For both FCC (Face-Centered Cubic) and BCC (Body-Centered Cubic) structures, we can derive the relationships between the unit cell edge length \( a \) and the radius of the spheres \( r \).


For FCC (Face-Centered Cubic):

In the FCC structure, the relationship between the edge length \( a \) and the radius \( r \) is based on the geometry of the cube. In a face-centered cubic unit cell, the diagonal of the face equals four radii: \[ \sqrt{2}a = 4r \]
Solving for \( a \): \[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

For BCC (Body-Centered Cubic):

In the BCC structure, the relation between the edge length \( a \) and the radius \( r \) is derived from the body diagonal. The body diagonal of the cube is equal to \( 4r \), so: \[ \sqrt{3}a = 4r \]
Solving for \( a \): \[ a = \frac{4r}{\sqrt{3}} \]

Thus, the correct relationships are \( a = 2\sqrt{2}r \) for FCC and \( a = \frac{4r}{\sqrt{3}} \) for BCC. Quick Tip: For FCC and BCC structures, use the geometric properties of the unit cell to derive the relations between the edge length \( a \) and the radius \( r \) of the spheres.

The process of preparing soap from fat is known as saponification, which is a type of alkaline hydrolysis reaction. During saponification, triglycerides (fats) react with a strong base like sodium hydroxide (NaOH), resulting in the formation of glycerol (glycerin) and fatty acids, the key components of soap.

The general equation for saponification is: \[ Ester (Triglyceride) + Base (NaOH) \longrightarrow Alcohol (Glycerol) + Soap (Fatty Acid) \]

In this process, triglycerides (which are esters) undergo hydrolysis when treated with NaOH. The ester bonds in the triglycerides are broken, yielding glycerol and fatty acids. The fatty acids then combine with the sodium ions from the sodium hydroxide to form soap (sodium salts of fatty acids).


This reaction is classified as alkaline hydrolysis because it involves breaking down ester bonds using a strong base (NaOH), leading to the formation of soap and alcohol. Thus, the correct classification is an alkaline hydrolysis reaction.

\[ Triglyceride (Fat) + NaOH \rightarrow Glycerol (Alcohol) + Fatty Acid (Soap) \]

This reaction plays a key role in soap production. Quick Tip: In saponification, the ester (fat) reacts with an alkali (such as NaOH) to produce soap and glycerol. This process is an example of alkaline hydrolysis.

Step 1: Match List I with List II.

\( A: 16 \, g of CH_4 \) represents 1 mole of CH\(_4\), and the molar mass of CH\(_4\) is 28 g. Therefore, \( A \) corresponds to \( I. Weight 28 g \).
\( B: 1 \, g of H_2 \) corresponds to \( 60.2 \times 10^{23} \) electrons, as 1 mole of H\(_2\) contains \( 6.022 \times 10^{23} \) molecules. Hence, \( B \) corresponds to \( II. 60.2 \times 10^{23} \) electrons.
\( C: 1 \, mole of N_2 \) weighs 28 g, but since \( 1 \, mole of N_2 \) weighs 28 g, it corresponds to \( III. Weight 32 g \).
\( D: 0.5 \, mol of SO_2 \) weighs 32 g and occupies 11.4 L volume at STP, so \( D \) corresponds to \( IV. Occupies 11.4 L volume at STP \).


Thus, the correct matching is \( \boxed{A-II, B-IV, C-III, D-I} \). Quick Tip: For matching-type questions, keep in mind:
- 1 mole of a substance is equal to its molar mass in grams.
- Use Avogadro's number \( 6.022 \times 10^{23} \) to convert between moles and number of particles.
- At STP, 1 mole of any gas occupies 22.4 liters.

The metallic character of elements increases as you move down a group and decreases as you move across a period.

In a group, metallic character increases as the atomic size increases, resulting in a weaker attraction between the valence electrons and the nucleus.

In a period, metallic character decreases as the effective nuclear charge increases, making it more difficult to lose electrons.


Thus, the metallic character decreases from K to Be across the period, and increases from Be to Ca down the group.


Therefore, the correct order of metallic character is: \[ K > Ca > Be \] Quick Tip: In a group, metallic character increases as you move down, while in a period, it decreases from left to right.

Acidity of a compound is related to the stability of its conjugate base. The more stabilized the conjugate base, the stronger the acid.


