Heating mixture of ${\rm{C}}{{\rm{u}}_2}{\rm{O}}\;{\rm{and}}\;{\rm{C}}{{\rm{u}}_2}{\rm{S}}$ will give

${\rm{Cu}} + {\rm{S}}{{\rm{O}}_2}$

${\rm{C}}{{\rm{u}}_2}{\rm{S}}{{\rm{O}}_3}$

${\rm{Cu}} + {\rm{S}}{{\rm{O}}_3}$

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

31 of 85Question permalink

Remembering +4 | -1

Fe ore is concentrated by:

View answer and explanation

Correct option: A

Magnetic treatment

Option A (Magnetic treatment) is correct because iron ore is concentrated using magnetic separation due to its magnetic properties.

Solution approach

Correct Answer + Core Reason: Magnetic treatment is used to concentrate Fe ore.

Step-by-Step Solution:

Step 1: Iron ores often contain magnetic impurities or are themselves magnetic (e.g., magnetite, $Fe_3O_4$ ).

Step 2: Magnetic separation involves using a magnetic field to separate the magnetic components of the ore from the non-magnetic gangue (impurities).

Step 3: This method is effective for concentrating iron ores.

Why other options are wrong:
- Option B: Froth floatation is used for sulfide ores, not typically for iron ores (unless they contain sulfide impurities).
- Option C: Electrolysis is used for refining metals, not for concentrating ores.
- Option D: Roasting is a process of heating ores to drive off volatile impurities, but not primarily for concentration.

Key Concepts:
- Concentration of ores
- Magnetic separation
- Magnetic properties of iron ores

Memory Aids:
- Iron is attracted to magnets

Quick Tips:
- Consider the properties of the ore when choosing a concentration method.

32 Medium

32 of 85Question permalink

Understanding +4 | -1

Which pair of elements can form alloy?

View answer and explanation

Correct option: C

Fe and C

Option C (Fe and C) is correct because iron and carbon form steel, which is a well-known alloy.

Solution approach

Correct Answer + Core Reason: Fe and C can form an alloy.

Step-by-Step Solution:

Step 1: An alloy is a mixture of two or more metals or a metal and a non-metal.

Step 2: Iron (Fe) and carbon (C) form steel, which is a common and important alloy.

Step 3: Zinc (Zn) and lead (Pb) do not readily form alloys due to differences in their crystal structures and atomic sizes.

Step 4: Iron (Fe) and mercury (Hg) do not typically form alloys; mercury forms amalgams with some metals, but not iron.

Step 5: Carbon (C) and platinum (Pt) can form compounds, but not a typical alloy.

Why other options are wrong:
- Option A: Zn and Pb - Do not readily form alloys.
- Option B: Fe and Hg - Mercury doesn't typically alloy with iron.
- Option D: C and Pt - Form compounds, not typical alloys.

Key Concepts:
- Alloys
- Composition of alloys

Memory Aids:
- Steel = Iron + Carbon

Quick Tips:
- Know common alloy compositions.

33 Medium

33 of 85Question permalink

Remembering +4 | -1

Aluminium is extracted by the electrolysis of:

View answer and explanation

Correct option: D

Alumina mixed with cryolite

Option D (Alumina mixed with cryolite) is correct because alumina (

\mathrm{Al_2O_3}

) is mixed with cryolite (

\mathrm{Na_3AlF_6}

) to lower its melting point and increase conductivity for efficient electrolysis.

Solution approach

Correct Answer + Core Reason: Alumina mixed with cryolite. This mixture is electrolyzed to extract aluminum.

Step-by-Step Solution:

Step 1: Recall the Hall-Héroult process: Aluminum is extracted by the electrolysis of alumina ( $\mathrm{Al_2O_3}$ ) dissolved in molten cryolite ( $\mathrm{Na_3AlF_6}$ ).

Step 2: Explain the role of cryolite: Cryolite lowers the melting point of alumina and increases the conductivity of the electrolyte.

Step 3: Evaluate the other options: Electrolyzing bauxite, alumina alone, or molten cryolite alone is not the standard method.

Step 4: Conclusion: Alumina mixed with cryolite is the correct answer.

Why other options are wrong:
- Option A (Bauxite): Bauxite is the ore of aluminum, but it needs to be processed to obtain alumina before electrolysis.
- Option B (Alumina): Alumina has a very high melting point, making direct electrolysis impractical.
- Option C (Molten cryolite): Cryolite alone does not contain aluminum; it's used as a solvent for alumina.

Key Concepts:
- Aluminum extraction
- Hall-Héroult process
- Role of cryolite

Memory Aids:
- Alumina + Cryolite = Aluminum Electrolysis

Quick Tips:
- Remember the Hall-Héroult process and the function of cryolite.

34 Medium

34 of 85Question permalink

Remembering +4 | -1

Which metal is used as a reducing agent in smelting?

View answer and explanation

Correct option: A

C

Option A (C) is correct because carbon (in the form of coke or charcoal) is a common reducing agent used in smelting to extract metals from their ores.

Solution approach

Correct Answer + Core Reason: C. Carbon is commonly used as a reducing agent in smelting.

Step-by-Step Solution:

Step 1: Recall the smelting process: Smelting involves using a reducing agent to extract a metal from its ore.

Step 2: Identify common reducing agents: Carbon (in the form of coke or charcoal) is a common and inexpensive reducing agent used in smelting.

Step 3: Evaluate the other options: Aluminum and zinc are also reducing agents, but carbon is more commonly used in large-scale smelting operations.

Step 4: Conclusion: Carbon is the correct answer.

