Step 1: Understanding the Concept:

To compare the total number of atoms across different substances, we first need to convert their given amounts (moles, mass, or volume) into the number of moles of molecules. Then, multiply by the atomicity (number of atoms per molecule) and Avogadro's number (\(N_A\)) to find the total atom count.


Step 2: Key Formula or Approach:

Moles from mass: \(n = \frac{W}{M_{w}}\)

Moles from volume at STP: \(n = \frac{V(in L)}{22.7}\) (or \(22.4\))

Total Atoms = \(n \times (Atoms per molecule) \times N_A\)


Step 3: Detailed Explanation:

Let's evaluate each option:


(A) 2 moles of cyclohexane

Chemical formula of cyclohexane: \(C_6H_{12}\)

Atoms per molecule = \(6 (C) + 12 (H) = 18\) atoms/molecule.

Number of moles = 2.

Total atoms = \(2 mol \times 18 atoms/molecule \times N_A = 36 N_A\) atoms.


(B) 684 g of sucrose

Chemical formula of sucrose: \(C_{12}H_{22}O_{11}\)

Molar mass of sucrose = \(12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g/mol\).

Atoms per molecule = \(12 + 22 + 11 = 45\) atoms/molecule.

Number of moles = \(\frac{684 g}{342 g/mol} = 2\) moles.

Total atoms = \(2 mol \times 45 atoms/molecule \times N_A = 90 N_A\) atoms.


(C) 90.8 L of dihydrogen at STP

Chemical formula of dihydrogen: \(H_2\)

Atoms per molecule = 2 atoms/molecule.

Using molar volume at STP as \(22.7 L/mol\) (standard IUPAC 1 bar condition):

Number of moles = \(\frac{90.8 L}{22.7 L/mol} = 4\) moles.

(If using old standard \(22.4 L/mol\), \(n = 4.05\) moles. It won't alter the relative order).

Total atoms = \(4 mol \times 2 atoms/molecule \times N_A = 8 N_A\) atoms.


Comparing the quantities:

(B) \(90 N_A\) > (A) \(36 N_A\) > (C) \(8 N_A\).

The correct descending order is B > A > C.


Step 4: Final Answer:

The correct order is B > A > C. Quick Tip: To avoid redundant calculations, don't multiply out Avogadro's number (\(6.022 \times 10^{23}\)). Keep values in terms of \(N_A\) when you only need to establish a comparative order.

Step 1: Understanding the Concept:

According to Bohr's model for hydrogen-like (single-electron) species, the radius of an orbit depends on the principal quantum number \(n\) and the atomic number \(Z\). Species with the same ratio of \(n^2 / Z\) will have identical orbital radii.


Step 2: Key Formula or Approach:

Bohr radius formula: \(r = 0.529 \frac{n^2}{Z} \AA\)

Therefore, \(r \propto \frac{n^2}{Z}\).


Step 3: Detailed Explanation:

Let's calculate the proportionality factor \(\frac{n^2}{Z}\) for each given species:

A. H (first orbit): \(Z = 1\), \(n = 1\)
\(\frac{n^2}{Z} = \frac{1^2}{1} = 1\)


B. He\(^+\) (first orbit): \(Z = 2\), \(n = 1\)
\(\frac{n^2}{Z} = \frac{1^2}{2} = 0.5\)


C. He\(^+\) (Second orbit): \(Z = 2\), \(n = 2\)
\(\frac{n^2}{Z} = \frac{2^2}{2} = \frac{4}{2} = 2\)


D. Li\(^{2+}\) (first orbit): \(Z = 3\), \(n = 1\)
\(\frac{n^2}{Z} = \frac{1^2}{3} = \frac{1}{3} \approx 0.33\)


E. Be\(^{3+}\) (Second orbit): \(Z = 4\), \(n = 2\)
\(\frac{n^2}{Z} = \frac{2^2}{4} = \frac{4}{4} = 1\)


Comparing the calculated values:

Species A (H) and E (Be\(^{3+}\)) both yield \(\frac{n^2}{Z} = 1\). Thus, their orbital radii are identical.


Step 4: Final Answer:

Options A and E have identical radii. Quick Tip: For comparative radius questions in Bohr's model, simply evaluating \(n^2/Z\) is sufficient. Don't waste time multiplying by \(0.529\) \AA\ unless the absolute numerical radius is explicitly requested.

Step 1: Understanding the Concept:

Molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). A \(\pi\) (pi) bond is formed by the sideways (lateral) overlap of \(p\)-orbitals. When they combine "out of phase" (subtractive overlap), they form a \(\pi^\) antibonding orbital, which features a node directly between the nuclei.


Step 2: Key Formula or Approach:

Antibonding Orbital (\(\Psi^\)) = \(\Psi_A - \Psi_B\).

A \(\pi^\) orbital is characterized by a nodal plane perpendicular to the internuclear axis (between the atoms) in addition to the nodal plane containing the internuclear axis itself.


Step 3: Detailed Explanation:

Let the internuclear axis be the z-axis. The \(p_x\) or \(p_y\) atomic orbitals are oriented perpendicular to this axis.

When two \(p_x\) (or \(p_y\)) orbitals approach each other out-of-phase (e.g., positive lobe of one approaches the negative lobe of the other), they repel and create a region of zero electron density (a nodal plane) exactly bisecting the internuclear axis.

The resulting shape consists of four distinct lobes pointing away from the center, with alternating phases (signs of the wave function) diagonally.

Looking at standard diagrams:
Image (Option A) shows an s-orbital overlapping out of phase, producing a \(\sigma^\) antibonding orbital.
Image (Option B) shows two p-orbitals overlapping sideways in-phase, producing a bonding \(\pi\) orbital (two large continuous lobes).
Image (Option C) shows two p-orbitals overlapping sideways out-of-phase, producing four distinct lobes separated by a central vertical nodal plane. This is the hallmark \(\pi^\) antibonding orbital.
Image (Option D) shows head-on overlap creating a \(\sigma\) bond or represents a d-orbital array, not a sideways overlap.

Thus, the third image accurately reflects the \(\pi^\) symmetry.


Step 4: Final Answer:

Option (C) correctly represents the \(\pi^\) antibonding molecular orbital. Quick Tip: To quickly identify an antibonding orbital graphically, look for a newly formed central node separating the two atoms. A \(\pi^\) specifically will have 4 separate lobes because it retains the original nodal plane of the p-orbitals along the axis.

Step 1: Understanding the Concept:

Osmosis involves the flow of solvent molecules across a semi-permeable membrane (SPM) from a region of lower solute concentration to a region of higher solute concentration. The true osmotic concentration is determined by the osmolarity (Molarity \(\times\) van 't Hoff factor \(i\)). The solution with the lower osmolarity is deemed "hypotonic".


Step 2: Key Formula or Approach:

Osmotic pressure: \(\pi = i C R T\)

Hypotonic means lower \(\pi\) (and thus lower \(i \times C\)).

Hypertonic means higher \(\pi\) (and thus higher \(i \times C\)).