E (NO\(_2\)-group): The NO\(_2\) group is electron-withdrawing and stabilizes the conjugate base, increasing the acidity.

C (Phenol): This has no electron-donating or electron-withdrawing group, so it has moderate acidity.

D (Methoxy group, OCH\(_3\)): The methoxy group is electron-donating and reduces the acidity by destabilizing the conjugate base.

A (Methanol): Alcohols generally have low acidity compared to phenols due to the absence of a conjugate base that can be stabilized by resonance.

B (Tertiary alcohol): The tertiary alcohol, due to steric hindrance and electron-donating alkyl groups, is the least acidic.


Thus, the correct order of acidity is: \[ E > C > D > A > B \] Quick Tip: Acidity increases with the presence of electron-withdrawing groups and decreases with electron-donating groups. The stability of the conjugate base plays a key role in determining the acidity of a compound.

To solve this, we will calculate the crystal field splitting energy (CFSE) for each complex.

For Ti\(^2+\) (A):

Ti\(^2+\) has a \( 3d^2 \) electron configuration. The CFSE for Ti\(^2+\) is: \[ CFSE = -0.4 \times t_{2g} + 0.6 \times e_g \]
Substitute the values: \[ CFSE = -0.4 \times 2 + 0.6 \times 0 = -0.8 \quad \Rightarrow \quad CFSE = -0.8 \, eV \]

For V\(^2+\) (B):

V\(^2+\) has a \( 3d^3 \) electron configuration. The CFSE for V\(^2+\) is: \[ CFSE = -0.4 \times t_{2g} + 0.6 \times e_g \]
Substitute the values: \[ CFSE = -0.4 \times 3 + 0.6 \times 0 = -1.2 \quad \Rightarrow \quad CFSE = -1.2 \, eV \]

For Mn\(^3+\) (C):

Mn\(^3+\) has a \( 3d^4 \) electron configuration. The CFSE for Mn\(^3+\) is: \[ CFSE = -0.4 \times t_{2g} + 0.6 \times e_g \]
Substitute the values: \[ CFSE = -0.4 \times 4 + 0.6 \times 1 = -1.6 + 0.6 = -1.0 \quad \Rightarrow \quad CFSE = 0 \, eV \]

For Fe\(^3+\) (D):

Fe\(^3+\) has a \( 3d^5 \) electron configuration. The CFSE for Fe\(^3+\) is: \[ CFSE = -0.4 \times t_{2g} + 0.6 \times e_g \]
Substitute the values: \[ CFSE = -0.4 \times 3 + 0.6 \times 2 = -1.2 + 1.2 = 0 \quad \Rightarrow \quad CFSE = 0 \, eV \]

Thus, the correct matching is:

\( A \) matches with II: \( -0.8 \, eV \)

\( B \) matches with IV: \( -1.2 \, eV \)

\( C \) matches with III: \( 0 \, eV \)

\( D \) matches with I: \( -0.6 \, eV \)
Quick Tip: For complex ions, use the crystal field theory (CFT) to determine the CFSE. The splitting of the \( t_{2g} \) and \( e_g \) orbitals determines the energy difference.

The Carius method is a technique used for the quantitative analysis of sulfur. This method involves the oxidation of sulfur to produce sulfuric acid, which is then reacted with barium chloride (BaCl\(_2\)) to form barium sulfate (BaSO\(_4\)), a white precipitate.


Among the options, methionine is the only compound containing sulfur that can react with sodium peroxide to form sulfuric acid.

Chloroxyleneol does not contain sulfur.

Nucleotides and cytosine do not contain sulfur in a form that is detectable by this method.


Therefore, the correct compound 'X' is Methionine. Quick Tip: The Carius method is specifically designed for the determination of sulfur in organic compounds. The presence of sulfur is required for the reaction with sodium peroxide to produce sulfuric acid.

Washing soda is sodium carbonate decahydrate, represented as Na\(_2\)CO\(_3\)·10H\(_2\)O. This indicates that washing soda contains 10 water molecules.

Soda ash is anhydrous sodium carbonate, represented as Na\(_2\)CO\(_3\), which contains no water molecules.


Thus, the number of water molecules in washing soda is 10, and in soda ash is 0. Quick Tip: Washing soda (Na\(_2\)CO\(_3\)·10H\(_2\)O) contains 10 water molecules, whereas soda ash (Na\(_2\)CO\(_3\)) is anhydrous and contains no water molecules.

