All questions of CSIR-UGC (NET) Chemical Science Mock Test for UGC NET Exam

BET is important theory that is used to analysis technique for measurements of specific surface area of material

Understanding the Problem
To solve the problem, we need to determine the normality (N) of the first solution mixed with the HCl solution. Given that the final normality of the mixed solution is 5 N, we can set up the problem using the concept of equivalents.
Calculating Equivalents
1. Normality and Volume of HCl
- HCl is a strong acid and completely dissociates, so its normality is equal to its molarity. Given that the HCl solution is 2 M, it is also 2 N.
- Volume of HCl = 2.5 L.
- Therefore, equivalents of HCl = Normality × Volume = 2 N × 2.5 L = 5 equivalents.
2. Normality and Volume of the Other Solution
- Let the normality of the other solution be N and the volume be 1.5 L.
- Therefore, equivalents of the other solution = N × 1.5 L.
Setting Up the Equation
- Since the total equivalents in the mixture should equal the final normality multiplied by the total volume of the resulting solution, we have:
Total Volume = 1.5 L + 2.5 L = 4 L.
- The equation becomes:
5 N = (N × 1.5) + 5.
Solving for N
- Rearranging the equation:
5N = 1.5N + 5
5N - 1.5N = 5
3.5N = 5
N = 5 / 3.5
N = 10.
Thus, the value of N is 10, confirming that the correct answer is option B.

Explanation:

Hybridization in H2O:
- In the water molecule (H2O), the central oxygen atom is sp3 hybridized.
- Oxygen has 6 electrons in its valence shell (2s2 2p4). To form 2 sigma bonds with the two hydrogen atoms, oxygen hybridizes one s and three p orbitals to form four sp3 hybrid orbitals.

Shape of H2O:
- The shape of the H2O molecule is angular or bent.
- The two lone pairs of electrons on the oxygen atom cause repulsion, leading to a bent molecular geometry.
- The bond angle in water is approximately 104.5 degrees.

Explanation of Options:
- Tetrahedral: This shape is seen in molecules with sp3 hybridization and no lone pairs, such as methane (CH4).
- Angular or Bent Structure: This is the correct shape for H2O due to the presence of lone pairs causing distortion.
- Trigonal Planar: This shape is seen in molecules with sp2 hybridization, like boron trifluoride (BF3).
- Pyramidal: This shape is observed in molecules with sp3 hybridization and one lone pair, such as ammonia (NH3).
Therefore, based on the concept of hybridization and lone pair repulsion in the H2O molecule, the correct shape is an angular or bent structure.

Explanation:

Ground State of Oxygen Atom:
- The ground state for the O atom is 3P1. This means that the oxygen atom has three unpaired electrons in its outer shell, making it paramagnetic.

Paramagnetism of Oxygen:
- Both the oxygen atom and the diatomic molecule are paramagnetic with two unpaired electrons. This is due to the presence of unpaired electrons in their electronic configurations.

Singlet Excited State of Dioxygen:
- The most readily accessible singlet excited state for dioxygen has an empty π* orbital. This empty orbital allows for the promotion of electrons to higher energy levels, resulting in the formation of excited states.
Therefore, the correct statements for oxygen are B and C. Oxygen exhibits paramagnetic properties due to the presence of unpaired electrons and has a singlet excited state with an empty π* orbital.

