Metals and Their Compounds Practice Questions — JAMB UTME

Question 1Question →

In the industrial extraction of sodium metal using the Downs cell, calcium chloride ( $\text{CaCl}_2$ ) is added to molten sodium chloride ( $\text{NaCl}$ ) primarily to:

lower the melting point of the sodium chloride electrolyte

act as a reducing agent for sodium ions

prevent chlorine gas from discharging at the graphite anode

increase the solubility of sodium metal in the molten mixture

Show answer & explanation

Answer: lower the melting point of the sodium chloride electrolyte

Answer

Calcium chloride is added to lower the melting point of the sodium chloride electrolyte.

Pure sodium chloride melts at

801^\circ\text{C}

. Adding calcium chloride lowers the melting point of the mixture to about

600^\circ\text{C}

, which conserves energy and minimizes evaporation of the liberated sodium metal during electrolysis in the Downs cell.

Step-by-Step Solution

1

Identify the physical properties of pure sodium chloride

Pure

\text{NaCl}

has a high melting point of approximately

801^\circ\text{C}

.

Maintaining such a high temperature industrially requires significant thermal energy.

2

Determine the effect of adding calcium chloride (

\text{CaCl}_2

)

Mixing

\text{CaCl}_2

with

\text{NaCl}

forms an electrolytic mixture with a reduced melting point of about

600^\circ\text{C}

.

Lowering the melting point makes the process economically viable and reduces heat loss and electrode wear.

Key Concept

Function of flux/admixture in the Downs process for sodium extraction

Estimated Time:45s

Question 2Question →

When a few drops of aqueous sodium hydroxide are added to a solution of aluminium chloride, a white gelatinous precipitate forms. Upon adding excess sodium hydroxide, the precipitate dissolves to produce a clear, colorless solution. Which chemical species is present in the final clear solution?

Tetrahydroxoaluminate(III) ion,

[Al(OH)_4]^-

Aluminium hydroxide,

Al(OH)_3

Aluminium oxide,

Al_2O_3

Hexaaquaaluminium(III) ion,

[Al(H_2O)_6]^{3+}

Show answer & explanation

Answer: Tetrahydroxoaluminate(III) ion, $[Al(OH)_4]^-$

Answer

The tetrahydroxoaluminate(III) ion,

[Al(OH)_4]^-

Aluminium hydroxide,

Al(OH)_3

, is an amphoteric hydroxide. It initially precipitates as a white gelatinous solid when alkali is added. When excess strong alkali like sodium hydroxide is added, it acts as an acid and reacts further with hydroxide ions to form the soluble complex ion, tetrahydroxoaluminate(III),

[Al(OH)_4]^-

, yielding a clear solution.

Step-by-Step Solution

1

Identify the initial precipitation reaction

Adding a few drops of

OH^-

ions precipitates white gelatinous aluminium hydroxide:

Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)

Aluminium ions react with hydroxide ions to form insoluble aluminium hydroxide.

2

Apply the amphoteric property of aluminium hydroxide with excess strong base

Adding excess sodium hydroxide causes

Al(OH)_3

to act as an acid, dissolving according to:

Al(OH)_3(s) + OH^-(aq) \rightarrow [Al(OH)_4]^-(aq)

Amphoteric hydroxides react with excess alkalis to form soluble complex aluminate ions.

Key Concept

Amphoteric nature of aluminium compounds

Estimated Time:1m 0s

Question 3Question →

Match each aluminium alloy or extraction reagent on the left with its correct composition, primary industrial application, or function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Duralumin matches with the composition of

Al

,

Cu

,

Mg

,

Mn

used in aircraft bodies; Magnalium matches with the composition of

Al

,

Mg

used in balance beams; Alnico matches with the composition of

Al

,

Ni

,

Co

,

Fe

used in permanent magnets; Cryolite matches with the molten solvent that lowers the melting point of alumina in extraction.

Each item correctly matches its specific chemical composition and technological usage in metallurgy.

Step-by-Step Solution

1

Identify the composition and application of Duralumin

Duralumin contains aluminium, copper, magnesium, and manganese, providing high strength and lightness for aircraft construction.

Alloying aluminium with copper and manganese enhances structural strength.

2

Identify the composition and application of Magnalium

Magnalium is an aluminium-magnesium alloy prized for low density and corrosion resistance in optical/scientific instruments.

Magnesium lowers density and improves machinability.

3

Identify the composition and application of Alnico

Alnico is composed of aluminium, nickel, cobalt, and iron, essential for permanent magnets.

The combination of ferromagnetic metals with aluminium yields high magnetic coercivity.

4

Identify the role of Cryolite in electrolysis

Cryolite acts as a molten solvent for alumina to reduce operating energy costs and enhance ionic conductivity.

Pure alumina melts at over

2000^\circ C

; cryolite lowers this operating temperature to around

950^\circ C

.