Why other options are wrong:
- Option B (Al): Aluminum is a strong reducing agent but is more expensive than carbon and used in specific cases like the Goldschmidt process.
- Option C (Zn): Zinc can be a reducing agent, but it is not as commonly used as carbon in smelting.
- Option D (None of these): Carbon is a commonly used reducing agent.

Key Concepts:
- Smelting
- Reducing agents
- Carbon as a reducing agent

Memory Aids:
- Carbon = Common Reducing Agent

Quick Tips:
- Know the common reducing agents used in metallurgy.

35 Medium

35 of 85Question permalink

Understanding +4 | -1

In the electrolysis of alumina, cryolite is added to:

View answer and explanation

Correct option: A

Lower the melting point of alumina and to increase the electrical conductivity

Option A (Lower the melting point of alumina and to increase the electrical conductivity) is correct because cryolite acts as a solvent for alumina and improves the conductivity of the electrolytic mixture.

Solution approach

Correct Answer + Core Reason: Lower the melting point of alumina and to increase the electrical conductivity. Cryolite acts as a solvent and increases conductivity.

Step-by-Step Solution:

Step 1: Describe the electrolysis of alumina. Alumina ( $\mathrm{Al}_2\mathrm{O}_3$ ) is electrolyzed to produce aluminum metal.

Step 2: Explain the problem with pure alumina. Pure alumina has a very high melting point (2072 $^\circ$ C), making electrolysis difficult and energy-intensive. It also has poor electrical conductivity.

Step 3: Explain the role of cryolite. Cryolite ( $\mathrm{Na}_3\mathrm{AlF}_6$ ) is added to lower the melting point of alumina to around 900-950 $^\circ$ C and to increase the electrical conductivity of the mixture.

Step 4: Explain how cryolite works. Cryolite dissolves alumina, forming a solution that is more conductive and has a lower melting point.

Why other options are wrong:
- Option B: Minimise the anode effect. While cryolite can help reduce the anode effect to some extent, its primary purpose is to lower the melting point and increase conductivity.
- Option C: Remove impurities from alumina. Cryolite does not primarily function to remove impurities from alumina. Its main role is to improve the electrolytic process.
- Option D: None of the above. Option A is the correct reason for adding cryolite.

Key Concepts:
- Electrolysis
- Alumina
- Cryolite
- Melting Point Depression
- Electrical Conductivity

Memory Aids:
- Cryolite: \"Cryo\" - think of \"lowering the temperature.\"

Quick Tips:
- Remember that cryolite is added to alumina to make the electrolysis process more efficient.

36 Medium

36 of 85Question permalink

Remembering +4 | -1

Which of the following is the chief ore of copper?

View answer and explanation

Correct option: C

\mathrm{CuFe}{\mathrm{S}_2}

Option C (

\mathrm{CuFe}{\mathrm{S}_2}

) is correct because Chalcopyrite is the most abundant and chief ore of copper.

Solution approach

Correct Answer + Core Reason: $\mathrm{CuFe}{\mathrm{S}_2}$ . Chalcopyrite is the most abundant copper ore.

Step-by-Step Solution:
Step 1: Identify the common ores of copper.
Step 2: Recall that Chalcopyrite ( $\mathrm{CuFe}{\mathrm{S}_2}$ ) is the most abundant and chief ore of copper.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{u}_2}\mathrm{S}$ (Copper glance or Chalcocite) is an ore of copper, but not the chief ore.
- Option B: $\mathrm{C}{\mathrm{u}_2}\mathrm{O}$ (Cuprite) is an ore of copper, but not the chief ore.
- Option D: $\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}$ (Malachite) is an ore of copper, but not the chief ore.

Key Concepts:
- Ores of copper

Memory Aids:
- Chalcopyrite is the chief ore of copper.

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

37 Medium

37 of 85Question permalink

Remembering +4 | -1

In the extraction of Zn, the formation of blue flame is due to the burning of:

View answer and explanation

Correct option: D

CO

Option D (CO) is correct because the blue flame observed during zinc extraction is due to the combustion of carbon monoxide.

Solution approach

Correct Answer + Core Reason: CO is responsible for the blue flame in Zn extraction.

Step-by-Step Solution:

Step 1: In the extraction of zinc, zinc oxide (ZnO) is reduced using carbon (coke).

Step 2: The reaction produces zinc vapor and carbon monoxide gas: $ZnO + C \rightarrow Zn + CO$

Step 3: The carbon monoxide gas then burns in air, producing a blue flame: $2CO + O_2 \rightarrow 2CO_2$

Why other options are wrong:
- Option A: ZnO is a reactant, not the cause of the flame.
- Option B: C (carbon) reacts to form CO, which then burns. Carbon itself doesn't produce a blue flame in this context.
- Option C: Zn (zinc) burns with a bluish-white flame, not a blue flame.
- Option D: CO is the correct answer.

Key Concepts:
- Extraction of zinc
- Reduction with carbon
- Combustion of carbon monoxide

Memory Aids:
- CO burns blue

Quick Tips:
- Remember the reactions involved in the extraction of zinc.

38 Medium

38 of 85Question permalink

Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a negative slope with respect to temperature.
Reason R: CO tends to get decomposed at higher temperature.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: B

A is correct but R is not correct

Option B (A is correct but R is not correct) is correct because the oxidation of carbon to carbon monoxide has a negative slope due to the increase in entropy (

\Delta S

is positive), and CO does not tend to decompose at higher temperatures.

Solution approach

Correct Answer + Core Reason: A is correct but R is not correct. The slope is negative due to entropy increase, and CO does not decompose at higher temperatures.