To calculate the wave number of the second scattered stokes line and second anti-stokes line in a Raman spectrometer, we need to consider the rotational constant of 16O, which is 10 cm^−1, and the wave number of the incident radiation, which is 20891 cm^−1.
- **Calculating the wave number of the second scattered stokes line:**
To find the wave number of the second scattered stokes line, we need to subtract the rotational constant from the incident wave number. The formula for the stokes line is:
Wave number of stokes line = Incident wave number - 2 * Rotational constant
Wave number of stokes line = 20891 cm^−1 - 2 * 10 cm^−1
Wave number of stokes line = 20891 cm^−1 - 20 cm^−1
Wave number of stokes line = 20871 cm^−1
- **Calculating the wave number of the second anti-stokes line:**
To find the wave number of the second anti-stokes line, we need to add the rotational constant to the incident wave number. The formula for the anti-stokes line is:
Wave number of anti-stokes line = Incident wave number + 2 * Rotational constant
Wave number of anti-stokes line = 20891 cm^−1 + 2 * 10 cm^−1
Wave number of anti-stokes line = 20891 cm^−1 + 20 cm^−1
Wave number of anti-stokes line = 20911 cm^−1
Therefore, the wave number of the second scattered stokes line is 20871 cm^−1 and the wave number of the second anti-stokes line is 20911 cm^−1.
Hence, the correct answer is option C: 20791 cm^−1, 20991 cm^−1.

First Ionization Energy Trend
- The first ionization energy is the energy required to remove the most loosely bound electron from an atom.
- The trend in first ionization energy generally increases across a period and decreases down a group in the periodic table.
- In this case, the correct order of first ionization energy follows Li < b="" />< be="" />< />
- This order is based on the atomic structure and effective nuclear charge of the elements.

Explanation of the Trend
- Lithium (Li) has the lowest first ionization energy in the given options because it has one electron in its outermost shell, which is relatively easy to remove compared to the other elements.
- Boron (B) has a higher first ionization energy than lithium because it has one more proton in its nucleus, leading to a stronger attraction to its electrons.
- Beryllium (Be) has a lower first ionization energy compared to boron because it has a filled 2s subshell, making it more stable and easier to remove an electron.
- Carbon (C) has the highest first ionization energy in the given options due to its half-filled p orbital, which provides extra stability and makes it harder to remove an electron compared to the other elements.

Conclusion
- The trend in first ionization energy among Li, B, Be, and C follows a logical pattern based on the atomic structure and electron configuration of the elements.
- Understanding these trends can help predict the behavior of elements in chemical reactions and their physical properties.

Effects of adding glycerine to water:
1. Freezing point depression:
When glycerine is added to water, it lowers the freezing point of water. This phenomenon is known as freezing point depression. Glycerine disrupts the formation of ice crystals, preventing water from freezing at its normal temperature.
2. Colligative property:
This effect is a result of a colligative property of solutions, where the presence of a solute (glycerine in this case) decreases the freezing point of the solvent (water). The more glycerine added, the lower the freezing point of the water.
3. Increased fluidity:
As the freezing point of water is lowered by the addition of glycerine, the resulting solution becomes more fluid and less viscous. This can be observed as the solution flows more easily compared to pure water.
In conclusion, adding glycerine to water results in the lowering of the freezing point of water due to the colligative property of solutions. This effect increases the fluidity of the solution and prevents it from freezing at its normal temperature.

Introduction to Superoxide Dismutase (SOD)
Superoxide dismutase is an essential enzyme that plays a critical role in cellular defense against oxidative stress. It catalyzes the dismutation of superoxide radicals into oxygen and hydrogen peroxide, thereby protecting cells from damage.
Metal Ion Composition of SOD
Superoxide dismutase requires metal ions for its enzymatic activity. The specific metal ions associated with SOD vary among different organisms and isoforms. The correct answer for the metal ions present in SOD is:

• Cu(II) and Zn(II)

Function of Metal Ions
The metal ions in SOD serve several critical functions:

• Copper (Cu(II)): Acts as a cofactor that facilitates the conversion of superoxide radicals into hydrogen peroxide and molecular oxygen.
• Zinc (Zn(II)): Provides structural stability to the enzyme and aids in the proper positioning of the active site.

Importance of Cu(II) and Zn(II) Combination
The combination of copper and zinc:

• Enhances Catalytic Efficiency: This duo is vital for the rapid dismutation of superoxide radicals.
• Protects Against Oxidative Damage: SOD activity is crucial in mitigating oxidative stress, which is linked to various diseases.

