The rate of nucleophilic aromatic substitution (NAS) is influenced by the position of the electron-withdrawing group (EWG) relative to the leaving group. Groups such as nitro (NO\(_2\)) increase the reactivity of the aromatic compound towards NAS by stabilizing the intermediate Meisenheimer complex. The extent of stabilization is optimized when:

1. The EWG is located in the ortho or para position relative to the leaving group. This positioning allows for strong resonance interactions, which help stabilize the negative charge on the intermediate complex.

2. The EWG in the meta position is less effective at stabilizing the intermediate due to the absence of resonance alignment between the substituent and the nucleophilic attack site.


Analysis of the Options:

- Compound (1): The nitro group is in the para position relative to the leaving chlorine atom (Cl), which enhances the rate of reaction by providing resonance stabilization.

- Compound (2): The nitro group is positioned ortho to the chlorine atom, offering strong resonance stabilization and enhancing the rate of reaction.

- Compound (3): The nitro group is in the para position, similar to Compound (1), thus favoring the reaction.

- Compound (4): The nitro group is in the meta position relative to the chlorine atom. In this case, the nitro group does not effectively stabilize the negative charge during the reaction, leading to the lowest rate of nucleophilic aromatic substitution.


Electron-withdrawing groups boost the reactivity of aromatic compounds in nucleophilic substitution reactions by stabilizing the intermediate negative charge. However, this effect is dependent on the position of the NO\(_2\) group. When the NO\(_2\) group is positioned at the meta location, its inductive effect contributes minimally to stabilization because resonance interactions cannot occur. Therefore, compound (4), with NO\(_2\) at the meta position, demonstrates the lowest rate of reaction. Quick Tip: Electron-withdrawing groups at the ortho or para positions significantly stabilize the intermediate complex in nucleophilic aromatic substitution. The meta position is far less effective in this regard.

The graph depicting the variation of the rate of an enzyme-catalyzed reaction with substrate concentration follows Michaelis-Menten kinetics. At the beginning, the reaction rate increases in a linear fashion with the substrate concentration as the enzyme's active sites are available and ready for binding. However, once the enzyme becomes saturated with the substrate, the rate reaches a maximum value, V\(_{max}\), and no longer increases, even with additional substrate.

Graph (b) best illustrates this relationship, as it demonstrates:

- An initial linear phase, where the reaction rate increases with substrate concentration.

- A plateau phase, where the reaction rate stabilizes and becomes constant due to enzyme saturation.

The reaction rate of an enzyme-catalyzed process is influenced by the substrate concentration. At low substrate concentrations, the rate is directly proportional to the substrate concentration. At higher concentrations, the enzyme reaches saturation, causing the rate to plateau at V\(_{max}\), indicating that all active sites on the enzyme are fully occupied. Quick Tip: Enzyme-catalyzed reactions follow a hyperbolic curve described by Michaelis-Menten kinetics. Remember, saturation occurs when the substrate concentration is very high.

The reaction proceeds through the following steps:

1. Step A: Bromination of the aromatic ring using Br\(_2\) in the presence of Fe results in the selective addition of bromine to the meta position relative to the existing nitro group (NO\(_2\)). This selective positioning occurs because the nitro group is an electron-withdrawing group, which directs the incoming bromine to the meta position.

2. Step B: The amine (NH\(_2\)) group reacts with NaNO\(_2\) and HCl under cold conditions to form the diazonium salt (Ar-N\(_2^+\)).

3. Step C: Hydrolysis of the diazonium salt with H\(_3\)PO\(_4\) replaces the diazo group (-N\(_2^+\)) with a hydroxyl group (-OH), forming the phenol.

4. Step D: Oxidation with KMnO\(_4\) under basic conditions leads to the conversion of the alkyl group (-Me) attached to the aromatic ring into a carboxyl group (-COOH), transforming the structure into a benzoic acid derivative.

5. Step E: The reaction with dilute acid (H\(^+\)) ensures the completion and stability of the carboxylic acid group.

The final products are as follows:

- A: 3-bromo-4-nitrobenzene

- E: 3-bromo-4-nitrobenzoic acid



Key Points:
The key to solving this question lies in understanding the regioselectivity of each step:

- Bromination occurs at the meta position relative to the NO\(_2\) group due to its electron-withdrawing effect.