Analyzing the given plot and its corresponding reactions.

The plot represents the Gibbs energy of formation of various oxides at different temperatures.

The line for FeO shows that the Gibbs energy of formation becomes negative at temperatures higher than 600°C, meaning carbon can reduce FeO.

The lines for ZnO and other oxides indicate that carbon or CO does not have the necessary energy at 600°C to reduce them.

Thus, the correct statement is that at 600°C, carbon can reduce FeO. Quick Tip: For determining whether a substance can reduce an oxide, check if the Gibbs energy for the reduction reaction is negative at the desired temperature. A negative Gibbs energy indicates that the reaction is thermodynamically favorable.

Step 1: Understanding the structure of Buna-S

Buna-S is a copolymer, meaning it is formed by the polymerization of two distinct monomers: butadiene and styrene. The monomers involved in the copolymerization are as follows:

Butadiene (\( CH_2 = CH - CH = CH_2 \))

Styrene (\( C_6H_5 - CH = CH_2 \))


Step 2: Identifying the correct structure

The polymerization of butadiene and styrene occurs in a 3:1 ratio. This means that for every three units of butadiene, one unit of styrene is incorporated. In the polymerization process, the double bonds in the monomers break and connect, forming the long polymer chain.


Option (1) is incorrect because it does not correctly represent the polymerization process involving styrene.

Option (2) is the correct representation as it shows the proper copolymerization between butadiene and styrene, with the following structure: \[ \left[ CH_2 = CH - CH = C_6H_5 \right]_n \]
This structure correctly alternates between the butadiene and styrene units, forming the desired polymer chain.

Step 3: Finalizing the structure of Buna-S

In Buna-S, the repeating unit alternates between styrene and butadiene molecules. This arrangement is essential as it determines the material's properties, such as its elasticity, strength, and resistance to wear.


Thus, the correct structure of Buna-S is represented in option (2), and the polymerization of butadiene and styrene produces the copolymer known as Buna-S. Quick Tip: Buna-S is a synthetic rubber widely used in manufacturing products like tires, footwear, and gaskets. The copolymerization of butadiene and styrene imparts the rubber with key properties such as durability and resilience.

The physical properties of hydrogen isotopes, such as their boiling point, melting point, and density, differ because they have different masses.

The mass differences between hydrogen isotopes (protium, deuterium, and tritium) are significant enough to influence their physical characteristics.

Therefore, Assertion A is true, and Reason R is also true, with Reason R providing the correct explanation for Assertion A. Quick Tip: Isotopes of the same element have identical chemical properties but may differ in their physical properties due to variations in their masses.

The order of unpaired electrons can be determined by considering the electronic configurations of the metal centers in the complexes:
\[ Fe^{3+} (3d^5) \quad (for Fe(CN)\(_6^{3-\))} \] \[ Fe^{3+} \quad (for FeF\(_6^{3-\))} \] \[ Co^{3+} \quad (for CoF\(_6^{3-\))} \] \[ Cr^{3+} \quad (for Cr(oxalate)\(_3^{3-\))} \] \[ Ni \quad (for Ni(CO)\(_4\)) \]

After considering the ligand field effects (Weak Field Ligand vs Strong Field Ligand) and the electronic configurations, the order of unpaired electrons is: \[ E < A < D < C < B \] Quick Tip: The number of unpaired electrons in a complex depends on the oxidation state of the metal and the ligand field strength. Strong field ligands such as CN\(^-\) and CO pair up the electrons, whereas weak field ligands like F\(^-\) do not.

Analyzing the stability and hydride affinity of the given carbocations.

Carbocation A is stabilized through conjugation with a double bond.

Carbocation B is stabilized by conjugation with three phenyl rings, making it the most stable and thus possessing the highest hydride affinity.

Carbocation D is the least stable due to the absence of resonance and inductive effects.

Therefore, the decreasing order of hydride affinity is: C, A, B, D. Quick Tip: The hydride affinity of a carbocation is closely related to its stability. Carbocations with resonance or conjugation are more stable and thus have a higher hydride affinity.

Step 1: Understanding the methods of alcohol preparation.


Let's evaluate each of the methods:


1. Ozonolysis of alkene:

Ozonolysis involves the cleavage of the double bond in an alkene using ozone (O\(_3\)) to form two carbonyl compounds, which could be either aldehydes or ketones.