Step-by-Step Solution:

Step 1: Understand the Ellingham diagram.\nIn an Ellingham diagram, the slope of the line representing the change in Gibbs free energy ( $\Delta G$ ) with respect to temperature (T) is related to the change in entropy ( $\Delta S$ ) by the equation:
$\Delta G = \Delta H - T\Delta S$
$\frac{{\partial (\Delta G)}}{{\partial T}} = -\Delta S$

Step 2: Analyze the oxidation of carbon to carbon monoxide.\nThe reaction is:
$2\mathrm{C}(s) + {\mathrm{O}_2}(g) \to 2\mathrm{CO}(g)$ \nIn this reaction, the number of moles of gas increases (1 mole of O2 to 2 moles of CO), so the change in entropy ( $\Delta S$ ) is positive. Therefore, the slope ( $\frac{{\partial (\Delta G)}}{{\partial T}}$ ) is negative.

Step 3: Analyze Reason R: CO tends to get decomposed at higher temperature.\nCarbon monoxide (CO) is a relatively stable molecule and does not readily decompose at higher temperatures. In fact, it is more likely to react with oxygen to form carbon dioxide (CO2).

Why other options are wrong:
- Option A: A is correct, so this is wrong.
- Option C: R is not correct, so this is wrong.
- Option D: R is not correct, so this is wrong.

Key Concepts:
- Ellingham Diagram: $\Delta G$ vs. T.
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$ .
- Entropy: A measure of disorder.

Memory Aids:
- Negative slope = Positive entropy change.

Quick Tips:
- The slope of the Ellingham diagram is related to the change in entropy of the reaction.

39 Medium

39 of 85Question permalink

Analyzing +4 | -1

Which of the following statement(s) is (are) incorrect?

View answer and explanation

Correct option: D

None

Option D (None) is correct because all the given statements about metallurgical processes are accurate.

Solution approach

Correct Answer + Core Reason: None. All statements are correct regarding the metallurgical processes.

Step-by-Step Solution:

Step 1: Analyze statement A: Serpeck's process involves removing silica from bauxite by heating it with coke in a nitrogen atmosphere at high temperatures ( $1800^{\circ} \mathrm{C}$ ). This is a correct description of the process.

Step 2: Analyze statement B: In the extraction of lead from galena, roasting and self-reduction occur in the same furnace under different conditions of temperature and air supply. This is also a correct description.

Step 3: Analyze statement C: Tin is indeed obtained by the carbon reduction of tin oxide. This statement is accurate as well.

Step 4: Since all the statements are correct, the incorrect statement is "None".

Why other options are wrong:
- Option A: This is wrong because the statement about Serpeck's process is correct.
- Option B: This is wrong because the statement about lead extraction is correct.
- Option C: This is wrong because the statement about tin extraction is correct.

Key Concepts:
- Serpeck's process
- Extraction of lead from galena
- Carbon reduction of tin oxide

Memory Aids:
- Serpeck's process: Bauxite + Coke + N2 -> Removal of Silica
- Lead Extraction: Roasting + Self-Reduction
- Tin Extraction: Carbon Reduction of Tin Oxide

Quick Tips:
- Remember the specific conditions and reactants for each metallurgical process.

40 Medium

40 of 85Question permalink

Remembering +4 | -1

The process of extraction of Au and Ag ores is based on their solubility in:

View answer and explanation

Correct option: D

KCN

Option D is correct because the extraction of Au and Ag ores is based on their solubility in KCN (potassium cyanide) solution, which forms soluble cyanide complexes.

Solution approach

Correct Answer + Core Reason: KCN. The process of extraction of Au and Ag ores is based on their solubility in KCN.

Step-by-Step Solution:

Step 1: Cyanide Leaching: Gold and silver ores are leached with a dilute solution of potassium cyanide (KCN).

Step 2: Complex Formation: Gold and silver react with cyanide ions to form soluble cyanide complexes, such as $[Au(CN)_2]^-$ and $[Ag(CN)_2]^-$ .

Step 3: Conclusion: The extraction is based on their solubility in KCN.

Why other options are wrong:
- Option A: Ammonia ( $NH_3$ ) is not used for the extraction of gold and silver.
- Option B: Hydrochloric acid (HCl) does not dissolve gold and silver ores in the way required for extraction.
- Option C: Nitric acid ( $HNO_3$ ) can dissolve gold and silver, but it's not the basis of the cyanide extraction process.

Key Concepts:
- Cyanide Process
- Gold and Silver Extraction
- Potassium Cyanide (KCN)

Memory Aids:
- Remember that KCN is used to dissolve gold and silver in the cyanide process.

Quick Tips:
- Know the reagents used in different metal extraction processes.

41 Medium

41 of 85Question permalink

Remembering +4 | -1

Thermite process is used in reduction of

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}

Option A (

\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}

) is correct because the thermite process, using aluminum as a reducing agent, is commonly used to reduce chromium(III) oxide.

Solution approach

Correct Answer + Core Reason: Thermite process reduces $Cr_2O_3$ .

Step-by-Step Solution:

Step 1: Understand the thermite process.

Step 2: The thermite process uses aluminum as a reducing agent to reduce metal oxides, such as $Cr_2O_3$ .

Step 3: The reaction is: $Cr_2O_3 + 2Al \rightarrow Al_2O_3 + 2Cr$

Why other options are wrong:
- Option B: $Al_2O_3$ is already an oxide of aluminum and cannot be reduced further by the thermite process.
- Option C: $PbO_2$ can be reduced by other methods, but the thermite process is less common for it.
- Option D: $CuO$ can be reduced by other methods, but the thermite process is less common for it.

Key Concepts:
- Thermite process
- Reduction of metal oxides

Memory Aids:
- Thermite is used for welding railway tracks, which often contain chromium steel.

Quick Tips:
- Remember that the thermite process uses aluminum as a strong reducing agent.