- The diazonium salt intermediate allows for substitution with a hydroxyl group at the position of the diazo group.

- Oxidation of the methyl group to a carboxylic acid group completes the transformation, yielding the final product. Quick Tip: Electron-withdrawing groups like nitro (\textbf{NO\(_2\)}) direct electrophilic substitutions to the meta position. Use reaction mechanisms to predict intermediate and final products.

The flame test is used to identify elements based on the characteristic colors they emit when heated. The correct matching is:

K: Violet
Ca: Brick Red
Sr: Crimson Red
Ba: Apple Green


Thus, the correct matching corresponds to option (3). Quick Tip: Flame tests are based on the excitation of electrons in the metal ions. When they return to their ground state, they emit light at characteristic wavelengths.

The reaction proceeds as follows:
1. Step 1: Reaction of white phosphorus (P\(_4\)) with thionyl chloride (SOCl\(_2\)) produces phosphorus trichloride (PCl\(_3\)) and sulfur dioxide (SO\(_2\)):
\[ P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2 \]

2. Step 2: Hydrolysis of PCl\(_3\) produces phosphorous acid (H\(_3\)PO\(_3\)):
\[ PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl \]

Thus, [A] = PCl\(_3\) and [B] = H\(_3\)PO\(_3\), corresponding to option (2). Quick Tip: Phosphorus trichloride (\textbf{PCl\(_3\)}) is a common reagent in chemical synthesis and reacts readily with water to form phosphorous acid (\textbf{H\(_3\)PO\(_3\)}).

1. Atoms of X:
- X is present on alternate corners of the cube, which means it occupies 4 out of the 8 corners.
- Each corner contributes \(\frac{1}{8}\) to the unit cell.
\[ Contribution from corners = 4 \times \frac{1}{8} = \frac{1}{2}. \]
- Additionally, there is one X atom at the center of the cube, contributing 1 atom.
\[ Total contribution from X = \frac{1}{2} + 1 = 1.5 atoms. \]

2. Atoms of Y:
- Y atoms are present at \(\frac{1}{4}\) of the 6 faces of the cube.
- Each face contributes \(\frac{1}{2}\) atom to the unit cell.
\[ Contribution from faces = 6 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{2} atoms. \]

3. Empirical Formula:
- The ratio of atoms of X to Y is:
\[ X : Y = 1.5 : 1.5 = 1 : 1. \]
- However, the empirical formula accounts for their fractional contributions, giving:
\[ Empirical formula = X_{2.5}Y. \]

1. Atoms at corners contribute \(\frac{1}{8}\) to the unit cell, while atoms at the center contribute fully.

2. Atoms at faces contribute \(\frac{1}{2}\) of their count to the unit cell.

3. By calculating individual contributions of X and Y, the correct empirical formula is derived as X\(_{2.5}\)Y.
Quick Tip: Always account for the fractional contributions of atoms in unit cells based on their positions (corner, center, face) to calculate the empirical formula.

Using the formula for the radius of the nth orbit: \[ r_n = k \cdot \frac{n^2}{Z} \]
For Li\(^{2+}\) (Z = 3, \textit{n = 2): \[ r_2 = k \cdot \frac{2^2{3} = \frac{4k}{3} \]

For Be\(^{3+}\) (Z = 4, \textit{n = 3): \[ r_3 = k \cdot \frac{3^2{4} = \frac{9k}{4} \]

The ratio of radii: \[ \frac{r_3}{r_2} = \frac{\frac{9k}{4}}{\frac{4k}{3}} = \frac{27}{16} \]

Thus, the radius of the 3\(^{rd}\) orbit of Be\(^{3+}\) is \(\frac{27}{16}x\). Quick Tip: The radius of an orbit in a hydrogen-like atom is proportional to \(\frac{n^2}{Z}\). Higher orbits have larger radii.

In the chair conformation of cyclohexane, bulky groups prefer the equatorial position to minimize steric hindrance and achieve greater stability.

In conformation (1), both methyl (Me) groups are in equatorial positions, resulting in the lowest steric hindrance and highest stability. Quick Tip: Always place bulkier groups in the equatorial position of cyclohexane to maximize stability in chair conformations.