Step 3: Detailed Explanation:

Let's evaluate the side 'x':

Solution is \(0.1 M K_4[Fe(CN)_6]\).

Dissociation: \(K_4[Fe(CN)_6] \to 4K^+ + [Fe(CN)_6]^{4-}\)

Number of ions \(i_x = 4 + 1 = 5\).

Effective concentration (Osmolarity) \(= i_x \cdot C = 5 \times 0.1 = 0.5 M\).


Let's evaluate the side 'y':

Solution is \(0.1 M FeCl_3\).

Dissociation: \(FeCl_3 \to Fe^{3+} + 3Cl^-\)

Number of ions \(i_y = 1 + 3 = 4\).

Effective concentration (Osmolarity) \(= i_y \cdot C = 4 \times 0.1 = 0.4 M\).


Now analyze the options:

(A) Blue color formation requires the reaction of \(Fe^{3+}\) with \([Fe(CN)_6]^{4-}\) to form Prussian blue. Since an SPM does not allow solute ions to pass through, they cannot mix. No color is formed.

(B) A semi-permeable membrane selectively allows solvent (water) to pass, completely blocking ionic solutes. False.

(C) Side 'y' has an osmolarity of \(0.4 M\), which is less than side 'x' (\(0.5 M\)). Thus, the solution on side 'y' is hypotonic relative to 'x'. True.

(D) Reverse osmosis requires applying an external pressure specifically greater than the osmotic pressure difference to the hypertonic side ('x'). "Any value" is technically incorrect; it must exceed the threshold \(\Delta \pi\).


Step 4: Final Answer:

The statement "Solution on side 'y' is hypotonic" is correct. Quick Tip: When assessing osmotic traits (hypertonic/hypotonic) of electrolytes, never look just at the molarity. Always multiply the given molarity by the van 't Hoff factor (\(i\)) to find the total particle concentration.

Step 1: Understanding the Concept:

When a weak acid is partially neutralized by a strong base, a buffer solution containing the weak acid and its conjugate base (salt) is formed. The pH of this buffer is calculated using the Henderson-Hasselbalch equation.


Step 2: Key Formula or Approach:

Neutralization reaction: \(CH_3COOH + NaOH \to CH_3COONa + H_2O\).

Henderson-Hasselbalch equation: \(pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)\).


Step 3: Detailed Explanation:

From the full neutralization data:
\(20 mL\) of acetic acid requires \(28.4 mL\) of \(0.1 M\) NaOH.

Milli-moles (mmol) of NaOH used for full neutralization = \(M \times V = 0.1 mol/L \times 28.4 mL = 2.84 mmol\).

Because NaOH and acetic acid react in a 1:1 molar ratio, the \(20 mL\) of acetic acid contains exactly \(2.84 mmol\) of acid.


Preparation of Solution (X):

We mix \(20 mL\) of this same acetic acid (containing \(2.84 mmol\)) with \(14.2 mL\) of \(0.1 M\) NaOH.

Milli-moles of NaOH added = \(0.1 \times 14.2 = 1.42 mmol\).


The NaOH will react with the acetic acid:

Initial: Acid = \(2.84 mmol\), NaOH = \(1.42 mmol\), Salt = \(0\)

Reacts: \(-1.42 mmol\) acid, \(-1.42 mmol\) NaOH, \(+1.42 mmol\) Salt

Final: Acid = \(2.84 - 1.42 = 1.42 mmol\). NaOH = \(0\). Salt = \(1.42 mmol\).


The resulting mixture is an acidic buffer containing equal amounts of the weak acid and its salt.

Apply the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right) \]
Since they are in the same total volume, the ratio of concentrations equals the ratio of milli-moles:
\[ pH = 4.75 + \log \left( \frac{1.42}{1.42} \right) \] \[ pH = 4.75 + \log(1) = 4.75 + 0 = 4.75 \]

Step 4: Final Answer:

The pH of solution (X) is 4.75. Quick Tip: If the volume of the strong base added is exactly half the volume required for full neutralization (the "half-equivalence point"), the resulting buffer always has \([Salt] = [Acid]\), meaning \(pH = pK_a\) instantaneously.

Step 1: Understanding the Concept:

Identify the fundamental reaction mechanism (substitution vs. addition, and electrophilic vs. nucleophilic vs. free radical) for each classic named organic reaction listed.


Step 2: Key Formula or Approach:

Review the reagent behaviors: Alkoxide ions are nucleophiles. Lewis acids generate electrophiles. Light (hv) initiates free radical cascades.


Step 3: Detailed Explanation:

Let's analyze each reaction:

A. Williamson Synthesis:

Reaction: \(R-X + R'-O^-Na^+ \to R-O-R' + NaX\).

Mechanism: The alkoxide ion acts as a nucleophile attacking the alkyl halide in an \(S_N2\) mechanism. This is Nucleophilic substitution. (A \(\to\) III)


B. Friedel Craft Reaction:

Reaction: Benzene \(+ R-Cl \xrightarrow{AlCl_3} Alkylbenzene + HCl\).

Mechanism: The Lewis acid \(AlCl_3\) generates an alkyl carbocation (an electrophile), which attacks the electron-rich aromatic ring. This is Electrophilic substitution. (B \(\to\) IV)


C. Bromination of vinyl benzene:

Reaction: Addition of \(Br_2\) across the \(C=C\) double bond of the vinyl group.

Mechanism: The pi electrons of the alkene attack the polarizable \(Br_2\) to form a bromonium ion, followed by bromide attack. Alkene additions are classically Electrophilic addition. (C \(\to\) I)


D. Chlorination of toluene in light:

Reaction: Toluene \(+ Cl_2 \xrightarrow{h\nu} Benzyl chloride\).

Mechanism: The presence of UV light (\(h\nu\)) causes homolytic cleavage of \(Cl_2\), producing chlorine free radicals that substitute the benzylic hydrogens. This is Free radical substitution. (D \(\to\) II)


Matching sequence: A-III, B-IV, C-I, D-II.


Step 4: Final Answer:

Option (C) is the correct match. Quick Tip: Light (\(h\nu\)), heat (\(\Delta\)), or peroxides are universally the triggering agents for Free Radical mechanisms. Spotting "in light" for Toluene chlorination instantly pairs it with "Free radical".

Step 1: Understanding the Concept:

Ionization enthalpy (IE) is the energy required to remove an electron from an isolated gaseous atom or ion. Successive ionization enthalpies always increase (\(IE_1 < IE_2 < IE_3 \dots\)) because removing an electron from a positively charged ion requires significantly more energy than removing it from a neutral atom. Additionally, ionization is strictly an endothermic process.


Step 2: Key Formula or Approach:

For any element: \(IE_2 > IE_1 > 0\).

Removing the second electron from a Group 2 metal usually takes roughly double the energy of the first.


Step 3: Detailed Explanation:

Given \(IE_1 = +737 kJ/mol\).

We must estimate \(IE_2\).

1. Ionization energy represents an input of energy, hence it must always be a positive value. This immediately eliminates options (A) \(-906 kJ/mol\) and (B) \(-856 kJ/mol\).