42 Medium

42 of 85Question permalink

Understanding +4 | -1

Presence of small impurity usually makes a metal quite hard because the impurities:

View answer and explanation

Correct option: B

Reduce the number of slide planes

Option B (Reduce the number of slide planes) is correct because impurities disrupt the regular arrangement of atoms, hindering the movement of dislocations and reducing the number of slide planes.

Solution approach

Correct Answer + Core Reason: Reduce the number of slide planes. Impurities disrupt the lattice structure, hindering dislocation movement.

Step-by-Step Solution:

Step 1: Understand the concept of metal hardness.\nMetal hardness is related to the ease with which dislocations (line defects) can move through the crystal lattice. The easier the dislocations move, the softer the metal.

Step 2: Consider the effect of impurities.\nImpurities disrupt the regular arrangement of atoms in the crystal lattice. This disruption hinders the movement of dislocations.

Step 3: Relate dislocation movement to slide planes.\nSlide planes are planes in the crystal lattice along which dislocations can easily move. When impurities are present, they reduce the number of these slide planes by impeding dislocation movement.

Why other options are wrong:
- Option A: While impurities can slightly alter the lattice structure, the primary effect is on dislocation movement, not a complete change in the structure.
- Option C: While some impurities can affect the number of mobile electrons, this is not the primary reason for increased hardness.
- Option D: Impurities may reduce crystal symmetry locally, but the main effect is on dislocation movement.

Key Concepts:
- Metal Hardness: Resistance to deformation.
- Dislocations: Line defects in the crystal lattice.
- Slide Planes: Planes along which dislocations move easily.

Memory Aids:
- Impurities "Immobilize" dislocations, making the metal harder.

Quick Tips:
- Hardness is related to the ease of dislocation movement. Impurities hinder this movement.

43 Medium

43 of 85Question permalink

Remembering +4 | -1

Titanium containing mineral found in our country is

View answer and explanation

Correct option: C

Elmanite

Option C (Elmanite) is correct because Ilmenite is a titanium-containing mineral found in many countries, including India.

Solution approach

Correct Answer + Core Reason: Elmanite. Ilmenite is a titanium-containing mineral.

Step-by-Step Solution:
Step 1: Identify the minerals listed.
Step 2: Recall that Ilmenite ( $\mathrm{FeTiO}_3$ ) is a titanium-containing mineral.

Why other options are wrong:
- Option A: Bauxite ( $\mathrm{Al}_2\mathrm{O}_3.\mathrm{xH}_2\mathrm{O}$ ) is an ore of aluminum, not titanium.
- Option B: Chalcopyrites ( $\mathrm{CuFeS}_2$ ) is an ore of copper, not titanium.
- Option D: Dolomite ( $\mathrm{CaMg(CO}_3\mathrm{)}_2$ ) is a mineral containing calcium and magnesium, not titanium.

Key Concepts:
- Minerals and their composition.

Memory Aids:
- Ilmenite contains titanium.

Quick Tips:
- Remember the composition of common minerals.

44 Medium

44 of 85Question permalink

Understanding +4 | -1

Purpose of smelting of an ore is

View answer and explanation

Correct option: C

To reduce it

Option C (To reduce it) is correct because smelting involves using a reducing agent to extract the metal from its oxide ore.

Solution approach

Correct Answer + Core Reason: To reduce it. Smelting uses a reducing agent to convert metal oxides into the metallic form.

Step-by-Step Solution:

Step 1: Define smelting. Smelting is a metallurgical process used to extract a metal from its ore by heating it with a reducing agent.

Step 2: Explain the role of the reducing agent. The reducing agent (e.g., carbon, carbon monoxide) removes oxygen from the metal oxide, converting it into the metallic form.

Step 3: Example of a smelting reaction. For example, $\mathrm{Fe}_2\mathrm{O}_3 + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO}_2$ .

Why other options are wrong:
- Option A: To oxidize it. Smelting is a reduction process, not an oxidation process. The metal oxide is reduced to the metal.
- Option B: To remove vaporisation impurities. While some volatile impurities may be removed during smelting, the primary purpose is reduction.
- Option D: To obtain an alloy. Smelting primarily aims to obtain the pure metal, not an alloy. Alloys may be produced in subsequent steps.

Key Concepts:
- Smelting
- Reduction
- Reducing Agent
- Metal Extraction

Memory Aids:
- Smelting: \"Smelt\" - to reduce the metal oxide.

Quick Tips:
- Remember that smelting is a reduction process used to extract metals from their ores.

45 Medium

45 of 85Question permalink

Remembering +4 | -1

The impurity of sulphur makes the iron:

View answer and explanation

Correct option: B

Red short

Option B (Red short) is correct because the presence of sulfur as an impurity in iron makes it brittle when hot, a condition known as red shortness.

Solution approach

Correct Answer + Core Reason: Sulphur impurity causes "Red short" in iron.

Step-by-Step Solution:

Step 1: Recall the effects of impurities on iron.

Step 2: Sulphur makes iron brittle when hot (red short).

Step 3: Phosphorus makes iron brittle when cold (cold short).

Why other options are wrong:
- Option A: Fibrous is not a term associated with sulphur impurity.
- Option C: Cold short is caused by phosphorus impurity.
- Option D: Malleable means easily shaped, the opposite of what sulphur does.

Key Concepts:
- Impurities in iron
- Red shortness
- Cold shortness

Memory Aids:
- Sulphur = Hot = Red short; Phosphorus = Cold = Cold short.

Quick Tips:
- Remember the specific impurities and their effects on iron.

46 Medium

46 of 85Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: C

Self reduction

Option C (Self reduction) is correct because after partial roasting, the remaining sulphide ore reacts with the formed oxide to reduce itself to the metal.