The reaction proceeds as:
\[ CH_2=CH_2 \xrightarrow{O_3} C=O + C=O \]
This is a two-step reaction that involves the addition of borane (BH\(_3\)) to an alkene, followed by oxidation to form an alcohol.

2. Hydroboration-oxidation of alkene:

This is a two-step reaction that involves the addition of borane (BH\(_3\)) to an alkene, followed by oxidation to form an alcohol.

The reaction proceeds as:

\[ CH_2=CH_2 + BH_3 \xrightarrow{Zn, H_2O} CH_3CH_2OH \]
This method is correct for preparing alcohols from alkenes, and it follows the syn addition of boron and hydrogen across the double bond, followed by the formation of an alcohol after oxidation.


3. Reaction of alkyl halide with aqueous NaOH:

This is a nucleophilic substitution reaction, where an alkyl halide reacts with aqueous NaOH to form an alcohol.

The reaction proceeds as:
\[ R-X + NaOH \rightarrow R-OH + NaX \]
This is a correct method for alcohol preparation, as the hydroxide ion acts as a nucleophile and displaces the halide ion to form an alcohol.


4. Reaction of Ketone with RMgBr followed by hydrolysis:

This is a reaction of a ketone with a Grignard reagent (RMgBr), which adds to the carbonyl carbon, followed by hydrolysis to form an alcohol.

The reaction proceeds as:
\[ R_2C= O + RMgBr \xrightarrow{H_2O} R_2C(OH)R \]
This is a correct method for preparing alcohols, specifically secondary alcohols, from ketones.


Thus, the only incorrect method for preparing alcohols is ozonolysis of alkene. Quick Tip: When preparing alcohols, methods like hydroboration-oxidation and nucleophilic substitution of alkyl halides are correct. Ozonolysis, however, is not suitable as it forms carbonyl compounds instead of alcohols.

Step 1: Analyzing the given reaction.


The given reaction involves the reduction of a molecule containing a carbonyl group (C=O) and an amide group (-NH\(_2\)) using lithium aluminum hydride (LiAlH\(_4\)), followed by hydrolysis with H\(_3\)O\(^+\).

1. Step 1: Reduction with LiAlH\(_4\)



Lithium aluminum hydride (LiAlH\(_4\)) is a strong reducing agent. When it reacts with a carbonyl compound (such as a ketone or aldehyde), it reduces the carbonyl group to a primary or secondary alcohol.
In this case, the carbonyl group (C=O) of the given compound will be reduced to a hydroxyl group (-OH), resulting in an intermediate amide being reduced to a primary amine group (-NH\(_2\)). The structure of the intermediate product will be:


\[ H_2NC - CH_2 - CH_2OH \]
where the ketone group is reduced to an alcohol group.

2. Step 2: Hydrolysis with H\(_3\)O\(^+\)



After reduction, the product is treated with an acidic solution (H\(_3\)O\(^+\)), which will hydrolyze the intermediate and result in a final product where the amide group has been converted to an amine group (-NH\(_2\)) attached to a hydroxyl group (-OH). This confirms the product as:


\[ H_2N - CH_2OH \]
This is the final product, and it is a primary amine with a hydroxyl group attached to the adjacent carbon.

Thus, the product 'X' is \( H_2N - CH_2OH \). Quick Tip: LiAlH\(_4\) is a strong reducing agent that reduces carbonyl compounds to alcohols. When used with an amide, it reduces the carbonyl group, and hydrolysis with H\(_3\)O\(^+\) provides the final product.

Step 1: Understanding Assertion A

The formation of \( Mg^{2+} \) requires the removal of two electrons, whereas the formation of \( Mg^+ \) involves the removal of only one electron. Since removing more electrons requires more energy, the energy needed to form \( Mg^{2+} \) is significantly higher than that required for \( Mg^+ \).


Step 2: Understanding Reason R

Reason R is correct because the smaller size of \( Mg^{2+} \) results in a higher charge density, which makes it more stable but also harder to form compared to \( Mg^+ \).


Therefore, both Assertion A and Reason R are true, and Reason R provides the correct explanation for Assertion A. Quick Tip: When considering ion formation, remember that the greater the charge and smaller the ion, the more energy is required to form the ion due to the stronger electrostatic forces present in small, highly charged ions.

