Step 1: Understanding the Concept:

We first determine the empirical formula of the hydrocarbon using the given mass ratio of Carbon to Hydrogen.

Using the fact that it contains exactly two carbon atoms, we find its exact molecular formula and molar mass.

Finally, we apply stoichiometry to the combustion reaction to find the mass of \(CO_2\) produced.


Step 2: Key Formula or Approach:

Moles = \(\frac{Given Mass}{Atomic/Molar Mass}\).

Balanced combustion equation: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\).


Step 3: Detailed Explanation:

The given mass ratio of C : H is 12 : 1.

Calculate the mole ratio by dividing by their respective atomic masses:

Moles of C = \(\frac{12}{12} = 1\).

Moles of H = \(\frac{1}{1} = 1\).

The empirical formula is CH.

Since the molecule contains exactly two carbon atoms, its molecular formula must be \(C_2H_2\) (Ethyne).

The molar mass of \(C_2H_2\) is \((2 \times 12) + (2 \times 1) = 26 g/mol\).

The amount of hydrocarbon burned is 3.38 g. Calculate the moles of \(C_2H_2\):
\[ Moles of C_2H_2 = \frac{3.38}{26} = 0.13 mol \]
Write the balanced combustion equation:
\[ C_2H_2 + \frac{5}{2}O_2 \rightarrow 2CO_2 + H_2O \]
From the stoichiometry, 1 mole of \(C_2H_2\) produces 2 moles of \(CO_2\).

So, 0.13 moles of \(C_2H_2\) will produce:
\[ Moles of CO_2 = 2 \times 0.13 = 0.26 mol \]
Now, calculate the mass of the produced \(CO_2\) (Molar mass of \(CO_2 = 44 g/mol\)):
\[ Mass of CO_2 = 0.26 mol \times 44 g/mol = 11.44 g \]

Step 4: Final Answer:

The weight of \(CO_2\) formed is \(11.44 g\).
Quick Tip: To avoid writing the full balanced combustion equation, apply the Principle of Atomic Conservation (POAC) on Carbon: Moles of C atoms in reactant = Moles of C atoms in product. \(2 \times n_{C_2H_2} = 1 \times n_{CO_2}\).

Step 1: Understanding the Concept:

The dissociation of the weak acid \(H_2A\) occurs in a solution already containing a strong acid (\(HCl\)).

Due to the common ion effect, the presence of the strong acid heavily suppresses the dissociation of the weak acid.

We can assume the concentration of \(H^+\) is determined entirely by the \(HCl\), and the concentration of un-ionized \(H_2A\) remains equal to its initial concentration.


Step 2: Key Formula or Approach:

The first dissociation step is: \(H_2A \rightleftharpoons H^+ + HA^-\).

The equilibrium constant expression is: \(K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}\).

Given values: \(K_{a1} = 8.1 \times 10^{-8}\), \([HCl] = 0.1 M\), Initial \([H_2A] = 0.1 M\).


Step 3: Detailed Explanation:

The strong acid completely dissociates:
\(HCl \rightarrow H^+ + Cl^-\)

So, the initial concentration of \(H^+\) is \(0.1 M\).

Let \(x\) be the amount of \(H_2A\) that dissociates. At equilibrium:
\([H_2A] = 0.1 - x \approx 0.1 M\) (since \(x\) is extremely small).
\([H^+] = 0.1 + x \approx 0.1 M\) (common ion effect).
\([HA^-] = x\).

Substitute these equilibrium concentrations into the \(K_{a1}\) expression:
\[ 8.1 \times 10^{-8} = \frac{(0.1) \cdot [HA^-]}{0.1} \]
The \(0.1\) terms cancel out perfectly:
\[ [HA^-] = 8.1 \times 10^{-8} M \]
(Note: The second dissociation step involving \(K_{a2}\) consumes a negligible amount of \(HA^-\) due to its extremely small constant \(1.0 \times 10^{-13}\), so this approximation holds true).


Step 4: Final Answer:

The concentration of \(HA^-\) is \(8.1 \times 10^{-8} M\).
Quick Tip: When a weak acid is placed in a strong acid of the same concentration, the concentration of the first conjugate base \([HA^-]\) is almost exactly equal to \(K_{a1}\).

Step 1: Understanding the Concept:

Two molecules are isostructural if they possess the exact same molecular geometry (shape).

We determine the shape using VSEPR theory by counting the number of bond pairs (bp) and lone pairs (lp) on the central atom.


Step 2: Key Formula or Approach:

Steric number (SN) = (Number of valence electrons on central atom + number of monovalent atoms - cationic charge + anionic charge) / 2.

Find SN and the number of lone pairs to determine hybridization and geometry.


Step 3: Detailed Explanation:

First, analyze the reference molecule \(SF_4\):

Sulfur (S) has 6 valence electrons. It forms 4 single bonds with F.

Remaining electrons = 2 (1 lone pair).

Total electron domains = 4 (bp) + 1 (lp) = 5. Hybridization is \(sp^3d\).

Geometry with 1 lone pair in a trigonal bipyramidal arrangement is "See-saw".


Now analyze the options:

A. \(BrF_4^-\): Bromine (Br) has 7 valence e⁻, +1 for negative charge = 8 e⁻.

4 bonds with F means 4 remaining e⁻ (2 lone pairs).

SN = 4 (bp) + 2 (lp) = 6 (\(sp^3d^2\)). Geometry is "Square planar".


B. \(CH_4\): Carbon (C) has 4 valence e⁻. 4 bonds, 0 lone pairs.

SN = 4 (bp). Geometry is "Tetrahedral".


C. \(IF_4^+\): Iodine (I) has 7 valence e⁻, -1 for positive charge = 6 e⁻.

4 bonds with F means 2 remaining e⁻ (1 lone pair).

SN = 4 (bp) + 1 (lp) = 5 (\(sp^3d\)). Geometry is "See-saw". (Matches!)


D. \(XeF_4\): Xenon (Xe) has 8 valence e⁻.

4 bonds with F means 4 remaining e⁻ (2 lone pairs).

SN = 4 (bp) + 2 (lp) = 6 (\(sp^3d^2\)). Geometry is "Square planar".


E. \(XeO_2F_2\): Xenon (Xe) has 8 valence e⁻.

It forms 2 double bonds with O (using 4 e⁻) and 2 single bonds with F (using 2 e⁻). Total 6 e⁻ used.

Remaining e⁻ = 2 (1 lone pair).

Electron domains = 4 (sigma bonds) + 1 (lp) = 5 (\(sp^3d\)). Geometry is "See-saw". (Matches!)


Both \(IF_4^+\) and \(XeO_2F_2\) have a see-saw structure identical to \(SF_4\).