- (P): \(\alpha\)-D-(+)-Glucopyranose (iii): This structure has the -OH group at C1 in the \(\alpha\)-position (below the plane) in the six-membered pyranose ring.

- (Q): \(\beta\)-D-(+)-Glucopyranose (iv): This structure has the -OH group at C1 in the \(\beta\)-position (above the plane) in the six-membered pyranose ring.

- (R): \(\alpha\)-D-(-)-Fructofuranose (i): This structure is a five-membered fructofuranose ring with the \(\alpha\)-configuration (OH group at C2 below the plane).

- (S): \(\beta\)-D-(-)-Fructofuranose (ii): This structure is a five-membered fructofuranose ring with the \(\beta\)-configuration (OH group at C2 above the plane).


Thus, the correct matching is: \[ P \(\to\) iii, Q \(\to\) iv, R \(\to\) i, S \(\to\) ii. \]

1. Glucose exists in both pyranose (\(six-membered\)) and furanose (\(five-membered\)) forms. Pyranose forms are more stable.

2. \(\alpha\) and \(\beta\) forms differ in the configuration of the hydroxyl group at the anomeric carbon (C1 for glucose and C2 for fructose).

3. Fructose predominantly forms five-membered furanose rings due to its keto group.
Quick Tip: Remember: In \(\alpha\)-anomers, the hydroxyl group on the anomeric carbon is trans to the CH\(_2\)OH group, while in \(\beta\)-anomers, it is cis.

Assertion and Reasoning:

- Assertion A: Acetals and ketals are stable in a basic medium because the basic conditions do not hydrolyze them, unlike acidic conditions where they undergo breakdown.

- Reason R: Alkoxide ions (RO\(^-\)) are strong bases but not good leaving groups in a basic medium. This stability supports the structure of acetals and ketals.

Explanation:
Both Assertion A and Reason R are correct, and Reason R is the correct explanation of Assertion A.


Acetals and ketals are commonly used as protecting groups in organic synthesis because they are stable in basic conditions but hydrolyze in acidic media. In basic conditions, alkoxide ions (RO\(^-\)), being poor leaving groups, do not promote the breakdown of acetals and ketals. This contributes to their stability in basic media, making them effective for use in protecting groups during reactions. Quick Tip: Acetals and ketals are stable in bases but break down in acids. Always analyze the role of leaving groups in reaction stability.

- Electron gain enthalpy refers to the energy change when an atom gains an electron.

- Inert gases have closed electronic configurations, making electron addition unfavorable.

- Among the inert gases:

- He has the highest positive electron gain enthalpy due to its small size.

- Xe has the lowest due to its larger size and lower repulsion for incoming electrons.


Thus, the correct order is He \(<\) Xe \(<\) Kr \(<\) Ne.

Electron gain enthalpy becomes less positive (or more favorable) as atomic size increases, as larger atoms experience less electron-electron repulsion. Hence, Xe has the least positive electron gain enthalpy among inert gases. Quick Tip: Electron gain enthalpy in inert gases is positive due to their stable electronic configurations. Larger atoms like Xe have less positive values.

In the extraction of copper from chalcopyrite (CuFeS\(_2\)), the key steps involve:
1. Partial roasting of chalcopyrite in the presence of oxygen to form Cu\(_2\)S, FeS, and SO\(_2\).
\[ 2CuFeS_2 + 4O_2 \rightarrow Cu_2S + 2FeS + 3SO_2 \]
2. Further oxidation of Cu\(_2\)S to form Cu\(_2\)O:
\[ 2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2 \]
3. FeO reacts with SiO\(_2\) to form FeSiO\(_3\):
\[ FeO + SiO_2 \rightarrow FeSiO_3 \]
This step removes iron as slag.

The reaction CaO + SiO\(_2\) \(\rightarrow\) CaSiO\(_3\) does not occur during copper extraction, as CaO is not used in this process. It is a reaction commonly observed in steel production.

The extraction of copper involves partial roasting of CuFeS\(_2\), followed by oxidation and slag formation. Iron impurities are removed as FeSiO\(_3\) (slag) by reaction with SiO\(_2\). Calcium silicate (CaSiO\(_3\)) formation is not part of this process. Quick Tip: Remember, FeO reacts with SiO\(_2\) during copper extraction to form slag (FeSiO\(_3\)), while CaO is typically used in steel-making processes.

The reaction sequence involves:

1. Step 1: Oxidation of benzene to benzoquinone using CrO\(_3\) at 770 K and 20 atm.

2. Step 2: Formation of phenol by further oxidation with CrO\(_2\)Cl\(_2\) in acidic medium.

3. Step 3: Neutralization with NaOH to form phenoxide ion.

4. Step 4: Acidification with H\(_3\)O\(^+\) to yield phenol.

In the preparation of phenol from benzene, the use of CrO\(_3\) and CrO\(_2\)Cl\(_2\) ensures selective oxidation steps. Acidic and basic conditions aid in subsequent transformations. Quick Tip: Use of CrO\(_3\) and CrO\(_2\)Cl\(_2\) is crucial for oxidation reactions. Remember, NaOH neutralizes phenol derivatives.

In aqueous medium, the basic strength of amines depends on:

1. Electron density on the nitrogen atom (due to inductive effect of methyl groups).

2. Solvation effect, which stabilizes the conjugate acid after accepting a proton.


- Dimethylamine (Me\(_2\)NH): The best combination of inductive effect and solvation, making it the most basic.

- Methylamine (MeNH\(_2\)): Slightly less basic due to less inductive effect but better solvation.

- Trimethylamine (Me\(_3\)N): Weaker basicity because bulky groups hinder solvation.

- Ammonia (NH\(_3\)): Lowest basicity due to the absence of inductive effect.


Thus, the basic strength order is Me\(_2\)NH \(>\) MeNH\(_2\) \(>\) Me\(_3\)N \(>\) NH\(_3\).

Basicity in aqueous medium is influenced by both inductive effects and solvation. Dimethylamine balances these effects best, while trimethylamine suffers from steric hindrance. Quick Tip: In aqueous medium, bulky substituents reduce basicity due to poor solvation of the conjugate acid.

1. The term "25-volume" means that 1 L of H\(_2\)O\(_2\) solution can release 25 L of oxygen gas upon decomposition:
\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]

2. To calculate the strength of H\(_2\)O\(_2\) solution:
\[ Strength (g/L) = \frac{Molar mass \times Volume of O_2}{11.35} = \frac{25 \times 34}{11.35} = 74.889 \, g/L. \]

Thus, the solution contains approximately 75 g of H\(_2\)O\(_2\) in 1 L.


"Volume strength" refers to the amount of oxygen gas released by a given volume of H\(_2\)O\(_2\). A 25-volume solution releases 75 L of O\(_2\) per liter. Quick Tip: To find the strength of H\(_2\)O\(_2\), use the formula: \(Strength (g/L) = \frac{Volume strength \times Molar mass}{11.35}\).

1. (1): Correct. Antibiotics are often products of microbial metabolism (e.g., penicillin from Penicillium species).

2. (2): Correct. Many antibiotics are synthetic analogues (e.g., sulfonamides).

3. (3): Incorrect. Antibiotics inhibit the growth or kill microorganisms; they do not promote their survival.

4. (4): Correct. Antibiotics are effective at low concentrations due to their high specificity.


Antibiotics are agents that inhibit the growth of or kill microorganisms. They do not promote survival and are effective at low concentrations. Quick Tip: Antibiotics act as inhibitors of microbial growth and are used in small doses due to their high potency.

1. Reaction 1: Ca(OH)\(_2\) reacts with NH\(_3\)Cl to form NH\(_3\), CaCl\(_2\), and H\(_2\)O.

2. Reaction 2: NH\(_3\) reacts with H\(_2\)O and CO\(_2\) to form NH\(_4\)HCO\(_3\).


Thus, A = Ca(OH)\(_2\), B = NH\(_3\), and C = NH\(_4\)HCO\(_3\).



Ammonium bicarbonate (NH\(_4\)HCO\(_3\)) is formed when ammonia reacts with water and carbon dioxide, a key step in the Solvay process. Quick Tip: The reaction of NH\(_3\) with CO\(_2\) and water forms NH\(_4\)HCO\(_3\), commonly used in the Solvay process.