Conclusion
In summary, superoxide dismutase primarily contains the metal ions Cu(II) and Zn(II), which are essential for its function in protecting cells from oxidative damage. Understanding the role of these metal ions helps in comprehending the enzyme's significance in biological systems and its potential therapeutic applications.

Calculation of Exit Pressure of the Compressor:
- Given:
Feed mixture composition: 95% air and 5% ammonia vapor
Recovery of ammonia by compressor: 40%
Partial pressure of ammonia vapor: 10 mm Hg

Calculating the partial pressure of ammonia in the feed mixture:
Partial pressure of ammonia = 5% of total pressure
Assuming total pressure of feed mixture = 1 atm = 760 mm Hg
Partial pressure of ammonia = 5% of 760 mm Hg = 0.05 * 760 = 38 mm Hg

Calculating the partial pressure of recovered ammonia:
Partial pressure of recovered ammonia = 40% of the partial pressure of ammonia in feed mixture
Partial pressure of recovered ammonia = 0.40 * 38 mm Hg = 15.2 mm Hg

Calculating the partial pressure of remaining ammonia in the exit stream:
Partial pressure of remaining ammonia = Partial pressure of ammonia in feed mixture - Partial pressure of recovered ammonia
Partial pressure of remaining ammonia = 38 mm Hg - 15.2 mm Hg = 22.8 mm Hg

Calculating the total exit pressure of the compressor:
Total exit pressure = Partial pressure of remaining ammonia + Partial pressure of air
Total exit pressure = 22.8 mm Hg + 760 mm Hg = 782.8 mm Hg
Therefore, the exit pressure of the compressor is 782.8 mm Hg, which is closest to option b) 326.8 mm Hg.

Initial Gas Mixture:
- H2: 20%
- N2: 80%
- Total pressure: P

Step 1: Calculate the partial pressure of N2 in the initial mixture
- Partial pressure of N2 = 80% of total pressure = 0.8P

Step 2: Add O2 to the mixture
- Since the same amount of O2 is added, the total pressure remains as P.

Step 3: Calculate the new partial pressure of N2 after adding O2
- Since the total pressure is still P, the partial pressure of N2 remains the same as before = 0.8P
Therefore, the new partial pressure of N2 after adding O2 to the gas mixture remains 0.8P. Hence, the correct answer is option 'B'.

Converting 101 to binary:
- To convert the decimal number 101 to binary, we need to find the binary representation of the number using the powers of 2.
- Break down 101 into powers of 2: 101 = 64 + 32 + 4 + 1.
- Representing 101 as a sum of powers of 2: 64 (2^6) + 32 (2^5) + 4 (2^2) + 1 (2^0).

Writing 101 in binary:
- Using the powers of 2 representation, we get 101 written in binary as 1100101.
- So, the binary representation of the decimal number 101 is 1100101.
- The correct option from the given choices is option 'D' which is 1100101.

Polyamines are small organic molecules that play important roles in various physiological processes in plants, including tissue culture systems. In tissue culture, polyamines have been found to have several effects on plant growth and development. However, one of these effects, promotion of tuber and bulb formation, is not considered a main effect of polyamines in the tissue culture system.

Somatic embryogenesis
One of the main effects of polyamines in the tissue culture system is the promotion of somatic embryogenesis. Somatic embryogenesis refers to the process of inducing the formation of embryos from somatic cells, bypassing the need for sexual reproduction. Polyamines have been shown to enhance the frequency and quality of somatic embryogenesis in various plant species. They can stimulate cell division and differentiation, leading to the formation of somatic embryos.

Adventitious root formation
Polyamines also play a crucial role in adventitious root formation in tissue culture. Adventitious roots are roots that develop from non-root tissues, such as stem or leaf cuttings. Polyamines have been found to enhance the formation and elongation of adventitious roots in tissue culture systems. They promote cell division and differentiation in the root meristem, leading to the development of new roots.