Step 4: Final Answer:

The correct choice is C and E Only.
Quick Tip: To quickly find isostructural species, identify the number of lone pairs. If two species have the same steric number AND the same number of lone pairs, they are virtually always isostructural.

Step 1: Understanding the Concept:

Internal energy (\(U\)) is a state function. This means the change in internal energy for a round trip (from X to Y and back to X) is exactly zero.

We apply the First Law of Thermodynamics to both the forward and reverse processes to find the unknown work value.


Step 2: Key Formula or Approach:

First Law of Thermodynamics (IUPAC convention): \(\Delta U = q + w\).

Here, \(q\) is positive if heat is absorbed, negative if evolved.
\(w\) is positive if work is done ON the gas, negative if work is done BY the gas.

For a cyclic process: \(\Delta U_{X \to Y} = -\Delta U_{Y \to X}\).


Step 3: Detailed Explanation:

Analyze the forward process (\(X \to Y\)):

Heat absorbed, \(q_1 = +10 J\).

Work done BY the gas means it expanded, so \(w_1 = -18 J\).

Calculate the change in internal energy:
\[ \Delta U_{X \to Y} = q_1 + w_1 = 10 J + (-18 J) = -8 J \]
Analyze the reverse process (\(Y \to X\)):

Since internal energy is a state function:
\[ \Delta U_{Y \to X} = -\Delta U_{X \to Y} = -(-8 J) = +8 J \]
Heat is evolved in this step, so \(q_2 = -6 J\).

Apply the First Law to find the work \(w_2\):
\[ \Delta U_{Y \to X} = q_2 + w_2 \] \[ 8 J = -6 J + w_2 \] \[ w_2 = 8 J + 6 J = +14 J \]
Since \(w_2\) is positive, it means \(14 J\) of work is done ON the gas by the surroundings.


Step 4: Final Answer:

14 J of the work is done on the gas 'A' by the surrounding.
Quick Tip: Strictly adhere to IUPAC sign conventions in Chemistry: Work done BY the system is negative (\(-w\)), Work done ON the system is positive (\(+w\)). This contrasts with standard Physics conventions.

Step 1: Understanding the Concept:

Osmotic pressure depends directly on the molar concentration (Molarity) of the solution.

If two solutions have the identical concentration and temperature, they will have identical osmotic pressures.

Furthermore, mixing two solutions of the exact same concentration will result in a final mixture with that same constant concentration.


Step 2: Key Formula or Approach:

Molarity \(C = \frac{moles of solute}{Volume of solution in L} = \frac{mass / Molar mass}{V}\).

Osmotic pressure \(\pi = CRT\).


Step 3: Detailed Explanation:

Analyze Solution A:

Mass \(m_A = 1 g\), Volume \(V_A = 0.5 L\).

Concentration \(C_A = \frac{1 / 50000}{0.5} = \frac{2}{50000} = 4 \times 10^{-5} mol/L\).

Calculate the osmotic pressure \(x\):
\[ x = \pi_A = C_A R T = (4 \times 10^{-5}) \times 0.083 \times 300 \] \[ x = 1200 \times 10^{-5} \times 0.083 = 99.6 \times 10^{-5} = 9.96 \times 10^{-4} bar \]
Analyze Solution B:

Mass \(m_B = 2 g\), Volume \(V_B = 1 L\).

Concentration \(C_B = \frac{2 / 50000}{1} = \frac{2}{50000} = 4 \times 10^{-5} mol/L\).

Since \(C_A = C_B\) and temperature is the same, the osmotic pressure is identical:
\[ y = \pi_B = 9.96 \times 10^{-4} bar \]
Analyze the Mixed Solution:

Total mass = \(1 g + 2 g = 3 g\).

Total volume = \(0.5 L + 1 L = 1.5 L\).

Concentration \(C_{mix} = \frac{3 / 50000}{1.5} = \frac{2}{50000} = 4 \times 10^{-5} mol/L\).

Since the final concentration remains identical, the final osmotic pressure \(z\) is also identical.
\[ z = \pi_{mix} = 9.96 \times 10^{-4} bar \]

Step 4: Final Answer:

The values are \(9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}\).
Quick Tip: Before doing any complex calculations, always compare the ratio of mass to volume for both solutions. If \(m_1/v_1 = m_2/v_2\), the concentrations are equal, meaning mixing them changes nothing about intensive properties like concentration or osmotic pressure.

Step 1: Understanding the Concept:

Part (a) is a simple weak acid pH calculation before any base is added.

Part (b) describes a partial neutralization. Since we add exactly half the volume of base needed to fully neutralize the acid, we create an equimolar buffer solution of the weak acid and its conjugate base.


Step 2: Key Formula or Approach:

For a weak acid with \(\alpha \ll 1\): \([H^+] = \sqrt{K_a \cdot C}\) and \(pH = \frac{1}{2}(pK_a - \log C)\).

For a buffer solution, Henderson-Hasselbalch equation: \(pH = pK_a + \log\frac{[Salt]}{[Acid]}\).


Step 3: Detailed Explanation:

(a) At the start of the titration:

Concentration of HX, \(C = 0.2 M\).

Calculate the \([H^+]\) concentration:
\[ [H^+] = \sqrt{K_a \cdot C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{1.0 \times 10^{-4}} = 10^{-2} M \] \[ pH = -\log(10^{-2}) = 2.0 \]
(b) After adding 10 mL of 0.2 M NaOH:

Initial millimoles of HX = \(20.0 mL \times 0.2 M = 4.0 mmol\).

Millimoles of NaOH added = \(10.0 mL \times 0.2 M = 2.0 mmol\).

The strong base reacts completely with the weak acid:
\(HX + NaOH \rightarrow NaX + H_2O\)

Remaining HX = \(4.0 - 2.0 = 2.0 mmol\).

Formed NaX (salt) = \(2.0 mmol\).

Since the millimoles of the weak acid and its conjugate base are equal (\([Salt] = [Acid]\)), this is the half-equivalence point.

Using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{2.0}{2.0}\right) = 3.3 + \log(1) = 3.3 + 0 = 3.3 \]

Step 4: Final Answer:

The pH values are 2.0 and 3.3 respectively.
Quick Tip: At the exact half-equivalence point of any weak acid-strong base titration, the pH is always perfectly equal to the \(pK_a\) of the weak acid. Identifying this saves you from doing buffer math.

Step 1: Understanding the Concept:

The order of a chemical reaction can be deduced directly from the units of its rate constant (\(k\)).

Once the order is identified, all standard kinetic properties (rate law, half-life formula, and graphical relationships) are fixed.


Step 2: Key Formula or Approach:

General unit of rate constant: \(M^{1-n}s^{-1}\) or \((mol L^{-1})^{1-n}s^{-1}\), where \(n\) is the order.