Key Concept

Aluminium Alloys and Extraction Metallurgy

Question 4Question →

During the industrial extraction of iron from hematite ( $\text{Fe}_2\text{O}_3$ ) in the blast furnace, which chemical species acts as the primary reducing agent responsible for converting the iron ore to iron in the upper region of the furnace?

Carbon(II) oxide (

\text{CO}

)

Solid carbon (

\text{C}

)

Carbon(IV) oxide (

\text{CO}_2

)

Calcium oxide (

\text{CaO}

)

Show answer & explanation

Answer: Carbon(II) oxide ( $\text{CO}$ )

Answer

Carbon(II) oxide (

\text{CO}

)

In the blast furnace, carbon(II) oxide (

\text{CO}

) gas is produced when carbon dioxide reacts with red-hot coke. Because it is a gas, carbon(II) oxide thoroughly mixes with and reduces the solid hematite (

\text{Fe}_2\text{O}_3

) to iron in the upper, cooler region of the furnace.

Step-by-Step Solution

1

Identify the chemical reactions occurring in the blast furnace.

Coke burns in oxygen to form

\text{CO}_2

, which then reacts with excess hot coke to produce

\text{CO}

:

\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g)

.

Gaseous

\text{CO}

is generated to serve as the gaseous reducing agent.

2

Determine which species reduces hematite in the upper zone (400°C – 700°C).

Gaseous carbon(II) oxide reduces

\text{Fe}_2\text{O}_3

:

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)

.

Gaseous

\text{CO}

provides intimate surface contact with the solid ore compared to solid coke.

Key Concept

Blast furnace reduction of iron ore by carbon(II) oxide gas

Question 5Question →

During the industrial extraction of iron in a blast furnace, hematite ( $Fe_2O_3$ ) is reduced to iron in the upper region of the furnace. Which chemical species acts as the primary reducing agent in this upper zone?

Carbon(II) oxide (

CO

)

Solid carbon (

C

)

Calcium oxide (

CaO

)

Carbon(IV) oxide (

CO_2

)

Show answer & explanation

Answer: Carbon(II) oxide ( $CO$ )

Answer

Carbon(II) oxide (

CO

)

In the blast furnace, carbon(II) oxide (

CO

) gas is produced when hot carbon dioxide reacts with excess coke. Rising

CO

gas encounters descending hematite (

Fe_2O_3

) in the upper, cooler zone of the furnace and reduces it to iron:

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

.

Step-by-Step Solution

1

Identify the chemical reactions occurring in the upper region of the blast furnace.

Near the top of the furnace (200°C–700°C), carbon(II) oxide gas reacts with descending hematite ore.

Gas-solid contact allows

CO

to easily reduce the porous iron ore.

2

Write the overall reduction equation.

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

This chemical equation confirms that

CO

gains oxygen (is oxidized) while reducing

Fe_2O_3

to metallic iron.

Key Concept

Role of gaseous carbon(II) oxide as the principal reducing agent in blast furnace iron extraction

Estimated Time:45s

Question 6Question →

Match each chemical substance or process associated with iron extraction and rust prevention on the left with its corresponding chemical role or function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Limestone (CaCO₃) matches with thermal decomposition to provide calcium oxide as a basic flux. Carbon monoxide (CO) matches with serving as the main reducing agent. Galvanization matches with sacrificial coating of iron using zinc. Calcium silicate (CaSiO₃) matches with forming molten slag that floats on molten iron.

Limestone decomposes into calcium oxide, which acts as a basic flux to neutralize silica impurities. Carbon monoxide is the main gaseous reducing agent reducing hematite to metallic iron. Galvanization applies a protective, sacrificial layer of zinc onto iron surfaces. Calcium silicate forms the molten slag layer that sits above molten iron to protect it from re-oxidation.

Step-by-Step Solution

1

Identify the role of limestone in the blast furnace.

Limestone undergoes endothermic decomposition to form calcium oxide (

\text{CaO}

), acting as a basic flux.

Flux is required to react with acidic impurities like silicon(IV) oxide.

2

Identify the primary reducing agent in iron extraction.

Carbon monoxide (

\text{CO}

) reduces hematite (

\text{Fe}_2\text{O}_3

) to iron in the upper and middle zones of the furnace.

At elevated blast furnace temperatures, gaseous carbon monoxide readily abstracts oxygen from iron ores.

3

Determine the role of zinc coating on iron (galvanization).

Galvanization provides sacrificial protection against corrosion.

Zinc oxidizes preferentially to iron because of its higher position in the electrochemical series.

4

Determine the identity and function of slag.

Calcium silicate (

\text{CaSiO}_3

) constitutes molten slag.

Slag is less dense than liquid iron, floating on top to prevent oxidation by incoming air blasts.