2. The second ionization energy (\(IE_2\)) involves overcoming the stronger electrostatic pull of the newly formed \(+1\) cation, ensuring that \(IE_2\) is strictly greater than \(IE_1\).
\(IE_2 > +737 kJ/mol\).

This eliminates option (D) \(+590 kJ/mol\).


The only physically viable option remaining that is both positive and greater than \(737 kJ/mol\) is \(+1450 kJ/mol\).


Step 4: Final Answer:

The estimated value is +1450 kJ/mol. Quick Tip: Never pick a negative value for Ionization Enthalpy; extracting electrons from a nucleus always requires you to supply energy (endothermic).

Step 1: Understanding the Concept:

Identify the element 'E' based on its group and electronegativity. Once 'E' is identified as Thallium (Tl) due to periodic trends, apply the concept of the Inert Pair Effect to determine the stability of its oxidation states and its resulting redox behavior.


Step 2: Key Formula or Approach:

Electronegativity trend in Group 13: B(2.0), Al(1.5), Ga(1.6), In(1.7), Tl(1.8).

Electronegativity of Ge = 1.8.

Inert Pair Effect stabilizes the lower oxidation state (\(+1\)) relative to the higher oxidation state (\(+3\)) in heavy post-transition metals like Tl.


Step 3: Detailed Explanation:

The group 13 element with an electronegativity of 1.8 (equal to Ge) is Thallium (Tl).

Let's evaluate the properties of Thallium (\(E = Tl\)):

Because Tl is at the very bottom of Group 13, its outermost s-electrons (\(6s^2\)) are strongly attracted to the nucleus due to poor shielding by internal d and f orbitals. This is known as the Inert Pair Effect.

Consequently, for Thallium, the \(+1\) oxidation state is significantly more stable than the \(+3\) oxidation state.


Let's check the given statements:

C. False. Tl\(^{3+}\) is much less stable than Tl\(^+\).

A. False. A reducing agent forces reduction by being oxidized itself. Tl\(^{3+}\) is already at its maximum group oxidation state and cannot be oxidized further.

B. True. Since Tl\(^+\) is much more stable than Tl\(^{3+}\), Tl\(^{3+}\) strongly desires to gain two electrons to reduce down to Tl\(^+\). Therefore, it acts as a strong oxidizing agent.

D. True. Because Tl\(^{3+}\) is highly eager to undergo reduction, the standard reduction potential \(E^\circ\) for Tl\(^{3+} + 3e^- \to Tl\) (and also Tl\(^{3+} \to\) Tl\(^+\)) is a highly positive value, indicating a spontaneous reduction process.


Thus, statements B and D are correct.


Step 4: Final Answer:

Options B and D Only are correct. Quick Tip: For heavy p-block elements (like Pb, Bi, Tl), the "Inert Pair Effect" is the culprit behind almost every anomaly. It makes their highest oxidation state unstable and highly oxidative.

Step 1: Understanding the Concept:

Write down the electronic configurations of the given Lanthanide and transition series elements using Aufbau principles and the known half-filled/fully-filled stability anomalies. Compare the specific electron count populating the \(4f\) subshell.


Step 2: Key Formula or Approach:

General Lanthanide config: \([Xe] 4f^{1-14} 5d^{0-1} 6s^2\).

Exceptions occur at half-filled (\(f^7\)) and fully-filled (\(f^{14}\)) states, where an electron occupies the \(5d\) orbital instead to preserve \(f\)-shell symmetry.


Step 3: Detailed Explanation:

Let's deduce the configuration for each element:

- Eu (63): \([Xe] 4f^7 6s^2\) \(\implies\) 7 electrons in \(4f\).

- Gd (64): \([Xe] 4f^7 5d^1 6s^2\) \(\implies\) 7 electrons in \(4f\). (Anomaly to maintain half-filled stability).

- Dy (66): \([Xe] 4f^{10} 6s^2\) \(\implies\) 10 electrons in \(4f\).

- Ho (67): \([Xe] 4f^{11} 6s^2\) \(\implies\) 11 electrons in \(4f\).

- Tm (69): \([Xe] 4f^{13} 6s^2\) \(\implies\) 13 electrons in \(4f\).

- Yb (70): \([Xe] 4f^{14} 6s^2\) \(\implies\) 14 electrons in \(4f\).

- Lu (71): \([Xe] 4f^{14} 5d^1 6s^2\) \(\implies\) 14 electrons in \(4f\). (Anomaly to maintain fully-filled stability).

- Hf (72): \([Xe] 4f^{14} 5d^2 6s^2\) \(\implies\) 14 electrons in \(4f\). (Post-lanthanide transition metal).


Check the given pairs:

A. (Eu and Gd): \(4f^7\) and \(4f^7\). Match! (Both have 7)

B. (Dy and Ho): \(4f^{10}\) and \(4f^{11}\). No match.

C. (Yb and Hf): \(4f^{14}\) and \(4f^{14}\). Match! (Both have 14)

D. (Lu and Tm): \(4f^{14}\) and \(4f^{13}\). No match.


Therefore, pairs A and C are the correct ones.


Step 4: Final Answer:

Options A and C Only are correct. Quick Tip: In the f-block, the electron count stalls at \(f^7\) (Eu, Gd) and \(f^{14}\) (Yb, Lu) because the incoming electron prefers the \(5d\) orbital rather than disrupting the highly stable half-filled or fully-filled \(4f\) state.

Step 1: Understanding the Concept:

Evaluate the oxidation state and d-electron count for Nickel in all three complexes. Use Crystal Field Theory (CFT) to distribute these electrons in octahedral and tetrahedral geometries to find unpaired electrons. Finally, use the spectrochemical series to order the splitting energy (\(\Delta\)), which is directly proportional to the frequency of absorbed light.


Step 2: Key Formula or Approach:

Frequency of absorption \(\nu \propto \Delta E\).

Splitting energy depends on geometry (\(\Delta_o > \Delta_t\)) and ligand strength (Spectrochemical series: \(en > NH_3 > Cl^-\)).
\(Ni^{2+}\) is a \(3d^8\) system.


Step 3: Detailed Explanation:

Unpaired Electrons Analysis:

In all three complexes, the metal ion is \(Ni^{2+}\), which has a \(3d^8\) configuration.

(A) \([Ni(en)_3]^{2+}\): Octahedral complex. The \(d^8\) configuration in an octahedral field (\(t_{2g}^6 e_g^2\)) always has exactly 2 unpaired electrons in the \(e_g\) orbitals, regardless of ligand strength.

(C) \([Ni(NH_3)_6]^{2+}\): Octahedral complex. Similar to the above, \(d^8\) in an octahedral field gives 2 unpaired electrons.

(B) \([NiCl_4]^{2-}\): Tetrahedral complex. Chloride is a weak field ligand. The \(d^8\) configuration in a tetrahedral field (\(e^4 t_2^4\)) yields exactly 2 unpaired electrons in the \(t_2\) orbitals.

So, the number of unpaired electrons is 2, 2, 2 respectively.