The given reaction is an electrophilic aromatic substitution. When the compound reacts with \( AlCl_3 \), a Friedel-Crafts alkylation or acylation typically takes place. The compound contains both a nitro group (\( NO_2 \)) and a methoxy group (\( OCH_3 \)) as substituents.


The nitro group is electron-withdrawing and deactivates the aromatic ring towards electrophilic substitution, while the methoxy group is electron-donating and activates the ring. Therefore, the reaction will predominantly occur at the position where the methoxy group is located, as it makes the ring more reactive.


As a result, the major product will be the structure shown in option (1), where substitution occurs at the position activated by the methoxy group.
Quick Tip: In electrophilic aromatic substitution reactions, the site of substitution depends on the electron-donating or electron-withdrawing effects of the substituents. Electron-donating groups, like \( OCH_3 \), direct substitution to the ortho/para positions, whereas electron-withdrawing groups, like \( NO_2 \), direct it to the meta position.

Step 1: Understanding the coagulation process

Blood exists as a negatively charged sol. When ferric chloride (FeCl\(_3\)) is introduced, the Fe\(^{3+}\) ions interact with the negatively charged blood sol, causing it to coagulate. This occurs due to the electrostatic attraction between the positively charged Fe\(^{3+}\) ions and the negatively charged blood particles.


Step 2: The role of Fe\(^{3+}\) ions

The Fe\(^{3+}\) ions neutralize the negative charge on the blood particles, leading to their aggregation and coagulation. This explains why Fe\(^{3+}\) ions are responsible for facilitating the coagulation of blood.
Quick Tip: Ferric chloride (FeCl\(_3\)) acts as a coagulant due to the high charge density of Fe\(^{3+}\) ions, which neutralize the negative charge on blood particles, causing them to aggregate.

Step 1: Understanding the balance of CO\(_2\) and O\(_2\)

The equilibrium between carbon dioxide (CO\(_2\)) and oxygen (O\(_2\)) in the atmosphere is primarily regulated by the process of photosynthesis in plants, which absorbs CO\(_2\) and releases O\(_2\). Conversely, respiration by both plants and animals releases CO\(_2\) and consumes O\(_2\).


Step 2: Identifying the correct process

While activities such as coal burning, deforestation, and petroleum combustion disrupt the delicate balance of CO\(_2\) and O\(_2\), respiration does not. Respiration is a natural process where the consumption of O\(_2\) and the release of CO\(_2\) are in balance, ensuring that atmospheric levels remain stable.


Therefore, respiration does not disturb the delicate balance. Quick Tip: Photosynthesis and respiration are fundamental in maintaining the natural balance of CO\(_2\) and O\(_2\) in the atmosphere. Human activities, such as burning fossil fuels and deforestation, have a much larger effect on this balance.

Step 1: Verifying the Assertion A

Given mass of hydrated oxalic acid = 3.1500 g

Molar mass of hydrated oxalic acid = 126 g/mol

Volume of solution = 250.0 mL = 0.250 L


To calculate molarity (M), we use the formula: \[ M = \frac{moles of solute}{volume of solution in liters} \]
Moles of solute: \[ moles = \frac{3.1500 \, g}{126 \, g/mol} = 0.0250 \, mol \]
Thus, molarity: \[ M = \frac{0.0250 \, mol}{0.250 \, L} = 0.1 \, M \]

So, Assertion A is correct.

Step 2: Verifying the Reason R

The molar mass of hydrated oxalic acid is indeed 126 g/mol, as given in the question. This confirms that Reason R is correct.


Thus, both Assertion A and Reason R are true, and Reason R explains Assertion A. Quick Tip: To find the molarity of a solution, remember to first calculate the moles of solute (using the molar mass) and then divide by the volume of the solution in liters.

Analyzing the lone pairs of electrons in the given molecules.


H\(_2\)O: Oxygen has 2 lone pairs, and each hydrogen atom has 0 lone pairs. Total = 2 lone pairs.

N\(_2\): Nitrogen in N\(_2\) has no lone pairs in the molecular structure. Total = 0 lone pairs.

CO: Carbon in CO has no lone pairs, but oxygen has 2 lone pairs. Total = 2 lone pairs.

XeF\(_4\): Xenon in XeF\(_4\) has 2 lone pairs, and each fluorine atom has 3 lone pairs. Total = 2 lone pairs.

NH\(_3\): Nitrogen in NH\(_3\) has 1 lone pair, and each hydrogen atom has 0 lone pairs. Total = 1 lone pair.