For 2nd order: Rate = \(k[X]^2\) and \(t_{1/2} = \frac{1}{k[X]_0}\).


Step 3: Detailed Explanation:

Analyze the given rate constant:
\(k = 1 \times 10^{-3} mol^{-1} L s^{-1} = 1 \times 10^{-3} M^{-1}s^{-1}\).

Equating units: \(1 - n = -1 \implies n = 2\). So, this is a second-order reaction.

Now evaluate the statements:

A. Rate = \(k[X]^2\). If \([X] \to 4[X]\), Rate \(\to k(4[X])^2 = 16k[X]^2\). The rate becomes 16 times. (TRUE)

B. As derived from the units, the reaction is a second-order reaction. (TRUE)

C. For a 2nd order reaction, \(t_{1/2} = \frac{1}{k[X]_0}\). It is inversely dependent on the initial concentration, not independent. (FALSE)

D. The thermal decomposition of \(N_2O_5\) is a well-known first-order reaction. (FALSE)

E. A plot of \(\ln([R]_0 / [R])\) vs time yielding a straight line represents the integrated rate law for a first-order reaction. For a 2nd order reaction, the linear plot is \(1/[R]\) vs time. (FALSE)

Therefore, only statements A and B are correct.


Step 4: Final Answer:

The correct choice is A and B Only.
Quick Tip: Always memorize the generic formula for rate constant units: \((concentration)^{1-n}(time)^{-1}\). It allows instant identification of the reaction order without any experimental data tables.

Step 1: Understanding the Concept:

Oxides are classified based on their acid-base characteristics:

- Basic: Typically metal oxides of groups 1 & 2 (e.g., \(Na_2O\), \(K_2O\)).

- Acidic: Typically non-metal oxides with high oxidation states (e.g., \(Cl_2O_7\), \(SO_3\)).

- Amphoteric: Certain metal/metalloid oxides that react with both acids and bases (e.g., \(Al_2O_3\), \(ZnO\), \(As_2O_3\)).

- Neutral: Non-metal oxides that react with neither (strictly \(CO\), \(NO\), \(N_2O\)).


Step 2: Key Formula or Approach:

We must systematically evaluate each option to find the one containing exactly one of each class.


Step 3: Detailed Explanation:

Let's analyze the options:

Option (A): \(Na_2O\) (Basic), \(K_2O\) (Basic), \(Al_2O_3\) (Amphoteric), \(As_2O_3\) (Amphoteric).

*Result: Missing Acidic and Neutral oxides.*

Option (B): \(Al_2O_3\) (Amphoteric), \(As_2O_3\) (Amphoteric), \(CO\) (Neutral), \(NO\) (Neutral).

*Result: Missing Basic and Acidic oxides.*

Option (C): \(K_2O\) (Basic), \(Cl_2O_7\) (Acidic), \(As_2O_3\) (Amphoteric), \(NO\) (Neutral).

*Result: Contains exactly one of each kind!*

Option (D): \(Na_2O\) (Basic), \(N_2O\) (Neutral), \(Al_2O_3\) (Amphoteric), \(CO\) (Neutral).

*Result: Missing Acidic oxides.*


Step 4: Final Answer:

The correct set is \(K_2O\), \(Cl_2O_7\), \(As_2O_3\) and \(NO\).
Quick Tip: There are only three neutral oxides you must strictly memorize for exams: \(CO\), \(NO\), and \(N_2O\). If a set asks for a neutral oxide, it MUST contain one of these three.

Step 1: Understanding the Concept:

Ionization enthalpy generally decreases down a group due to increased atomic size.

However, deviations occur in the p-block due to poor shielding by d and f-orbitals (d-block contraction and lanthanide contraction), increasing the effective nuclear charge (\(Z_{eff}\)) on valence electrons.


Step 2: Key Formula or Approach:

Evaluate Group 13 second IE anomalies (B, Al, Ga).

Evaluate Group 14 first IE anomalies (Si, Ge, Sn, Pb).


Step 3: Detailed Explanation:

Evaluate Statement I:

We are looking at the Second Ionization Energy (IE\(_2\)) of Group 13 elements.

B (Group 13, Period 2): \(B^+\) is \(1s^2 2s^2\). Removing an electron from the small \(2s\) orbital requires very high energy.

Al (Group 13, Period 3): \(Al^+\) is \([Ne] 3s^2\).

Ga (Group 13, Period 4): \(Ga^+\) is \([Ar] 3d^{10} 4s^2\).

Because the \(3d\) electrons in Gallium shield the nucleus very poorly, the \(4s\) electrons experience a significantly higher effective nuclear charge compared to Aluminum. Thus, it is harder to remove the \(4s\) electron from \(Ga^+\) than the \(3s\) electron from \(Al^+\).

The actual order of IE\(_2\) is \(B > Ga > Al\).

The statement claims \(B > Al > Ga\), which is FALSE.


Evaluate Statement II:

We are looking at the First Ionization Energy (IE\(_1\)) of Group 14 elements.

The general trend is a decrease down the group: \(C > Si > Ge > Sn\).

However, for Lead (Pb), the presence of filled \(4f\) and \(5d\) orbitals causes severe lanthanide contraction. The very poor shielding heavily increases \(Z_{eff}\), making Pb's IE\(_1\) slightly higher than Sn's.

The actual order of IE\(_1\) is \(C > Si > Ge > Pb > Sn\).

The statement claims \(Si < Ge < Pb < Sn\), which asserts a completely reversed and incorrect trend.

Therefore, Statement II is FALSE.


Step 4: Final Answer:

Both Statement I and Statement II are false.
Quick Tip: Always watch out for Gallium (Ga) and Lead (Pb) in ionization energy trend questions. The poor shielding of d-orbitals makes Ga abnormally high compared to Al, and f-orbitals make Pb abnormally high compared to Sn.

Step 1: Understanding the Concept:

Statement I relies on the stability of electronic configurations. The Third Ionization Energy (IE\(_3\)) involves removing an electron from a \(+2\) cation. High stability of the \(+2\) ion means a high IE\(_3\).

Statement II relies on Crystal Field Stabilization Energy (CFSE). CFSE depends on the oxidation state of the central metal and the field strength of the ligands.


Step 2: Key Formula or Approach:

Write electronic configurations for the \(+2\) ions to assess IE\(_3\).

For CFSE (\(\Delta_o\)): It increases with higher oxidation state of the central metal ion and with stronger field ligands (Spectrochemical series).


Step 3: Detailed Explanation:

Evaluate Statement I:

Sc: \([Ar] 3d^1 4s^2 \implies Sc^{2+}\) is \([Ar] 3d^1\). Removing the 3rd electron leaves a highly stable noble gas core (\(Ar\)). Thus, its IE\(_3\) is exceptionally low (lowest in the 3d series).