Promotion of shoot formation
Polyamines are known to promote shoot formation in tissue culture. Shoot formation refers to the regeneration of shoots from explants, such as leaf or stem segments. Polyamines stimulate cell division and differentiation in the shoot meristem, leading to the formation of new shoots. They also play a role in the regulation of shoot development by influencing the balance between shoot proliferation and shoot elongation.

Promotion of tuber and bulb formation (Not a main effect)
In contrast to the other effects, the promotion of tuber and bulb formation is not considered a main effect of polyamines in the tissue culture system. While polyamines have been shown to affect the development of storage organs like tubers and bulbs in plants, their role in this process is not as extensively studied or well-established as their effects on somatic embryogenesis, adventitious root formation, and shoot formation. Therefore, it is not considered a main effect of polyamines in the tissue culture system.

In conclusion, polyamines have various effects in the tissue culture system, including the promotion of somatic embryogenesis, adventitious root formation, and shoot formation. However, the promotion of tuber and bulb formation is not considered a main effect of polyamines in the tissue culture system.

Concentration factors are used to express the amount of solute present in a given amount of solvent or solution. These factors include molarity, molality, mole fraction, and weight fraction. Among these, molarity is the concentration factor that is affected by changes in temperature.

Molarity, also known as molar concentration, is defined as the number of moles of solute present in one liter of solution. It is expressed in moles per liter (mol/L). The molarity of a solution can change with temperature because the volume of the solution is dependent on temperature.

When the temperature of a solution increases, the volume of the solution generally expands. This expansion leads to an increase in the total volume of the solution, while the amount of solute remains constant. As a result, the molarity of the solution decreases.

Similarly, when the temperature of a solution decreases, the volume of the solution generally contracts. This contraction leads to a decrease in the total volume of the solution, while the amount of solute remains constant. As a result, the molarity of the solution increases.

Therefore, it can be concluded that molarity is affected by changes in temperature because the volume of the solution changes with temperature, which in turn affects the concentration of the solution.

On the other hand, molality, mole fraction, and weight fraction are concentration factors that are not affected by changes in temperature.

Molality is defined as the number of moles of solute present in one kilogram of solvent. It is expressed in moles per kilogram (mol/kg). Since molality is based on the mass of the solvent, it does not depend on the temperature-dependent volume changes.

Mole fraction is a dimensionless quantity that represents the ratio of the number of moles of a component to the total number of moles in the solution. It is not affected by changes in temperature because it is a ratio of moles and does not involve volume or mass.

Weight fraction, also known as mass fraction, is the ratio of the mass of solute to the total mass of the solution. Like mole fraction, it is a ratio of masses and is not affected by changes in temperature.

In summary, molarity is the concentration factor that is affected by changes in temperature. Molality, mole fraction, and weight fraction are not affected by changes in temperature as they are based on ratios of moles or masses and do not involve volume changes.

Understanding p1d1 Configuration
The p1d1 configuration refers to the arrangement of electrons in the atomic orbitals. The specific arrangement in question allows us to derive the possible terms.
Terms Derived from p1d1 Configuration
To analyze the terms, we consider the following:
- p Orbital (1 electron): The p orbital can have a maximum of 3 orbitals (px, py, pz) and holds one electron in this case. The possible angular momentum (L) states are:
- L = 1 (p state)
- d Orbital (1 electron): The d orbital can have a maximum of 5 orbitals (dxy, dyz, dzx, dx2-y2, dz2) and also holds one electron here. The possible angular momentum states are:
- L = 2 (d state)
Combining the Angular Momentum States
When combining the p and d states, we use the vector coupling of angular momentum:
- The total angular momentum (L) can be calculated as:
- L can range from |L1 - L2| to |L1 + L2|, where L1 is from the p orbital (1) and L2 is from the d orbital (2).
Thus, the possible values for the total angular momentum (L) are:
- L = 1 (p) + L = 2 (d):
- Possible values: L = 1, 2, 3 (which corresponds to terms: P, D, F)
Possible Terms and Their Multiplicities
- 3F: This term arises from the maximum multiplicity due to two unpaired electrons.
- 1D: This term arises when the total spin is paired.
- 1F: This is a singlet state with a total spin of S = 0.
Given these combinations, the possible terms from the p1d1 configuration can be summarized as:
- 3F
- 1D
Conclusion
Therefore, the correct answer from the options provided is 3F and 1D, which corresponds to option 'C'.