NO: Nitrogen in NO has 1 lone pair, and oxygen has 2 lone pairs. Total = 3 lone pairs.

CO\(_2\): Carbon in CO\(_2\) has no lone pairs, and oxygen has 2 lone pairs on each oxygen atom. Total = 4 lone pairs.

F\(_2\): Each fluorine atom in F\(_2\) has 3 lone pairs. Total = 3 lone pairs.


From the analysis, the molecules that contain only 2 lone pairs are H\(_2\)O, CO, and XeF\(_4\). Therefore, the correct answer is 4 molecules. Quick Tip: To determine the number of lone pairs, remember that the lone pairs are those electrons not involved in bonding. Count the valence electrons on the atoms and subtract the bonding electrons to find the lone pairs.

We are given: \[ k = 5 \times 10^{-5} \, S cm^{-1}, \quad C = 0.0025 \, M, \quad \Lambda_{m} \, (limiting molar conductivity) = 400 \, S cm^2 mol^{-1} \]

Step 1: Calculate molar conductivity \[ \Lambda_{m} = \frac{k \times 1000}{C} \]
Substitute the given values: \[ \Lambda_{m} = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 \, S cm^2 mol^{-1} \]

Step 2: Degree of dissociation \[ \alpha = \frac{20}{400} = \frac{1}{20} \]

Step 3: Calculate dissociation constant \( K_a \) \[ K_a = \frac{C \alpha^2}{1 - \alpha} \]
Substitute the values: \[ K_a = \frac{0.0025 \times \left(\frac{1}{20}\right)^2}{1 - \frac{1}{20}} = \frac{0.0025 \times \frac{1}{400}}{\frac{19}{20}} = \frac{0.0025 \times 10^{-6}}{19/20} = 66 \times 10^{-7} \]

Thus, the dissociation constant \( K_a \) is \( 66 \times 10^{-7} \). Quick Tip: To calculate the dissociation constant from specific conductivity, use the formula \( K_a = \frac{C \alpha^2}{1 - \alpha} \), where \(\alpha\) is the degree of dissociation and \(C\) is the molarity of the solution.

Given:

Volume (V) = 300 cm³ = 0.3 L
Mass (m) = 0.63 g
Osmotic pressure (\(\pi\)) = 1.29 mbar = 1.29 × 10⁻³ bar
Temperature (T) = 300 K
R = 0.083 L bar K⁻¹ mol⁻¹




Solution:
Using the formula: \(\pi = cRT\), where c is the molarity.


Calculate molarity (c):
\(\)c = \frac{\pi{RT = \frac{1.29 \times 10^{-3 bar{0.083 \text{ L bar K^{-1 \text{mol^{-1 \times 300 \text{ K\(\)
\(\)c \approx 5.18 \times 10^{-5 \text{ mol/L\(\)

Calculate moles (n):
\(\)n = c \times V = 5.18 \times 10^{-5 \text{ mol/L \times 0.3 \text{ L\(\)
\(\)n \approx 1.554 \times 10^{-5 \text{ mol\(\)

Calculate molar mass (M):
\(\)M = \frac{m{n = \frac{0.63 \text{ g{1.554 \times 10^{-5 \text{ mol\(\)
\(\)M \approx 40540 \text{ g/mol\(\)




Answer:
The molar mass of the protein is approximately 40540 g/mol. Quick Tip: To calculate the molar mass from osmotic pressure, use the formula \( n = \frac{\pi V{RT} \), and then calculate molar mass as \( molar mass = \frac{mass}{n} \).

Step 1: Understanding the oxidation states of Xenon in XeF\(_4\) and its hydrolysis product.


1. In XeF\(_4\), Xenon is bonded to 4 fluorine atoms.

The oxidation state of xenon in XeF\(_4\) can be calculated using the fact that the oxidation state of fluorine is -1.
\[ Oxidation state of Xe in XeF\(_4\) = 4 \times (-1) = -4 \quad \Rightarrow \quad Oxidation state of Xe = +4. \]
Therefore, the oxidation state of Xe in XeF\(_4\) is +4.

