Non-Metals and Their Compounds Practice Questions — JAMB UTME — Page 5

Match each noble gas on the left with its corresponding primary application or characteristic property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches with filling weather balloons and deep-sea diving gas mixtures; Neon matches with advertising signs producing reddish-orange glow; Argon matches with inert shielding in arc welding and light bulbs; Radon matches with radioactive cancer treatment.

Each noble gas possesses specific physical characteristics: Helium's light weight and low solubility suit balloons and diving gas; Neon's electrical discharge color suits signage; Argon's abundance and chemical inertness suit welding and lighting; Radon's radioactivity suits cancer treatment.

Step-by-Step Solution

1

Identify the key physical and chemical properties of each Group 0 (noble gas) element.

Helium is the lightest non-flammable gas with minimal blood solubility; Neon exhibits characteristic light emission; Argon is an abundant inert gas; Radon is radioactive.

Matching noble gases requires aligning their unique electronic stability and physical properties with industrial and medical uses.

2

Pair each gas to its correct industrial or medical application.

Helium pairs with weather balloons/diving gas; Neon pairs with advertising glow lamps; Argon pairs with welding/bulbs; Radon pairs with radiotherapy.

Each application specifically relies on the unique physical state or reactivity profile of that element.

Key Concept

Noble gases are unreactive Group 8/0 elements with stable octet (or duplet) electron configurations whose distinct physical properties dictate specific industrial and medical uses.

Question 82Question →

When solid lead(II) trioxonitrate(V), $Pb(NO_3)_2$ , is heated strongly in a dry test tube, it decomposes to yield a yellow solid residue, oxygen gas, and a reddish-brown gas. Which formula represents this reddish-brown gas, and what is the oxidation state of nitrogen in it?

NO_2

with an oxidation state of

+4

N_2O

with an oxidation state of

+1

NO

with an oxidation state of

+2

N_2O_5

with an oxidation state of

+5

Show answer & explanation

Answer: $NO_2$ with an oxidation state of $+4$

Answer

The reddish-brown gas is nitrogen(IV) oxide (

NO_2

), in which nitrogen has an oxidation state of

+4

.

Heating lead(II) trioxonitrate(V) decomposes it into lead(II) oxide (

PbO

), nitrogen(IV) oxide (

NO_2

), and oxygen (

O_2

). Nitrogen(IV) oxide is a characteristic reddish-brown gas, and assigned oxidation state calculations give

+4

for nitrogen in

NO_2

.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of lead(II) trioxonitrate(V).

2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)

Heavy metal nitrates decompose on heating to yield the metal oxide, nitrogen(IV) oxide gas, and oxygen gas.

2

Identify the physical property of the gaseous products.

PbO

is a yellow solid residue (when hot/cold depending on form),

O_2

is a colorless gas, and

NO_2

is a distinctive reddish-brown acidic gas.

Nitrogen(IV) oxide (

NO_2

) is the only brown oxide of nitrogen produced in this reaction.

3

Calculate the oxidation state of nitrogen in

NO_2

.

Let

x

be the oxidation state of N.

x + 2(-2) = 0 \implies x = +4

.

Oxygen has an oxidation number of

-2

in neutral covalent oxides.

Key Concept

Thermal Decomposition of Metal Nitrates and Oxides of Nitrogen

Question 83Question →

During the industrial isolation of noble gases from liquid air, argon is collected in a fraction between nitrogen and oxygen. What physical property explains why argon distills over after nitrogen but before oxygen?

Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Argon has a lower boiling point than nitrogen but a higher boiling point than oxygen.

Argon has a higher vapor density than oxygen but a lower vapor density than nitrogen.

Argon has a lower molar mass than nitrogen but a higher molar mass than oxygen.

Show answer & explanation

Answer: Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Answer

Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Fractional distillation of liquefied air separates gases according to their boiling points. Nitrogen distills first at

-196\ ^{\circ}\text{C}

(

77\text{ K}

), argon distills next at

-186\ ^{\circ}\text{C}

(

87\text{ K}

), and oxygen distills last at

-183\ ^{\circ}\text{C}

(

90\text{ K}

). Therefore, argon distills over after nitrogen but before oxygen because its boiling point is higher than that of nitrogen and lower than that of oxygen.

Step-by-Step Solution

1

Identify the separation principle of liquid air distillation.

Components of liquid air are separated based on differences in their boiling points during fractional distillation.

Fractional distillation separates liquids with different boiling points as the liquid mixture is warmed.

2

Compare the boiling points of nitrogen, argon, and oxygen.

Nitrogen boils at

-196\ ^{\circ}\text{C}

, argon boils at

-186\ ^{\circ}\text{C}

, and oxygen boils at

-183\ ^{\circ}\text{C}

.

Lower boiling point components boil and distill off first as vapor.

3

Deduce the sequence of distillation for argon.

Since

-196\ ^{\circ}\text{C} < -186\ ^{\circ}\text{C} < -183\ ^{\circ}\text{C}

, nitrogen distills off first, followed by argon, while oxygen remains liquid longest.

Argon distills after nitrogen because its boiling point is higher than nitrogen's, and before oxygen because its boiling point is lower than oxygen's.

