced and deposited at the cathode. The zinc ions remain in the solution.

Why other options are wrong:
- Option A: Pure $\mathrm{Zn}$ Pure $\mathrm{Cu}$ . This is incorrect because the impure sample must be the anode, and the pure metal (Cu) must be the cathode.
- Option B: Impure sample Pure $\mathrm{Cu}$ . This is incorrect because the impure sample must be the anode, not the cathode.
- Option C: Impure $\mathrm{Zn}$ Impure sample. This is incorrect because the pure metal (Cu) must be the cathode, and the impure sample is the anode.

Key Concepts:
- Electrolytic Refining
- Anode
- Cathode
- Electrolysis

Memory Aids:
- Anode: \"A\"node - where the \"impure\" metal dissolves.
- Cathode: \"C\"athode - where the \"pure\" metal collects.

Quick Tips:
- Remember that in electrolytic refining, the impure metal is always the anode and the pure metal is always the cathode.

25 Medium

25 of 85Question permalink

Understanding +4 | -1

In which of the following method of purification, metal is converted to its volatile compound which is decomposed to give pure metal?
(a) Heating with stream of carbon monoxide
(b) Heating with iodine (c) Liquation (d) Distillation
Correct options are

View answer and explanation

Correct option: A

a & b are correct

Option A (a & b are correct) is correct because both Mond's process (using CO) and Van Arkel method (using iodine) involve volatile compound formation and decomposition.

Solution approach

Correct Answer + Core Reason: a & b are correct. Both methods involve converting the metal to a volatile compound and then decomposing it to obtain the pure metal.

Step-by-Step Solution:

Step 1: Analyze each method of purification.

Step 2: (a) Heating with a stream of carbon monoxide (Mond's Process): This is used for refining nickel. Nickel reacts with CO to form volatile Ni(CO)4, which is then decomposed to give pure Ni.

Step 3: (b) Heating with iodine (Van Arkel Method): This is used for refining zirconium and titanium. The metal reacts with iodine to form a volatile iodide, which is then decomposed on a hot filament to give pure metal.

Step 4: (c) Liquation: This method is based on differences in melting points of the metal and impurities. It does not involve forming a volatile compound.

Step 5: (d) Distillation: This method is based on differences in boiling points. It does not necessarily involve forming a volatile compound of the metal itself, although it relies on the metal being volatile.

Why other options are wrong:
- Option B: Liquation (c) does not involve forming a volatile compound.
- Option C: Liquation (c) does not involve forming a volatile compound.
- Option D: Distillation is a general purification technique based on boiling points, not necessarily volatile compound formation of the metal.

Key Concepts:
- Mond's Process (Ni purification)
- Van Arkel Method (Zr, Ti purification)
- Liquation
- Distillation

Memory Aids:
- Mond = CO, Van Arkel = Iodine

Quick Tips:
- Remember the specific metals purified by Mond's and Van Arkel methods.

26 Medium

26 of 85Question permalink

Remembering +4 | -1

Percentage of gold in Fool’s gold is

View answer and explanation

Correct option: A

Zero

Option A is correct because Fool's gold (pyrite,

FeS_2

) is an iron sulfide mineral and does not contain any gold.

Solution approach

Correct Answer + Core Reason: Zero. Fool’s gold (pyrite) contains no gold.

Step-by-Step Solution:

Step 1: Fool's Gold Composition: Fool's gold is iron pyrite ( $FeS_2$ ).

Step 2: Gold Content: Iron pyrite does not contain any gold.

Step 3: Conclusion: The percentage of gold in Fool's gold is zero.

Why other options are wrong:
- Option B: Fool's gold contains no gold, so the percentage is not 8.
- Option C: Fool's gold contains no gold, so the percentage is not 16.
- Option D: Fool's gold contains no gold, so the percentage is not 30.

Key Concepts:
- Mineral Composition
- Fool's Gold (Pyrite)

Memory Aids:
- Remember that Fool's gold is iron pyrite and contains no gold.

Quick Tips:
- Know the composition of common minerals.

27 Medium

27 of 85Question permalink

Remembering +4 | -1

Plumbo-solvency refers to:

View answer and explanation

Correct option: C

Dissolution of lead in water containing air

Option C (Dissolution of lead in water containing air) is correct because plumbo-solvency refers to the ability of water, especially when aerated, to dissolve lead from pipes.

Solution approach

Correct Answer + Core Reason: Dissolution of lead in water containing air. Plumbo-solvency refers to the dissolution of lead in water containing air.

Step-by-Step Solution:

Step 1: Define plumbo-solvency: Plumbo-solvency is the ability of water to dissolve lead.

Step 2: Explain the process: Lead can dissolve in water, especially when the water is aerated (contains dissolved oxygen and carbon dioxide). This is a concern when lead pipes are used for drinking water.

Step 3: Evaluate the other options: The other options do not accurately describe plumbo-solvency.

Step 4: Conclusion: Dissolution of lead in water containing air is the correct answer.

Why other options are wrong:
- Option A (Oxidation of lead to lead oxide): While oxidation is involved, plumbo-solvency specifically refers to the dissolution process.
- Option B (Oxidation of lead to red lead): Red lead formation is not directly related to plumbo-solvency.
- Option D (Making lead wires by forcing heated metal through a die): This is a manufacturing process, not related to plumbo-solvency.

Key Concepts:
- Environmental chemistry
- Lead contamination
- Plumbo-solvency

Memory Aids:
- Plumbo = Lead, Solvency = Dissolving

Quick Tips:
- Understand the environmental concerns related to lead in drinking water.