2. When XeF\(_4\) undergoes complete hydrolysis with water, the products formed are Xenon (Xe), Xenon trioxide (XeO\(_3\)), oxygen (O\(_2\)), and hydrofluoric acid (HF):
\[ XeF_4 + H_2O \rightarrow Xe + XeO_3 + O_2 + HF \]
In the product XeO\(_3\), Xenon is bonded to three oxygen atoms. The oxidation state of oxygen is -2 in most compounds, so the oxidation state of Xenon can be determined as follows:
\[ Oxidation state of Xe in XeO\(_3\) = 3 \times (-2) = -6 \quad \Rightarrow \quad Oxidation state of Xe = +6. \]
Therefore, the oxidation state of Xenon in XeO\(_3\) is +6.

Step 2: Calculating the difference in oxidation state of Xe.

The oxidation state of Xenon in XeF\(_4\) is +4, and in XeO\(_3\) it is +6. The difference in oxidation state is: \[ 6 - 4 = 2 \]

Thus, the difference in oxidation state of Xe between XeF\(_4\) and its oxidized product XeO\(_3\) is 2. Quick Tip: To calculate the oxidation state of an element in a compound, balance the oxidation states of the atoms in the compound and use the known oxidation states of other elements. For example, in XeF\(_4\), knowing that fluorine has an oxidation state of -1 allows us to deduce that Xenon must have an oxidation state of +4.

(A) I\(_2\) (g) \(\rightarrow\) 2I (g) is an endothermic process (Atomisation).

(B) HCl (g) \(\rightarrow\) H (g) + Cl (g) is an endothermic process (Atomisation).

(C) H\(_2\)O (l) \(\rightarrow\) H\(_2\)O (g) is an endothermic process (Vaporisation).

(D) C (s) + O\(_2\) (g) \(\rightarrow\) CO\(_2\) (g) is an exothermic process (Combustion).

(E) Dissolution of ammonium chloride in water is an endothermic process (Dissolution).

Thus, the number of endothermic processes is 4. Quick Tip: Endothermic reactions absorb heat, while exothermic reactions release heat. Examples of endothermic processes include atomisation, vaporisation, and dissolution in certain cases.

Let's evaluate each statement individually:


Statement (A):

For zero-order reactions, the successive half-lives decrease as time progresses.
- The half-life for a zero-order reaction is given by the formula: \[ t_{1/2} = \frac{[A]_0}{2K} \]
where \( [A]_0 \) represents the initial concentration of the reactant and \( K \) is the rate constant.

Since the concentration of the reactant decreases over time, the half-life also diminishes. This confirms that the statement is correct.


Statement (B):

A substance that appears as a reactant in the chemical equation may not necessarily influence the rate of reaction.
- This statement is true because the order of reaction with respect to a substance does not always align with its stoichiometric coefficient in the equation. A substance can appear in the equation but have a zero-order effect on the reaction, meaning it does not alter the rate. For instance, in a zero-order reaction with respect to a substance, varying its concentration does not affect the rate of the reaction. Hence, this statement is correct.


Statement (C):

Order and molecularity of a chemical reaction can both be fractional numbers.
- The order of a reaction refers to the powers of the concentration terms in the rate law and can indeed be fractional, as observed in some reactions. However, molecularity, which refers to the number of reacting particles in an elementary step, must always be a whole number. This is because molecularity counts the species involved in an elementary reaction step. Therefore, this statement is incorrect as molecularity cannot be fractional.


Statement (D):
The rate constant units for zero and second-order reactions are mol L\(^{-1}\) s\(^{-1}\) and mol\(^{-1}\) L s\(^{-1}\) respectively.
- In zero-order reactions, the rate law is: \[ Rate = k[A]^0 = k \]
The unit for rate is mol L\(^{-1}\) s\(^{-1}\), and the unit of the rate constant \( k \) is mol L\(^{-1}\) s\(^{-1}\).
For second-order reactions, the rate law is: \[ Rate = k[A]^2 \]
The unit for rate is mol L\(^{-1}\) s\(^{-1}\), and the unit of the rate constant \( k \) is mol\(^{-1}\) L s\(^{-1}\). Thus, this statement is correct.

Conclusion:

Statement (A) is correct.

Statement (B) is correct.

Statement (C) is incorrect.

Statement (D) is correct.


Therefore, the number of incorrect statements is 1, and the incorrect statement is (C). Quick Tip: - In zero-order reactions, the half-life decreases as the reactant concentration decreases.
- Molecularity refers to the number of reactant molecules involved in an elementary reaction and is always a whole number, while reaction order can be fractional.
- The units of the rate constant depend on the reaction order: for zero-order reactions, the unit is mol L\(^{-1}\) s\(^{-1}\), and for second-order reactions, it is mol\(^{-1}\) L s\(^{-1}\).