Key Concept

Isolation of noble gases from liquid air by fractional distillation

Estimated Time:1m 0s

Question 84Question →

What is the oxidation number of chlorine in potassium trioxochlorate(V), $\text{KClO}_3$ ?

+5

+3

+7

-1

Show answer & explanation

Answer: $+5$

Answer

The oxidation number of chlorine in

\text{KClO}_3

is

+5

.

In potassium trioxochlorate(V),

\text{KClO}_3

, potassium contributes

+1

and three oxygen atoms contribute a total of

-6

. For the compound to be electrically neutral, chlorine must have an oxidation state of

+5

.

Step-by-Step Solution

1

Assign known oxidation numbers to potassium and oxygen

Potassium (Group 1 metal) has an oxidation state of

+1

. Oxygen has an oxidation state of

-2

.

Standard rules assign

+1

to alkali metals and

-2

to oxygen in oxoacids/oxoanions.

2

Formulate an equation for the neutral molecule

\text{KClO}_3

(+1) + x + 3(-2) = 0

The sum of all oxidation numbers in a neutral chemical compound must equal zero.

3

Solve for the unknown oxidation state

x

of chlorine

+1 + x - 6 = 0 \implies x - 5 = 0 \implies x = +5

Solving the linear algebraic equation yields

+5

.

Key Concept

Oxidation state calculation of halogens in oxoacids and oxoacid salts

Question 85Question →

When chlorine gas is bubbled into a hot, concentrated solution of potassium hydroxide (

\text{KOH}

), it undergoes disproportionation according to the chemical equation:

3\text{Cl}_2\text{(g)} + 6\text{KOH(aq)} \rightarrow 5\text{KCl(aq)} + \text{KClO}_3\text{(aq)} + 3\text{H}_2\text{O(l)}

What mass of potassium trioxochlorate(V) (

\text{KClO}_3

) is produced when

6.72\text{ dm}^3

of chlorine gas measured at s.t.p. reacts completely?
[Molar volume of gas at s.t.p. =

22.4\text{ dm}^3\text{mol}^{-1}

; Molar masses:

\text{K} = 39\text{ g/mol}

,

\text{Cl} = 35.5\text{ g/mol}

,

\text{O} = 16\text{ g/mol}

]

12.25\text{ g}

36.75\text{ g}

11.43\text{ g}

9.05\text{ g}

Show answer & explanation

Answer: $12.25\text{ g}$

Answer

The correct mass of potassium trioxochlorate(V) produced is

12.25\text{ g}

.

The option specifying

12.25\text{ g}

is correct because

6.72\text{ dm}^3

of

\text{Cl}_2

gas at s.t.p. corresponds to

0.30\text{ mol}

. Based on the 3:1 stoichiometric ratio from the balanced equation (

3\text{Cl}_2 \rightarrow 1\text{KClO}_3

),

0.10\text{ mol}

of

\text{KClO}_3

is produced. Multiplying

0.10\text{ mol}

by the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

) yields

12.25\text{ g}

.

Step-by-Step Solution

1

Calculate the amount of chlorine gas in moles at s.t.p.

n(\text{Cl}_2) = \frac{6.72\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.30\text{ mol}

At s.t.p.,

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

.

2

Determine the moles of potassium trioxochlorate(V) (

\text{KClO}_3

) formed using stoichiometric ratios.

n(\text{KClO}_3) = \frac{1}{3} \times 0.30\text{ mol} = 0.10\text{ mol}

From the balanced equation,

3\text{ moles}

of

\text{Cl}_2

produce

1\text{ mole}

of

\text{KClO}_3

.

3

Calculate the molar mass of

\text{KClO}_3

.

\text{Molar Mass} = 39 + 35.5 + (3 \times 16) = 122.5\text{ g/mol}

Summing the atomic masses of one potassium, one chlorine, and three oxygen atoms.

4

Calculate the mass of

\text{KClO}_3

produced.

\text{Mass} = 0.10\text{ mol} \times 122.5\text{ g/mol} = 12.25\text{ g}

Mass is obtained by multiplying the number of moles by the molar mass.

Key Concept

Disproportionation reactions of halogens in hot concentrated alkalis and gas stoichiometry at s.t.p.

Question 86Question →

When concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ) is added to sucrose crystals, a black mass of carbon and steam are produced. What property of the acid does this reaction demonstrate?

Dehydrating action

Oxidizing action

Reducing action

Precipitating action

Show answer & explanation

Answer: Dehydrating action

Answer

Dehydrating action

Concentrated tetraoxosulfate(VI) acid has a very high affinity for water. When added to carbohydrates such as sucrose (

C_{12}H_{22}O_{11}

), it removes hydrogen and oxygen atoms in a 2:1 ratio as water molecules, leaving behind a charred black mass of elemental carbon.

Step-by-Step Solution

1

Analyze the chemical transformation in the given reaction

Sucrose (

C_{12}H_{22}O_{11}

) reacts with concentrated

H_2SO_4

to yield elemental carbon (

12C

) and water vapor (

11H_2O

).

The acid extracts hydrogen and oxygen in the exact ratio of water from the carbohydrate molecule.

