d and f Block Elements — Page 27 | JEE Chemistry

Q261

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : In the Ellingham diagram, a sharp change in slope of the line is observed for

\mathrm{Mg} \rightarrow \mathrm{MgO}

at

\sim 1120^{\circ} \mathrm{C}

Reason R: There is a large change of entropy associated with the change of state In the light of the above statements, choose the correct answer from the options given below

A Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B

\mathbf{A}

is false but

\mathbf{R}

is true

C Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option A

Solution

In the Ellingham diagram, a sharp change in slope of the line is observed for $\mathrm{Mg}-\mathrm{MgO}$ at $\sim 1120^{\circ} \mathrm{C}$ because that is the boiling point of magnesium.

There is a large increase in entropy associated with the change of state of magnesium.

So, both Assertion (A) and Reason (R) are true and (R) is the correct explanation of $(A)$ .

Q262

Given below are two statements : Statement I : In the metallurgy process, sulphide ore is converted to oxide before reduction. Statement II : Oxide ores in general are easier to reduce. In the light of the above statements, choose the most appropriate answer from the options given below:

A Both Statement I and Statement II are correct

B Both Statement I and Statement II are incorrect

C Statement I is correct but Statement II is incorrect

D Statement I is incorrect but Statement II is correct

Correct Answer

Option A

Solution

Option A is the correct answer.

Statement I : In the metallurgy process, sulphide ore is converted to oxide before reduction.

This is correct.

In metallurgy, it is common to first convert sulphide ores to oxides because it's typically easier to reduce oxides to extract the metal.

This process is known as roasting, where the sulphide ore is heated in the presence of air to convert it into an oxide.

2 \mathrm{ZnS}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{ZnO}+2 \mathrm{SO}_2

Statement II : Oxide ores in general are easier to reduce.

This is also correct.

Oxides are generally easier to reduce than most other types of compounds.

This is because the bond between metal and oxygen in metal oxides is often weaker than the bond between metal and other elements in various types of ores.

For example, the C-O bond in CO2 (a product of the reduction of a metal oxide) is weaker and more easily broken than the C-S bond in CS2 (a product of the reduction of a metal sulphide), making CO2 more volatile and easier to remove from the system.

So, both Statement I and Statement II are correct, making Option A the right choice.

Q263

The major components in "Gun Metal” are :

A Al, Cu, Mg and Mn

B Cu, Ni and Fe

C Cu, Sn and Zn

D Cu, Zn and Ni

Correct Answer

Option C

Solution

Composition of Gun metal $\to$ 90% Cu + 8% Sn + 2% Zn Note : .tg .tg Alloys Composition Brass 60% Cu + 40% Zn Bronze 75-90% Cu + 10-25% Sn Gun-metal 90% Cu + 8% Sn + 2% Zn Aluminium bronze 90% Cu + 10% Al Monel metal 30% Cu + 67% Ni + 3% (Fe + Mn) German Silver 56% Cu + 24% Zn + 20% Ni Manganin 78-86% Cu + 13-18% Mn + 1-4% Ni Bell metal 80% Cu + 20% Sn Constantan 60% Cu + 40% Ni

Q264

In the extraction process of copper, the product obtained after carrying out the reactions (i)

2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}+2 \mathrm{SO}_{2}

(ii)

2 \mathrm{Cu}_{2} \mathrm{O}+\mathrm{Cu}_{2} \mathrm{~S} \rightarrow 6 \mathrm{Cu}+\mathrm{SO}_{2}

is called

A Copper matte

B Blister copper

C Reduced copper

D Copper scrap

Correct Answer

Option B

Solution

The sequence of reactions stated here is part of the copper extraction process, specifically the smelting and conversion stages.

Initially, copper sulfide (Cu $_2$ S) reacts with oxygen (O $_2$ ) to form copper(I) oxide (Cu $_2$ O) and sulfur dioxide (SO $_2$ ), according to the equation :

2 \mathrm{Cu}_2 \mathrm{S} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Cu}_2 \mathrm{O} + 2 \mathrm{SO}_2.

Subsequently, the resulting copper(I) oxide then reacts with remaining copper sulfide to yield copper (Cu) and additional sulfur dioxide :

2 \mathrm{Cu}_2 \mathrm{O} + \mathrm{Cu}_2 \mathrm{S} \rightarrow 6 \mathrm{Cu} + \mathrm{SO}_2.

The output of this process, which contains about 98-99% copper, is known as "blister copper".

This name derives from the blister-like appearance of the copper surface, which results from the release of sulfur dioxide gas during the reaction.

Q265

In Hall - Heroult process, the following is used for reducing

\mathrm{Al}_{2} \mathrm{O}_{3}

:-

A Magnesium

B

\mathrm{CaF}_{2}

C Graphite

D

\mathrm{Na}_{3} \mathrm{AlF}_{6}

Correct Answer

Option C

Solution

In the Hall-Héroult process, graphite anodes are used to reduce alumina (Al₂O₃).

During the electrolytic reduction, Al³⁺ ions are reduced at the cathode to form aluminum, while at the anode, oxide ions are oxidized to form oxygen gas.