Absorption Frequency Analysis:

The frequency of light absorbed (\(\nu\)) corresponds to the crystal field splitting energy (\(\Delta\)). \(\Delta E = h\nu = \Delta_o\) (or \(\Delta_t\)).

1. Octahedral fields split orbitals much more than tetrahedral fields: \(\Delta_o \approx \frac{9}{4} \Delta_t\). So complexes (A) and (C) absorb at a much higher frequency than complex (B).

2. Between the two octahedral complexes (A) and (C), we look at ligand strength. Ethylenediamine (en) is a bidentate ligand and is stronger than Ammonia (NH\(_3\)).

According to the spectrochemical series: \(en > NH_3\).

Thus, \(\Delta_o (en) > \Delta_o (NH_3)\).

Bringing it all together: \(\Delta_E(A) > \Delta_E(C) > \Delta_E(B)\).

Therefore, the frequency order is \(\nu_{(A)} > \nu_{(C)} > \nu_{(B)}\).


Step 4: Final Answer:

The correct option is 2, 2, 2 and (A) > (C) > (B). Quick Tip: For a \(d^8\) metal ion like \(Ni^{2+}\), it will always have exactly 2 unpaired electrons in any high-spin or low-spin octahedral or tetrahedral geometry. It physically cannot be forced to pair completely in these symmetrical fields!

Step 1: Understanding the Concept:

The bond length of a carbon-oxygen bond depends on its bond order. A pure single bond is the longest, a pure double bond is shorter, and intermediate bond orders (due to resonance or aromaticity) fall in between. We evaluate the resonance structures of each given species to determine their C-O bond orders.


Step 2: Key Formula or Approach:

Bond Length \(\propto \frac{1}{Bond Order}\)

Identify the dominant resonance contributors for each structure to estimate the bond order.


Step 3: Detailed Explanation:

Let's analyze the three structures:

1. Structure (x) [Tropone]: Tropone is a seven-membered ring with a ketone group. To achieve aromaticity (a stable \(6\pi\) electron system, following Hückel's rule), the pi electrons of the C=O bond strongly shift towards the highly electronegative oxygen atom, creating a tropylium cation structure (\(C^+ - O^-\)). This dipolar resonance structure is the major contributor, meaning the C-O bond has predominantly single bond character (bond order \(\approx 1\)). Thus, it is the longest.


2. Structure (z) [Acetate ion]: The acetate ion (\(CH_3COO^-\)) exhibits perfect equivalent resonance between the two oxygen atoms. The pi bond is delocalized equally over both C-O bonds. Therefore, the bond order is exactly 1.5. This makes it shorter than a single bond but longer than a double bond.


3. Structure (y) [Acetone]: Acetone (\(CH_3COCH_3\)) is a simple ketone with no significant resonance delocalization extending the pi bond outside of the C=O group. It is a pure double bond (bond order = 2). Thus, it has the shortest bond length.


Ordering the bond lengths from longest to shortest (lowest bond order to highest):
\(x\) (B.O. \(\approx 1\)) \(> z\) (B.O. \(= 1.5\)) \(> y\) (B.O. \(= 2\)).


Step 4: Final Answer:

The correct order of bond length is \(x > z > y\). Quick Tip: Always look for aromaticity driven by polarization in cyclic ketones (like tropone or cyclopropenone). The desire to form an aromatic ring drastically increases the single-bond character of the exocyclic double bond.

Step 1: Understanding the Concept:

We first need to identify the highest oxidation states of Manganese (Mn) when combined with Fluorine and Oxygen. Due to oxygen's ability to form multiple pi-bonds, it can stabilize a higher oxidation state than fluorine. After finding the difference \(|x|\), we determine the electron configurations of the given transition metal ions to count their unpaired electrons.


Step 2: Key Formula or Approach:

Highest Mn Fluoride: \(MnF_4\) (Oxidation state = +4).

Highest Mn Oxide: \(Mn_2O_7\) (Oxidation state = +7).

Unpaired electrons = number of electrons in singly occupied d-orbitals.


Step 3: Detailed Explanation:

The highest known fluoride of manganese is \(MnF_4\), where Mn is in the \(+4\) oxidation state.

The highest known oxide of manganese is \(Mn_2O_7\), where Mn is in the \(+7\) oxidation state.

The difference is \(|x| = |4 - 7| = |-3| = 3\).

So we are looking for ions with exactly 3 unpaired electrons.


Let's write the electronic configurations for the given ions:

A. \(Sc^{3+}\): Neutral Sc is \([Ar] 3d^1 4s^2\). \(Sc^{3+}\) is \([Ar] 3d^0\). (0 unpaired electrons)

B. \(Zn^{2+}\): Neutral Zn is \([Ar] 3d^{10} 4s^2\). \(Zn^{2+}\) is \([Ar] 3d^{10}\). (0 unpaired electrons)

C. \(V^{2+}\): Neutral V is \([Ar] 3d^3 4s^2\). \(V^{2+}\) is \([Ar] 3d^3\). (3 unpaired electrons)

D. \(Fe^{2+}\): Neutral Fe is \([Ar] 3d^6 4s^2\). \(Fe^{2+}\) is \([Ar] 3d^6\). (4 unpaired electrons)

E. \(Co^{2+}\): Neutral Co is \([Ar] 3d^7 4s^2\). \(Co^{2+}\) is \([Ar] 3d^7\). (3 unpaired electrons)


The ions containing exactly 3 unpaired electrons are \(V^{2+}\) and \(Co^{2+}\) (C and E).


Step 4: Final Answer:

Options C and E Only are correct. Quick Tip: Oxygen stabilizes higher oxidation states better than fluorine because it can form multiple bonds (\(p\pi - d\pi\) bonding), distributing the high positive charge density of the central metal more effectively.

Step 1: Understanding the Concept:

The graph plots the concentration of reactants \([R]\) against time \(t\). The steepness of the curve at any given time represents the rate of the reaction. A steeper drop indicates a faster reaction, which inherently corresponds to a higher rate constant \(k\), assuming all other factors (like reaction order and initial concentration) are held constant.


Step 2: Key Formula or Approach:

For any order reaction, the rate is given by \(-\frac{d[R]}{dt} = k[R]^n\).

If the order \(n\) is identical across multiple reactions, the rate of concentration decay \(-\frac{d[R]}{dt}\) is directly proportional to the rate constant \(k\).


Step 3: Detailed Explanation:

Let's analyze the options based on the visual properties of the curves:

- Curve 3 decreases the most rapidly, meaning the reactant is consumed the fastest.

- Curve 2 decreases at an intermediate pace.

- Curve 1 decreases the slowest.


If we assume the reactions all have the same order (as proposed in option B), then the reaction that drops fastest must necessarily have the highest rate constant \(k\).

Since Curve 3 decays much faster than Curve 2, the rate constant \(k_3\) must be greater than \(k_2\). Therefore, statement (B) is entirely correct.


Evaluating other options briefly:

(A) We cannot definitively prove they are all the same order just from the visual curves without log plots.

(C) If reaction 1 is zero-order (which it visually resembles as a nearly straight line), the unit of \(k\) would be \(M \cdot s^{-1}\), not \(s^{-1}\) (which is for 1st order).