Solution approach

Correct Answer + Core Reason: Self reduction. After partial roasting, the remaining sulphide ore reacts with the oxide formed to give the metal.

Step-by-Step Solution:

Step 1: Sulphide ores are often subjected to partial roasting to convert a portion of the sulfide to oxide or sulfate.

Step 2: In the case of copper, the following reactions occur:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Step 3: The remaining sulfide ore then reacts with the oxide formed to produce the metal. This is known as self-reduction or auto-reduction.

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

Key Concepts:
- Self-reduction (auto-reduction)
- Roasting of sulfide ores
- Copper extraction

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

Quick Tips:
- Remember that self-reduction involves the reaction of a sulfide with its oxide.

47 Medium

47 of 85Question permalink

Remembering +4 | -1

Bessemer’s converter is used in the manufacture of:

View answer and explanation

Correct option: C

Steel

Option C (Steel) is correct because the Bessemer converter is a steel-making process that removes impurities from pig iron by oxidation.

Solution approach

Correct Answer + Core Reason: Bessemer's converter is used in the manufacture of steel.

Step-by-Step Solution:

Step 1: Recall the function of Bessemer's converter.

Step 2: Bessemer's converter is used to remove impurities from pig iron by oxidation, converting it into steel.

Why other options are wrong:
- Option A: Cast iron is a product of the blast furnace, not the Bessemer converter.
- Option B: Pig iron is the raw material fed into the Bessemer converter.
- Option D: Wrought iron is produced by other methods, not the Bessemer converter.

Key Concepts:
- Steel manufacturing
- Bessemer process

Memory Aids:
- Think of Bessemer as "Better Steel Maker".

Quick Tips:
- Understand the different processes involved in iron and steel production.

48 Medium

48 of 85Question permalink

Understanding +4 | -1

The extraction of nickel involves:

View answer and explanation

Correct option: C

The formation and thermal decomposition of

\mathrm{Ni}{\left( {\mathrm{CO}} \right)_4}

Option C (The formation and thermal decomposition of

\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}

) is correct because the Mond process involves forming a volatile nickel carbonyl complex and then decomposing it at a higher temperature to obtain pure nickel.

Solution approach

Correct Answer + Core Reason: The formation and thermal decomposition of $\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}$ . This describes the Mond process.

Step-by-Step Solution:

Step 1: The Mond process is used for the extraction and purification of nickel.

Step 2: Impure nickel is reacted with carbon monoxide at a relatively low temperature (around 50-60°C) to form volatile nickel tetracarbonyl, $\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}$ :
${\mathrm{Ni}} + 4{\mathrm{CO}} \rightleftharpoons {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}}$

Step 3: The gaseous nickel tetracarbonyl is then separated from the impurities.

Step 4: The pure nickel tetracarbonyl is then heated to a higher temperature (around 180-200°C) to decompose it, yielding pure nickel and carbon monoxide:
${\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \rightleftharpoons {\mathrm{Ni}} + 4{\mathrm{CO}}$

Why other options are wrong:
- Option A: The formation of $\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}$ is only part of the process.
- Option B: The decomposition of $\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}$ is only part of the process.
- Option D: The formation and catalytic decomposition of $\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}$ is incorrect because the decomposition is thermal, not catalytic.

Key Concepts:
- Mond process
- Extraction of nickel
- Coordination complexes
- Thermal decomposition

Memory Aids:
- Mond process: Nickel carbonyl formation and decomposition.

Quick Tips:
- Remember the Mond process involves a volatile carbonyl complex.

49 Medium

49 of 85Question permalink

Analyzing +4 | -1

Heating mixture of

\mathrm{C}{\mathrm{u}_2}\mathrm{O}\;\mathrm{and}\;\mathrm{C}{\mathrm{u}_2}\mathrm{S}

will give

View answer and explanation

Correct option: D

\mathrm{Cu} + \mathrm{S}{\mathrm{O}_2}

Option D is correct because when

Cu_2O

and

Cu_2S

are heated together, they react to form copper metal and sulfur dioxide.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cu} + \mathrm{S}{\mathrm{O}_2}$ . Heating a mixture of $Cu_2O$ and $Cu_2S$ results in the formation of copper metal and sulfur dioxide.

Step-by-Step Solution:

Step 1: Write the reaction between $Cu_2O$ and $Cu_2S$ : $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Step 2: Identify the products: The products of the reaction are copper (Cu) and sulfur dioxide ( $SO_2$ ).

Why other options are wrong:
- Option A: $Cu_2SO_3$ is not formed in this reaction.
- Option B: $CuO + CuS$ are not the products of this reaction. The reaction proceeds to form copper metal and sulfur dioxide.
- Option C: $Cu + SO_3$ is incorrect. Sulfur trioxide ( $SO_3$ ) is not formed in this reaction; sulfur dioxide ( $SO_2$ ) is the correct product.

Key Concepts:
- Extraction of metals
- Roasting process
- Chemical reactions of metal oxides and sulfides

Memory Aids:
- Remember that copper sulfide and copper oxide react to produce copper and sulfur dioxide.

Quick Tips:
- Balance the chemical equation to correctly identify the products.

50 Medium

50 of 85Question permalink

Remembering +4 | -1

Which among the following is a carbonate ore ?

View answer and explanation

Correct option: C

Siderite

Option C (Siderite) is correct because Siderite (

FeCO_3

) is a carbonate ore.

Solution approach

Correct Answer + Core Reason: Siderite. Siderite is a carbonate ore of iron ( $FeCO_3$ ).

Step-by-Step Solution:

Step 1: Define Carbonate Ore: A carbonate ore contains the carbonate anion ( $CO_3^{2-}$ ).