The oxygen gas then reacts with the carbon in the graphite anodes to produce carbon dioxide.

This means that graphite is consumed during the process, and the anodes have to be replaced periodically.

So, in the context of the question, graphite (Option C) is the correct answer, as it is the material of the anode used for the electrolytic reduction of Al₂O₃ in the Hall-Héroult process.

Q266

Which of the following metals can be extracted through alkali leaching technique?

A

\mathrm{Sn}

B Cu

C

\mathrm{Au}

D

\mathrm{Pb}

Correct Answer

Option C

Solution

The alkali leaching technique is a method to extract metals from their ores by using an alkaline solution, typically a sodium hydroxide (NaOH) solution, sometimes in the presence of other substances like a cyanide for gold and silver ores.

Among the options given, the metal that can be extracted through alkali leaching technique is: Option C:

\mathrm{Au}

(Gold) Gold can be extracted using a process known as cyanidation or cyanide leaching.

In this process, finely ground gold-bearing ore is treated with a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN) in the presence of oxygen to dissolve the gold.

The chemical reaction for the dissolution of gold in the cyanide solution is given by:

\mathrm{4Au + 8NaCN + O_2 + 2H_2O \rightarrow 4Na[Au(CN)_2] + 4NaOH}

The gold is then recovered from the solution by various methods, such as adsorption onto activated carbon, followed by elution and eventually electro-winning or by precipitation with zinc dust (the Merrill-Crowe process).

Because of environmental and safety concerns, alternatives to cyanide leaching have been researched, yet cyanidation remains the primary method for the recovery of gold from its ores.

As for the other options: Option A:

\mathrm{Sn}

(Tin) is generally extracted through pyrometallurgical processes such as smelting, rather than alkali leaching.

Option B: Copper (Cu) is often extracted using hydrometallurgical processes like sulfuric acid leaching rather than alkali leaching.

Copper can also be extracted using the process called solvent extraction and electrowinning (SX-EW) after the leaching step.

Option D:

\mathrm{Pb}

(Lead) is typically extracted using pyrometallurgical methods, including smelting and refining, and not alkali leaching.

Q267

The purification method based on the following physical transformation is :

\underset{(X)}{\text{ Solid }} \xrightarrow[]{\text{ Heat }} \underset{(X)}{\text{ Vapour }} \xrightarrow{\text{ Cool }} \underset{(X)}{\text{ Solid }}

A Extraction

B Crystallization

C Distillation

D Sublimation

Correct Answer

Option D

Solution

The purification method described by the transformation:

\underset{(X)}{\text{Solid}} \xrightarrow[]{\text{Heat}} \underset{(X)}{\text{Vapour}} \xrightarrow[]{\text{Cool}} \underset{(X)}{\text{Solid}}

is known as Sublimation.

In sublimation, a solid substance directly transforms into its gaseous phase upon heating and reverts back to the solid phase upon cooling.

This technique is commonly used to purify substances, particularly those that sublime more easily than others under the same conditions.

Sublimation is effective for separating substances with different volatility, like separating ammonium chloride from a mixture with sodium chloride, as ammonium chloride sublimates while sodium chloride does not.

Q268

When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3 ; a dark green product is formed which disproportioates in acidic solution to afforda dark purple solution. X is :

A Ti

B V

C Cr

D Mn

Correct Answer

Option D

Solution

X is Mn.

MnO2 + 2KOH + KNO3 $\to$ K2MnO4 + KNO2 + H2O Here K2MnO4 is dark green.

3MnO4 $-$ 2 + 4H+ $\to$ 2MnO4 $-$ + MnO2 + 2H2O In acidic solution, K2MnO4 changes to dark purple solution of KMnO4.

Q269

Match the refining methods (Column I) with metals (Column II). .tg .tg Column I (Refining Methods) Column II (Metals)

List - I		List - II
(a)	Zr	(I)	Liquation
(b)	Ni	(II)	Zone Refining
(c)	Sn	(III)	Mond Process
(d)	Ga	(IV)	Van Arkel Method

A (I)-(c) ; (II)-(a) ; (III)-(b) ; (IV)-(d)

B (I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)

C (I)-(b) ; (II)-(d) ; (III)-(a) ; (IV)-(c)

D (I)-(b) ; (II)-(c) ; (III)-(d) ; (IV)-(a)

Correct Answer

Option B

Solution

Liquation is used for Sn.

Zone refining is used for Ga.

Van Arkel method is used for Zr.

Mond’s process is used for refining of Ni.

Q270

Match List I with List II .tg .tg List-I List-II A. Reverberatory furnace I. Pig Iron B. Electrolytic cell II. Aluminium C. Blast furnace III. Silicon D. Zone Refining furnace IV. Copper Choose the correct answer from the options given below :

A A-III, B-IV, C-I, D-II

B A-IV, B-II, C-I, D-III

C A-I, B-IV, C-II, D-III

D A-I, B-III, C-II, D-IV

Correct Answer

Option B

Solution

(A) Reverberatory furnance is used for extraction of copper.

(B) Electrolytic cell is used for obtaining highly reactive metals like aluminium.

(C) Blast furnace is used for pig iron.

(D) Zone refining is used for purification of silicon.