(D) Thermal decomposition of HI on a gold surface is a classic zero-order reaction. A zero-order reaction yields a perfectly straight line on a \([R]\) vs \(t\) plot, which resembles Curve 1, not Curve 2.


Step 4: Final Answer:

The correct statement is that the rate constant of reaction 3 is larger than reaction 2 if their orders are the same. Quick Tip: When comparing decay curves starting from the same initial concentration, the curve closest to the origin (steepest initial descent) invariably represents the fastest kinetics and the highest rate constant.

Step 1: Understanding the Concept:

We must trace the multi-step organic synthesis starting from Styrene (Compound X). The sequence involves electrophilic addition, double dehydrohalogenation to form a terminal alkyne, nucleophilic substitution to extend the carbon chain, and finally, a Birch reduction to form an alkene.


Step 2: Key Formula or Approach:

1. \(Br_2/CHCl_3\): Bromination of alkene.

2. Excess \(NaNH_2\): Double \(E2\) elimination forming an alkyne, followed by deprotonation to form an acetylide ion.

3. \(CH_3I\): \(S_N2\) alkylation of the acetylide ion.

4. \(Na/NH_3(l)\): Birch reduction yielding a trans-alkene.

Molar mass \(M_w = \sum (atomic masses)\).


Step 3: Detailed Explanation:

Let's map out the reaction intermediates:

Step (i): Compound (X) is Styrene (\(C_6H_5-CH=CH_2\)).

Reacting with \(Br_2/CHCl_3\) adds bromine across the double bond, yielding 1,2-dibromoethylbenzene:
\(C_6H_5-CHBr-CH_2Br\).


Step (ii): Treatment with excess \(NaNH_2\) causes double dehydrohalogenation. The first elimination forms a bromoalkene, and the second forms a terminal alkyne (phenylacetylene). Because \(NaNH_2\) is a very strong base and is in excess, it immediately deprotonates the acidic terminal alkyne to form Sodium phenylacetylide:
\(C_6H_5-C\equiv C^- Na^+\).


Step (iii): Adding Methyl iodide (\(CH_3I\)) leads to a straightforward \(S_N2\) reaction where the acetylide ion attacks the methyl group, kicking off iodine. This forms 1-phenylpropyne:
\(C_6H_5-C\equiv C-CH_3\).


Step (iv): Reacting the internal alkyne with Sodium in liquid Ammonia (\(Na/NH_3(l)\)) is the Birch reduction condition. It reduces internal alkynes exclusively to trans-alkenes. Thus, the major product (Y) is trans-1-phenylpropene:
\(C_6H_5-CH=CH-CH_3\).


Determine the molar mass of Product (Y):

Molecular formula of \(C_6H_5-CH=CH-CH_3\) is \(C_9H_{10}\).

Molar Mass \(= (9 \times 12) + (10 \times 1) = 108 + 10 = 118 g/mol\).


Step 4: Final Answer:

The molar mass of product (Y) is 118. Quick Tip: When an alkyne is treated with \(NaNH_2\), always look for the word "excess". It indicates that after elimination, the terminal alkyne is deprotonated into an excellent nucleophile ready for chain elongation.

Step 1: Understanding the Concept:

To classify stereoisomers, one must first locate any stereocenters (chiral carbons). A carbon is chiral only if it is bonded to four completely distinct groups. If a molecule lacks a chiral center (and has no other forms of chirality), it is achiral.


Step 2: Key Formula or Approach:

Identify the 4 substituents attached to the central carbon atom in both given structures. If any two substituents are identical, the carbon is achiral. Any two 3D representations of the same achiral molecule are simply identical molecules.


Step 3: Detailed Explanation:

Let's analyze the groups attached to the central carbon atom in the provided visual structures:

- Top bond: Bromine (Br)

- Right bond: Chlorine (Cl)

- Bottom bond: Methyl group (Me)

- Left bond: Methyl group (\(CH_3\))


The notation "Me" is standard chemical shorthand for the methyl group (\(CH_3\)). Therefore, the central carbon is attached to:

Br, Cl, \(CH_3\), and \(CH_3\).


Because the central carbon is attached to two identical groups (two methyl groups), it does not have 4 distinct substituents. Therefore, the central carbon is not a chiral center.

Since the molecule is achiral, it does not have enantiomers, diastereomers, or meso forms. Any rotational variation or 3D sketch of this specific connectivity represents the exact same achiral molecule.


Therefore, the two structures are identical molecules.


Step 4: Final Answer:

The structures are identical molecules. Quick Tip: Examiners often use mixed nomenclature like writing "\(CH_3\)" on one branch and "Me" on another to trick you into assuming they are different groups. Always mentally expand abbreviations first!

Step 1: Understanding the Concept:

Acidity is determined by the stability of the conjugate base formed after releasing an \(H^+\) ion. Resonance (-M) and inductive (-I) electron-withdrawing groups stabilize the negative charge, increasing acidity. Electron-donating groups (+M, +I) destabilize the conjugate base, decreasing acidity. Furthermore, carboxylic acids are inherently far more acidic than phenols.


Step 2: Key Formula or Approach:

Acidity order: Carboxylic acids \(\gg\) Phenols.

Substituent effects: \(-M, -I\) increase acidity; \(+M, +I\) decrease acidity.


Step 3: Detailed Explanation:

Let's categorize the given molecules:

A. Phenol

B. p-Nitrophenol (contains strongly electron-withdrawing \(-NO_2\) group; \(-M, -I\))

C. p-Methoxyphenol (contains electron-donating \(-OCH_3\) group; \(+M, -I\))

D. p-Nitrobenzoic acid (Carboxylic acid with \(-NO_2\) group; \(-M, -I\))

E. Benzoic acid (Unsubstituted carboxylic acid)


First, separate the carboxylic acids (D, E) from the phenols (A, B, C). Carboxylic acids are much stronger acids because their conjugate base (carboxylate) features equivalent resonance over two highly electronegative oxygen atoms.

So, {D, E > {A, B, C.


Evaluate the Carboxylic Acids (D, E):

The \(NO_2\) group in D withdraws electron density, heavily stabilizing the carboxylate anion.

Thus, D (p-nitrobenzoic acid) > E (benzoic acid).


Evaluate the Phenols (A, B, C):

- B has a \(-NO_2\) group (-M effect), drastically stabilizing the phenoxide ion \(\implies\) Most acidic phenol.

- A is the baseline phenol.

- C has an \(-OCH_3\) group. Though it has a -I effect, its resonance donating effect (+M) heavily outweighs it at the para position, destabilizing the phenoxide ion \(\implies\) Least acidic phenol.

Thus, B > A > C.


Combining the two series:

D > E > B > A > C.


Step 4: Final Answer:

The descending order is D > E > B > A > C. Quick Tip: Always split mixed acidity/basicity lists by their primary functional groups first (e.g., sulfonic acids > carboxylic acids > phenols > alcohols). Only apply M/I effect comparisons within identical functional groups to save time.