Step 2: Consider Corundum: Corundum is aluminum oxide ( $Al_2O_3$ ).

Step 3: Consider Copper Glance: Copper glance is copper sulfide ( $Cu_2S$ ).

Step 4: Consider Siderite: Siderite is iron(II) carbonate ( $FeCO_3$ ).

Step 5: Consider Magnetite: Magnetite is an iron oxide ( $Fe_3O_4$ ).

Why other options are wrong:
- Option A: Corundum is an oxide ore, not a carbonate ore.
- Option B: Copper glance is a sulfide ore, not a carbonate ore.
- Option D: Magnetite is an oxide ore, not a carbonate ore.

Key Concepts:
- Carbonate Ores
- Ore Types (Oxide, Sulfide, Carbonate)

Memory Aids:
- Siderite: $FeCO_3$ (Iron Carbonate)

Quick Tips:
- Memorize common ore formulas and their classifications.
- Recognize the carbonate anion ( $CO_3^{2-}$ ) in carbonate ores.

51 Medium

51 of 85Question permalink

Analyzing +4 | -1

Which of the following statements about the advantage of roasting sulphide ore before reduction is true?

a. The

\Delta_{{f}} {G}^{\mathrm{o}}

of the sulphide is greater than that for

{CS}_2

and

{H}_2 {S}

b. The

\Delta {G}

is negative for roasting sulphide ore to oxide.

c. Roasting sulphide to oxide is thermodynamically feasible

d. Carbon and hydrogen are suitable reducing agents for metal sulphides

View answer and explanation

Correct option: B

a, b, and c

Option B (a, b, and c) is correct because all three statements accurately describe the advantages of roasting sulfide ore before reduction.

Solution approach

Correct Answer + Core Reason: a, b, and c. These statements correctly describe the thermodynamic favorability of roasting sulfide ores.

Step-by-Step Solution:

Step 1: Analyze statement a: The $\Delta_{{f}} {G}^{\mathrm{o}}$ of the sulfide is greater than that for ${CS}_2$ and ${H}_2 {S}$ . This implies that the formation of $CS_2$ and $H_2S$ is more favorable than the formation of the metal sulfide, making roasting advantageous.

Step 2: Analyze statement b: The $\Delta {G}$ is negative for roasting sulfide ore to oxide. A negative $\Delta G$ indicates that the roasting process is thermodynamically favorable.

Step 3: Analyze statement c: Roasting sulfide to oxide is thermodynamically feasible. This is a direct consequence of the negative $\Delta G$ value.

Step 4: Analyze statement d: Carbon and hydrogen are suitable reducing agents for metal sulfides. This is generally not true; they are better suited for reducing metal oxides.

Step 5: Therefore, statements a, b, and c are correct.

Why other options are wrong:
- Option A: Missing statement c, which is also correct.
- Option C: Includes statement d, which is incorrect.
- Option D: Only includes statement c, missing a and b.

Key Concepts:
- Gibbs Free Energy ( $\Delta G$ )
- Thermodynamics of Roasting
- Ellingham Diagrams

Memory Aids:
- Negative $\Delta G$ = Spontaneous/Feasible Reaction
- Roasting: Sulfide -> Oxide + $SO_2$

Quick Tips:
- Understand the relationship between $\Delta G$ and the feasibility of a reaction.
- Remember that carbon and hydrogen are generally used to reduce oxides, not sulfides.

52 Medium

52 of 85Question permalink

Understanding +4 | -1

Electrolysis of fused carnallite gives:

View answer and explanation

Correct option: D

\mathrm{K},\mathrm{Mg }\;\mathrm{and}\;\mathrm{ C}{\mathrm{l}_2}

Option D (K, Mg and Cl₂) is correct because carnallite (KCl.MgCl₂.6H₂O) upon electrolysis yields potassium, magnesium, and chlorine gas.

Solution approach

Correct Answer + Core Reason: $\mathrm{K},\mathrm{Mg }\;\mathrm{and}\;\mathrm{ C}{\mathrm{l}_2}$ . Electrolysis of fused carnallite produces potassium, magnesium, and chlorine gas.

Step-by-Step Solution:

Step 1: Identify the components of carnallite. Carnallite is a double salt with the formula KCl.MgCl₂.6H₂O.

Step 2: Understand the electrolysis process. Electrolysis involves the decomposition of a substance by passing an electric current through it in a molten or dissolved state.

Step 3: Determine the products at the cathode (reduction). At the cathode, the metal ions (K⁺ and Mg²⁺) are reduced to their respective metals. Since magnesium has a higher reduction potential than potassium, both will be produced.

Step 4: Determine the product at the anode (oxidation). At the anode, chloride ions (Cl⁻) are oxidized to chlorine gas (Cl₂).

Step 5: Write the half-cell reactions:
* Cathode: K⁺ + e⁻ → K
* Cathode: Mg²⁺ + 2e⁻ → Mg
* Anode: 2Cl⁻ → Cl₂ + 2e⁻

Step 6: Combine the products. The products are K, Mg, and Cl₂.

Why other options are wrong:
- Option A: This option only lists Mg, which is incorrect as K and Cl₂ are also produced.
- Option B: This option only lists K, which is incorrect as Mg and Cl₂ are also produced.
- Option C: This option lists K and CO₂, which is incorrect. CO₂ is not a product of carnallite electrolysis.
- Option D: This option correctly identifies all the products of the electrolysis of fused carnallite.

Key Concepts:
- Electrolysis
- Reduction potential
- Oxidation and reduction reactions
- Electrolytic cell

Memory Aids:
- AN OX (Anode Oxidation), RED CAT (Reduction Cathode)

Quick Tips:
- Remember to consider all ions present in the electrolyte when predicting electrolysis products.
- Understand the difference between electrolysis of molten salts and aqueous solutions.