Key Concept

Industrial extraction of iron in the blast furnace and sacrificial protection mechanisms against iron rusting.

Question 7Question →

Match each transition metal or transition metal compound listed on the left with its corresponding industrial catalytic process on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Finely divided iron matches the Haber process for manufacturing ammonia; Vanadium(V) oxide matches the Contact process for manufacturing tetraoxosulfate(VI) acid; Nickel matches the hydrogenation of vegetable oils to margarine; Platinum matches the Ostwald process for manufacturing trioxonitrate(V) acid.

Transition metals and their oxides serve as effective industrial catalysts due to their partially filled d-orbitals, variable oxidation states, and ability to adsorb reactant molecules onto their surfaces. Finely divided iron is the standard catalyst in the Haber process for ammonia synthesis, vanadium(V) oxide catalyzes sulfur dioxide oxidation in the Contact process, nickel catalyzes the hydrogenation of unsaturated vegetable oils, and platinum catalyzes the catalytic oxidation of ammonia in the Ostwald process.

Step-by-Step Solution

1

Identify the industrial reaction associated with finely divided iron.

Finely divided iron catalyzes

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

in the Haber process.

Iron provides a surface for nitrogen and hydrogen molecules to adsorb and react efficiently.

2

Identify the catalyst used in the Contact process.

Vanadium(V) oxide (

V_2O_5

) catalyzes

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

.

Vanadium exhibits variable oxidation states (

V^{5+}

and

V^{4+}

) allowing intermediate redox steps.

3

Identify the catalyst used in organic hydrogenation.

Nickel (

Ni

) catalyzes the conversion of unsaturated vegetable oils to saturated fats.

Finely divided nickel adsorbs hydrogen gas and liquid oil to facilitate addition across carbon-carbon double bonds.

4

Identify the catalyst used in the Ostwald process.

Platinum (

Pt

) catalyzes the oxidation of ammonia (

NH_3

) to nitrogen(II) oxide (

NO

).

Platinum gauze provides high surface area and stability at elevated temperatures required for ammonia oxidation.

Key Concept

Industrial Catalytic Applications of Transition Metals

Question 8Question →

Copper is a transition metal with an atomic number of $29$ . What is the ground-state electronic configuration of the copper(II) ion ( $Cu^{2+}$ ) present in copper compounds such as copper(II) tetraoxosulfate(VI)?

[Ar] 3d^9

[Ar] 3d^8 4s^1

[Ar] 3d^7 4s^2

[Ar] 3d^{10} 4s^1

Show answer & explanation

Answer: $[Ar] 3d^9$

Answer

The ground-state electronic configuration of the copper(II) ion (

Cu^{2+}

) is

[Ar] 3d^9

.

The correct answer specifies

[Ar] 3d^9

. Neutral copper has a ground-state configuration of

[Ar] 3d^{10} 4s^1

. Ionization to form

Cu^{2+}

requires losing two electrons; the first comes from the outermost

4s

orbital, and the second comes from the

3d

subshell, leaving

[Ar] 3d^9

.

Step-by-Step Solution

1

Determine the electronic configuration of a neutral copper atom (

Cu, Z=29

).

Neutral copper has the ground-state configuration

[Ar] 3d^{10} 4s^1

(anomalous filling for extra stability of a full d-subshell).

A completely filled

3d^{10}

subshell provides lower overall energy than a

3d^9 4s^2

arrangement.

2

Identify which electrons are removed when forming the

Cu^{2+}

ion.

Two electrons must be removed: 1 electron from the

4s

orbital and 1 electron from the

3d

orbital.

Electrons in the outermost principal quantum level (

n=4

) are always removed first during cation formation.

3

Write the final electronic configuration of

Cu^{2+}

.

[Ar] 3d^9

Subtracting

1

electron from

4s

and

1

electron from

3d^{10}

leaves

9

electrons in the

3d

subshell and

0

in

4s

.

Key Concept

Electronic configuration of transition metal cations and anomalous electron filling in copper.

Estimated Time:1m 0s

Question 9Question →

Match each metallic alloy listed in Column A with its corresponding elemental composition and primary application in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Duralumin matches with Aluminum, Copper, Magnesium, and Manganese (aircraft construction); Brass matches with Copper and Zinc (musical instruments and fittings); Stainless Steel matches with Iron, Carbon, Chromium, and Nickel (cutlery and surgical tools); Bronze matches with Copper and Tin (statues, bearings, and medals).

Each alloy is paired strictly according to its constituent metals and major application: Duralumin (

\text{Al}+\text{Cu}+\text{Mg}+\text{Mn}

) for light aircraft structures, Brass (

\text{Cu}+\text{Zn}

) for acoustic instruments, Stainless Steel (

\text{Fe}+\text{C}+\text{Cr}+\text{Ni}

) for rust-proof tools, and Bronze (

\text{Cu}+\text{Sn}

) for low-friction bearings and statues.