Pressure does not affect the rate of reaction:
Pressure does not affect the rate of a reaction that does not involve gases because pressure only affects the rate of reactions involving gases. For reactions that do not involve gases, the pressure of the system does not impact the rate at which the reaction occurs.

Factors affecting the rate of reaction:
- Temperature: Increasing the temperature generally increases the rate of reaction as it provides more energy for the molecules to react.
- Concentration: Higher concentration of reactants usually leads to a faster rate of reaction as there are more molecules colliding with each other.
- Catalyst: Catalysts can speed up a reaction by providing an alternative pathway with lower activation energy, but they do not participate in the reaction itself.
Therefore, in the absence of gases in a reaction, pressure does not play a role in determining the rate of reaction. It is important to consider the specific factors that influence the rate of a reaction in order to accurately predict and control the rate at which it occurs.

A particular wavelength can be influenced by various factors, including the concentration of the absorbing substance, the path length of the light through the sample, and the presence of other substances that may interfere with the absorption. The absorption at a specific wavelength is typically measured using a spectrophotometer, which can provide information about the concentration of the absorbing substance in a sample. By comparing the absorption at different wavelengths, scientists can determine the unique absorption spectrum of a particular substance, which can be used for identification and quantification purposes.

The Concept of Interplanar Spacing
In crystallography, the interplanar spacing (d) is a measure of the distance between parallel planes of atoms in a crystal lattice. For cubic crystals, the interplanar spacing can be determined using Miller indices.
Miller Indices
- Miller indices are a notation system in crystallography for planes in crystal lattices.
- For cubic crystals, the indices (hkl) correspond to specific planes.
- The notation d_hkl indicates the spacing between the planes indexed by (hkl).
Given Values
- d_111 = 325.6 pm (picometers)
- We need to find d_333.
Calculation of d_333
The relationship between d_hkl is given by the formula:
d_hkl = a / sqrt(h^2 + k^2 + l^2)
Where 'a' is the lattice parameter (the distance between two points in the crystal), and (h, k, l) are the Miller indices.
For d_111:
- We have (h, k, l) = (1, 1, 1).
- Thus, d_111 = a / sqrt(1^2 + 1^2 + 1^2) = a / sqrt(3).
For d_333:
- We have (h, k, l) = (3, 3, 3).
- Thus, d_333 = a / sqrt(3^2 + 3^2 + 3^2) = a / sqrt(27) = a / (3*sqrt(3)).
Using the relationship between d_111 and d_333:
- d_333 = d_111 / 3 = 325.6 pm / 3 = 108.5 pm.
Conclusion
Thus, the value of d_333 is indeed 108.5 pm, which corresponds to option 'C'.

Initial Conditions
- Initial amount of water: 10 liters
- Water flow in rate: 8 liters/second
- Water flow out rate: 5 liters/second
Net Flow Rate
- Net flow rate = Flow in rate - Flow out rate
- Net flow rate = 8 liters/s - 5 liters/s = 3 liters/s
Time Duration
- Duration: 10 seconds
Change in Water Amount
- Change in water amount = Net flow rate x Time duration
- Change in water amount = 3 liters/s x 10 seconds = 30 liters
Total Amount of Water After 10 Seconds
- Total amount of water = Initial amount + Change in water amount
- Total amount of water = 10 liters + 30 liters = 40 liters
Conclusion
- After 10 seconds, the amount of water in the vessel will be 40 liters. Thus, the correct answer is option 'D'.

Explanation:

C) 22/7:
- 22/7 is an approximation for the mathematical constant π (pi), which represents the ratio of a circle's circumference to its diameter.
- It is an irrational number and its decimal representation is approximately 3.14159.
- Unlike the other options provided, 22/7 is a specific numerical approximation for π, rather than a geometric or mathematical concept.

A) Ratio of the circumference of a circle to its diameter:
- The ratio of a circle's circumference to its diameter is a constant value known as π (pi).
- It is approximately equal to 3.14159 and is the same for all circles, making it a universal constant in geometry.

B) Sum of the three angles of a plane triangle expressed in radians:
- The sum of the three angles of any plane triangle is always equal to π radians or 180 degrees.
- This is a fundamental property of triangles in geometry and is independent of the specific dimensions or shape of the triangle.

D) Net volume of a hemisphere and a cone:
- The net volume of a hemisphere with a unit radius and a cone with a unit radius and height can be calculated using specific formulas for each shape.
- These values are determined by mathematical formulas and are not dependent on approximations or constants like π.
In conclusion, option C (22/7) stands out as different from the other options because it represents a specific numerical approximation for π, while the rest are based on fundamental geometric or mathematical concepts.

Electrolysis of 50% H2SO4
The electrolysis of 50% H2SO4 involves the decomposition of sulfuric acid using an electric current. This process results in the formation of various products depending on the concentration of the sulfuric acid and the conditions of the electrolysis.

Products of Electrolysis
- When 50% H2SO4 is electrolyzed, the main products that are generally formed are hydrogen gas (H2) and oxygen gas (O2).
- The reaction at the anode is 2H2O → O2 + 4H+ + 4e-, leading to the formation of oxygen gas.
- The reaction at the cathode is 2H+ + 2e- → H2, resulting in the generation of hydrogen gas.
- Therefore, the electrolysis of 50% H2SO4 mainly produces H2 and O2 as the final products.

Other Possible Products
- In some cases, the electrolysis of sulfuric acid can also lead to the formation of hydrogen peroxide (H2O2) as a byproduct. This is due to the presence of oxygen and hydrogen ions in the electrolyte.
- The formation of hydrogen peroxide can occur through various intermediate reactions involving the reduction of oxygen at the cathode.

Correct Answer
- In the given options, the correct answer is H2O2 (hydrogen peroxide) as one of the products of the electrolysis of 50% H2SO4.
- While hydrogen gas (H2) and water (H2O) are the primary products, the presence of oxygen can lead to the formation of hydrogen peroxide (H2O2) as an additional product.
In conclusion, the electrolysis of 50% H2SO4 produces hydrogen gas, oxygen gas, and hydrogen peroxide as the main products.

Incorrect statement: In the emission spectrum, transition takes place with the absorption of a photon of energy hv.

Explanation:
The emission spectrum is a set of wavelengths of electromagnetic radiation emitted by an atom when its electrons transition from higher energy levels (excited state) to lower energy levels (ground state). The emission spectrum is a characteristic property of each element and can be used to identify the presence of specific elements.

Statement (a): Emission spectrum results when an electron in an atom undergoes a transition from an excited state to the ground state.
This statement is correct. When an electron in an atom jumps from a higher energy state to a lower energy state, it emits a photon of specific energy, resulting in the emission spectrum.

Statement (b): In the emission spectrum, transition takes place with the emission of a photon of energy hv.
This statement is correct. In the emission spectrum, the transition of an electron from a higher energy level to a lower energy level results in the emission of a photon of specific energy. The energy of the emitted photon is given by the equation E = hv, where E is the energy of the photon, h is Planck's constant, and v is the frequency of the photon.

Statement (c): In the emission spectrum, transition takes place with the absorption of a photon of energy hv.
This statement is incorrect. In the emission spectrum, the transition of an electron occurs with the emission, not absorption, of a photon. When an electron drops from a higher energy level to a lower energy level, it releases energy in the form of a photon.

Therefore, the incorrect statement is option (c). The emission spectrum is characterized by the emission of photons, not the absorption of photons.

Explanation:

Identifying the correct statement:
- Statement B is the correct statement about Infrared (IR) spectroscopy.
- IR spectroscopy is commonly used to identify the functional group(s) present in a compound based on the characteristic absorption bands in the infrared region.

Explanation of Incorrect Statements:

Statement A:
- IR spectroscopy is not typically used to determine the band gap, band structure, or charge carrier concentration of a compound.
- These properties are more commonly studied using techniques such as UV-Vis spectroscopy, X-ray crystallography, or Hall effect measurements.

Statement C:
- IR spectroscopy is used to characterize different stretching and bending modes of vibration in molecules.
- This is one of the key applications of IR spectroscopy, as different functional groups exhibit characteristic IR absorption bands corresponding to specific vibrational modes.

Statement D:
- Heteronuclear diatomic molecules are typically not IR-active.
- IR absorption in diatomic molecules is primarily observed in homonuclear molecules due to changes in dipole moment during vibration.

Therefore, the correct statement is B: "It is used to identify the functional group(s) of a compound."

Saponification Explained
Saponification is a chemical reaction that plays a crucial role in the production of soap. It involves the transformation of fats or oils (which are typically esters) into soap and glycerin. Here’s a detailed breakdown of why option 'A' is the correct answer:
Definition of Saponification
- Saponification specifically refers to the hydrolysis of esters in the presence of a strong base.
- In the case of fats and oils, they are triglycerides, which are esters formed from glycerol and fatty acids.
Process of Saponification
- When a triglyceride undergoes saponification, it reacts with a strong base like sodium hydroxide (NaOH) or potassium hydroxide (KOH).
- This reaction cleaves the ester bonds, resulting in the formation of carboxylate salts (the soap) and glycerol.
Key Reaction Components
- Ester Cleavage: The reaction breaks the ester bonds, leading to the production of carboxylic acids and alcohol.
- Alkaline Hydrolysis: The presence of a base facilitates the hydrolysis process, effectively breaking down the ester molecules.
Conclusion
- Therefore, option 'A' accurately describes saponification as it highlights the cleaving of ester molecules into carboxylic acids and alcohols, making it the correct choice among the options provided.
Understanding saponification is essential for grasping the chemistry behind soap production and its applications in everyday life.

Let's analyze each statement to determine its correctness.

Statement A: they are involved in O2 transport in biological systems
- This statement is correct. Both deoxy-hemerythrin and deoxy-hemocyanin are involved in the transport of oxygen in biological systems.

Statement B: they contain two metal ions in their active site
- This statement is incorrect. Deoxy-hemerythrin contains two iron ions in its active site, while deoxy-hemocyanin contains two copper ions in its active site. Therefore, this statement is only true for deoxy-hemerythrin, not for deoxy-hemocyanin.

Statement C: active site metal centers are bridged by amino acid residues
- This statement is correct. Both deoxy-hemerythrin and deoxy-hemocyanin have metal centers in their active sites that are bridged by amino acid residues. In deoxy-hemerythrin, the iron ions are bridged by histidine residues, while in deoxy-hemocyanin, the copper ions are bridged by histidine residues.

Statement D: they prefer to bind only one O2 per active site
- This statement is incorrect. Deoxy-hemerythrin can bind two oxygen molecules per active site, while deoxy-hemocyanin can bind only one oxygen molecule per active site. Therefore, this statement is only true for deoxy-hemocyanin, not for deoxy-hemerythrin.


Based on the analysis above, the correct statements are:

- A. they are involved in O2 transport in biological systems
- B. they contain two metal ions in their active site (only true for deoxy-hemerythrin)
- D. they prefer to bind only one O2 per active site (only true for deoxy-hemocyanin)

Therefore, the correct answer is option 'A' (A, B, and D).