Step 1: Understanding the Concept:

According to Brønsted-Lowry acid-base theory, the strength of a conjugate acid is inversely proportional to the strength of its corresponding base. Therefore, the "strongest conjugate acid" will be formed by protonating the "weakest base" among the options.


Step 2: Key Formula or Approach:

Strongest Conjugate Acid \(\iff\) Weakest Base.

Basicity of anilines depends on the availability of the lone pair on the nitrogen atom. Electron-withdrawing groups (EWG) pull electron density away from the ring via -M and -I effects, decreasing lone pair availability and reducing basicity.


Step 3: Detailed Explanation:

Let's analyze the given options (aniline derivatives):

- Image a: Aniline. The baseline aromatic amine.

- Imageb: p-Anisidine (\(OCH_3\) group at para position). The \(-OCH_3\) group is a strongly electron-donating group (+M effect), which increases electron density on the nitrogen, making it the strongest base.

- Image c: p-Nitroaniline (\(NO_2\) group at para position). The \(-NO_2\) group is a powerful electron-withdrawing group (-M and -I effects). It heavily delocalizes the nitrogen lone pair into the aromatic ring and towards the oxygen atoms, making the lone pair highly unavailable for protonation. This makes it the weakest base.

- Image c: p-Toluidine (\(CH_3\) group at para position). The \(-CH_3\) group is weakly electron-donating (+I and hyperconjugation), making it a slightly stronger base than aniline.


Order of basicity: p-Anisidine > p-Toluidine > Aniline > p-Nitroaniline.

Since p-Nitroaniline (Option C) is the weakest base, its protonated form (the conjugate acid) will be the strongest acid.


Step 4: Final Answer:

Option (C) represents the weakest base, thus yielding the strongest conjugate acid. Quick Tip: When asked for the "strongest conjugate acid," immediately cross it out and write "weakest base" to prevent mental mix-ups while evaluating inductive and mesomeric effects.

Step 1: Understanding the Concept:

Oxidation of an aldose with hot nitric acid (\(HNO_3\)) converts both the top aldehyde (\(-CHO\)) group and the bottom primary alcohol (\(-CH_2OH\)) group into carboxylic acid (\(-COOH\)) groups, yielding an aldaric acid. If the resulting aldaric acid is optically inactive, it must possess an internal plane of symmetry (a meso compound).


Step 2: Key Formula or Approach:

For a 4-carbon aldaric acid to have a plane of symmetry, the chiral centers at C2 and C3 must be exact mirror reflections of each other across the horizontal center plane of the Fischer projection. This means both \(-OH\) groups must lie on the exact same side.


Step 3: Detailed Explanation:

Let's analyze the properties required:

1. It is a D-sugar. By definition, in a Fischer projection, the \(-OH\) group on the highest numbered chiral carbon (C3 for an aldotetrose) must be on the right side.

2. It yields a meso aldaric acid. After \(HNO_3\) oxidation, the top and bottom groups are both \(-COOH\). To have a plane of symmetry, the \(-OH\) on C2 must be on the exact same side as the \(-OH\) on C3.

Since the C3 \(-OH\) is on the right (D-sugar), the C2 \(-OH\) must also be on the right.


Therefore, the structure of the D-aldotetrose must have:

- C1: \(-CHO\) (top)

- C2: \(-OH\) on the right

- C3: \(-OH\) on the right

- C4: \(-CH_2OH\) (bottom)

This specific sugar is known as D-erythrose.


Let's evaluate the options provided in the images:

- Image a : Both \(-OH\) groups are on the left. This is L-erythrose.

- Image b : C2 \(-OH\) on right, C3 \(-OH\) on left. This is D-threose (yields optically active aldaric acid).

- Image c : Both \(-OH\) groups are on the right. This is D-erythrose.

- Image d : C2 \(-OH\) on left, C3 \(-OH\) on right. This is L-threose.


Step 4: Final Answer:

Option (C) is correct as it represents D-erythrose. Quick Tip: Mnemonic for 4-carbon sugars: "Erythrose has groups on the Earth's Equator (same side). Threose has them Thrown apart (opposite sides)."

Step 1: Understanding the Concept:

In the systematic qualitative analysis of inorganic cations, different metal ions are precipitated sequentially using specific group reagents. We must identify which of the given metals specifically belongs to the analytical group whose reagent is \(H_2S\) gas passed in an alkaline medium (\(NH_4OH\)). After identifying the metal, we state its highest known oxidation state.


Step 2: Key Formula or Approach:

Group II reagent: \(H_2S\) in acidic medium (\(HCl\)). Precipitates \(Pb^{2+}, Cu^{2+}\).

Group III reagent: \(NH_4OH\) in presence of \(NH_4Cl\). Precipitates \(Fe^{3+}\) as \(Fe(OH)_3\).

Group IV reagent: \(H_2S\) in basic medium (\(NH_4OH\)). Precipitates \(Mn^{2+}, Zn^{2+}, Ni^{2+}, Co^{2+}\).


Step 3: Detailed Explanation:

Let's classify the given ions into their analytical groups:

1. \(Pb^{2+}\): Belongs to Group I (precipitates as chloride) and Group II (precipitates as sulfide in acidic medium).

2. \(Cu^{2+}\): Belongs to Group II. It precipitates as \(CuS\) when \(H_2S\) is passed in an acidic medium.

3. \(Fe^{3+}\): Belongs to Group III. Its specific group reagent is \(NH_4OH\) (with \(NH_4Cl\)), precipitating it as \(Fe(OH)_3\).

4. \(Mn^{2+}\): Belongs to Group IV. The specific group reagent for Group IV is \(H_2S\) passed in an alkaline medium provided by \(NH_4OH\). This precipitates manganese as Manganese(II) sulfide (\(MnS\)).


Thus, the metal that is specifically precipitated by \(H_2S\) in the presence of \(NH_4OH\) as its group reagent is Manganese (Mn).


The question asks for the highest possible oxidation state of this metal.

Manganese has the electronic configuration \([Ar] 3d^5 4s^2\). By losing all of its 4s and 3d valence electrons, it can achieve a maximum oxidation state of +7 (commonly seen in the permanganate ion, \(MnO_4^-\)).


Step 4: Final Answer:

The highest possible oxidation state is +7. Quick Tip: While \(H_2S\) in basic medium would technically precipitate Group II and III metals if they were still present, the phrase "as group reagent" restricts the target strictly to Group IV cations.

Step 1: Understanding the Concept:

We must identify the functional groups of the chemical structures provided in List-I and match them with the classic qualitative organic laboratory tests in List-II designed to detect those specific functional groups.


Step 2: Key Formula or Approach:

- Alcohols react with Lucas Reagent (\(ZnCl_2 / HCl\)).

- Amines react with Hinsberg's Reagent (Benzenesulfonyl chloride).

- Aldehydes react with Tollen's Reagent (Ammoniacal Silver Nitrate).

- Phenols react with Phthalic anhydride to give the Phthalein dye test.