Step-by-Step Solution

1

Analyze the composition and primary characteristic of Duralumin.

Duralumin consists of

\text{Al} + \text{Cu} + \text{Mg} + \text{Mn}

. Its key physical property is low density combined with high tensile strength.

Aluminum is the base metal in lightweight aviation alloys.

2

Differentiate between the copper-based alloys: Brass and Bronze.

Brass is an alloy of copper and zinc (

\text{Cu} + \text{Zn}

), whereas Bronze is an alloy of copper and tin (

\text{Cu} + \text{Sn}

).

Zinc is added to copper to make brass; tin is added to copper to make bronze.

3

Identify the alloying elements that impart corrosion resistance to steel.

Stainless steel contains chromium and nickel added to iron and carbon.

Chromium reacts with oxygen to form a thin, unreactive passivation layer of chromium(III) oxide (

\text{Cr}_2\text{O}_3

).

Key Concept

Compositions, physical property modifications, and industrial applications of major metallic alloys.

Question 10Question →

During the industrial extraction of iron in the blast furnace, limestone ( $\text{CaCO}_3$ ) is decomposed by heat to form calcium oxide ( $\text{CaO}$ ), which then reacts with silica ( $\text{SiO}_2$ ) impurities present in the ore. Which of the following chemical formulas represents the molten slag produced from this reaction?

\text{CaSiO}_3

\text{FeSiO}_3

\text{CaCO}_3

\text{CaO}

Show answer & explanation

Answer: $\text{CaSiO}_3$

Answer

\text{CaSiO}_3

(Calcium trioxosilicate(IV))

In the lower region of the blast furnace, limestone (

\text{CaCO}_3

) decomposes into calcium oxide (

\text{CaO}

). The basic

\text{CaO}

reacts with acidic silica (

\text{SiO}_2

) impurities present in hematite to form molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

), commonly known as slag.

Step-by-Step Solution

1

Identify thermal decomposition of limestone

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

High temperatures in the blast furnace break down limestone into basic calcium oxide.

2

Combine basic oxide flux with acidic silica impurity

\text{CaO}(s) + \text{SiO}_2(s) \rightarrow \text{CaSiO}_3(l)

Calcium oxide acts as a basic flux that neutralizes acidic sand/silica impurities to form molten slag.

Key Concept

Slag Formation in Iron Extraction

Estimated Time:45s

Question 11Question →

During the electrolytic refining of copper, a steady current of $5.0\text{ A}$ is passed through an aqueous copper(II) tetraoxosulfate(VI) ( $CuSO_4$ ) solution for $965\text{ seconds}$ . What mass of copper, in grams, is deposited at the cathode? (Faraday's constant $F = 96,500\text{ C mol}^{-1}$ ; Molar mass of $Cu = 64\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 1.6

Answer

1.6 g

Passing a

5.0\text{ A}

current for

965\text{ s}

transfers

4825\text{ C}

of electric charge, corresponding to

0.05\text{ mol}

of electrons. Because the deposition of copper from

CuSO_4

follows

Cu^{2+} + 2e^- \rightarrow Cu

, every

2\text{ moles}

of electrons deposit

1\text{ mole}

of copper metal. Thus,

0.025\text{ mol}

of copper is deposited, which corresponds to

0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

.

Step-by-Step Solution

1

Calculate the total quantity of electricity (

Q

) passed through the electrolyte.

Q = I \times t = 5.0\text{ A} \times 965\text{ s} = 4825\text{ C}

Electric charge is the product of current in amperes and time in seconds.

2

Calculate the moles of electrons transferred.

n(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.05\text{ mol}

of electrons

One mole of electrons carries a charge equivalent to 1 Faraday (

96,500\text{ C}

).

3

Use the cathode half-equation to find the moles of deposited copper.

Cathode reaction:

Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

. Moles of

Cu = \frac{0.05\text{ mol}}{2} = 0.025\text{ mol}

Reduction of one mole of copper(II) ions requires two moles of electrons.

4

Convert the moles of deposited copper into mass.

Mass of

Cu = n \times M = 0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

Multiplying the chemical amount of copper by its molar mass yields the mass in grams.

Key Concept

Quantitative electrolysis of copper(II) ions using Faraday's laws of electrolysis

Question 12Question →

What is the ground-state electronic configuration of the iron(III) ion, $Fe^{3+}$ ? (Atomic number of $Fe = 26$ )

[Ar] 3d^5

[Ar] 3d^3 4s^2

[Ar] 3d^4 4s^1

[Ar] 3d^6

Show answer & explanation

Answer: $[Ar] 3d^5$

Answer

The correct ground-state electronic configuration of the iron(III) ion is

[Ar] 3d^5

.