28 Medium

28 of 85Question permalink

Remembering +4 | -1

The metal which is refined by liquation is

View answer and explanation

Correct option: B

Sn

Option B (Sn) is correct because tin (Sn) is refined by liquation due to its low melting point.

Solution approach

Correct Answer + Core Reason: Sn. Tin is refined by liquation due to its relatively low melting point compared to impurities.

Step-by-Step Solution:

Step 1: Define Liquation: Liquation is a process of separating a metal from impurities by heating the impure metal to a temperature just above the melting point of the metal but below the melting point of the impurities. The molten metal flows away, leaving the solid impurities behind.

Step 2: Consider Zn: Zinc is typically refined by distillation or electrolytic refining.

Step 3: Consider Sn: Tin has a relatively low melting point and can be effectively refined by liquation.

Step 4: Consider In: Indium is often refined by electrolytic methods.

Step 5: Consider Ni: Nickel is refined by the Mond process.

Why other options are wrong:
- Option A: Zn is refined by distillation or electrolytic refining.
- Option C: In is often refined electrolytically.
- Option D: Ni is refined by the Mond process.

Key Concepts:
- Liquation
- Refining Methods

Memory Aids:
- Liquation: Low melting point metal separates from higher melting point impurities.

Quick Tips:
- Remember that liquation is suitable for metals with low melting points.
- Know the refining methods for common metals.

29 Medium

29 of 85Question permalink

Analyzing +4 | -1

Mixture of chlorides of copper, cadmium, chromium iron and aluminium was dissolved in water acidified with

{HCl}

and

{H}_2{S}

gas was passed for a sufficient time. It was filtered boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filtrate would give test for

View answer and explanation

Correct option: B

sodium and aluminium

Option B (sodium and aluminium) is correct because after the given treatment, only sodium and aluminium ions remain in the filtrate.

Solution approach

Correct Answer + Core Reason: sodium and aluminium. After precipitation of other metal ions, sodium and aluminium remain in the filtrate.

Step-by-Step Solution:

Step 1: Dissolve the mixture in water acidified with HCl: $Cu^{2+}$ , $Cd^{2+}$ , $Cr^{3+}$ , $Fe^{3+}$ , and $Al^{3+}$ ions are now in solution.

Step 2: Pass $H_2S$ gas: $Cu^{2+}$ and $Cd^{2+}$ precipitate as sulfides (CuS and CdS) because their sulfide salts have low $K_{sp}$ values in acidic medium.

Step 3: Filter, boil, and add nitric acid: This step oxidizes $Fe^{2+}$ to $Fe^{3+}$ to ensure complete precipitation in the next step.

Step 4: Add ammonium chloride and sodium hydroxide: $Fe^{3+}$ precipitates as $Fe(OH)_3$ and $Cr^{3+}$ precipitates as $Cr(OH)_3$ . $Al^{3+}$ also precipitates as $Al(OH)_3$ .

Step 5: Filter: The filtrate contains only $Na^+$ ions from NaOH and possibly $AlO_2^-$ (aluminate ion) if excess NaOH was added, which can be considered as $Al^{3+}$ remaining in solution in a different form.

Step 6: Identify the ions in the filtrate: The filtrate contains sodium ions ( $Na^+$ ) and aluminium ions (as $AlO_2^-$ or $Al^{3+}$ ).

Why other options are wrong:
- Option A: Iron precipitates as $Fe(OH)_3$ and is not present in the filtrate.
- Option C: Iron precipitates as $Fe(OH)_3$ and is not present in the filtrate.
- Option D: Iron and cadmium precipitate as sulfides and hydroxides, respectively, and are not present in the filtrate.

Key Concepts:
- Selective precipitation
- Solubility product ( $K_{sp}$ )
- Qualitative analysis

Memory Aids:
- Remember the order of precipitation based on $K_{sp}$ values.

Quick Tips:
- Understand the conditions required for precipitation of different metal ions.

30 Medium

30 of 85Question permalink

Analyzing +4 | -1

Assertion (A) Carbonate and hydroxide ores are concentrated by frothfloatation process.

Reason (R) In froth floatation process, mineral oil is used due to itpreferentially wets the gangue particles.

View answer and explanation

Correct option: D

If both Assertion and Reason and false.

Option D is correct because both the Assertion and Reason are false. Carbonate and hydroxide ores are generally not concentrated by froth floatation, and mineral oil preferentially wets the ore particles, not the gangue.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason and false. Carbonate and hydroxide ores are not concentrated by froth floatation, and mineral oil wets the ore particles, not the gangue.

Step-by-Step Solution:

Step 1: Analyze the assertion: Carbonate and hydroxide ores are concentrated by froth floatation process.
- This is generally false. Froth floatation is more commonly used for sulfide ores.

Step 2: Analyze the reason: In froth floatation process, mineral oil is used due to it preferentially wets the gangue particles.
- This is false. Mineral oil preferentially wets the ore particles, not the gangue particles.

Step 3: Determine the correctness of both statements.

Why other options are wrong:
- Option A: Both statements are false.
- Option B: Both statements are false.
- Option C: Both statements are false.

Key Concepts:
- Froth floatation
- Ore concentration methods

Memory Aids:
- Froth floatation is primarily for sulfide ores.
- Oil wets ore, water wets waste.

Quick Tips:
- Remember the types of ores that are commonly concentrated by froth floatation.

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