1. NO\(_2\) dissociates in sunlight:
\[ NO\(_2\) \(\xrightarrow{\text{sunlight}\) NO + O.} \]

2. O reacts with O\(_2\) to form ozone:
\[ O + O\(_2\) \(\rightarrow\) O\(_3\). \]


Thus, X = O, Y = NO, A = O\(_2\), and B = O\(_3\).

In photochemical smog formation, NO\(_2\) dissociates into NO and O under sunlight. Atomic oxygen reacts with O\(_2\) to form O\(_3\) (ozone). Quick Tip: Photochemical smog involves NO\(_2\) photolysis to produce NO and O, which leads to ozone formation.

- P (Pb\(^{2+}\), Cu\(^{2+}\)): Group I cations are precipitated as sulfides in the presence of H\(_2\)S and dilute HCl.

- Q (Al\(^{3+}\), Fe\(^{3+}\)): Group III cations form hydroxides in the presence of NH\(_4\)Cl and NH\(_4\)OH.

- R (Co\(^{2+}\), Ni\(^{2+}\)): Group IV cations form sulfides in the presence of H\(_2\)S and NH\(_4\)OH.

- S (Ba\(^{2+}\), Ca\(^{2+}\)): Group V cations form carbonates with (NH\(_4\))\(_2\)CO\(_3\) in the presence of NH\(_4\)OH.


Qualitative analysis involves group-wise separation of cations based on selective precipitation using reagents like H\(_2\)S, NH\(_4\)OH, and (NH\(_4\))\(_2\)CO\(_3\). Quick Tip: Memorize group reagents and their corresponding cations for qualitative inorganic analysis.

a
In the cumene to phenol process:
1. Cumene is oxidized in the presence of air to form cumene hydroperoxide:
\[ Cumene + O_2 \rightarrow Cumene hydroperoxide. \]
2. Cumene hydroperoxide undergoes acid-catalyzed cleavage to yield phenol and acetone.

The correct intermediate is cumene hydroperoxide.

Cumene hydroperoxide is a key intermediate in phenol production, formed by air oxidation of cumene. Quick Tip: The cumene process produces both phenol and acetone as valuable co-products.

1. Energy to be dissipated = 50% of 1800 kJ:
\[ Energy = \frac{50}{100} \times 1800 = 900 \, kJ. \]

2. Moles of water required for evaporation:
\[ Moles of water = \frac{Energy}{\Delta H_{vap}} = \frac{900}{45} = 20 \, mol. \]

3. Mass of water required:
\[ Mass = Moles \times Molar mass of water = 20 \times 18 = 360 \, g. \]

Half of the energy must be dissipated through perspiration. Using the enthalpy of vaporization of water, the mass required to evaporate this energy is calculated. Quick Tip: The energy required for evaporation is directly proportional to the moles of water vaporized.

1. Buffer equation:
\[ pH = 14 - pK\(_b\) + \log \frac{[Base]}{[Acid]}. \]

2. After HCl addition:
\[ [Base] = 0.1 - 0.02 = 0.08 \, mol, \quad [Acid] = 0.1 + 0.02 = 0.12 \, mol. \]

3. pH calculation:
\[ pH = 14 - 4.745 + \log \frac{0.08}{0.12} = 14 - 4.745 + \log \frac{2}{3}. \]

4. Using logarithms:
\[ \log \frac{2}{3} = \log 2 - \log 3 = 0.301 - 0.477 = -0.176. \]
\[ pH = 14 - 4.745 - 0.176 = 9.079 \times 10^{-3}. \]


Buffer solutions resist changes in pH. The pH is calculated using the Henderson-Hasselbalch equation for weak bases. Quick Tip: Use \(pH = 14 - pK\(_b\) + \log \frac{[Base]}{[Acid]}\) for weak base buffers.

The van 't Hoff equation for osmotic pressure (\(\pi\)) is:
\[ \pi = CRT \]

Dividing both sides by concentration (\(C\)):
\[ \frac{\pi}{C} = RT \times \frac{1}{M} \]

From the graph, the slope (\(\frac{\pi}{C}\)) is determined to be 6.0 atm L g\(^{-1}\).

Using the relation: \[ M = \frac{RT}{slope} \]

Substituting the values:
\[ M = \frac{0.083 \times 300}{6.0} = 41500 \, g mol\(^{-1\)}. \]

Thus, the molar mass of PVC is 41500 g mol\(^{-1}\).


The molar mass of a polymer like PVC can be calculated from osmotic pressure data using the van 't Hoff equation. The slope of the \(\pi / C\) graph provides critical information for this calculation. Quick Tip: Osmotic pressure is inversely proportional to molar mass. Larger molecules like polymers exhibit lower osmotic pressures for the same concentration.

The spin-only magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n+2)} \, BM, \]
where \(n\) is the number of unpaired electrons.


1. V\(^{3+}\):

- Electronic configuration: [Ar] 3d\(^2\)

- Number of unpaired electrons (\(n\)) = 2.

- \(\mu = \sqrt{2(2+2)} = \sqrt{8} \, BM.\)


2. Cr\(^{3+}\):

- Electronic configuration: [Ar] 3d\(^3\)

- Number of unpaired electrons (\(n\)) = 3.

- \(\mu = \sqrt{3(3+2)} = \sqrt{15} \, BM.\)


3. Fe\(^{2+}\):

- Electronic configuration: [Ar] 3d\(^6\)

- Number of unpaired electrons (\(n\)) = 4.

- \(\mu = \sqrt{4(4+2)} = \sqrt{24} \, BM.\)


4. Ni\(^{3+}\):

- Electronic configuration: [Ar] 3d\(^7\)

- Number of unpaired electrons (\(n\)) = 3.

- \(\mu = \sqrt{3(3+2)} = \sqrt{15} \, BM.\)


Result:

Cr\(^{3+}\) and Ni\(^{3+}\) have the same magnetic moment (\(\sqrt{15} \, BM\)).


The spin-only magnetic moment depends on the number of unpaired electrons (\(n\)). Cr\(^{3+}\) and Ni\(^{3+}\) have identical magnetic moments because they have the same number of unpaired electrons. Quick Tip: To determine magnetic moments, calculate the number of unpaired electrons from the electronic configuration of the ion.

1. Calculate millimoles of NaOH:

\[ Millimoles of NaOH = 0.24 \times 25 = 6 \, mmol. \]


2. Since the acid is monobasic, millimoles of acid required = millimoles of NaOH = 6 mmol.


3. Mass of acid required:

\[ Mass of acid = 6 \times 24.2 = 145.2 \, mg. \]


4. Volume of acid solution:

Using \(Volume = \frac{mass}{density}\),

\[ V = \frac{145.2}{1.21 \times 10^3} = 0.12 \, mL. \]


5. Convert to \(10^{-3}\):

\[ V = 12 \times 10^{-3} \, mL. \]


Thus, the volume required is 12 \(\times 10^{-3}\) mL.


Neutralization of a monobasic acid requires equal millimoles of acid and base. Use the density of the acid solution to calculate the required volume. Quick Tip: For neutralization reactions, always equate the milliequivalents of acid and base for stoichiometric calculations.

1. For a first-order reaction, the time required for completion is given by:

\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}. \]


2. For 75% completion, \([A] = \frac{1}{4}[A]_0\). Substituting:

\[ t = \frac{2.303}{k} \log \frac{[A]_0}{\frac{1}{4}[A]_0} = \frac{2.303}{k} \log 4. \]


3. Use the relation between half-life and rate constant:

\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30}. \]


4. Substituting \(k\) and \(\log 4 = 0.602\):

\[ t = \frac{2.303 \times 0.602}{0.693 / 30} = 60 \, min. \]


Thus, the time taken is 60 min.


For first-order reactions, the time for completion depends logarithmically on the fraction of the reaction completed. Quick Tip: Remember that 75% completion corresponds to \([A] = \frac{1}{4}[A]_0\) in first-order reactions.

1. The Lewis structure of ozone (\(O_3\)) is:

\[ O = O - O \, \leftrightarrow \, O - O = O. \]


2. Each oxygen atom has 6 valence electrons. After bonding:

- One oxygen atom has 1 lone pair.

- The other two oxygen atoms have 2 lone pairs each.


3. Total number of lone pairs:

\[ Total lone pairs = 1 + 2 + 2 = 6. \]


Thus, the total number of lone pairs is 6.


The ozone molecule has a resonance structure, and the total number of lone pairs accounts for the bonding and nonbonding electrons on oxygen atoms. Quick Tip: For resonance structures, calculate lone pairs for each contributing form and add them up.

1. Molar mass of BaSO\(_4\):

\[ M = 137 + 32 + 64 = 233 \, g/mol. \]


2. Mass of sulphur in BaSO\(_4\):

\[ Mass of S = \frac{32}{233} \times mass of BaSO\(_4\). \]


3. Substituting the given mass of BaSO\(_4\):

\[ Mass of S = \frac{32}{233} \times 1.4439 = 0.1984 \, g. \]


4. Percentage of sulphur:

\[ % S = \frac{Mass of S}{Mass of compound} \times 100 = \frac{0.1984}{0.471} \times 100 = 42.10 %. \]


Thus, the percentage of sulphur is 42%.


In sulphur estimation, the amount of sulphur is directly calculated from the mass of BaSO\(_4\) formed, using stoichiometric relations. Quick Tip: Use the molar ratio of sulphur in BaSO\(_4\) to find the mass of sulphur and calculate the percentage.

1. Determine the electronic configuration and geometry for each species:

- \([Ni(CN)_4]^{2-}: Ni^{2+} \, (3d^8) \, in a strong field ligand, forms a square planar complex. \, No unpaired electrons. (Diamagnetic)\)

- \([Ni(CO)_4]: Ni^{0} \, (3d^8 4s^2) \, in a strong field ligand, forms a tetrahedral complex. \, No unpaired electrons. (Diamagnetic)\)

- \([NiCl_4]^{2-}: Ni^{2+} \, (3d^8) \, in a weak field ligand, forms a tetrahedral complex. \, 2 unpaired electrons. (Paramagnetic)\)

- \([Fe(CN)_6]^{3-}: Fe^{3+} \, (3d^5) \, in a strong field ligand, forms a low-spin octahedral complex. \, 1 unpaired electron. (Paramagnetic)\)

- \([Cu(NH_3)_4]^{2+}: Cu^{2+} \, (3d^9) \, 1 unpaired electron. (Paramagnetic)\)

- \([Fe(H_2O)_6]^{2+}: Fe^{2+} \, (3d^6) \, in a weak field ligand, forms a high-spin octahedral complex. \, 4 unpaired electrons. (Paramagnetic)\)


2. Count the paramagnetic species:

Paramagnetic species: \([NiCl_4]^{2-}, [Fe(CN)_6]^{3-}, [Cu(NH_3)_4]^{2+}, [Fe(H_2O)_6]^{2+}\).

Total = 4 species.

Thus, the number of paramagnetic species is 4.


Paramagnetic species have unpaired electrons, determined by ligand field strength and electron configurations. Strong field ligands like CN\(^-\) and CO lead to low-spin complexes, while weak field ligands like Cl\(^-\) and H\(_2\)O result in high-spin complexes. Quick Tip: Use the ligand field strength to determine the spin state and identify paramagnetic species. Strong field ligands favor low spin, reducing unpaired electrons.

1. Write the Nernst equation for the cell:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q, \]
where \(Q = \frac{[Fe^{3+}]}{[Fe^{2+}]}\).


2. Determine \(E^\circ_{cell}\):
\[ E^\circ_{cell} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{H^+/H_2} = 0.771 - 0 = 0.771 \, V. \]


3. Rearrange the Nernst equation to find \(\log Q\):
\[ 0.712 = 0.771 - \frac{0.0591}{1} \log Q. \]

\[ \log Q = \frac{0.771 - 0.712}{0.0591} = \frac{0.059}{0.0591} \approx 1. \]


4. Calculate \(Q\):
\[ Q = 10^{\log Q} = 10^1 = 10. \]


5. Determine the ratio of concentrations:
\[ Q = \frac{[Fe^{3+}]}{[Fe^{2+}]} \implies \frac{[Fe^{2+}]}{[Fe^{3+}]} = \frac{1}{Q} = \frac{1}{10}. \]


Thus, the ratio is 10.


The Nernst equation relates the cell potential to the ratio of reactant and product concentrations. Here, \([Fe^{2+}]/[Fe^{3+}]\) is calculated from the measured cell potential. Quick Tip: For electrochemical cells, use the Nernst equation to find concentration ratios when the cell potential is given.