Neutral iron (

Z=26

) has the ground-state electron configuration

[Ar] 3d^6 4s^2

. When transition metal atoms form cations, electrons are removed first from the outermost principal energy level (

4s

) before removing electrons from the inner

3d

subshell. Ionization to

Fe^{3+}

involves removing the two

4s

electrons and one

3d

electron, resulting in a stable half-filled

d

-subshell configuration of

[Ar] 3d^5

.

Step-by-Step Solution

1

Determine the electronic configuration of the neutral iron atom (

Fe

,

Z = 26

).

Neutral

Fe

has 26 electrons:

1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2

or abbreviated as

[Ar] 3d^6 4s^2

.

The

4s

orbital is filled before

3d

in neutral atoms according to the Aufbau principle.

2

Apply the cation ionization rule for transition metals to form

Fe^{3+}

.

Remove 3 electrons in total: first remove 2 electrons from the outermost

4s

orbital, then remove 1 electron from the

3d

orbital.

Electrons in the outermost shell (

n=4

) are lost before inner

(n-1)d

electrons during ionization.

3

Write the resulting electronic configuration for

Fe^{3+}

.

[Ar] 3d^5

Removing two

4s

electrons leaves

[Ar] 3d^6

, and removing one more

3d

electron yields

[Ar] 3d^5

.

Key Concept

Electronic configuration of transition metal cations

Question 13Question →

Match each aluminium extraction component, compound, or alloy listed on the left with its accurate chemical role, structural behavior, or application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Molten Cryolite matches with its role as a solvent lowering the melting point of alumina and improving conductivity; reaction of bauxite with concentrated alkali matches with the formation of soluble sodium tetrahydroxoaluminate(III); anhydrous aluminium chloride below 400 °C matches with its existence as a covalent dimer Al2Cl6 with dative bonds; Duralumin matches with the alloy composed of Al, Cu, Mn, Mg used in aircraft structure.

Cryolite serves as a flux and solvent reducing alumina's melting point from

2050^\circ\text{C}

to

950^\circ\text{C}

. The reaction of bauxite with concentrated sodium hydroxide exploits aluminium's amphoteric property to produce soluble tetrahydroxoaluminate(III). Anhydrous aluminium chloride forms a coordinate-bonded dimer (

\text{Al}_2\text{Cl}_6

) at low temperatures. Duralumin is an aluminium-copper-manganese-magnesium alloy key to aerospace applications due to its high strength-to-weight ratio.

Step-by-Step Solution

1

Analyze the role of cryolite in industrial extraction

Cryolite lowers the melting temperature of

\text{Al}_2\text{O}_3

from

2050^\circ\text{C}

to

950^\circ\text{C}

and increases conductivity.

Pure

\text{Al}_2\text{O}_3

has an extremely high melting point and poor conductivity in the solid state; cryolite provides a suitable molten electrolyte mixture.

2

Examine the Bayer process chemical separation of amphoteric aluminium oxide

Dissolution in concentrated

\text{NaOH}

forms soluble complex ion

\text{Na}[\text{Al}(\text{OH})_4]

.

Aluminium oxide is amphoteric and reacts with strong base, whereas basic impurities like

\text{Fe}_2\text{O}_3

remain insoluble.

3

Evaluate the molecular structure of anhydrous aluminium chloride

Forms

\text{Al}_2\text{Cl}_6

dimer via coordinate bonds.

Aluminium in monomeric

\text{AlCl}_3

has only 6 valence electrons; dimerization allows each aluminium atom to complete its octet.

4

Identify the composition and application of Duralumin

Alloy of

\text{Al}

,

\text{Cu}

,

\text{Mn}

,

\text{Mg}

used in aircraft.

Addition of copper and magnesium to aluminium dramatically increases mechanical strength without significantly compromising low density.

Key Concept

Chemical principles of aluminium extraction (Bayer and Hall-Héroult processes), amphoteric reactivity, dimerization of aluminium chloride, and metallurgical properties of aluminium alloys.

Question 14Question →

Match each transition metal complex species on the left with its corresponding IUPAC name and oxidation state of the central metal ion on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are:

[Fe(CN)_6]^{3-}

matches Hexacyanoferrate(III) ion (

Fe

in

+3

state);

[Cu(NH_3)_4]^{2+}

matches Tetraamminecopper(II) ion (

Cu

in

+2

state);

[Co(H_2O)_6]^{2+}

matches Hexaaquacobalt(II) ion (

Co

in

+2

state); and

[Ni(CO)_4]

matches Tetracarbonylnickel(0) (

Ni

in

0

state).

Each complex is correctly paired based on the number and charge of its ligands, the resulting oxidation state of the central transition metal, and IUPAC nomenclature rules for cationic, neutral, and anionic complexes.

Step-by-Step Solution

1

Determine the charge of each ligand present in the transition metal complex.