53 Medium

53 of 85Question permalink

Remembering +4 | -1

One of the following metals is obtained by leaching its ore with dilute cyanide solution. Identify it.

View answer and explanation

Correct option: C

Silver

Option C is correct because silver is extracted by leaching its ore with a dilute cyanide solution, forming a soluble silver cyanide complex.

Solution approach

Correct Answer + Core Reason: Silver. Silver is obtained by leaching its ore with dilute cyanide solution.

Step-by-Step Solution:

Step 1: Cyanide Leaching: Silver ore is leached with a dilute solution of cyanide ions (CN-).

Step 2: Complex Formation: Silver reacts with cyanide ions to form a soluble silver cyanide complex, $[Ag(CN)_2]^-$ .

Step 3: Conclusion: Silver is extracted using cyanide leaching.

Why other options are wrong:
- Option A: Titanium is not extracted using cyanide leaching.
- Option B: Vanadium is not extracted using cyanide leaching.
- Option D: Zinc is not extracted using cyanide leaching.

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

Memory Aids:
- Remember that silver is extracted using the cyanide process.

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

54 Medium

54 of 85Question permalink

Understanding +4 | -1

Roasting of HgS in air produces:

View answer and explanation

Correct option: D

Hg

Option D (Hg) is correct because roasting HgS in air produces mercury and sulfur dioxide.

Solution approach

Correct Answer + Core Reason: Hg. Roasting HgS in air produces mercury and sulfur dioxide.

Step-by-Step Solution:

Step 1: Roasting is a metallurgical process involving heating a sulfide ore in the presence of air.

Step 2: When HgS (cinnabar) is roasted in air, it undergoes oxidation to form mercury and sulfur dioxide.

Step 3: The reaction is: $HgS(s) + O_2(g) \rightarrow Hg(l) + SO_2(g)$

Why other options are wrong:
- Option A: HgO is not the direct product of roasting HgS in air.
- Option B: $HgSO_3$ is not formed during roasting.
- Option C: $HgSO_4$ is not formed during roasting.

Key Concepts:
- Roasting
- Metallurgy of Mercury
- Oxidation reactions

Memory Aids:
- Roasting sulfides produces metal and $SO_2$ .

Quick Tips:
- Roasting is typically used for sulfide ores.

55 Medium

55 of 85Question permalink

Analyzing +4 | -1

Which is wrongly matched for refining of metal elements?

View answer and explanation

Correct option: D

\mathrm{Zn}, \mathrm{Cd}

from impurities : Liquation with high boiling points

Option D (

\mathrm{Zn}, \mathrm{Cd}

from impurities : Liquation with high boiling points) is wrongly matched because liquation is suitable for metals with low melting points, not high boiling points.

Solution approach

Correct Answer + Core Reason: $\mathrm{Zn}, \mathrm{Cd}$ from impurities : Liquation with high boiling points. Liquation is used for metals with low melting points, not high boiling points.

Step-by-Step Solution:

Step 1: Define zone refining. Zone refining is a method used to purify solid materials, such as semiconductors, by repeatedly passing a molten zone through the material.

Step 2: Define cupellation. Cupellation is a process used to purify precious metals, such as silver and gold, by oxidizing base metals and removing them as slag.

Step 3: Define carbon reduction. Carbon reduction is a method used to extract metals from their oxides by heating them with carbon.

Step 4: Define liquation. Liquation is a process used to separate metals with different melting points. The mixture is heated to a temperature above the melting point of the lower-melting metal, which then flows away from the higher-melting metal.

Why other options are wrong:
- Option A: $\mathrm{Si}, \mathrm{Ge}$ from impurities : Zone refining method. This is correctly matched. Zone refining is commonly used for purifying silicon and germanium.
- Option B: Purification of $\mathrm{Ag}$ : Cupellation. This is correctly matched. Cupellation is a standard method for purifying silver.
- Option C: Extraction of $\mathrm{Sn}$ from $\mathrm{SnO}_2: \mathrm{C}$ reduction. This is correctly matched. Tin is extracted from tin oxide by carbon reduction.
- Option D: $\mathrm{Zn}, \mathrm{Cd}$ from impurities : Liquation with high boiling points. This is wrongly matched. Liquation is suitable for metals with low melting points, not high boiling points. Also, boiling point is not relevant to Liquation.

Key Concepts:
- Zone Refining
- Cupellation
- Carbon Reduction
- Liquation
- Metallurgical Processes

Memory Aids:
- Liquation: \"Liquid\" - think of separating metals based on their melting points.

Quick Tips:
- Remember the specific applications of each refining method.

56 Medium

56 of 85Question permalink

Remembering +4 | -1

Pig iron is converted into steel by reducing the amount of carbon contained in it, in a:

View answer and explanation

Correct option: C

Bessemer’s converter

Option C (Bessemer’s converter) is correct because it is the equipment used to reduce the carbon content in pig iron to produce steel.

Solution approach

Correct Answer + Core Reason: Bessemer’s converter is used for converting pig iron into steel by reducing the carbon content.

Step-by-Step Solution:

Step 1: Pig iron contains a high percentage of carbon and other impurities.

Step 2: The Bessemer process involves blowing air through molten pig iron to oxidize and remove carbon and other impurities.

Step 3: The Bessemer converter is specifically designed for this process.

Why other options are wrong:
- Option A: Blast furnace is used for the initial extraction of iron from ore, not for converting pig iron to steel.
- Option B: Pyrite burner is used to produce sulfur dioxide for sulfuric acid production, not for steelmaking.
- Option D: None of these is incorrect because Bessemer's converter is the correct answer.