Step 3: Detailed Explanation:

Let's analyze the visual structures and map them:

A. Cyclohexanol: This is an alicyclic ring with an \(-OH\) group. It is a secondary alcohol. Alcohols are identified using the Lucas test, where secondary alcohols produce turbidity after 5-10 minutes. (A \(\to\) III)


B. Cyclohexylamine: This is an alicyclic ring with an \(-NH_2\) group. It is a primary aliphatic amine. Amines are famously identified using Hinsberg's reagent test (forming a clear solution that precipitates upon acidification). (B \(\to\) I)


C. Cyclohexanecarbaldehyde: This is an alicyclic ring attached to a \(-CHO\) group. It is an aldehyde. Aldehydes uniquely reduce Tollen's reagent to form a silver mirror. (C \(\to\) IV)


D. Phenol: This is a benzene ring (indicated by the conjugated circle inside the hexagon) attached directly to an \(-OH\) group. Phenols undergo condensation with phthalic anhydride in the presence of concentrated \(H_2SO_4\) to form distinct indicator dyes (like phenolphthalein). This is the Phthalein dye test. (D \(\to\) II)


Matching sequence: A-III, B-I, C-IV, D-II.


Step 4: Final Answer:

Option (A) is the correct match. Quick Tip: Carefully distinguish between Cyclohexanol (no double bonds in the ring) and Phenol (benzene ring). While both have \(-OH\) groups, phenols fail the Lucas test and uniquely answer the Phthalein dye or \(FeCl_3\) tests.

Step 1: Understanding the Concept:

A bomb calorimeter measures the heat of reaction at constant volume, which corresponds to the change in internal energy (\(\Delta U_c^\circ\)). We must convert this experimental data into molar internal energy, then use the relation \(\Delta H = \Delta U + \Delta n_g RT\) to find the standard enthalpy of combustion (\(\Delta H_c^\circ\)). Finally, Hess's Law relates combustion enthalpy to the enthalpies of formation.


Step 2: Key Formula or Approach:

1. Moles \(n = \frac{Mass}{Molar Mass}\)

2. \(\Delta U_c^\circ = \frac{Heat}{moles}\)

3. \(\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT\)

4. \(\Delta H_c^\circ = \sum \Delta H_f^\circ (products) - \sum \Delta H_f^\circ (reactants)\)


Step 3: Detailed Explanation:

Molar mass of ethanol (\(C_2H_5OH\)) = \((2 \times 12) + (6 \times 1) + 16 = 46 g/mol\).

Moles of ethanol burnt \(n = \frac{3.365 g}{46 g/mol} = 0.073152 mol\).


Heat produced at constant volume (\(\Delta U\)) for these moles is \(-99.472 kJ\) (negative because heat is released).

Molar internal energy of combustion:
\(\Delta U_c^\circ = \frac{-99.472}{0.073152} = -1359.8 kJ/mol \approx -1360 kJ/mol\).


Write the balanced combustion equation for liquid ethanol:
\(C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)\)

Change in gaseous moles (\(\Delta n_g\)) = Moles of gaseous products - Moles of gaseous reactants
\(\Delta n_g = 2 (CO_2) - 3 (O_2) = -1\).


Convert \(\Delta U_c^\circ\) to \(\Delta H_c^\circ\):
\(\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT\)
\(\Delta H_c^\circ = -1360 kJ/mol + (-1) \times (8.314 \times 10^{-3} kJ/K\cdotmol) \times (298.15 K)\)
\(\Delta H_c^\circ = -1360 - 2.478 = -1362.478 kJ/mol\).


Use Hess's Law to find the enthalpy of formation of ethanol:
\(\Delta H_c^\circ = [2 \Delta H_f^\circ (CO_2) + 3 \Delta H_f^\circ (H_2O)] - [\Delta H_f^\circ (C_2H_5OH) + 3 \Delta H_f^\circ (O_2)]\)

We know \(\Delta H_f^\circ (O_2) = 0\).

The enthalpy of formation of \(CO_2\) is identical to the combustion of graphite = \(-393.5 kJ/mol\).
\(-1362.478 = [2(-393.5) + 3(-285.8)] - \Delta H_f^\circ (C_2H_5OH)\)
\(-1362.478 = [-787.0 - 857.4] - \Delta H_f^\circ (C_2H_5OH)\)
\(-1362.478 = -1644.4 - \Delta H_f^\circ (C_2H_5OH)\)
\(\Delta H_f^\circ (C_2H_5OH) = -1644.4 + 1362.478 = -281.922 kJ/mol\).


Rounding to the nearest integer, we get \(-282 kJ/mol\).


Step 4: Final Answer:

The value is -282. Quick Tip: "Bomb calorimeter" is the universal code word for "Constant Volume". Any heat value given for a bomb calorimeter directly provides \(\Delta U\), not \(\Delta H\). You must always manually add the \(\Delta n_g RT\) correction term.

Step 1: Write equilibrium moles


For reaction
\[ 2N_2O_5 \rightleftharpoons 2N_2O_4 + O_2 \]

Assume initially \(2\) moles of \(N_2O_5\).

Degree of dissociation:
\[ \alpha=0.5 \]

Hence equilibrium moles are
\[ N_2O_5 = 2(1-\alpha)=2(1-0.5)=1 \]
\[ N_2O_4 = 2\alpha = 1 \]
\[ O_2=\alpha=0.5 \]

Total moles:
\[ n_{total}=1+1+0.5=2.5 \]



Step 2: Find partial pressures


Given total pressure
\[ P=2\ atm \]

Using
\[ P_i = X_iP \]

For \(N_2O_5\):
\[ P_{N_2O_5}=\frac{1}{2.5}\times 2=0.8\ atm \]

For \(N_2O_4\):
\[ P_{N_2O_4}=\frac{1}{2.5}\times 2=0.8\ atm \]

For \(O_2\):
\[ P_{O_2}=\frac{0.5}{2.5}\times 2=0.4\ atm \]



Step 3: Calculate \(K_p\)

\[ K_p= \frac{(P_{N_2O_4})^2(P_{O_2})} {(P_{N_2O_5})^2} \]
\[ = \frac{(0.8)^2(0.4)}{(0.8)^2} \]
\[ =0.4 \]



Step 4: Use free energy relation


Formula:
\[ \Delta G^\circ=-RT\ln K_p \]

Temperature:
\[ T=50+273=323\ K \]

So,
\[ \Delta G^\circ = -8.314\times 323\times \ln(0.4) \]

Now
\[ \ln(0.4)=\ln\left(\frac{4}{10}\right) \]
\[ =\ln 4-\ln 10 \]

Using
\[ \ln 4 = 2\ln 2 \]

and
\[ \ln 2 = 2.303\log 2 = 2.303(0.30)=0.6909 \]
\[ \ln 4=2(0.6909)=1.3818 \]

Hence
\[ \ln(0.4)=1.3818-2.303=-0.9212 \]

Therefore
\[ \Delta G^\circ = -8.314\times 323\times (-0.9212) \]
\[ =2470\ J mol^{-1} \]



Final Answer:
\[ \boxed{2470\ J mol^{-1}} \] Quick Tip: For dissociation reactions with stoichiometric coefficients like \(2A \to 2B + C\), choosing the initial moles equal to the coefficient of the reactant (i.e., \(2\) instead of \(1\)) completely prevents dealing with fractions during the ICE table setup.

Step 1: Understanding the Concept:

By identifying the anode and cathode based on standard reduction potentials, we can establish the standard cell potential \(E^\circ_{cell}\). Using the Nernst equation alongside the given non-standard \(E_{cell}\) and the reaction quotient \(Q\), we can back-calculate \(n\), the total number of electrons exchanged. Analyzing the balanced redox equation reveals the specific oxidation state \(x\).


Step 2: Key Formula or Approach:

Cathode (Reduction) = Higher \(E^\circ_{red}\).

Anode (Oxidation) = Lower \(E^\circ_{red}\).
\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)

Nernst Equation: \(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q\)


Step 3: Detailed Explanation:

Identify the half-reactions:
\(M^{x+}/M\) has higher reduction potential (\(+0.15 V\)), so it acts as the Cathode (Reduction).
\(Fe^{3+}/Fe\) has lower reduction potential (\(-0.036 V\)), so it acts as the Anode (Oxidation).


Calculate \(E^\circ_{cell}\):
\(E^\circ_{cell} = 0.15 V - (-0.036 V) = 0.186 V\).


Use the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \]
Substitute the given values (\(E_{cell} = 0.2057 V\), \(Q = 10^{-2}\)):
\[ 0.2057 = 0.186 - \frac{0.059}{n} \log(10^{-2}) \] \[ 0.2057 - 0.186 = -\frac{0.059}{n} (-2) \] \[ 0.0197 = \frac{0.118}{n} \] \[ n = \frac{0.118}{0.0197} \approx 5.989 \approx 6 \]

The total number of electrons transferred in the balanced cell reaction is 6.

Let's construct the balanced reaction.

Anode: \(Fe \to Fe^{3+} + 3e^-\)

Cathode: \(M^{x+} + xe^- \to M\)


To balance the electrons, we multiply the anode reaction by \(x\) and the cathode reaction by \(3\):

Total electrons transferred \(n = 3x\).

Since we calculated \(n = 6\):
\(3x = 6 \implies x = 2\).

This is consistent with "M is a p-block metal" since elements like Lead (Pb) or Tin (Sn) commonly exhibit a \(+2\) oxidation state.


Step 4: Final Answer:

The value of \(x\) is 2. Quick Tip: If \(E_{cell} > E^\circ_{cell}\), it immediately implies that the reaction quotient \(Q < 1\). The logarithmic term \(\log Q\) becomes negative, mathematically adding to the standard potential.

Step 1: Understanding the Concept:

For a first-order reaction, the time required for the concentration to drop by a specific ratio is constant regardless of the starting concentration. We can find the rate constant \(k\) from the \(t=20\) data point, and then use it to find the time \(x\) corresponding to the intermediate drop. Alternatively, a simple logarithmic ratio scaling gives the answer directly.


Step 2: Key Formula or Approach:

First-order kinetics equation: \(k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]_t}\right)\)

Rearranging for time: \(t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)\)


Step 3: Detailed Explanation:

Let's analyze the concentration drops:

Initial concentration \([A]_0 = 0.6500 M\).


At \(t = 20 min\), \([A]_{20} = 0.00065 M\).

The ratio of concentrations is: \(\frac{[A]_0}{[A]_{20}} = \frac{0.6500}{0.00065} = 1000\).

So, \(k = \frac{1}{20} \ln(1000) = \frac{1}{20} \ln(10^3) = \frac{3 \ln 10}{20} min^{-1}\).


At \(t = x min\), \([A]_x = 0.0650 M\).

The ratio of concentrations is: \(\frac{[A]_0}{[A]_x} = \frac{0.6500}{0.0650} = 10\).

So, \(x = \frac{1}{k} \ln(10)\).


Substitute the value of \(k\) into the equation for \(x\):
\[ x = \frac{1}{\frac{3 \ln 10}{20}} \times \ln 10 \] \[ x = \frac{20}{3 \ln 10} \times \ln 10 \]
The \(\ln 10\) terms cancel out perfectly:
\[ x = \frac{20}{3} = 6.666\dots min \]

Rounding to the nearest integer, we get 7.


Step 4: Final Answer:

The value of \(x\) is 7. Quick Tip: Recognize powers of 10 in concentration drops! Dropping to \(1/10\)th takes time \(t_{90%}\). Dropping to \(1/1000\)th takes \(3 \times t_{90%}\). Since 3 cycles take 20 mins, one cycle takes \(20/3 = 6.67\) mins. No logs needed!

Step 1: Understanding the Concept:

This problem uses the Carius method for the quantitative estimation of sulfur. All the sulfur present in the known mass of the organic compound is converted quantitatively into a precipitable sulfate (\(BaSO_4\)). By weighing the precipitate, we determine the exact mass of sulfur, which is then expressed as a percentage of the original compound's mass.


Step 2: Key Formula or Approach:

Moles of S = Moles of \(BaSO_4 = \frac{Mass of BaSO_4}{Molar mass of BaSO_4}\)

Mass of S = Moles of S \(\times\) Atomic mass of S (32 g/mol)

Percentage of S = \(\frac{Mass of S}{Mass of compound X} \times 100\)


Step 3: Detailed Explanation:

First, find the actual mass of the organic compound (X) used in the experiment:

Mass of X = Moles of X \(\times\) Molar mass of X

Mass of X = \((2.0 \times 10^{-3} mol) \times 76 g/mol = 0.152 g\).


Next, find the mass of sulfur recovered from the \(BaSO_4\) precipitate:

Moles of \(BaSO_4\) produced = \(\frac{0.4813 g}{233 g/mol} = 0.0020657 mol\).

Since each mole of \(BaSO_4\) contains exactly one mole of Sulfur atoms:

Moles of S = \(0.0020657 mol\).

Mass of S = \(0.0020657 mol \times 32 g/mol = 0.0661 g\).


Now, calculate the percentage of Sulfur in the compound:
\[ % S = \frac{0.0661 g}{0.152 g} \times 100 \] \[ % S = 0.434868 \dots \times 100 = 43.4868\dots % \]

The question asks for the answer in the format \(\alpha \times 10^{-1} %\).
\(43.4868% = 434.868 \times 10^{-1} %\).


Rounding to the nearest integer, we get 435.


(Self-check alternative reasoning): Since \(0.002\) moles of compound yielded roughly \(0.002\) moles of S, there is exactly 1 Sulfur atom per molecule of X.

Theoretical percentage = \(\frac{Mass of 1 S atom}{Molar mass of X} \times 100 = \frac{32}{76} \times 100 = 42.1%\).

The experimental yield data gives \(43.5%\), which means the question specifically tests the calculation from the empirical \(BaSO_4\) data rather than the theoretical molecular formula deduction. Always trust the empirical data given.


Step 4: Final Answer:

The value is 435. Quick Tip: In quantitative estimation problems, ignore theoretical deductions (like assuming 1 atom of S per molecule) if direct empirical data (precipitate mass) is provided. Always calculate the percentage based exclusively on the given precipitate mass to avoid trap discrepancies.