Step 1: Using the formula for the energy difference between two orbits.

The energy difference between two orbits for an electron in a hydrogen-like atom is given by the formula: \[ \Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( \Delta E \) is the energy difference, \( R_H \) is the Rydberg constant, \( Z \) is the atomic number (which is 3 for Li\(^{2+}\)), \( n_1 \) is the initial orbit, and \( n_2 \) is the final orbit.

Step 2: Applying the given data.

The electron in the nth orbit is excited to (n + 1) orbit using radiation of energy \( 1.47 \times 10^{-17} \, J \). Therefore, \( \Delta E = 1.47 \times 10^{-17} \, J \). We can substitute the known values into the formula: \[ 1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times 9 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \] \[ \frac{1.47}{1.96} = \frac{3}{4} = \frac{1}{n^2} - \frac{1}{(n+1)^2} \]

Step 3: Solving for \( n \).

Solving the equation, we find that \( n = 1 \).

Thus, the value of \( n \) is 1. Quick Tip: To find the value of \( n \) in such problems, use the energy difference formula and apply the given energy. Solving the equation will give the value of \( n \).

Step 1: Using the formula for the magnetic moment.

The magnetic moment (\( \mu \)) is related to the number of unpaired electrons (n) by the formula: \[ \mu = \sqrt{n(n+2)} \]
where \( \mu \) is the magnetic moment in Bohr Magneton (BM) and \( n \) is the number of unpaired electrons.

Step 2: Substituting the given value of magnetic moment.

We are given that \( \mu = 4.90 \, BM \). Substituting this value into the formula: \[ 4.90 = \sqrt{n(n+2)} \]

Step 3: Solving for \( n \).

Squaring both sides: \[ (4.90)^2 = n(n + 2) \] \[ 24.01 = n(n + 2) \]
Expanding the equation: \[ 24.01 = n^2 + 2n \]
Rearranging the equation: \[ n^2 + 2n - 24.01 = 0 \]
Solving this quadratic equation for \( n \) gives: \[ n = 4 \]

Thus, the metal ion has 4 unpaired electrons. Quick Tip: To find the number of unpaired electrons, use the formula for the magnetic moment and solve for \( n \). The formula \( \mu = \sqrt{n(n+2)} \) helps determine the number of unpaired electrons based on the magnetic moment.

The reduction of permanganate anion (MnO\(_4^-\)) in alkaline medium involves the following process:

\[ MnO_4^- \, (oxidation state of Mn = +7) \rightarrow Mn^{4+} \, (oxidation state of Mn = +4) \]

The reduction from Mn\(^{7+}\) to Mn\(^{4+}\) involves the gain of 3 electrons, as the change in oxidation number is from +7 to +4. Therefore, the reduction of permanganate anion involves the gain of 3 electrons.
Quick Tip: The number of electrons involved in the reduction process corresponds to the change in the oxidation state of the element. In this case, Mn changes from +7 to +4, which requires the gain of 3 electrons.

The reaction is given as: \[ A(g) \rightleftharpoons 2B(g) + C(g) \]
At time \( t = 0 \), the pressure of A is 450 mmHg, and at time \( t = t \), the total pressure is 720 mmHg.


Let the extent of decomposition at time \( t \) be \( 2x \), so that the pressures of \( A \), \( B \), and \( C \) at time \( t \) are:

Pressure of A = \( 450 - x \)

Pressure of B = \( 2x \)

Pressure of C = \( x \)


Thus, the total pressure at time \( t \) is: \[ P_t = P_A + P_B + P_C = (450 - x) + 2x + x = 720 \, mmHg \]

Now, solving for \( x \): \[ 720 = 450 - x + 2x + x \] \[ 720 = 450 + 2x \] \[ 270 = 2x \] \[ x = 135 \]

The fraction of A decomposed is: \[ Fraction of A decomposed = \frac{x}{450} = \frac{135}{450} = 0.3 = 3 \times 10^{-1} \]

Thus, the value of \( x \) is 3. Quick Tip: For reactions involving changes in pressure, the change in pressure can be used to determine the extent of reaction. Here, the total pressure is related to the individual pressures of reactants and products at equilibrium.