2

Identify the specific property corresponding to removing elements of water

This process is classified as dehydration.

Dehydration refers specifically to the removal of chemically combined water or its constituent elements from a substance.

Key Concept

Dehydrating action of concentrated tetraoxosulfate(VI) acid

Estimated Time:45s

Question 87Question →

In the industrial manufacture of tetraoxosulfate(VI) acid via the Contact Process, several crucial chemical and physical steps are carried out in a specific sequence to maximize yield and prevent efficiency loss. Arrange the following steps of the Contact Process in the correct sequential order from first to last.

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Answer

The correct sequence of the Contact Process is: Combustion/roasting to produce

SO_2

gas

\rightarrow

Purification of

SO_2

gas to remove catalyst poisons

\rightarrow

Catalytic conversion of

SO_2

to

SO_3

over

V_2O_5

catalyst

\rightarrow

Absorption of

SO_3

gas in concentrated

H_2SO_4

to form oleum

\rightarrow

Controlled dilution of oleum with water to yield

H_2SO_4

.

The Contact Process must follow a rigorous order: first generating raw

SO_2

gas, purifying it to prevent catalyst poisoning by impurities like

As_2O_3

, catalytically oxidizing

SO_2

to

SO_3

over

V_2O_5

, absorbing

SO_3

in 98%

H_2SO_4

to form oleum (

H_2S_2O_7

), and finally diluting oleum with water to produce concentrated

H_2SO_4

.

Step-by-Step Solution

1

Identify the initial feedstock generation stage.

Combustion of sulfur or roasting of sulfide ores produces

SO_2

gas.

Sulfur(IV) oxide is the essential chemical precursor required for the process.

2

Determine the necessary gas purification stage prior to catalysis.

Passing the

SO_2

and air mixture through scrubbers and precipitators removes dust particles and arsenic(III) oxide (

As_2O_3

).

Arsenic compounds act as catalyst poisons, permanently deactivating the vanadium(V) oxide catalyst if not removed first.

3

Identify the catalytic oxidation step.

Purified

SO_2

reacts with

O_2

over a

V_2O_5

catalyst at 450 °C and 1–2 atm to form

SO_3

.

This exothermic equilibrium reaction converts sulfur(IV) oxide to sulfur(VI) oxide.

4

Determine the absorption stage for sulfur(VI) oxide.

SO_3

gas is absorbed into concentrated (98%)

H_2SO_4

to form oleum (

H_2S_2O_7

).

Direct hydration of

SO_3

with water is extremely exothermic and produces a fine acid fog that cannot be easily condensed industrially.

5

Identify the final dilution/hydration step.

Oleum (

H_2S_2O_7

) is diluted with a calculated volume of water to form concentrated

H_2SO_4

.

Reacting oleum with water yields high-purity tetraoxosulfate(VI) acid safely and efficiently.

Key Concept

Sequential chemical and industrial stages of the Contact Process for tetraoxosulfate(VI) acid production

Estimated Time:2m 0s

Question 88Question →

During the laboratory preparation and isolation of nitrogen gas from atmospheric air, a student must pass atmospheric air through several reagents in a specific sequence to remove impurities. What is the correct order of steps to isolate nitrogen gas from atmospheric air, starting from the initial removal of carbon(IV) oxide to the final collection of the gas?

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Answer

The correct sequence to isolate nitrogen gas from atmospheric air is: 1. Pass air through concentrated caustic alkali solution (

KOH

/

NaOH

) to remove

CO_2

. 2. Pass the gas through concentrated

H_2SO_4

to remove water vapor. 3. Pass dry gas over red-hot copper turnings to remove

O_2

. 4. Collect the remaining nitrogen gas over water.

Atmospheric air consists primarily of

N_2

(78%),

O_2

(21%),

CO_2

(0.03%), water vapor, and noble gases. To isolate nitrogen, impurities are removed according to chemical reactivity:

CO_2

is removed first by neutralisation with an alkali (

KOH

), moisture is absorbed by a dehydrating agent (concentrated

H_2SO_4

), oxygen is removed by reduction of red-hot copper turnings to

CuO

, and the remaining nitrogen gas (mixed with trace noble gases like argon) is collected over water.

Step-by-Step Solution

1

Identify the acidic gas impurity in air and select its removal agent.

Carbon(IV) oxide (

CO_2

) is acidic and must be scrubbed first using concentrated

KOH

or

NaOH

solution:

2KOH_{(aq)} + CO_{2(g)} \rightarrow K_2CO_{3(aq)} + H_2O_{(l)}

.

Removing

CO_2

first prevents it from contaminating subsequent drying and heating apparatus.

2

Dry the remaining gas mixture.

Passing the remaining gases (

O_2

,

N_2

, water vapor, noble gases) through concentrated

H_2SO_4

absorbs moisture.

Gas must be thoroughly dried before passing over hot copper turnings to prevent thermal shock and unwanted reactions.

3

Remove oxygen gas chemically.

Passing dry air over red-hot copper turnings removes

O_2

:

2Cu_{(s)} + O_{2(g)} \rightarrow 2CuO_{(s)}

.