Zn: \([Ar] 3d^{10} 4s^2 \implies Zn^{2+}\) is \([Ar] 3d^{10}\). This is a completely filled, extremely stable d-subshell. Disrupting this full shell requires enormous energy. Thus, its IE\(_3\) is exceptionally high (highest in the 3d series).

Statement I is TRUE.


Evaluate Statement II:

We compare the Crystal Field Splitting Energy (\(\Delta_o\)) of three complexes.

1) \([Co(H_2O)_6]^{2+}\): Cobalt is in +2 oxidation state. \(H_2O\) is a weak field ligand.

2) \([Co(H_2O)_6]^{3+}\): Cobalt is in +3 oxidation state. A higher oxidation state strongly pulls the ligands closer, significantly increasing the splitting energy (\(\Delta_o\)) compared to the +2 state, even with the same weak ligand.

3) \([Co(en)_3]^{3+}\): Cobalt is in +3 oxidation state, but Ethylenediamine (\(en\)) is a much stronger field ligand than \(H_2O\) (as per the spectrochemical series). This results in the highest splitting energy.

Therefore, the order of CFSE is: \([Co(H_2O)_6]^{2+} < [Co(H_2O)_6]^{3+} < [Co(en)_3]^{3+}\).

Statement II is TRUE.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: A \(d^0\), \(d^5\), or \(d^{10}\) configuration signifies an impending massive jump in ionization energy. For CFSE, remember the priority: Oxidation state of central metal affects \(\Delta_o\) more dramatically than moderate ligand changes.

Step 1: Understanding the Concept:

Coordination isomerism arises from the interchange of ligands between the cationic and anionic entities of different metal ions present in a complex salt.

To form a valid coordination isomer, exchanging the ligands must result in a new pair of complex cation and complex anion without forming a neutral (non-salt) complex.


Step 2: Key Formula or Approach:

Analyze each given pair:

1) Ensure that swapping ligands doesn't result in a neutral molecule (which would stop it from being a complex salt).

2) Identify complexes with different metal centers, as they represent the clearest and most standard examples of coordination isomerism (e.g., swapping all ligands between two different metals gives a distinctly new compound).


Step 3: Detailed Explanation:

Let's evaluate the given complexes:

A. \([Ag(NH_3)_2][Ag(CN)_2]\): Exchanging one ligand gives \([Ag(NH_3)(CN)]\), which is a neutral molecule. Therefore, it cannot exist as a complex salt isomer.

B. \([Co(NH_3)_6][Cr(CN)_6]\): Exchanging all ligands gives \([Cr(NH_3)_6][Co(CN)_6]\), which is a clear coordination isomer. (Valid)

C. \([Co(NH_3)_6][Co(CN)_6]\): While it can form partial exchange isomers like \([Co(NH_3)_5(CN)][Co(NH_3)(CN)_5]\), standard textbook definitions heavily emphasize the interchange of ligands between different metal ions for classical coordination isomerism.

D. \([Fe(NH_3)_6][Co(CN)_6]\): Exchanging ligands gives \([Co(NH_3)_6][Fe(CN)_6]\). (Valid)

E. \([Co(NH_3)_6][Fe(CN)_6]\): This is exactly the coordination isomer of D. If D shows it, E also inherently represents a system capable of it. (Valid)

Since D and E are coordination isomers of each other, they must appear together in the correct option. The only option pairing D, E, and another valid complex with differing metals is B, D, and E.


Step 4: Final Answer:

The complexes that show coordination isomerism strictly are B, D, and E.
Quick Tip: If two options (like D and E) are literally coordination isomers of each other, they must always be grouped together in the correct multiple-choice option. Finding this pair immediately narrows down your choices.

Step 1: Understanding the Concept:

In Liebig's combustion method, all the carbon in the organic compound is converted to \(CO_2\), and all the hydrogen is converted to \(H_2O\).

By knowing the mass of \(CO_2\) produced, we can calculate the exact mass of carbon present. Using the given percentage of carbon, we can then find the total mass \(X\) of the original compound.


Step 2: Key Formula or Approach:

Mass of Carbon = \(\frac{12}{44} \times Mass of CO_2\).

Percentage of Carbon = \(\left( \frac{Mass of C}{X} \right) \times 100\).

Rearranging gives \(X = \frac{Mass of C}{Percentage of C} \times 100\).


Step 3: Detailed Explanation:

First, calculate the mass of carbon in the 0.25 g of \(CO_2\) produced:
\[ Mass of C = \frac{12}{44} \times 0.25 g \] \[ Mass of C = \frac{3}{11} \times \frac{1}{4} = \frac{3}{44} g \approx 0.06818 g \]
We are given that the percentage of carbon in the original compound is 25%.
\[ 25 = \left( \frac{3/44}{X} \right) \times 100 \] \[ \frac{1}{4} = \frac{3}{44X} \] \[ 44X = 12 \implies X = \frac{12}{44} = \frac{3}{11} g \]
Convert this value into decimal form:
\[ X \approx 0.272727\dots g \]
The question asks for the answer in the format \(X = blank \times 10^{-3} g\).
\[ 0.2727 g = 272.7 \times 10^{-3} g \]
Rounding to the nearest integer, we get 273.

(Verification with Hydrogen: Mass of H = \(\frac{2}{18} \times 0.12 = 0.0133 g\). %H = \(\frac{0.0133}{0.2727} \times 100 \approx 4.89%\), which perfectly matches the given data).


Step 4: Final Answer:

The value of X is \(273 \times 10^{-3} g\).
Quick Tip: You only need to use one of the elemental percentages (Carbon or Hydrogen) to find the total mass. Doing the calculation with Carbon is mathematically cleaner because \(25%\) simplifies nicely to \(1/4\).

Step 1: Understanding the Concept:

Carbocations (electron-deficient species) are highly stabilized by electron-donating groups (+R or +I effects).

Carbanions (electron-rich species) are highly stabilized by electron-withdrawing groups (-R or -I effects).

We must evaluate the nature of the \(-OCH_3\) and \(-NO_2\) groups attached to the aromatic rings.


Step 2: Key Formula or Approach:
\(-OCH_3\) has lone pairs on oxygen, allowing it to donate electron density into the ring via resonance (+R effect).
\(-NO_2\) is strongly electronegative and has a formal positive charge on nitrogen, allowing it to withdraw electron density from the ring via resonance (-R effect).


Step 3: Detailed Explanation:

Evaluate Statement I:

The molecule features a central carbocation (\(CH^+\)). The para-methoxy group (\(-OCH_3\)) pushes electron density towards the carbocation through the benzene ring via the +R (resonance) effect. This delocalizes the positive charge and stabilizes the carbocation. Statement I is TRUE.

