JEE Chemistry Environmental Chemistry MCQs Set 17

Chemistry Objective Questions and Answers: Environmental Chemistry

Access targeted multiple-choice questions for Environmental Chemistry designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Environmental Chemistry MCQs with Answers

Access the complete set of multiple-choice questions for Environmental Chemistry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In which region of the atmosphere does the temperature increase with altitude because of absorption of ultraviolet radiation by ozone?
(a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Exosphere
Show Answer & Explanation

Answer: (b) Stratosphere

Explanation:
1. In the troposphere (0–11 km) the temperature falls with height, and in the mesosphere it falls again.
2. In the stratosphere (11–50 km) ozone absorbs UV radiation and releases heat, so the temperature rises from about −56 °C to about −2 °C.
3. Answer: (b)

Teacher's Note:
1. Stratosphere holds the ozone layer.
2. Troposphere is the region of weather.

Question 2: The species present mainly in the thermosphere (ionosphere) are:
(a) N2, O2, CO2, water vapour
(b) O3 and O2
(c) O2+, O+, NO+ and free electrons
(d) CH4 and NH3
Show Answer & Explanation

Answer: (c) O2+, O+, NO+ and free electrons

Explanation:
1. In the thermosphere (85–500 km) high energy UV radiation ionises the gases, so the main species are O2+, O+, NO+ and electrons.
2. That is why the region is called the ionosphere.
3. Answer: (c)

Teacher's Note:
1. Troposphere: N2, O2, CO2, H2O.
2. Stratosphere: N2, O2, O3.

Question 3: In the destruction of ozone by chlorofluorocarbons, the chain carrier (catalyst) is:
(a) ClO·
(b) Cl·
(c) F·
(d) CF2Cl·
Show Answer & Explanation

Answer: (b) Cl·

Explanation:
1. UV radiation breaks CF2Cl2 to give Cl·.
2. Cl· + O3 → ClO· + O2 and ClO· + O → Cl· + O2; since Cl· is regenerated, it acts as a catalyst and one Cl· can destroy thousands of ozone molecules.
3. Answer: (b)

Teacher's Note:
1. Cl· is regenerated in the second step.
2. C–F bonds are too strong to break by stratospheric UV.

Question 4: The compound that temporarily acts as a chlorine reservoir and terminates the ozone-destroying chain in the stratosphere is:
(a) ClONO2
(b) CF2Cl2
(c) CH4
(d) HF
Show Answer & Explanation

Answer: (a) ClONO2

Explanation:
1. ClO· reacts with NO2 to form chlorine nitrate, ClO· + NO2 → ClONO2.
2. This ties up reactive chlorine temporarily and acts as a reservoir; the other species are not chlorine reservoirs in this chain.
3. Answer: (a)

Teacher's Note:
1. Chlorine nitrate is the main reservoir.
2. Polar stratospheric clouds release the chlorine again.

Question 5: The ozone layer protects life on earth because it absorbs:
(a) infrared radiation
(b) harmful ultraviolet radiation
(c) visible light
(d) cosmic rays only
Show Answer & Explanation

Answer: (b) harmful ultraviolet radiation

Explanation:
1. Ozone absorbs the harmful short-wavelength UV radiation of the sun.
2. Without this shield UV would reach the surface and cause skin cancer, cataract and damage to plants.
3. Answer: (b)

Teacher's Note:
1. Ozone absorbs UV, CO2 absorbs IR.
2. Depletion increases UV at the surface.

Question 6: The ozone hole is generally defined as a region in the stratosphere where the total ozone column falls below:
(a) 50 Dobson units
(b) 220 Dobson units
(c) 500 Dobson units
(d) 1000 Dobson units
Show Answer & Explanation

Answer: (b) 220 Dobson units

Explanation:
1. Ozone is measured in Dobson units (1 DU = 0.01 mm thickness of pure ozone at STP).
2. A region with a column below 220 DU is called an ozone hole; the normal global average is about 300 DU.
3. Answer: (b)

Teacher's Note:
1. Normal ozone column is about 300 DU.
2. Antarctic spring shows the deepest hole.