Hot copper chemically binds oxygen, leaving only unreactive nitrogen and traces of noble gases.

4

Collect the purified nitrogen gas.

Nitrogen gas is collected over water.

Nitrogen has very low solubility in water, making water displacement ideal for gas collection.

Key Concept

Laboratory Isolation of Nitrogen from Atmospheric Air

Estimated Time:1m 30s

Question 89Question →

Match each chlorine oxoacid listed on the left with its corresponding chlorine oxidation state and defining chemical characteristics on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypochlorous acid (HClO) matches with oxidation state +1 and bleaching/germicidal properties; Chlorous acid (HClO2) matches with oxidation state +3 and dioxochlorate(III) salt formation; Chloric acid (HClO3) matches with oxidation state +5 and trioxochlorate(V) salt formation; Perchloric acid (HClO4) matches with oxidation state +7 and being the strongest oxoacid.

Each chlorine oxoacid is correctly paired based on the oxidation state of chlorine (ranging from +1 in hypochlorous acid to +7 in perchloric acid) and its associated chemical behavior, where acid strength and oxidizing power in concentrated form increase with increasing oxygen content.

Step-by-Step Solution

1

Calculate the oxidation number of chlorine in each oxoacid using standard oxidation states (H = +1, O = -2).

HClO: 1 + Cl + (-2) = 0 → Cl = +1. HClO2: 1 + Cl + 2(-2) = 0 → Cl = +3. HClO3: 1 + Cl + 3(-2) = 0 → Cl = +5. HClO4: 1 + Cl + 4(-2) = 0 → Cl = +7.

Determining oxidation states is the first step in differentiating chlorine oxoacids.

2

Correlate the oxidation states with acid strength trends in halogen oxoacids.

Acid strength increases as the number of oxygen atoms increases (HClO < HClO2 < HClO3 < HClO4). Thus, HClO4 is the strongest oxoacid.

Additional oxygen atoms pull electron density away from the O-H bond, weakening it and stabilizing the resulting oxoanion.

3

Pair each acid with its systematic IUPAC nomenclature and chemical properties.

HClO (+1) is hypochlorous acid (oxochlorate(I)), HClO2 (+3) is chlorous acid (dioxochlorate(III)), HClO3 (+5) is chloric acid (trioxochlorate(V)), and HClO4 (+7) is perchloric acid (tetraoxochlorate(VII)).

This establishes the exact matching pairs between left and right items.

Key Concept

Oxidation States and Acid Strength Trends of Chlorine Oxoacids

Estimated Time:2m 0s

Question 90Question →

Match each chlorine oxoacid or oxoanion listed in Column I with its corresponding oxidation state, IUPAC designation, or chemical property in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypochlorous acid (HClO) matches with the weak, unstable acid (+1 oxidation state) decomposing in sunlight to release O2 gas; Chloric(V) acid (HClO3) matches with the strong oxidizing acid (+5 oxidation state) prepared from barium chlorate and dilute H2SO4; Perchloric acid (HClO4) matches with the strongest oxoacid (+7 oxidation state); Oxochlorate(I) anion (ClO-) matches with the active bleaching conjugate base formed in cold aqueous NaOH.

Each chlorine species is accurately paired according to oxidation state calculations, resonance stability of conjugate bases, and established laboratory synthesis routes.

Step-by-Step Solution

1

Determine the oxidation state of chlorine in each specified oxoacid and oxoanion species

In

\text{HClO}

, chlorine is

+1

. In

\text{HClO}_3

, chlorine is

+5

. In

\text{HClO}_4

, chlorine is

+7

. In

\text{ClO}^-

, chlorine is

+1

.

Oxidation numbers dictate IUPAC nomenclature and help categorize chemical reactivity.

2

Analyze acid strength trends among chlorine oxoacids

Acid strength increases with increasing number of terminal oxygen atoms:

\text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4

. Thus,

\text{HClO}_4

is the strongest oxoacid.

Electronegative terminal oxygen atoms withdraw electron density from the

\text{O-H}

bond, stabilizing the conjugate base via resonance.

3

Correlate specific preparation methods and stability characteristics to their respective species

\text{HClO}

decomposes into

\text{HCl}

and

\text{O}_2

.

\text{HClO}_3

is synthesized via

\text{Ba(ClO}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4\downarrow + 2\text{HClO}_3

.

\text{ClO}^-

is generated in cold alkaline chlorination:

\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O}

.

Matching unique reaction mechanisms and industrial/laboratory preparation routes identifies each chlorine compound.

Key Concept

Oxoacids of chlorine, oxidation states, relative acid strengths, and chemical preparation methods.

Question 91Question →

Match each noble gas listed on the left with its corresponding primary industrial application or characteristic use on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches with weather balloons and diving gas mixtures; Neon matches with advertising signs; Argon matches with arc welding inert atmosphere; Radon matches with cancer radiotherapy.

Each noble gas is matched to its unique application based on its physical and chemical properties: Helium for low density and low blood solubility, Neon for glowing light discharge, Argon for unreactive shielding during welding, and Radon for cancer radiotherapy.