Cyano (

CN^-

) carries a charge of

-1

, while ammine (

NH_3

), aqua (

H_2O

), and carbonyl (

CO

) are neutral (

0

charge).

Knowing ligand charges is required to set up the algebraic equation for the metal's oxidation state.

2

Calculate the oxidation state of the central metal ion in each complex.

For

[Fe(CN)_6]^{3-}

:

x + 6(-1) = -3 \Rightarrow x = +3

. For

[Cu(NH_3)_4]^{2+}

:

x + 4(0) = +2 \Rightarrow x = +2

. For

[Co(H_2O)_6]^{2+}

:

x + 6(0) = +2 \Rightarrow x = +2

. For

[Ni(CO)_4]

:

x + 4(0) = 0 \Rightarrow x = 0

.

The sum of oxidation numbers of the central atom and ligands equals the total net charge of the complex species.

3

Apply standard IUPAC naming rules for complex species.

Name ligands with multiplicative prefixes (tetra-, hexa-), follow with the metal name (using '-ate' suffix for anionic complexes like ferrate), and indicate the oxidation state in Roman numerals.

Anionic complexes modify the metal name root, whereas cationic and neutral complexes retain the standard metal element name.

Key Concept

IUPAC Nomenclature and Oxidation State Calculations for Transition Metal Complex Ions

Question 15Question →

When copper metal turnings are heated with concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ), a dense gas is evolved alongside the formation of copper(II) tetraoxosulfate(VI) ( $CuSO_4$ ) and water ( $H_2O$ ). Which of the following correctly identifies the gaseous product evolved and the specific role of concentrated $H_2SO_4$ in this chemical reaction?

Sulfur(IV) oxide (

SO_2

); concentrated

H_2SO_4

acts as an oxidizing agent.

Hydrogen gas (

H_2

); concentrated

H_2SO_4

acts as a typical acid reacting with a metal.

Sulfur(VI) oxide (

SO_3

); concentrated

H_2SO_4

acts as a dehydrating agent.

Hydrogen sulfide (

H_2S

); concentrated

H_2SO_4

acts as a reducing agent.

Show answer & explanation

Answer: Sulfur(IV) oxide ( $SO_2$ ); concentrated $H_2SO_4$ acts as an oxidizing agent.

Answer

Sulfur(IV) oxide (

SO_2

); concentrated

H_2SO_4

acts as an oxidizing agent.

Copper is an unreactive transition metal located below hydrogen in the electrochemical series. Consequently, it does not react with dilute acids to liberate hydrogen gas. However, when heated with concentrated tetraoxosulfate(VI) acid, the acid behaves as a strong oxidizing agent rather than a typical acid. Copper is oxidized to copper(II) tetraoxosulfate(VI) (

CuSO_4

), while the sulfur in

H_2SO_4

is reduced from oxidation state

+6

to

+4

, liberating sulfur(IV) oxide gas (

SO_2

) along with water.

Step-by-Step Solution

1

Analyze the position of copper in the electrochemical activity series.

Copper is less electropositive than hydrogen (

E^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}

), meaning it cannot displace

H_2

gas from dilute or non-oxidizing acids.

Direct single replacement of hydrogen by copper is thermodynamically unfavorable.

2

Determine the chemical action of hot concentrated tetraoxosulfate(VI) acid on copper.

Hot concentrated

H_2SO_4

is a strong oxidizing agent. It oxidizes copper from oxidation state

0

to

+2

(

Cu^{2+}

).

Oxidizing acids react with unreactive metals via redox pathways rather than acid-base displacement.

3

Formulate the balanced thermochemical equation.

Cu_{(s)} + 2H_2SO_{4(aq)} \rightarrow CuSO_{4(aq)} + 2H_2O_{(l)} + SO_{2(g)}

Sulfur in

H_2SO_4

(oxidation state

+6

) undergoes reduction to form sulfur(IV) oxide (

SO_2

, oxidation state

+4

).

Key Concept

Chemical properties of copper and oxidizing behavior of concentrated tetraoxosulfate(VI) acid

Estimated Time:2m 0s

Question 16Question →

Arrange the following sequential industrial steps of the Bayer process used in refining bauxite ore into pure alumina (aluminium oxide) in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological order of the Bayer process is: Digestion of bauxite in concentrated NaOH

\rightarrow

Filtration to remove red mud

\rightarrow

Seeding to precipitate hydrated aluminium hydroxide

\rightarrow

High-temperature calcination to yield pure anhydrous alumina.

The Bayer process refines crude bauxite into pure alumina through four distinct chemical phases: initial amphoteric dissolution in hot concentrated alkali (digestion), removal of solid iron oxide impurities via filtration (red mud removal), controlled crystallization of pure hydroxide (seeding/precipitation), and thermal decomposition (calcination) to produce anhydrous

\text{Al}_2\text{O}_3

.