Evaluate Statement II:

The molecule features a central carbanion (\(CH^-\)). The para-nitro group (\(-NO_2\)) pulls the excess electron density away from the carbanion into its own oxygen atoms via the -R (resonance) effect. This delocalizes the negative charge and stabilizes the carbanion. Statement II is TRUE.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: Always remember: "Like stabilizes like." Electron-donating groups (+R) stabilize positive charges, while electron-withdrawing groups (-R) stabilize negative charges.

Step 1: Understanding the Concept:

We must identify a hydrocarbon that undergoes three specific standard reactions: Isomerization, Aromatization, and Cyclization.

These are hallmark reactions of straight-chain alkanes (paraffins) in the petroleum industry (reforming processes).


Step 2: Key Formula or Approach:

Isomerization: Normal alkanes heat with \(AlCl_3/HCl\) to form branched alkanes.

Aromatization: Alkanes with 6 or more carbons heat over catalysts (\(Cr_2O_3, V_2O_5, Mo_2O_3\)) at high temp/pressure to form aromatic rings.

Carbon counting: The products dictate exactly how many carbon atoms are in the parent chain.


Step 3: Detailed Explanation:

Let's analyze the products to determine the carbon count of (X):

(i) 2,4-dimethyl pentane has \(5\) (pentane) \(+ 2\) (dimethyl) \(= 7\) carbon atoms.

(ii) Toluene (\(C_6H_5CH_3\)) has \(6 + 1 = 7\) carbon atoms.

(iii) Methyl cyclohexane (\(C_6H_{11}CH_3\)) has \(6 + 1 = 7\) carbon atoms.

Since all products contain 7 carbon atoms, the starting compound (X) must be a 7-carbon straight-chain alkane.

The presence of reagents like \(AlCl_3/HCl\) for branching definitively points to an alkane. Alkenes and trienes would undergo addition reactions rather than standard skeletal isomerization under these specific conditions.

Therefore, compound (X) is strictly n-heptane.


Step 4: Final Answer:

The correct name of the compound is Heptane.
Quick Tip: Catalytic reforming (aromatization) converts n-hexane to benzene and n-heptane to toluene. Knowing this standard industrial reaction provides an instant shortcut to the answer.

Step 1: Understanding the Concept:

Evaluate each statement based on standard principles of haloalkane and haloarene chemistry, specifically focusing on substitution vs. elimination conditions, carbocation stability, and phenol reactivity.


Step 2: Key Formula or Approach:

- Aqueous KOH favors Nucleophilic Substitution (\(S_N\)) to form alcohols. Alcoholic KOH favors Elimination (E2) to form alkenes.

- \(S_N1\) reactivity depends solely on the stability of the intermediate carbocation.

- Aryl and vinyl halides are unreactive towards \(S_N\) due to partial double bond character from resonance.


Step 3: Detailed Explanation:

A. TRUE. Alcoholic KOH provides the ethoxide ion (\(C_2H_5O^-\)), which acts as a strong base rather than a nucleophile due to the less polar nature of the alcohol solvent, heavily favoring dehydrohalogenation (elimination).

B. FALSE. The carbocation formed from \(C_6H_5-CHCl-C_6H_5\) is diphenylmethyl (\(Ph_2CH^+\)), which is stabilized by extensive resonance from two benzene rings. It is much more stable than the benzyl carbocation (\(PhCH_2^+\)). Thus, the reactivity order is reversed.

C. FALSE. Aryl halides are highly unreactive towards nucleophilic substitution compared to alkyl halides because the C-X bond acquires partial double bond character due to resonance with the benzene ring.

D. FALSE. Vinyl chloride (\(CH_2=CHCl\)) is a haloalkene. However, allyl chloride (\(CH_2=CH-CH_2Cl\)) is also a haloalkene, not a haloalkyne (which would require a triple bond).

E. TRUE. Aliphatic alcohols (\(R-OH\)) readily react with \(SOCl_2\) to form alkyl chlorides. Phenols (\(Ar-OH\)) do not react with \(SOCl_2\) because the C-O bond in phenol has partial double bond character and cannot be easily cleaved.


The only correct statements are A and E.


Step 4: Final Answer:

The correct choice is A and E Only.
Quick Tip: A classic distinction to memorize: Aq. KOH = Substitution (Alcohols). Alc. KOH = Elimination (Alkenes). Also, Phenols never undergo direct nucleophilic substitution on the C-OH bond.

Step 1: Understanding the Concept:

We first determine the empirical and molecular formula of compound "x" using its molar ratios and molar mass.

Once the structure of "x" is identified, we react it with 50% KOH, which is the classic condition for the Cannizzaro reaction (for aldehydes lacking alpha-hydrogens).


Step 2: Key Formula or Approach:

If molar ratios are equal, the empirical formula is CHO.

Empirical mass = \(12 (C) + 1 (H) + 16 (O) = 29 g/mol\).
\(n = \frac{Molar Mass}{Empirical Mass} = \frac{58}{29} = 2\).

Molecular formula = \(C_2H_2O_2\).


Step 3: Detailed Explanation:

The only logical stable structure for \(C_2H_2O_2\) is Glyoxal: \(OHC-CHO\).

Glyoxal is a dialdehyde with no alpha-hydrogens. When treated with concentrated (50%) base like KOH, it undergoes an intramolecular Cannizzaro reaction.

In an intramolecular Cannizzaro reaction, one aldehyde group is oxidized to a carboxylic acid salt, and the other is reduced to a primary alcohol.

Reaction:
\(OHC-CHO + KOH \rightarrow HOCH_2-COOK\) (Potassium glycolate)

Subsequent acidification replaces the potassium ion with a proton:
\(HOCH_2-COOK + H^+ \rightarrow HOCH_2-COOH\) (Glycolic acid)

This structure matches option C: \(O=C(OH)-CH_2-OH\).


Step 4: Final Answer:

The structure of "y" is \(O=C(OH)-CH_2-OH\).
Quick Tip: Any aldehyde lacking \(\alpha\)-hydrogens (like formaldehyde, benzaldehyde, or glyoxal) will immediately undergo the Cannizzaro reaction in the presence of concentrated base (\(>50%\)).

Step 1: Understanding the Concept:

The given molecule (X) is an ester: isopropyl acetate (\(CH_3COOCH(CH_3)_2\)).

Hydrolysis of an ester under acidic conditions breaks the ester bond to yield a carboxylic acid and an alcohol.

We need to identify these products and evaluate their physical and chemical properties against the given statements.


Step 2: Key Formula or Approach:

Ester Hydrolysis: \(R-COO-R' + H_2O \xrightarrow{H^+} R-COOH + R'-OH\).

Here, \(R = CH_3\) and \(R' = CH(CH_3)_2\).


Step 3: Detailed Explanation:

Perform the hydrolysis on isopropyl acetate:
\(CH_3COOCH(CH_3)_2 + H_2O \rightarrow CH_3COOH\) (Acetic acid) + \(CH_3CH(OH)CH_3\) (Isopropanol).

Let (Y) be Acetic acid and (Z) be Isopropanol (or vice versa).

Now, evaluate the statements:

A. Molar mass of Acetic acid (\(C_2H_4O_2\)) = \(2(12) + 4(1) + 2(16) = 60 g/mol\).

Molar mass of Isopropanol (\(C_3H_8O\)) = \(3(12) + 8(1) + 16 = 60 g/mol\).

They have the exact same molar mass. Statement A is TRUE.

B. \(NaHCO_3\) is a weak base used to test for carboxylic acids. Acetic acid will react to release effervescence of \(CO_2\) gas. Isopropanol (an alcohol) will not react. They can be distinguished. Statement B is TRUE.

C. HCN reacts with carbonyl groups (aldehydes and ketones) via nucleophilic addition. Neither acetic acid nor isopropanol behaves as a typical aldehyde/ketone for HCN addition. Statement C is FALSE.

D. 2,4-Dinitrophenylhydrazine (Brady's reagent) is a test for aldehydes and ketones. Neither the carboxylic acid nor the alcohol will undergo addition with 2,4-DNP. Statement D is FALSE.

Therefore, only statements A and B are correct.


Step 4: Final Answer:

The correct choice is A and B Only.
Quick Tip: It is a classic competitive exam trick to pair acetic acid and propanol (or isopropanol) because they coincidentally share the exact same molar mass (\(60 g/mol\)) despite belonging to entirely different functional groups.

Step 1: Understanding the Concept:

We must trace a multi-step organic synthesis starting from p-ethylnitrobenzene.

The steps involve electrophilic aromatic substitution, reduction of a nitro group, diazotization, deamination, and strong oxidation of an alkyl side chain.


Step 2: Key Formula or Approach:

1. \(Br_2/AlBr_3\): Electrophilic bromination. Ortho/para directing groups (alkyl) dominate over meta directing groups (nitro).

2. \(Sn/HCl\): Reduces \(-NO_2\) to \(-NH_2\).

3. \(NaNO_2/HCl\): Converts \(-NH_2\) to diazonium \(-N_2^+Cl^-\).

4. \(C_2H_5OH\): Mild reducing agent that replaces the diazonium group with a Hydrogen atom.

5. \(KMnO_4/H_3O^+\): Oxidizes any alkyl side chain with benzylic hydrogens directly into a \(-COOH\) group.


Step 3: Detailed Explanation:

Start with p-ethylnitrobenzene. The ethyl group is at position 1, and the nitro group is at position 4.

Reaction 1: Bromination. The ethyl group is activating and ortho/para directing. The nitro group is deactivating and meta directing. Both groups direct the incoming Bromine to the exact same position: ortho to the ethyl group (which is meta to the nitro group).

Thus, Compound A is 2-bromo-1-ethyl-4-nitrobenzene.

Reaction 2: \(Sn/HCl\) reduces the \(-NO_2\) group. Compound B is 4-amino-2-bromo-1-ethylbenzene.

Reaction 3: Diazotization. Compound C is the diazonium salt at position 4.

Reaction 4: Ethanol (\(C_2H_5OH\)) reduces the diazonium salt, removing the nitrogen group entirely and replacing it with H. Compound D is simply 2-bromo-1-ethylbenzene (or o-bromoethylbenzene).

Reaction 5: Strong oxidation with \(KMnO_4\). The ethyl side chain is completely oxidized to a carboxylic acid group. The bromine atom is unaffected.

Thus, Compound E is 2-bromobenzoic acid.


Step 4: Final Answer:

Compound A is 2-bromo-1-ethyl-4-nitrobenzene and Compound E is 2-bromobenzoic acid.
Quick Tip: Ethanol (\(CH_3CH_2OH\)) and Hypophosphorous acid (\(H_3PO_2\)) are the two standard reagents used to completely remove a diazonium group from a benzene ring (deamination).

Step 1: Understanding the Concept:

We must identify the amino acid that acts as the biological precursor for thyroid hormones and match it to its correct one-letter biochemical abbreviation.


Step 2: Key Formula or Approach:

The hormone produced by the thyroid gland that contains iodine is Thyroxine (T4).

Thyroxine is synthesized in the body through the iodination of the amino acid Tyrosine.

The one-letter codes for amino acids: Tyrosine is 'Y'. Threonine is 'T'.


Step 3: Detailed Explanation:

The hormone is explicitly described as an iodinated derivative. Insulin is a large peptide hormone and is not an iodinated derivative of a single amino acid. Thyroxine, however, is formed by the addition of iodine to the phenolic ring of Tyrosine.

Thus, Hormone (B) = Thyroxine.

The precursor amino acid (A) = Tyrosine.

Now, check the one-letter codes. Since 'T' is already taken by Threonine, Tyrosine is assigned the letter 'Y' (based on the second letter of its name to distinguish it).

Therefore, Amino acid (A) is Y.


Step 4: Final Answer:

The correct pair is Y and Thyroxine.
Quick Tip: One-letter amino acid codes are frequently tested. Remember the tricky ones: Y = Tyrosine, W = Tryptophan, F = Phenylalanine, D = Aspartic Acid, E = Glutamic Acid.

Step 1: Understanding the Concept:

The borax bead test is a qualitative analytical method used to identify transition metals. Only ions with unpaired d-electrons (colored ions) give a positive test.

The ionization enthalpy refers to the energy required to remove an electron from the specific ion listed.


Step 2: Key Formula or Approach:

Check electron configurations for unpaired electrons to confirm the borax bead test.

Evaluate the stability of the electron configuration to determine which ion has the highest ionization enthalpy (i.e., which is hardest to oxidize further).


Step 3: Detailed Explanation:

First, evaluate the borax bead test constraint:
\(Zn^{2+}\) has a \([Ar] 3d^{10}\) configuration. Since it has a completely filled d-subshell, it is diamagnetic and colorless, meaning it does not respond to the borax bead test. We can eliminate \(Zn^{2+}\).

The remaining ions (\(Fe^{2+}\), \(Fe^{3+}\), \(Cr^{2+}\)) all have partially filled d-orbitals, are colored, and give positive borax bead tests.

Next, evaluate the ionization enthalpy constraint:

We must find the energy required to remove one electron from each ion:

- \(Fe^{2+} \rightarrow Fe^{3+} + e^-\): Going from \(3d^6\) to \(3d^5\). Removing this electron achieves a highly stable half-filled state, so the energy required is relatively low.

- \(Cr^{2+} \rightarrow Cr^{3+} + e^-\): Going from \(3d^4\) to \(3d^3\). The \(3d^3\) state is exceptionally stable in aqueous/complex forms (half-filled \(t_{2g}\)), so this ionization is also relatively easy.

- \(Fe^{3+} \rightarrow Fe^{4+} + e^-\): \(Fe^{3+}\) has a \([Ar] 3d^5\) configuration. This is an exactly half-filled d-subshell, making it spherically symmetrical and exceptionally stable. Removing an electron from this perfectly stable \(d^5\) core requires an enormous amount of energy.

Therefore, \(Fe^{3+}\) has the highest ionization enthalpy among the active candidates.


Step 4: Final Answer:

The ion is \(Fe^{3+}\).
Quick Tip: Any ion with a \(d^5\) (like \(Fe^{3+}\), \(Mn^{2+}\)) or \(d^{10}\) configuration represents an energy "wall". Removing an electron from these states will always require the highest ionization energy among peers.

Step 1: Understanding the Concept:

This problem uses Einstein's Photoelectric Equation, which states that the total energy of an incident photon is split into the work function (binding energy) of the metal and the maximum kinetic energy of the ejected electron.


Step 2: Key Formula or Approach:

Photoelectric equation: \(E_{photon} = W + KE\).

Energy of a photon: \(E = \frac{hc}{\lambda}\).

We must convert all energy values to a common unit (Joules) before solving for the wavelength \(\lambda\).


Step 3: Detailed Explanation:

First, convert the work function \(W\) from eV to Joules:
\[ W = 2.3 eV \times (1.6 \times 10^{-19} J/eV) = 3.68 \times 10^{-19} J \]
The kinetic energy is given as:
\[ KE = 2.8 \times 10^{-20} J = 0.28 \times 10^{-19} J \]
Calculate the total energy of the incident photon:
\[ E = W + KE = (3.68 \times 10^{-19}) + (0.28 \times 10^{-19}) \] \[ E = 3.96 \times 10^{-19} J \]
Now, use the photon energy formula to find the wavelength \(\lambda\):
\[ E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E} \]
Substitute the given constants:
\[ hc = (6.6 \times 10^{-34} J s) \times (3.0 \times 10^8 m/s) = 19.8 \times 10^{-26} J m \] \[ \lambda = \frac{19.8 \times 10^{-26}}{3.96 \times 10^{-19}} \] \[ \lambda = 5 \times 10^{-7} m \]
Convert the wavelength to nanometers (\(1 m = 10^9 nm\)):
\[ \lambda = 5 \times 10^{-7} \times 10^9 nm = 500 nm \]
The question asks for the value in the format \(x \times 10^2 nm\).
\[ 500 nm = 5 \times 10^2 nm \]
Comparing this to the format, \(x = 5\).


Step 4: Final Answer:

The value of \(x\) is \(5\).
Quick Tip: Always align the powers of 10 before adding numbers in scientific notation. Converting \(2.8 \times 10^{-20}\) to \(0.28 \times 10^{-19}\) prevents a very common alignment error during addition.

Step 1: Understanding the Concept:

We are dealing with a gas-phase reaction where we track the progress using the total pressure of the system.

At time \(t = \infty\), the reaction is assumed to have gone to complete conversion. This allows us to calculate the initial pressure of the reactant.

Using the data at \(t = 30\) min, we can construct an ICE (Initial, Change, Equilibrium) table to find the partial pressures of individual components.


Step 2: Key Formula or Approach:

Let initial pressure of A be \(P_0\).

At \(t = \infty\), \(P_A = 0\), and the products B and C exert pressure based on stoichiometry.

At \(t = 30\), set up the change in pressure as \(-2x\) for A, \(+4x\) for B, and \(+x\) for C.

Total pressure = Sum of partial pressures.


Step 3: Detailed Explanation:

Let's analyze the state at \(t = \infty\):

Initial moles of A gives 0 moles of A, and 4/2 moles of B and 1/2 moles of C.

So, if initial pressure is \(P_0\), the final pressure is entirely due to products.
\[ P_{\infty} = P_B + P_C = \left(\frac{4}{2}\right)P_0 + \left(\frac{1}{2}\right)P_0 = 2P_0 + 0.5P_0 = 2.5P_0 \]
We are given \(P_{\infty} = 600 mm Hg\).
\[ 2.5P_0 = 600 \implies P_0 = \frac{600}{2.5} = 240 mm Hg \]
Now, set up the ICE table for \(t = 30\) min:
\(\quad\quad 2A(g) \rightarrow 4B(g) + C(g)\)

Initial: \(240 \quad\quad 0 \quad\quad 0\)

Change: \(-2x \quad +4x \quad +x\)

At 30m: \(240-2x \quad 4x \quad x\)

The total pressure at 30 minutes is the sum of these partial pressures:
\[ P_{total} = (240 - 2x) + 4x + x = 240 + 3x \]
We are given \(P_{total} = 300 mm Hg\) at \(t = 30\) min.
\[ 240 + 3x = 300 \] \[ 3x = 60 \implies x = 20 mm Hg \]
The question asks for the partial pressure of \(C(g)\) at 30 minutes.

From the ICE table, \(P_C = x\).

Therefore, \(P_C = 20 mm Hg\).


Step 4: Final Answer:

The pressure of C(g) at 30 minutes is \(20\).
Quick Tip: The "time = \(\infty\)" data point in chemical kinetics always gives you the theoretical 100% yield state, which is the most reliable way to reverse-calculate the exact initial concentration or pressure of the reactants.

Step 1: Understanding the Concept:

A fuel cell derives electrical work from a spontaneous chemical reaction. We first find the standard cell potential \(E^\circ_{cell}\).

Using this, we calculate the maximum possible thermodynamic work (\(\Delta G^\circ\)).

We factor in the cell's efficiency to find the actual usable work, and then apply this work to the physics formula for mechanical compression of a gas to find the volume change.


Step 2: Key Formula or Approach:

Overall Cell Potential: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\).

Maximum electrical work: \(W_{max} = -\Delta G^\circ = n F E^\circ_{cell}\).

Actual work: \(W_{actual} = Efficiency \times W_{max}\).

Mechanical work of compression: \(W = P_{ext} \Delta V\).


Step 3: Detailed Explanation:

Identify the anode and cathode based on standard reduction potentials. The reaction with the higher potential (\(1.23 V\)) will be the reduction (cathode). The other (\(0.02 V\)) will be reversed to oxidation (anode).
\[ E^\circ_{cell} = 1.23 V - 0.02 V = 1.21 V \]
The oxidation of 1 mole of \(CH_3OH\) involves 6 electrons, so \(n = 6\).

Calculate the maximum theoretical work for 1 mole of methanol:
\[ W_{max} = n F E^\circ_{cell} = 6 \times 96500 C \times 1.21 V \] \[ W_{max} = 579000 \times 1.21 = 700590 J \]
The fuel cell operates at 80% efficiency, so the actual usable work is:
\[ W_{actual} = 0.80 \times 700590 J = 560472 J \]
This work is used to compress a gas against an external pressure of \(1 kPa = 1000 Pa\).
\[ W_{actual} = P_{ext} \Delta V \] \[ 560472 J = 1000 Pa \times \Delta V \] \[ \Delta V = \frac{560472}{1000} = 560.472 m^3 \]
Rounding to the nearest integer gives 560.


Step 4: Final Answer:

The change in the volume of the gas is \(560\).
Quick Tip: For fuel cell thermodynamics, remember that \(n\) is the total number of electrons exchanged per mole of the primary fuel (here, methanol). Balancing the entire equation isn't necessary if you already see "6e-" in the methanol half-reaction.

Step 1: Understanding the Concept:

An ion is paramagnetic if it possesses one or more unpaired electrons in its atomic orbitals.

An ion is diamagnetic if all of its electrons are paired (typically occurring when subshells are completely empty, like \(d^0, f^0\), or completely full, like \(d^{10}, f^{14}\)).


Step 2: Key Formula or Approach:

Write out the electronic configuration for each neutral atom using the nearest noble gas core.

Remove electrons to form the ion (remember to remove from the outermost \(s\)-orbital first, then \(d\), then \(f\)).

Count the number of unpaired electrons.


Step 3: Detailed Explanation:

Let's evaluate each ion:

1. \(Mn^{2+} (Z=25)\): Neutral is \([Ar] 3d^5 4s^2\). Ion is \([Ar] 3d^5\). 5 unpaired e⁻ \(\rightarrow\) Paramagnetic.

2. \(Cu^{2+} (Z=29)\): Neutral is \([Ar] 3d^{10} 4s^1\). Ion is \([Ar] 3d^9\). 1 unpaired e⁻ \(\rightarrow\) Paramagnetic.

3. \(Zn^{2+} (Z=30)\): Neutral is \([Ar] 3d^{10} 4s^2\). Ion is \([Ar] 3d^{10}\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

4. \(Yb^{2+} (Z=70)\): Neutral is \([Xe] 4f^{14} 6s^2\). Ion is \([Xe] 4f^{14}\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

5. \(Sc^{3+} (Z=21)\): Neutral is \([Ar] 3d^1 4s^2\). Ion is \([Ar] 3d^0\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

6. \(La^{3+} (Z=57)\): Neutral is \([Xe] 5d^1 6s^2\). Ion is \([Xe] 4f^0 5d^0\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

7. \(Gd^{3+} (Z=64)\): Neutral is \([Xe] 4f^7 5d^1 6s^2\). Ion is \([Xe] 4f^7\). 7 unpaired e⁻ \(\rightarrow\) Paramagnetic.

8. \(Lu^{3+} (Z=71)\): Neutral is \([Xe] 4f^{14} 5d^1 6s^2\). Ion is \([Xe] 4f^{14}\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

9. \(Ti^{4+} (Z=22)\): Neutral is \([Ar] 3d^2 4s^2\). Ion is \([Ar] 3d^0\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

10. \(Ce^{4+} (Z=58)\): Neutral is \([Xe] 4f^1 5d^1 6s^2\). Ion is \([Xe] 4f^0\). 0 unpaired e⁻ \(\rightarrow\) Diamagnetic.

The paramagnetic ions are \(Mn^{2+}\), \(Cu^{2+}\), and \(Gd^{3+}\).

Total count is 3.


Step 4: Final Answer:

The number of paramagnetic ions is \(3\).
Quick Tip: For f-block elements, +3 is the most common oxidation state. Ions like \(La^{3+}\), \(Lu^{3+}\), and \(Ce^{4+}\) are explicitly stable because they achieve empty (\(f^0\)) or completely full (\(f^{14}\)) configurations, making them completely diamagnetic.

Step 1: Understanding the Concept:

We first decipher the organic reaction sequence to find the exact molecular structure of the final product (P).

Then, we apply the stoichiometry of the Carius method, which quantitatively converts all halogen atoms in an organic compound into a silver halide precipitate (\(AgBr\)).


Step 2: Key Formula or Approach:

Step (i): Reduction of nitro to amine.

Step (ii): Acetylation of amine for protection.

Step (iii): Electrophilic aromatic bromination.

Step (iv): Deprotection (hydrolysis) of the amide back to an amine.

Carius method stoichiometry: 1 mole of P (containing 1 Br) yields 1 mole of \(AgBr\).

Mass of \(AgBr\) = \(\frac{Mass of P}{Molar Mass of P} \times Molar Mass of AgBr\).


Step 3: Detailed Explanation:

Trace the reactions starting from p-Nitrotoluene:

(i) \(Sn/HCl\) followed by \(OH^-\) reduces the \(p\)-\(NO_2\) group to an \(-NH_2\) group. The product is p-toluidine.

(ii) \((CH_3CO)_2O\) acetylates the amine to protect it, forming \(p\)-methylacetanilide. This dampens the extreme activating effect of the amine group to prevent poly-bromination.

(iii) \(Br_2/AlBr_3\) is a bromination step. The \(-NHCOCH_3\) group is highly activating and ortho/para directing. The \(-CH_3\) group is weakly activating and ortho/para directing. The stronger \(-NHCOCH_3\) group controls the direction. Since the para position is blocked by the methyl group, bromination occurs at the ortho position relative to the \(-NHCOCH_3\) group.

(iv) \(H_3O^+\) hydrolyzes the protective amide group back to an amine.

The final product (P) is 2-bromo-4-methylaniline.

The molecular formula of (P) is \(C_6H_3(NH_2)(Br)(CH_3)\) which combines to \(C_7H_8BrN\).

Calculate the molar mass of (P):
\(M_P = (7 \times 12) + (8 \times 1) + 80 + 14 = 84 + 8 + 80 + 14 = 186 g/mol\).

Calculate the molar mass of \(AgBr\):
\(M_{AgBr} = 108 + 80 = 188 g/mol\).

According to the Carius method, 1 mole of P (186 g) yields 1 mole of \(AgBr\) (188 g).

For 1.0 g of P, the mass of \(AgBr\) produced is:
\[ Mass = \frac{188}{186} \times 1.0 g = 1.0107 g \]
Rounding to the nearest integer, we get 1 g.


Step 4: Final Answer:

The mass of the precipitate produced is \(1\).
Quick Tip: Acetylation of an aniline derivative before halogenation is a classic technique to guarantee mono-halogenation. It effectively steps down the intense activating power of the \(-NH_2\) group, preventing the formation of 2,4,6-tribromo products.