Step-by-Step Solution

1

Identify the primary application of Helium.

Helium's low density and minimal blood solubility pair it with weather balloons and deep-sea diving mixtures.

Prevents decompression sickness in divers and provides buoyancy in balloons.

2

Identify the primary application of Neon.

Neon emits a distinct reddish-orange light in electrical discharge tubes used for advertising signs.

Excited neon gas emits characteristic light when electrical discharge occurs.

3

Identify the primary application of Argon.

Argon serves as an inert shielding gas in electric arc welding.

Prevents atmospheric oxygen and nitrogen from reacting with hot metals being welded.

4

Identify the primary application of Radon.

Radon is radioactive and used in cancer radiotherapy.

Radiation emitted during radioactive decay destroys targeted cancer cells.

Key Concept

Industrial applications and chemical inertness of noble gases

Question 92Question →

In the Contact Process for the manufacture of tetraoxosulfate(VI) acid, $44.8\text{ dm}^3$ of sulfur(IV) oxide ( $SO_2$ ) gas measured at standard temperature and pressure (STP) is reacted with excess oxygen over a vanadium(V) oxide ( $V_2O_5$ ) catalyst. If the catalytic conversion efficiency of $SO_2$ to sulfur(VI) oxide ( $SO_3$ ) is $85\%$ , and all produced $SO_3$ is subsequently absorbed in concentrated $H_2SO_4$ and hydrated to form pure $H_2SO_4$ , what mass of pure $H_2SO_4$ in grams is produced? (Molar mass of $H_2SO_4 = 98\text{ g mol}^{-1}$ , Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 166.6

Answer

The mass of pure H2SO4 produced is 166.6 g.

The correct answer of 166.6 g is derived by converting 44.8 dm³ of SO2 at STP to 2.0 moles, taking 85% of that value to find the actual 1.70 moles of SO3 produced, and multiplying by the molar mass of H2SO4 (98 g/mol).

Step-by-Step Solution

1

Calculate the moles of SO2 gas supplied at STP

Moles of SO2 = 2.0 mol

Dividing the gas volume at STP (44.8 dm³) by the molar gas volume (22.4 dm³/mol) gives the molar quantity.

2

Determine theoretical yield of SO3

Theoretical moles of SO3 = 2.0 mol

From the stoichiometric mole ratio in 2SO2 + O2 -> 2SO3, 2 moles of SO2 produce 2 moles of SO3.

3

Calculate actual moles of SO3 produced considering catalytic efficiency

Actual moles of SO3 = 1.70 mol

Multiplying the theoretical yield (2.0 mol) by the 85% conversion efficiency gives the actual yield of 1.70 mol.

4

Calculate mass of H2SO4 produced from the actual SO3 formed

Mass of H2SO4 = 166.6 g

Overall absorption and hydration converts SO3 to H2SO4 in a 1:1 mole ratio (SO3 + H2O -> H2SO4). Multiplying 1.70 mol by molar mass 98 g/mol yields 166.6 g.

Key Concept

Stoichiometry of the Contact Process involving molar volume at STP and percentage conversion efficiency

Question 93Question →

Match each nitrogen oxide or nitrogen cycle component in Column I with its correct physical property or biological role in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dinitrogen monoxide (

N_2O

) matches with 'A sweet-smelling, neutral gas that relights a glowing splint'. Nitrogen dioxide (

NO_2

) matches with 'A reddish-brown, acidic gas that dissolves in water to form a mixture of two acids'. Nitrosomonas bacteria matches with 'Converts soil ammonium ions (

NH_4^+

) into trioxonitrate(III) ions (

NO_2^-

)'. Nitrobacter bacteria matches with 'Converts soil trioxonitrate(III) ions (

NO_2^-

) into trioxonitrate(V) ions (

NO_3^-

)'.

Each nitrogen oxide and nitrifying bacterium is correctly associated with its characteristic chemical behavior or distinct biochemical pathway in the nitrogen cycle.

Step-by-Step Solution

1

Differentiate between the physical and chemical properties of the nitrogen oxides.

N_2O

is neutral and sweet-smelling while supporting combustion.

NO_2

is acidic, reddish-brown, and forms

HNO_2

and

HNO_3

upon reaction with water.

Oxides of nitrogen vary in color, acidity, and combustion-supporting capabilities depending on the oxidation state of nitrogen.

2

Differentiate the roles of nitrifying bacteria in the nitrogen cycle.

Nitrosomonas oxidizes ammonium to nitrite (

NO_2^-

), whereas Nitrobacter oxidizes nitrite (

NO_2^-

) to nitrate (

NO_3^-

).

Nitrification proceeds in two distinct enzymatic steps mediated by specialized microbial species.

Key Concept

Properties of Nitrogen Oxides and Biological Nitrification Stages

Question 94Question →

When copper turnings are heated with concentrated tetraoxosulfate(VI) acid (

H_2SO_4

), a gas (

SO_2

) that turns acidified potassium heptaoxodichromate(VI) solution from orange to green is evolved according to the equation:

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

Which property of concentrated tetraoxosulfate(VI) acid is demonstrated in this reaction?

Dehydrating agent

Oxidizing agent

Reducing agent

Drying agent

Show answer & explanation

Answer: Oxidizing agent

Answer

Oxidizing agent

In this reaction, elemental copper (

Cu

) with an oxidation state of

0

is oxidized to

Cu^{2+}

(

+2

oxidation state in

CuSO_4

). Concentrated tetraoxosulfate(VI) acid acts as the oxidizing agent because it causes this oxidation while itself being reduced to sulfur(IV) oxide (

SO_2

), in which sulfur has an oxidation state of

+4

.

Step-by-Step Solution

1

Analyze the oxidation state changes of copper and sulfur in the given balanced equation

Copper (

Cu

) goes from oxidation state

0

to

+2

in

CuSO_4

(loss of electrons / oxidation). Sulfur in

H_2SO_4

goes from

+6

to

+4

in

SO_2

(gain of electrons / reduction).

Determining oxidation number changes identifies which reactant is oxidized and which acts as the oxidizing agent.

2

Identify the role of concentrated tetraoxosulfate(VI) acid

Since

H_2SO_4

causes copper to be oxidized to

Cu^{2+}

ions while itself being reduced to

SO_2

, concentrated

H_2SO_4

is acting as an oxidizing agent.

An oxidizing agent accepts electrons and undergoes reduction during a redox reaction.

Key Concept

Oxidizing property of concentrated tetraoxosulfate(VI) acid with metals

Question 95Question →

Arrange the following sequential stages involved in the industrial manufacture of tetraoxosulfate(VI) acid via the Contact Process in their correct chronological order from initial raw material processing to final acid product formation:

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Show answer & explanation

Answer

The correct sequence of stages in the Contact Process is: (1) Combustion of sulfur to produce

SO_2

, (2) Purification of

SO_2

gas to remove catalyst poisons, (3) Catalytic oxidation of

SO_2

to

SO_3

over

V_2O_5

, (4) Absorption of

SO_3

into concentrated

H_2SO_4

to form oleum (

H_2S_2O_7

), and (5) Controlled dilution of oleum with water to produce concentrated

H_2SO_4

.

The Contact Process progresses in five logical stages: sulfur combustion to generate

SO_2

, gas purification to protect the catalyst, catalytic oxidation of

SO_2

to

SO_3

over

V_2O_5

, absorption of

SO_3

into concentrated

H_2SO_4

to produce oleum (

H_2S_2O_7

), and finally hydration of oleum with water to yield pure

H_2SO_4

.

Step-by-Step Solution

1

Identify the initial feedstock generation phase

Elemental sulfur is burned in excess dry air to form

SO_2

(

S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}

).

Sulfur(IV) oxide gas must be produced first before subsequent catalytic oxidation can occur.

2

Determine the necessary gas conditioning and purification step

The

SO_2

stream is washed, dried, and passed through electrostatic precipitators to eliminate impurities like

As_2O_3

.

Arsenic impurities deactivate (poison) the vanadium(V) oxide catalyst if not removed prior to entering the catalytic converter.

3

Identify the core catalytic conversion reaction

SO_2

reacts reversibly with

O_2

over

V_2O_5

at

450^\circ\text{C}

and

1-2\text{ atm}

to form

SO_3

(

2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}

).

This key reversible exothermic step converts sulfur(IV) oxide to sulfur(VI) oxide under optimal yield conditions.

4

Identify the absorption mechanism for

SO_3

SO_3

gas is dissolved in 98% concentrated

H_2SO_4

to form oleum (

SO_{3(g)} + H_2SO_{4(l)} \rightarrow H_2S_2O_{7(l)}

).

Direct addition of

SO_3

to water generates enormous heat, causing the water to vaporize and create a fog of acid mist that will not condense easily.

5

Determine the final product hydration stage

Oleum is diluted with water to generate

H_2SO_4

of desired concentration (

H_2S_2O_{7(l)} + H_{2}O_{(l)} \rightarrow 2H_2SO_{4(l)}

).

Diluting oleum produces pure tetraoxosulfate(VI) acid safely without fog or mist formation.

Key Concept

Sequential chemical steps, conditions, and process rationale of the industrial Contact Process for tetraoxosulfate(VI) acid production

Estimated Time:2m 0s

Question 96Question →

Concentrated tetraoxosulfate(VI) acid reacts with copper metal as an oxidizing agent according to the balanced chemical equation:

Cu(s) + 2H_2SO_4(aq) \rightarrow CuSO_4(aq) + 2H_2O(l) + SO_2(g)

If

31.75\text{ g}

of copper turnings react completely with an excess of concentrated tetraoxosulfate(VI) acid, what volume of sulfur(IV) oxide (

SO_2

) gas, in

\text{dm}^3

, is evolved at standard temperature and pressure (STP)? [Molar mass of

Cu = 63.5\text{ g mol}^{-1}

; Molar volume of gas at STP =

22.4\text{ dm}^3\text{ mol}^{-1}

]

Show answer & explanation

Answer: 11.2

Answer

The volume of sulfur(IV) oxide gas evolved at STP is

11.2\text{ dm}^3

.