Step-by-Step Solution

1

Identify the chemical extraction reaction

Crude bauxite (

\text{Al}_2\text{O}_3\cdot x\text{H}_2\text{O}

) is treated with hot concentrated

\text{NaOH}

solution under pressure to dissolve aluminium amphoterically:

\text{Al}_2\text{O}_3(s) + 2\text{NaOH}(aq) + 3\text{H}_2\text{O}(l) \rightarrow 2\text{NaAl(OH)}_4(aq)

.

Aluminium oxide is amphoteric and forms soluble aluminate ions, whereas impurities like

\text{Fe}_2\text{O}_3

do not react.

2

Separate insoluble residues

The dense, insoluble residue known as 'red mud' (containing

\text{Fe}_2\text{O}_3

, silica, and titania) is filtered out.

Filtration purifies the liquid stream so that subsequent precipitates are free from iron contamination.

3

Precipitate hydrated aluminium hydroxide

The clear sodium tetrahydroxoaluminate filtrate is cooled and seeded with pure

\text{Al(OH)}_3

crystals to induce precipitation:

\text{NaAl(OH)}_4(aq) \rightarrow \text{Al(OH)}_3(s) + \text{NaOH}(aq)

.

Cooling and seeding shifts the equilibrium back toward solid

\text{Al(OH)}_3

formation.

4

Dehydrate the precipitate via heating

The collected

\text{Al(OH)}_3

is washed and calcined in rotary kilns at

1000^\circ\text{C}

:

2\text{Al(OH)}_3(s) \xrightarrow{\Delta} \text{Al}_2\text{O}_3(s) + 3\text{H}_2\text{O}(g)

.

Calcination removes all chemically bound water, producing pure dry alumina feed for the Hall-Héroult electrolytic cell.

Key Concept

Bayer Process for Bauxite Purification

Question 17Question →

Soft solder is an alloy commonly used in electrical wiring and plumbing to join metal components. Which pair of metals constitutes soft solder, and what structural mechanism explains why solder is harder than its pure constituent metals?

Lead and tin; foreign atoms of different sizes disrupt the regular metallic lattice, making it harder for layers of metal ions to slide past one another.

Copper and zinc; foreign atoms release extra delocalized electrons that increase electrostatic repulsion between layers of atoms.

Copper and tin; foreign atoms form localized covalent cross-links between metal ions in the crystal lattice.

Iron and carbon; foreign atoms decrease the density of the metal lattice, causing atoms to pack more loosely.

Show answer & explanation

Answer: Lead and tin; foreign atoms of different sizes disrupt the regular metallic lattice, making it harder for layers of metal ions to slide past one another.

Answer

Soft solder consists of lead and tin. Alloying introduces atoms of a different atomic size into the metallic lattice, creating distortion that restricts the movement of atomic layers, resulting in enhanced hardness.

Soft solder is an alloy of lead and tin. The addition of foreign atoms of differing atomic radii disrupts the regular, symmetrical arrangement of the pure metal lattice. This lattice distortion prevents atomic layers from sliding easily over one another, rendering the alloy harder than pure lead or pure tin.

Step-by-Step Solution

1

Identify the elemental composition of soft solder.

Soft solder is a binary alloy composed of lead (

Pb

) and tin (

Sn

).

Memorization and recognition of core industrial alloy compositions specified in the JAMB Chemistry syllabus.

2

Analyze the physical property changes resulting from metallic alloying.

The incorporation of foreign atoms disrupts the uniform layers of the host metal lattice.

Atoms of different sizes introduce lattice strain and distortion, making it significantly more difficult for atomic planes to slide over one another when shear stress is applied.

Key Concept

Composition of solder and structural explanation of enhanced hardness in alloys due to lattice distortion

Question 18Question →

Duralumin, an alloy primarily composed of aluminium along with copper, magnesium, and manganese, exhibits higher tensile strength than pure aluminium because the presence of atoms of different sizes distorts the metallic lattice and hinders the slipping of crystal planes.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. Duralumin consists of aluminium alloyed with copper, magnesium, and manganese. The introduction of atoms of different sizes distorts the regular metallic lattice of aluminium, which impedes the movement of atomic planes and increases overall tensile strength.

The statement is true because Duralumin is formed by alloying aluminium with copper, magnesium, and manganese. The differing atomic sizes of these constituent elements cause structural lattice distortion, which prevents atomic layers from sliding easily past each other and enhances tensile strength.

Step-by-Step Solution

1

Verify the elemental composition of Duralumin.

Duralumin is an aluminium-based alloy containing approximately 94% aluminium, 4% copper, 1% magnesium, and 0.5-1% manganese.

Confirming the constituent metals verifies the composition stated in the stem.

2

Analyze the structural effect of alloying on the metallic crystal lattice.