Question 7: The pH of normal rain water is about 5.6, not 7, because:
(a) it contains dissolved SO2
(b) it contains dissolved CO2, forming carbonic acid
(c) it contains dissolved NO2
(d) it contains dissolved O2
Show Answer & Explanation

Answer: (b) it contains dissolved CO2, forming carbonic acid

Explanation:
1. Atmospheric CO2 dissolves in rain water to give H2CO3, which ionises to H+ and HCO3−.
2. This makes unpolluted rain slightly acidic with pH about 5.6, and rain with pH below 5.6 is called acid rain.
3. Answer: (b)

Teacher's Note:
1. Carbonic acid lowers pH to 5.6.
2. Acid rain has pH below 5.6.

Question 8: Damage to marble monuments such as the Taj Mahal by acid rain is represented by:
(a) CaCO3 + H2SO4 → CaSO4 + H2O + CO2
(b) CaO + H2O → Ca(OH)2
(c) CaSO4 → CaO + SO3
(d) CaCO3 → CaO + CO2
Show Answer & Explanation

Answer: (a) CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Explanation:
1. Marble is calcium carbonate, which reacts with sulphuric acid of acid rain to give calcium sulphate, water and carbon dioxide.
2. Calcium sulphate is more soluble and is washed away, so the surface erodes.
3. Answer: (a)

Teacher's Note:
1. Acid + carbonate gives salt, water and CO2.
2. Marble cancer is another name for this damage.

Question 9: In the atmosphere sulphur dioxide is converted to sulphuric acid by the sequence:
(a) SO2 → SO3 (oxidation) then SO3 + H2O → H2SO4
(b) SO2 + H2 → H2S
(c) SO2 → S + O2
(d) SO2 + CO2 → CS2
Show Answer & Explanation

Answer: (a) SO2 → SO3 (oxidation) then SO3 + H2O → H2SO4

Explanation:
1. SO2 is oxidised to SO3 (for example 2SO2 + O2 → 2SO3, catalysed by particulates or oxidants such as O3 and H2O2).
2. SO3 dissolves in water to give H2SO4, which falls with rain.
3. Answer: (a)

Teacher's Note:
1. Oxidation first, then hydration.
2. Particulate matter acts as catalyst.

Question 10: The radiation absorbed by greenhouse gases and responsible for the greenhouse effect is:
(a) ultraviolet
(b) infrared
(c) X-rays
(d) gamma rays
Show Answer & Explanation

Answer: (b) infrared

Explanation:
1. The earth absorbs sunlight and re-emits heat as infrared radiation.
2. Greenhouse gases absorb part of this infrared radiation and radiate it back, keeping the lower atmosphere warm.
3. Answer: (b)

Teacher's Note:
1. Heat radiation is infrared.
2. Increased CO2 traps more heat.

Question 11: The atmospheric 'sink' for carbon monoxide is:
(a) ozone
(b) micro-organisms in soil
(c) plants through photosynthesis
(d) oceans through dissolution
Show Answer & Explanation

Answer: (b) micro-organisms in soil

Explanation:
1. Although large quantities of CO enter the air, its level does not rise much because soil bacteria convert it to CO2.
2. Soil micro-organisms are therefore the main sink of CO.
3. Answer: (b)

Teacher's Note:
1. Sink means a process that removes a pollutant.
2. CO is oxidised to CO2.

Question 12: A catalytic converter fitted in the exhaust of an automobile contains:
(a) Pt, Pd and Rh, which convert CO, hydrocarbons and NOx to harmless products
(b) Ni and Cu, which convert CO2 to CO
(c) Fe and Mo, which convert N2 to NH3
(d) Ag and Au, which convert CO to CH4
Show Answer & Explanation

Answer: (a) Pt, Pd and Rh, which convert CO, hydrocarbons and NOx to harmless products

Explanation:
1. Catalytic converters use the noble metals platinum, palladium and rhodium.
2. They oxidise CO to CO2 and unburnt hydrocarbons to CO2 and H2O, and reduce nitrogen oxides to N2.
3. Answer: (a)

Teacher's Note:
1. Three-way catalyst: Pt, Pd, Rh.
2. Converts CO, HC and NOx.

Question 13: Leaded petrol cannot be used in vehicles fitted with catalytic converters because:
(a) lead is expensive
(b) lead deactivates (poisons) the catalyst
(c) lead raises the octane number too much
(d) lead forms PbO that burns
Show Answer & Explanation

Answer: (b) lead deactivates (poisons) the catalyst

Explanation:
1. Lead compounds from tetraethyl lead coat the catalyst surface and make it inactive.
2. Therefore only unleaded petrol is used in vehicles with catalytic converters.
3. Answer: (b)

Teacher's Note:
1. Catalyst poisoning by Pb.
2. Unleaded petrol is mandatory for converters.

Question 14: Classical (London) smog differs from photochemical smog in that it:
(a) is oxidising and occurs in sunny weather
(b) is reducing, contains SO2 and smoke, and occurs in cool humid weather
(c) contains PAN
(d) contains ozone as the main component
Show Answer & Explanation

Answer: (b) is reducing, contains SO2 and smoke, and occurs in cool humid weather

Explanation:
1. Classical smog results from coal smoke and SO2 with fog in cool, humid weather.
2. It is a reducing smog, unlike the oxidising photochemical smog.
3. Answer: (b)

Teacher's Note:
1. SO2 makes it a reducing smog.
2. It is a winter phenomenon.

Question 15: Peroxyacetyl nitrate (PAN), a component of photochemical smog, has the formula:
(a) CH3COOONO2
(b) CH3CH2NO2
(c) CH3COONH2
(d) CH3ONO2
Show Answer & Explanation

Answer: (a) CH3COOONO2

Explanation:
1. PAN is CH3–CO–O–O–NO2, formed from hydrocarbons, oxygen and nitrogen oxides in sunlight.
2. It is a secondary pollutant that irritates the eyes and damages plants.
3. Answer: (a)

Teacher's Note:
1. PAN is a secondary pollutant.
2. It is not emitted directly.

Question 16: In the formation of photochemical smog the first photochemical step is:
(a) NO2 + hν → NO + O
(b) O3 + NO → NO2 + O2
(c) N2 + O2 → 2NO
(d) SO2 + O2 → SO3
Show Answer & Explanation

Answer: (a) NO2 + hν → NO + O

Explanation:
1. Nitrogen dioxide absorbs sunlight and splits into NO and an oxygen atom.
2. The oxygen atom reacts with O2 to give O3, and the ozone and NO then react with hydrocarbons to form formaldehyde, acrolein and PAN.
3. Answer: (a)

Teacher's Note:
1. NO2 photolysis starts the smog chain.
2. The oxygen atom forms ozone.

Question 17: Eutrophication of a lake is caused mainly by an excess of:
(a) heavy metals
(b) phosphates and nitrates from fertilisers and detergents
(c) chlorides
(d) carbon dioxide
Show Answer & Explanation

Answer: (b) phosphates and nitrates from fertilisers and detergents

Explanation:
1. Excess phosphates and nitrates act as nutrients and cause rapid growth of algae.
2. When the algae die, their decomposition uses up dissolved oxygen, and fish and other aquatic life die.
3. Answer: (b)

Teacher's Note:
1. Nutrient enrichment causes algal bloom.
2. Dissolved oxygen falls.

Question 18: For the growth and survival of fish the minimum amount of dissolved oxygen in water should be about:
(a) 1 ppm
(b) 6 ppm
(c) 20 ppm
(d) 40 ppm
Show Answer & Explanation

Answer: (b) 6 ppm

Explanation:
1. Fish need dissolved oxygen of at least about 6 ppm.
2. Below this level the growth of fish slows, and many species cannot survive.
3. Answer: (b)

Teacher's Note:
1. Oxygen solubility in water is low.
2. Warm or polluted water holds less dissolved oxygen.

Question 19: Minamata disease in Japan was caused by pollution of water with:
(a) lead
(b) mercury
(c) cadmium
(d) arsenic
Show Answer & Explanation

Answer: (b) mercury

Explanation:
1. Minamata disease resulted from eating fish contaminated with methylmercury released by an industrial plant.
2. It damages the nervous system.
3. Answer: (b)

Teacher's Note:
1. Minamata: Hg.
2. Itai-itai: Cd.

Question 20: Itai-itai disease, which causes severe bone and kidney damage, is caused by:
(a) mercury
(b) lead
(c) cadmium
(d) nitrates
Show Answer & Explanation

Answer: (c) cadmium

Explanation:
1. Itai-itai disease was due to cadmium contamination of rice and water.
2. Cadmium accumulates in the body and damages the kidneys and softens bones.
3. Answer: (c)

Teacher's Note:
1. Itai-itai: cadmium.
2. Heavy metals are cumulative poisons.

Question 21: Drinking water is fluoridated to about 1 ppm. The mass of fluoride ions present in 10 litres of such water (taking 1 ppm as 1 mg per litre) is:
(a) 1 mg
(b) 10 mg
(c) 100 mg
(d) 1 g
Show Answer & Explanation

Answer: (b) 10 mg

Explanation:
1. 1 ppm means 1 mg of fluoride per litre of water.
2. In 10 L the mass is 10 × 1 = 10 mg.
3. Answer: (b)

Teacher's Note:
1. ppm in water: mg per litre.
2. Above 10 ppm fluoride damages bones and teeth.

Question 22: DDT is a serious pollutant because:
(a) it is rapidly biodegradable
(b) it is non-biodegradable and accumulates along the food chain (biomagnification)
(c) it is a greenhouse gas
(d) it dissolves readily in water
Show Answer & Explanation

Answer: (b) it is non-biodegradable and accumulates along the food chain (biomagnification)

Explanation:
1. DDT is stable and fat soluble, so it is not decomposed by micro-organisms.
2. It accumulates in fatty tissues and its concentration increases at each step of the food chain.
3. Answer: (b)

Teacher's Note:
1. Non-biodegradable pollutants accumulate.
2. Biomagnification harms top predators.

Question 23: Which of the following is an example of green chemistry?
(a) Bleaching paper with chlorine
(b) Dry cleaning with tetrachloroethene
(c) Using liquefied CO2 (with a suitable detergent) in dry cleaning
(d) Burning fossil fuel for synthesis
Show Answer & Explanation

Answer: (c) Using liquefied CO2 (with a suitable detergent) in dry cleaning

Explanation:
1. Green chemistry reduces pollution at its source by using safer reagents and processes.
2. Replacing tetrachloroethene (a possible carcinogen and groundwater pollutant) by liquid CO2 in dry cleaning, or chlorine by H2O2 in bleaching paper, are standard examples.
3. Answer: (c)

Teacher's Note:
1. Prevent pollution rather than treat it.
2. Use of benign solvents is typical.

Question 24: Thermal pollution of a river by hot water from a power plant harms aquatic life mainly because:
(a) it increases the dissolved oxygen
(b) it decreases the solubility of oxygen and increases the oxygen demand of organisms
(c) it increases the pH to 14
(d) it adds phosphates
Show Answer & Explanation

Answer: (b) it decreases the solubility of oxygen and increases the oxygen demand of organisms

Explanation:
1. The solubility of gases in water decreases as temperature increases.
2. The warm water holds less dissolved oxygen, and the metabolic rate of organisms rises so they need more of it.
3. Answer: (b)

Teacher's Note:
1. Gas solubility falls on heating.
2. Fish die from oxygen deficiency.

Question 25: Fine particulates (smaller than about 2.5 micrometres) are particularly harmful because they:
(a) are removed by the nose
(b) penetrate deep into the lungs
(c) dissolve in blood only
(d) are completely biodegradable
Show Answer & Explanation

Answer: (b) penetrate deep into the lungs

Explanation:
1. Particles larger than about 5 micrometres lodge in the nasal passages, but the smaller ones travel deep into the lungs.
2. Long exposure may cause fibrosis and other lung diseases.
3. Answer: (b)

Teacher's Note:
1. Smaller size means deeper penetration.
2. Some fine particles are carcinogenic.

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FAQs

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