Concentrated tetraoxosulfate(VI) acid acts as an oxidizing agent when heated with copper metal, being reduced to sulfur(IV) oxide gas. From the balanced reaction equation,

1\text{ mole}

(

63.5\text{ g}

) of copper yields

1\text{ mole}

(

22.4\text{ dm}^3

at STP) of

SO_2

gas. Therefore,

31.75\text{ g}

(

0.5\text{ moles}

) of copper yields

0.5 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3

of

SO_2

gas.

Step-by-Step Solution

1

Calculate the moles of copper metal reacted

n(Cu) = \frac{31.75\text{ g}}{63.5\text{ g mol}^{-1}} = 0.5\text{ mol}

Converting the given mass of copper to moles allows stoichiometric comparison with the reaction products.

2

Determine the moles of sulfur(IV) oxide (

SO_2

) gas produced

n(SO_2) = 0.5\text{ mol}

From the balanced chemical equation,

1\text{ mole}

of

Cu

reacts to produce

1\text{ mole}

of

SO_2

gas.

3

Calculate the volume of

SO_2

gas evolved at STP

V(SO_2) = 0.5\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

At STP,

1\text{ mole}

of any ideal gas occupies a standard molar volume of

22.4\text{ dm}^3

.

Key Concept

Oxidizing action of concentrated tetraoxosulfate(VI) acid on metals and gas volume stoichiometry at STP

Question 97Question →

When sulfur(IV) oxide ( $SO_2$ ) gas is bubbled into an acidified solution of potassium tetraoxomanganate(VII) ( $KMnO_4$ ), a distinct chemical change occurs. Which statement correctly describes the chemical role of sulfur(IV) oxide and the observed color change?

Sulfur(IV) oxide acts as a reducing agent, decolorizing the purple solution.

Sulfur(IV) oxide acts as an oxidizing agent, changing the purple solution to green.

Sulfur(IV) oxide acts as a dehydrating agent, forming a dark brown precipitate.

Sulfur(IV) oxide acts as an allotropic catalyst, depositing yellow solid sulfur.

Show answer & explanation

Answer: Sulfur(IV) oxide acts as a reducing agent, decolorizing the purple solution.

Answer

Sulfur(IV) oxide acts as a reducing agent, decolorizing the purple potassium tetraoxomanganate(VII) solution.

In the reaction with acidified potassium tetraoxomanganate(VII), sulfur(IV) oxide (

SO_2

) is oxidized to tetraoxosulfate(VI) ions (

SO_4^{2-}

), with sulfur's oxidation state increasing from +4 to +6. Because it undergoes oxidation,

SO_2

acts as a reducing agent. Consequently, the purple tetraoxomanganate(VII) ions (

MnO_4^-

) are reduced to colorless manganese(II) ions (

Mn^{2+}

).

Step-by-Step Solution

1

Determine the oxidation state of sulfur in

SO_2

Sulfur has an oxidation state of +4 in

SO_2

.

Since oxygen has an oxidation state of -2, solving

S + 2(-2) = 0

gives

S = +4

.

2

Analyze the redox reaction between

SO_2

and acidified

KMnO_4

SO_2

is oxidized to

SO_4^{2-}

(oxidation state of S increases from +4 to +6), while

MnO_4^-

is reduced to

Mn^{2+}

.

A substance that undergoes oxidation by losing electrons functions as a reducing agent.

3

Identify the visual observation associated with the reduction of

MnO_4^-

to

Mn^{2+}

The intense purple color of the solution turns colorless.

Tetraoxomanganate(VII) ions (

MnO_4^-

) impart a purple color to the solution, whereas reduced manganese(II) ions (

Mn^{2+}

) are colorless.

Key Concept

Reducing action of sulfur(IV) oxide on acidified potassium tetraoxomanganate(VII)

Question 98Question →

What volume of nitrogen(IV) oxide gas, measured at standard temperature and pressure (STP), is evolved when $12.7\text{ g}$ of copper completely reacts with excess concentrated trioxonitrate(V) acid according to the equation below?

Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}

(Relative atomic mass: $Cu = 63.5$ ; Molar volume of gas at STP = $22.4\text{ dm}^3\text{/mol}$ )

8.96\text{ dm}^3

9.60\text{ dm}^3

4.48\text{ dm}^3

17.92\text{ dm}^3

Show answer & explanation

Answer: $8.96\text{ dm}^3$

Answer

The volume of nitrogen(IV) oxide gas evolved at STP is

8.96\text{ dm}^3

.

According to the balanced chemical equation,

1\text{ mol}

of copper reacts with excess concentrated trioxonitrate(V) acid to produce

2\text{ mol}

of nitrogen(IV) oxide gas. Since

12.7\text{ g}

of copper corresponds to

0.2\text{ mol}

, the reaction generates

0.4\text{ mol}

of

NO_2

. At STP,

0.4\text{ mol}

occupies

0.4 \times 22.4 = 8.96\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount of copper in moles

\text{Moles of } Cu = \frac{12.7\text{ g}}{63.5\text{ g/mol}} = 0.2\text{ mol}

Dividing given mass by relative atomic mass yields moles of reactant.

2

Determine moles of

NO_2

gas produced using stoichiometric coefficients

\text{Moles of } NO_2 = 0.2\text{ mol } Cu \times \left(\frac{2\text{ mol } NO_2}{1\text{ mol } Cu}\right) = 0.4\text{ mol}

The balanced chemical equation shows that

1\text{ mol}

of

Cu

yields

2\text{ mol}

of

NO_2

gas.

3

Calculate the volume of

NO_2

gas evolved at STP

\text{Volume of } NO_2 = 0.4\text{ mol} \times 22.4\text{ dm}^3\text{/mol} = 8.96\text{ dm}^3

One mole of any gas occupies

22.4\text{ dm}^3

at standard temperature and pressure.

Key Concept

Redox stoichiometric calculation of gas volumes produced by trioxonitrate(V) acid reactions

Question 99Question →

Potassium tetraoxomanganate(VII),

\text{KMnO}_4

, reacts with excess concentrated hydrochloric acid according to the redox equation:

2\text{KMnO}_4(s) + 16\text{HCl}(aq) \rightarrow 2\text{KCl}(aq) + 2\text{MnCl}_2(aq) + 8\text{H}_2\text{O}(l) + 5\text{Cl}_2(g)

Calculate the volume of chlorine gas (in

\text{dm}^3

) produced at s.t.p. when

15.8\text{ g}

of

\text{KMnO}_4

reacts completely. [Molar mass of

\text{KMnO}_4 = 158\text{ g mol}^{-1}

, molar volume of gas at s.t.p. =

22.4\text{ dm}^3\text{mol}^{-1}

]

Show answer & explanation

Answer: 5.6

Answer

The volume of chlorine gas liberated at s.t.p. is

5.6\text{ dm}^3

.

Converting

15.8\text{ g}

of

\text{KMnO}_4

gives

0.10\text{ mol}

. According to the balanced equation,

2\text{ mol}

of

\text{KMnO}_4

yields

5\text{ mol}

of

\text{Cl}_2

, giving

0.25\text{ mol}

of

\text{Cl}_2

. Multiplying

0.25\text{ mol}

by the molar volume at s.t.p. (

22.4\text{ dm}^3\text{mol}^{-1}

) yields

5.6\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the moles of potassium tetraoxomanganate(VII),

\text{KMnO}_4

\text{Moles of KMnO}_4 = \frac{15.8\text{ g}}{158\text{ g mol}^{-1}} = 0.10\text{ mol}

Converting the given mass of reactant to moles allows stoichiometric comparison.

2

Use the balanced redox equation to determine the mole ratio between

\text{KMnO}_4

and

\text{Cl}_2

\text{Moles of Cl}_2 = 0.10\text{ mol} \times \frac{5}{2} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

of

\text{KMnO}_4

produce

5\text{ moles}

of

\text{Cl}_2

gas.

3

Calculate the volume of

\text{Cl}_2

gas produced at s.t.p.

\text{Volume} = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 5.6\text{ dm}^3

At s.t.p.,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

.

Key Concept

Stoichiometric calculations in the laboratory preparation of chlorine from oxidation of hydrochloric acid

Question 100Question →

Helium, neon, and argon are Group 0 (Group 18) elements known for their extreme chemical unreactivity. Which statement accurately explains why helium exhibits this noble behavior despite having only two valence electrons ( $1s^2$ ), unlike the octet configuration ( $ns^2 np^6$ ) of other noble gases?

Its single energy level (

n=1

) holds a maximum of two electrons, forming a completely filled and highly stable duplet shell.

Its

1s^2

valence electrons easily form covalent bonds with electronegative non-metals to satisfy octet stability.

It readily ionizes under ambient conditions to produce a stable

He^{2+}

cation with a completely empty electron orbital.

Its exceptionally low boiling point during fractional distillation allows it to gain electrons from surrounding atmospheric gases.

Show answer & explanation

Answer: Its single energy level ( $n=1$ ) holds a maximum of two electrons, forming a completely filled and highly stable duplet shell.

Answer

Helium achieves chemical inertness because its single energy level (

n=1

) is completely filled with two electrons (

1s^2

), forming a stable duplet shell.

Helium has an atomic number of 2, placing its two electrons in the

1s

orbital (

1s^2

). Because the first principal energy level (

n=1

) can hold at most two electrons, this shell is completely filled, conferring maximum thermodynamic and chemical stability (duplet rule) without needing an octet.

Step-by-Step Solution

1

Analyze the electronic configuration of helium (

Z=2

).

Helium has an electronic configuration of

1s^2

.

The principal quantum number

n=1

contains only the

s

subshell, which holds a maximum of 2 electrons (

2n^2 = 2(1)^2 = 2

).

2

Compare the stability criteria for

n=1

versus higher energy levels (

n \ge 2

).

For

n=1

, two electrons create a completely filled valence shell (duplet stability). For

n \ge 2

, eight electrons (

ns^2 np^6

) are required for a full valence shell (octet stability).

Chemical unreactivity is determined by having a completely filled valence shell, which minimizes chemical potential energy.

Key Concept

Duplet vs Octet Stability in Noble Gases

Estimated Time:1m 0s

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