Key Concepts:
- Steelmaking
- Bessemer process
- Pig iron

Memory Aids:
- Bessemer = Better Steel

Quick Tips:
- Remember the specific processes and equipment used in metallurgy.

57 Medium

57 of 85Question permalink

Remembering +4 | -1

In the electrolytic purification of copper some gold is found in the:

View answer and explanation

Correct option: C

Anode mud

Option C (Anode mud) is correct because during electrolytic refining, less reactive metals like gold collect as anode mud.

Solution approach

Correct Answer + Core Reason: Anode mud contains gold during electrolytic purification of copper.

Step-by-Step Solution:

Step 1: Electrolytic refining involves using an impure copper anode and a pure copper cathode in a copper sulfate solution.

Step 2: During electrolysis, copper ions from the anode dissolve into the solution and are deposited on the cathode.

Step 3: Impurities less reactive than copper, such as gold and silver, do not dissolve and collect at the bottom of the anode as anode mud.

Why other options are wrong:
- Option A: Cathode is where pure copper is deposited, not where gold is found.
- Option B: Cathode mud is not a standard term in this process. Gold collects at the anode.
- Option D: None of these is incorrect because anode mud is the correct answer.

Key Concepts:
- Electrolytic refining
- Anode mud
- Electrochemistry

Memory Aids:
- Anode = Accumulation of less reactive metals

Quick Tips:
- Remember that less reactive metals collect at the anode during electrolytic refining.

58 Medium

58 of 85Question permalink

Remembering +4 | -1

Which metal adsorbs hydrogen?

View answer and explanation

Correct option: A

Pd

Option A (Pd) is correct because Palladium (Pd) is well-known for its ability to adsorb large volumes of hydrogen gas.

Solution approach

Correct Answer + Core Reason: Pd. Palladium is known for its ability to adsorb hydrogen.

Step-by-Step Solution:

Step 1: Define Adsorption: Adsorption is the adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface.

Step 2: Consider Pd: Palladium has a unique ability to adsorb large volumes of hydrogen gas within its crystal structure.

Step 3: Consider K, Al, and Zn: These metals do not have a significant capacity for hydrogen adsorption compared to palladium.

Why other options are wrong:
- Option B: K is an alkali metal and doesn't readily adsorb hydrogen.
- Option C: Al is a metal that forms a protective oxide layer but doesn't significantly adsorb hydrogen.
- Option D: Zn is a metal used in galvanization but doesn't significantly adsorb hydrogen.

Key Concepts:
- Adsorption
- Hydrogen Adsorption
- Palladium

Memory Aids:
- Palladium: Good at adsorbing Hydrogen

Quick Tips:
- Remember that transition metals like palladium are often used as catalysts due to their adsorption properties.

59 Medium

59 of 85Question permalink

Remembering +4 | -1

The material used for the lining of Bessemer’s converter in the extraction of copper is:

View answer and explanation

Correct option: B

Lime

Option B (Lime) is correct because lime (CaO) is used as a lining in Bessemer converters, especially when dealing with acidic impurities.

Solution approach

Correct Answer + Core Reason: Lime is used for the lining of Bessemer’s converter in the extraction of copper.

Step-by-Step Solution:

Step 1: The Bessemer converter is used to remove impurities from molten copper.

Step 2: The lining of the converter needs to withstand high temperatures and react with impurities.

Step 3: Lime (CaO) is a basic oxide that reacts with acidic impurities like silica ( $SiO_2$ ) to form slag, which can be removed.

Why other options are wrong:
- Option A: Silica is an acidic oxide and would be used to remove basic impurities, not in copper extraction where lime is more common.
- Option C: Iron is the metal being refined, not used as a lining material.
- Option D: Cu (copper) is the metal being extracted, not used as a lining material.

Key Concepts:
- Bessemer process
- Lining materials
- Acid-base reactions in metallurgy

Memory Aids:
- Lime neutralizes acidic impurities

Quick Tips:
- Consider the nature of the impurities when choosing a lining material.

60 Medium

60 of 85Question permalink

Remembering +4 | -1

Which method of purification is represented by the following equation:

Ti\left( s \right) + 2{I_2}\left( g \right)Ti{I_4}\left( g \right)Ti\left( s \right) + 2{I_2}\left( g \right)

View answer and explanation

Correct option: D

Van Arkel

Option D (Van Arkel) is correct because the Van Arkel method involves the formation of a volatile metal iodide, its decomposition on a hot filament, and deposition of pure metal.

Solution approach

Correct Answer + Core Reason: Van Arkel. The equation represents the Van Arkel method for purification.

Step-by-Step Solution:

Step 1: Recognize the reaction type: The given equation shows a metal reacting with iodine to form a volatile iodide, which then decomposes to release the pure metal.

Step 2: Recall the Van Arkel method: The Van Arkel method is a purification technique that uses this principle. The metal is heated with iodine to form a volatile iodide, which is then decomposed on a hot filament to deposit pure metal.

Step 3: Evaluate the other options: Zone refining, cupellation, and polling are different purification methods.

Step 4: Conclusion: Van Arkel is the correct answer.

Why other options are wrong:
- Option A (Zone refining): Zone refining involves melting a small zone of a solid and moving it along the material to concentrate impurities.
- Option B (Cupellation): Cupellation is used to separate noble metals from base metals by oxidation.
- Option C (Polling): Polling is used to reduce metal oxides in molten metals by stirring with green wood.

Key Concepts:
- Metal purification
- Van Arkel method
- Zone refining
- Cupellation
- Polling

Memory Aids:
- Van Arkel = Volatile Iodide

Quick Tips:
- Know the principles behind different metal purification techniques.

Your Progress

General principles and processes of isolation of elements practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice