Environmental Chemistry MCQ Quiz - Objective Question with Answer for Environmental Chemistry - Download Free PDF

The correct answer is: “A, C and D only”

Key Points

• Infrared (IR) Spectroscopy – Molecular vibrations and rotational states:
  • This is correctly matched.
  • IR spectroscopy involves the interaction of infrared radiation with matter, leading to excitation of vibrational and some rotational transitions in molecules.
  • This technique is widely used for identifying functional groups and characterizing molecular structures.
  • The vibrational modes are specific to different chemical bonds, making IR a fingerprint technique.
• Circular Dichroism (CD) – Protein structure via circularly polarized light:
  • This is correctly matched.
  • CD spectroscopy measures the differential absorption of left and right circularly polarized light by optically active (chiral) molecules.
  • It is especially useful in studying the secondary and tertiary structures of proteins (like α-helices and β-sheets).
  • The CD spectra provide insights into conformational changes and folding behavior of biomolecules.
• Nuclear Magnetic Resonance (NMR) Spectroscopy – Nuclear spin interactions:
  • This is correctly matched.
  • NMR spectroscopy explores the interaction of atomic nuclei with external magnetic fields and radiofrequency pulses.
  • It provides detailed information about the electronic environment and structure of organic molecules based on nuclear spin behaviors (especially of ¹H and ¹³C).
  • NMR is fundamental in structural elucidation of organic compounds, conformational studies, and reaction mechanisms.

Additional Information

• Optical Rotary Dispersion (ORD) – Magnetic moments of unpaired electrons:
  • This is incorrectly matched.
  • ORD deals with the rotation of plane-polarized light as it passes through a chiral substance.
  • It is not used for studying magnetic moments of unpaired electrons—that is instead the domain of Electron Paramagnetic Resonance (EPR) or Electron Spin Resonance (ESR).
  • ORD, like CD, is used for determining the stereochemistry and optical activity of chiral molecules, especially in organic and biochemical systems.

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The correct answer is: "Chlorophyll absorbs red and blue wavelengths, transmitting green"

Key Points

• Why chlorophyll appears green:
  • Chlorophyll is the primary pigment involved in photosynthesis in green plants, and it plays a major role in light absorption.
  • Chlorophyll strongly absorbs light in the blue (~430 nm) and red (~660 nm) regions of the visible light spectrum.
  • However, it does not efficiently absorb green wavelengths (around 500–550 nm). Instead, it reflects or transmits these wavelengths.
  • As a result, green light is what primarily reaches our eyes, and leaves containing chlorophyll appear green to us.
  • This selective absorption and reflection is fundamental to the energy-harvesting efficiency of photosynthesis.

Additional Information

• "Chlorophyll reflects only red wavelengths":
  • This is incorrect because chlorophyll primarily reflects green light, not red.
  • Red light is one of the main absorption peaks for chlorophyll, which is used to drive photosynthesis.
• "Chlorophyll reflects green and blue wavelengths":
  • This is incorrect as well. Chlorophyll absorbs blue light very effectively—it does not reflect it.
  • Only green light is poorly absorbed and thus reflected, giving leaves their green appearance.
• "Green light is absorbed most strongly by chlorophyll":
  • This is the opposite of what actually happens. Chlorophyll absorbs least in the green region of the spectrum.
  • Because green light is poorly absorbed, it is mostly reflected or transmitted, making leaves appear green.

Gibb's free energy :

where,

= Gibb's free energy

= enthalpy of the reaction

= entropy change

= temperature

Under standard condition:

For a spontaneous process, must be negative.

-----(1)

can or if the condition of eqn (1) is satisfied.

The correct answer is: 2) II and III

Key Points

• Atmosphere and Free Radicals:
  • The atmosphere contains various gases, such as nitrogen, oxygen, carbon dioxide, and noble gases.
  • Free radicals are highly reactive atoms or molecules with unpaired electrons.
  • While free radicals do exist in the atmosphere, primarily coming from natural and anthropogenic sources, describing their amount as "negligible" is not accurate. They play significant roles in atmospheric chemistry and processes such as the formation and breakdown of ozone.
• Aerosols as Atmospheric Particles:
  • Aerosols are tiny particles or droplets suspended in the atmosphere and can originate from natural sources (e.g., sea spray, volcanic ash) or human activities (e.g., combustion engines, industrial processes).
  • They can affect climate by scattering and absorbing sunlight and can also act as nuclei for cloud formation.
  • Aerosols play a crucial role in atmospheric physics and chemistry, impacting weather patterns, air quality, and human health.
• Particulates from Primary and Secondary Sources:
  • Primary particulates are directly emitted into the atmosphere from sources such as dust storms, sea spray, volcanic activity, and human activities like industrial operations and vehicle exhaust.
  • Secondary particulates form in the atmosphere due to chemical reactions between different kinds of gases, such as sulfur dioxide (SO<sub>2</sub>) and nitrogen oxides (NO<sub>x</sub>), which then react with other substances to form particulate matter like sulfates and nitrates.
  • These particles can impact human health, visibility, and climate.

Additional Information

• Role of Free Radicals:
  • Free radicals in the atmosphere are involved in various important processes, including the formation of secondary pollutants such as ozone and organic aerosols.
  • They are also central to the degradation of certain pollutants, acting as agents that facilitate their breakdown and removal from the atmosphere.
• Aerosols and Human Health:
  • Aerosol particles can have significant health impacts, particularly fine particulate matter (PM<sub>2.5</sub>), which can penetrate deep into the respiratory system causing chronic respiratory diseases and cardiovascular problems.
  • Control and monitoring of aerosols are crucial for public health, air quality management, and regulatory standards related to environmental quality.
• Environmental Impact of Particulates:
  • Particulate matter significantly influences both weather and climate change through its interactions with sunlight, as well as with clouds and precipitation processes.
  • They can also contribute to environmental issues such as acid rain and visibility reduction.

The correct option is 'A is true, but R is false'.
Key Points

• The thermodynamic function which determines the spontaneity of a process is the free energy. For a process to be spontaneous, the change in free energy must be negative.
  • This assertion is true.
  • Gibbs Free Energy (G) determines the spontaneity of a process at constant temperature and pressure.
  • The criterion for spontaneity is that the change in Gibbs Free Energy (ΔG) must be negative, i.e., ΔG
  • This indicates that the process can occur without the input of external energy.
• The change in free energy is related to the change in enthalpy and change in entropy. The change in entropy for a process must always be positive if it is spontaneous.
  • This reason is false.
  • The change in Gibbs Free Energy (ΔG) is indeed related to the change in enthalpy (ΔH) and the change in entropy (ΔS) by the equation: ΔG = ΔH - TΔS, where T is the absolute temperature.
  • However, the statement that the change in entropy (ΔS) must always be positive for a process to be spontaneous is incorrect.
  • For a process to be spontaneous, the condition ΔG
• Thus, while Statement (A) is true, Statement (R) is false.
  • Statement (R) provides an incorrect condition for spontaneity regarding the change in entropy.
  • The change in entropy does not necessarily need to be positive for the process to be spontaneous; only the free energy change must be negative.

Additional Information

• The concept of Gibbs Free Energy is essential to understanding the spontaneity of chemical reactions and processes.
• In exergonic reactions, ΔG is negative, indicating the process releases energy and can proceed spontaneously.
• In contrast, in endergonic reactions, ΔG is positive, indicating the need for an input of energy for the process to occur.

The correct answer is apex.Key Points

• The ecological pyramid is a graphical representation of the flow of energy and matter in an ecosystem, with different trophic levels arranged in a pyramid shape.
• The base of the ecological pyramid represents the primary producers, such as plants, which convert sunlight into organic matter through photosynthesis.
• As we move up the pyramid, each level represents organisms that consume the lower level, such as herbivores, carnivores, and top predators.
• Therefore, the base of the ecological pyramid is usually broad, as it supports a large number of organisms at the primary level, and it narrows towards the apex, which represents the top predator or the highest trophic level.
• Option 3, "apex," is the correct answer, as it correctly describes the narrowing of the ecological pyramid towards the top predator or the highest trophic level.

Additional Information

• Option 1, "bottom," is incorrect, as it refers to the base of the pyramid, which is broad and supports a large number of organisms.
• Option 2, "center," is incorrect, as there is no central point in the ecological pyramid, and the trophic levels are arranged in a hierarchical order.
• Option 4, "sides," is incorrect, as it does not describe the shape or structure of the ecological pyramid, which is a pyramid shape with a broad base and a narrow apex.
• The ecological pyramid is an important tool for understanding the flow of energy and matter in an ecosystem and the interdependence of different species.
• It helps us to understand the impact of human activities on the environment and the need for conservation and sustainable use of natural resources.

The correct answer is ~71.4 mg/l.

 Key PointsGiven: 

Volume of wastewater = 21 ml

The volume of wastewater plus dilution water = 300 ml

DO<sub>1</sub> = 10 mg/l

DO<sub>5</sub> = 5 mg/l

Using the formula: 

 = 0.07

 = ~71.4 mg/l

Therefore, the correct answer is ~71.4 mg/l.

The correct answer is ~104.7 μg/m3.

Key Points Given: 

Concentration of SO<sub>2</sub> = 40 ppb

Temperature = 25°C

The molecular weight of SO2 = 64

Using the formula: 

 = ~104.7 μg/m3

Therefore, the correct answer is ~104.7 μg/m3.

Correct answer: 1) 

Concept:

• Dinitrogen and dioxygen are the stable forms of nitrogen and oxygen atoms respectively.
• Dinitrogen has a very high bond enthalpy as it has triple bond.
• Due to this reason, it is practically inert at room temperature.
• However, the reactivity of dinitrogen increases as the temperature is increased.
• Dioxygen is a very reactive gas but due to presence of double bond the bond enthalpy of dioxygen is very high.

Explanation:

• Dinitrogen and dioxygen are main constituents of air (N2=78.08%,O2=20.95%" id="MathJax-Element-4-Frame" role="presentation" tabindex="0">N<sub>2</sub>=78.08%,O<sub>2</sub>=20.95%)
• They do not react with each other to form oxides of nitrogen.
• This is because dinitrogen has a triple bond between the two nitrogen atoms and for dinitrogen and dioxygen to react, this triple bond needs to be broken.
• But the bond dissociation energy that is the energy required to break the bond is very high for dinitrogen.
• This high amount of energy cannot be found in the atmosphere.
• Therefore, for this reaction to take place, it would require very high temperatures to generate enough energy to break the triple bond in dinitrogen.

Conclusion:

Thus, dinitrogen and dioxygen are main constituents of air but these do not react with each other to form oxides of nitrogen because the reaction is endothermic and requires very high temperature.

Additional Information

The correct answer is (A) is false but (R) is true.

 Key Points

• Organophosphorus pesticides vary in chemical structures and toxicities.
• The main groups are phosphate, phosphorothioate, O-alkyl phosphorothioate and phosphorodithioate.
• A phosphorthioate compound such as parathion is much more toxic than a phosphorodithioate compound like malathion. 
• According to the WHO classification of pesticides, the acute oral toxicity of parathion is 3–6 mg/kg (in class I), while malathion is unclassified with an acute toxicity to mammals of 1400 mg/kg.

Thus the assertion is false.

• The organophosphorus insecticides malathion and parathion are inactive in their parent state but can be metabolized to the active ChE inhibitors malaoxon and paraoxon, respectively.
• Malathion is relatively safe in mammals because it is hydrolyzed rapidly by plasma carboxylesterases.
• This detoxification occurs much more rapidly in birds and mammals than in insects.

Thus the reason is true.

Therefore, the correct answer is (A) is false but (R) is true.

The correct answer is They decompose and metabolize slowly.

 Key Points Organophosphates:

• Organophosphates are chemical substances produced by the process of esterification between phosphoric acid and alcohol.
• Organophosphates can undergo hydrolysis with the liberation of alcohol from the ester bond.
• These chemicals are the main components of herbicides, pesticides, and insecticides.
• Acute or chronic exposure to organophosphates can produce varying toxicity levels in humans, animals, plants, and insects.
• Organophosphates are classified as non-persistent pesticides.
• They do not bioaccumulate appreciably in humans and are rapidly metabolized and excreted in the urine.
• They are rapidly broken down into other chemicals so they do not build up in the environment.
• They are not likely to build up to high or dangerous levels in animal or plant foods that you might eat.
• Long-term exposure to organophosphates can cause confusion, anxiety, loss of memory, loss of appetite, disorientation, depression, and personality changes.
• Examples: Parathion, Chlorpyrifos, Diazinon, Dichlorvos, Phosmet, Fenitrothion, Tetrachlorvinphos, Azamethiphos etc.

Therefore, the correct answer is they decompose and metabolize slowly.

The correct answer is 2.0.

Key Points

• The ratio of the distance travelled by the analyte to the distance travelled by the solvent front on a chromatogram is known as the retardation or retention factor (R<sub>f</sub>) value.
• R<sub>f</sub> = Distance travelled by analyte / Distance travelled by solvent front.
• In chromatographic techniques, the mobility of analytes with mobile solvents is different from the mobility of analytes added to stationary phases (phases).
• This discrepancy is explained by the different affinities of analytes for stationary and mobile solvents.
• The longer an analyte remains in a stationary phase, the lower the R<sub>f</sub> value it will have, and the opposite is also true.
• R<sub>f</sub> values are the ratios of solute (analyte) migration distances to solvent fronts, they are always less than one.
• Because solutes must have some attractive qualities with stationary phases, the solvent front always travels more than the solute front.
• R<sub>f</sub> values will always be between 0 and 1 because the denominator has a greater value.

Therefore, the correct answer is 2.0.

The correct answer is ∼ 100 keV.

Key Points

• Nuclear cross-section is used to describe the probability that a nuclear reaction will occur.
• The Deuterium (D) – Tritium (T) reaction has the largest cross section and also the largest Q-value (i.e. the released energy of a reaction) of varieties of fusion reactions.
• It produces an alpha particle (or Helium-4), a neutron and releases 17.6 MeV of energy in the form of kinetic energy of the products (3.5 MeV to alpha particle and 14.1 MeV to neutron).
• At a plasm a ion temperature of 5 keV, the largest fraction of the D- T reaction rate occurs at deuteron energies of ~ 20-15 0 keV.
• At energies above ~ 100 keV, the uncertainties in the D-T cross section introduces errors into the calculated reaction rate of ~ 10.
• At energie s below ~ 100 keV, the fit to the cross section used in calculating the reaction rate is based on a single experiment , so the sensitivity of the reaction rate to the D-T cross section may be quite uncertain.

Therefore, the maximum cross-section for the fusion reaction  +  is reached at energies of nuclei at ∼ 100 keV.

The correct answer is A, B, C, D and E.

Key Points

• Atomic absorption spectroscopy, or AAS, is a technique for measuring the concentrations of metallic elements in different materials.
• As an analytical technique, it uses electromagnetic wavelengths, coming from a light source.
• Distinct elements will absorb these wavelengths differently. 
• Elements exist on an electromagnetic spectrum, and their atoms will absorb wavelengths of light that relate to their particular characteristics.
• Every element has its unique electronic structure.
• Therefore, the radiation absorbed represents a unique property of each element.
• So in atomic absorption spectroscopy, only a single element can be analyzed at a time.
• The dynamic range, also known as the linear dynamic range or the linear range, is the range over which a response is a well-defined (usually linear) function of the analyte concentration.
• The top of the linear range for most elements is between 0.20 and 0.30 absorbance units.
• Since mercury exists as mercury vapour at room temperature.
• Therefore, only cold vapour atomic absorption spectroscopy can be used for the analysis of mercury.
• Hydride generation was first used by Holak for the total determination of arsenic by atomic absorption spectroscopy.
• It is now well-established for elements such as arsenic, bismuth, antimony, selenium, tin, germanium and tellurium.
• AAS measurements are based on the famous Beer-Lambert law which states that at a selected wavelength the absorbance by absorbing species is in direct proportion to its concentration and path length of the sample cell.

Therefore, the correct answer is A, B, C, D and E.

The correct answer is (b) and (c) only.

Key PointsHalf-life of different radioactive species are:

Therefore, the correct answer is (b) and (c) only.

Additional Information

• Half-life (t½) is the time required for a quantity (of substance) to reduce to half of its initial value.
• The term is commonly used to describe how quickly unstable atoms undergo radioactive decay or how long stable traces survive.
• Half-life is constant over the lifetime of an exponentially decaying quantity, and it is a characteristic unit for the exponential decay equation. 
• The relationship between the half-life, T<sub>1/2</sub>, and the decay constant is given by T<sub>1/2</sub> = 0.693/λ.