Copper, magnesium, and manganese atoms have different atomic sizes compared to aluminium atoms, causing localized distortion in the metallic lattice.

Disrupting the uniform arrangement of identical host atoms alters the physical behavior of the metal matrix.

3

Relate lattice distortion to mechanical property modifications (tensile strength).

Lattice distortion creates stress fields that impede the movement of dislocations and slipping of crystal layers under applied force, resulting in higher tensile strength.

Interference with layer sliding directly explains why alloys are stronger and harder than their pure constituent metals.

Key Concept

Alloy composition and strengthening mechanism via lattice distortion

Question 19Question →

Match each chemical phenomenon or process involving iron and its compounds listed on the left with its corresponding chemical principle or characteristic observation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Galvanizing iron structural beams matches with providing sacrificial cathodic protection due to higher electropositivity of zinc.
2. Addition of aqueous sodium hydroxide to iron(II) tetraoxosulfate(VI) matches with forming a dirty-green precipitate that turns reddish-brown in air.
3. Accumulation of molten slag at the blast furnace hearth matches with floating on molten iron to prevent re-oxidation.
4. Reaction of aqueous iron(II) ions with acidified potassium tetraoxomanganate(VII) matches with decolorizing the purple solution via redox reaction.

Each pair correctly connects an iron chemical phenomenon with its true underlying property: zinc sacrificial protection relies on standard electrode potential differences;

\text{Fe}^{2+}

precipitation produces dirty-green

\text{Fe(OH)}_2

that oxidizes to brown

\text{Fe(OH)}_3

; slag (

\text{CaSiO}_3

) protects extracted molten iron from re-oxidation at the furnace base; and

\text{Fe}^{2+}

reduces purple

\text{MnO}_4^-

to colorless

\text{Mn}^{2+}

.

Step-by-Step Solution

1

Analyze the principle of rusting prevention via galvanization

Zinc is more reactive (more electropositive) than iron, so it corrodes preferentially in an electrochemically sacrificial manner.

Protective coatings composed of metals above iron in the electrochemical series function sacrificially.

2

Identify qualitative test reactions for iron(II) ions with strong bases

Adding

\text{OH}^-

ions to

\text{Fe}^{2+}

forms insoluble dirty-green

\text{Fe(OH)}_2

, which oxidizes in air to hydrated iron(III) oxide/hydroxide.

Iron(II) compounds undergo atmospheric oxidation rapidly in alkaline media.

3

Evaluate the industrial function of slag in the blast furnace hearth

Molten

\text{CaSiO}_3

forms an immiscible layer above liquid iron due to density differences, preventing oxygen in incoming air blasts from re-oxidizing the extracted metal.

Physical separation of hot molten iron from oxidative gases is crucial to preserve yield.

4

Examine redox properties of iron(II) species with standard oxidizing agents

\text{Fe}^{2+}

is oxidized to

\text{Fe}^{3+}

, while purple

\text{MnO}_4^-

is reduced to colorless

\text{Mn}^{2+}

in acidic solution.

Potassium tetraoxomanganate(VII) is a strong oxidizing agent used to confirm reducing species like

\text{Fe}^{2+}

.

Key Concept

Chemical reactivity, industrial extractions, qualitative identification, and corrosion mechanisms of iron and its compounds

Question 20Question →

In the industrial extraction of iron from haematite ( $Fe_2O_3$ ) in a blast furnace, which chemical species acts as the primary reducing agent for reducing iron(III) oxide in the upper zone of the furnace?

Carbon(II) oxide (

CO

)

Coke (

C

)

Calcium oxide (

CaO

)

Carbon(IV) oxide (

CO_2

)

Show answer & explanation

Answer: Carbon(II) oxide ( $CO$ )

Answer

Carbon(II) oxide (

CO

)

Gaseous carbon(II) oxide (

CO

) is the primary reducing agent in the upper, cooler region of the blast furnace. It reduces iron(III) oxide (

Fe_2O_3

) to iron according to the reaction equation

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe + 3CO_2(g)

.

Step-by-Step Solution

1

Identify the chemical reactions producing reducing species in the blast furnace.

Coke reacts with oxygen to form carbon(IV) oxide, which is subsequently reduced by excess hot coke higher up in the furnace to yield carbon(II) oxide (

CO_2 + C \rightarrow 2CO

).

Gaseous carbon(II) oxide rises through the furnace and comes into intimate contact with descending iron ore.

2

Analyze the reduction step of haematite in the upper zone.

Gaseous

CO

reduces

Fe_2O_3

via the reaction

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s/l) + 3CO_2(g)

.

Carbon(II) oxide is a powerful gaseous reducing agent at the moderate temperatures (400°C–700°C) found near the top of the furnace.

Key Concept

General Principles of Metallurgy and Metal Extraction - Reduction of Iron Ore in the Blast Furnace

Estimated Time:1m 0s

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