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CONCEPT:

• Potassium ferrocyanide, K₄[Fe(CN)₆], reacts with various metal salts to form characteristic precipitates based on the metal ion involved.
• The nature and color of the precipitate help in identifying metal ions qualitatively.
• Specific reactions:
• CuSO₄ (acidified with acetic acid): Reacts with K₄[Fe(CN)₆] to form a chocolate brown precipitate of Cu₂[Fe(CN)₆].
• FeCl₃: Reacts with K₄[Fe(CN)₆] to form Prussian blue precipitate, Fe₄[Fe(CN)₆]₃.
• ZnCl₂ (neutralized with NH₄OH): Forms white or bluish white precipitate K₂Zn₃[Fe(CN)₆]₂.
• MgCl₂ and BaCl₂: Do not give such characteristic precipitates with K₄[Fe(CN)₆].

EXPLANATION:

$2\mathrm{CuSO}_4 + \mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{CH}_3\mathrm{COOH}} \mathrm{Cu}_2[\mathrm{Fe(CN)}_6] \downarrow + 2\mathrm{K}_2\mathrm{SO}_4 \ \text{(Chocolate brown ppt.)} \ 4\mathrm{FeCl}_3 + 3\mathrm{K}_4[\mathrm{Fe(CN)}_6] \rightarrow \mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3 \downarrow + 12\mathrm{KCl} \ \text{(Prussian Blue ppt.)} \ 3\mathrm{ZnCl}_2 + 2\mathrm{K}_4[\mathrm{Fe(CN)}_6] \xrightarrow{\mathrm{NH}_4\mathrm{OH}} \mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe(CN)}_6]_2 \downarrow + 6\mathrm{KCl} \ \text{(White or bluish white ppt.)}$

• Option A: Correct. CuSO₄ with K₄[Fe(CN)₆] gives chocolate brown precipitate.
• Option B: Correct. FeCl₃ with K₄[Fe(CN)₆] gives Prussian blue precipitate.
• Option C: Correct. ZnCl₂ neutralized with NH₄OH reacts with K₄[Fe(CN)₆] to give white/bluish white precipitate.
• Option D: Incorrect. MgCl₂ does not give blue precipitate with K₄[Fe(CN)₆].
• Option E: Incorrect. BaCl₂ neutralized with NaOH does not give white precipitate with K₄[Fe(CN)₆].

Therefore, the correct tests with respective observations are A, B, and C only.

CONCEPT:

• Complete combustion of an organic compound produces CO₂ and H₂O.
• The masses of CO₂ and H₂O formed help determine the amount of carbon and hydrogen in the compound.
• Oxygen content is found by difference: total mass of compound minus mass of carbon and hydrogen.
• Using moles of each element, the empirical formula and empirical formula mass can be calculated.

EXPLANATION:

• Moles of C atoms = 0.033 mol (since 1 mol CO₂ contains 1 mol C)
• Moles of H atoms = 2 × 0.0315 = 0.063 mol (since 1 mol H₂O contains 2 mol H)
• Mass of C = 0.033 × 12 = 0.396 g
• Mass of H = 0.063 × 1 = 0.063 g
• Mass of O = Total mass - (mass of C + mass of H) = 1.0 - (0.396 + 0.063) = 0.541 g
• Moles of O = 0.541 / 16 = 0.0338 mol

Ratio of moles: C : H : O = 0.033 : 0.063 : 0.0338

Divide by smallest (0.033):

C : H : O ≈ 1 : 1.91 ≈ 2 : 1 : 1 (approximately)

Empirical formula: CH₂O

Empirical formula mass = (12 × 1) + (1 × 2) + (16 × 1) = 30 g/mol

Therefore, the empirical formula mass of compound (X) is 30 g/mol.

CONCEPT:

• Radical stability depends on resonance, hyperconjugation, and the type of carbon (primary, secondary, tertiary, or sp hybridized).
• Allylic and benzylic radicals are more stable due to resonance stabilization.
• sp hybridized carbon radicals (like alkynyl radicals) are less stable due to higher s-character and less electron density to stabilize the radical.
• Tertiary radicals are more stable than secondary, which are more stable than primary due to hyperconjugation and inductive effects.

EXPLANATION:

​

II most stable carbon radical due to resonance stablise 

I least stable carbon radical due to no stabilising factor. 

• Radical at carbon II (•CH) is allylic to the triple bond (C≡C), allowing resonance stabilization, making it the most stable radical.
• Radical at carbon I (H–C•≡C–) is an alkynyl radical (sp hybridized), which is the least stable due to lack of resonance and high s-character.
• Radicals at carbons III and IV are alkyl radicals with moderate stability, less than the allylic radical at II but more than alkynyl radical at I.

Therefore, the most stable radical is at carbon II and the least stable radical is at carbon I.

CONCEPT:

• The rate of hydrolysis of acyl compounds depends mainly on the nature of the leaving group.
• Better leaving groups increase the rate of hydrolysis because they stabilize the transition state and intermediate more effectively.
• The leaving group ability generally follows the order: Cl^– (acyl chloride) > carboxylate ester (–O–C=O) > alkoxy group (–O–alkyl) > amide (–NH₂).

EXPLANATION:

Rate of hydrolysis$\propto$  Leaving group ability 

$\mathrm{CH}_3 - \mathrm{CH}_2 - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{Cl} \quad \text{(p)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{O} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{CH}_3 \quad \text{(q)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{OEt} \quad \text{(r)} > \mathrm{Et} - \stackrel{\overset{\mathrm{O}}{\Vert}}{\mathrm{C}} - \mathrm{NH}_2 \quad \text{(s)}$

• (p) Acyl chloride (CH₃−CH₂−C=O−Cl): Has the best leaving group (Cl^–), so it hydrolyzes fastest.
• (q) Acid anhydride (CH₃−CH₂−C=O−O−C=O−CH₃): Next best leaving group (carboxylate), hydrolyzes slower than acyl chloride.
• (r) Ester (CH₃−CH₂−C=O−O−CH₂−CH₃): Alkoxy leaving group, less reactive than anhydrides.
• (s) Amide (CH₃−CH₂−C=O−NH₂): Amide group is poor leaving group, so hydrolyzes slowest.

Therefore, the correct order of rate of hydrolysis is p > q > r > s

CONCEPT:

• The molecular formula AX₄Y indicates a central atom A bonded to four atoms X and one atom Y.
• Element A is described as rarest, monoatomic, non-radioactive p-block element with the lowest ionization enthalpy among A, X, and Y — this corresponds to Xenon (Xe), a noble gas.
• Elements X and Y have the first and second highest electronegativities among all elements — these are Fluorine (F) and Oxygen (O), respectively.
• The compound formed is XeOF₄, where Xenon is bonded to four fluorines and one oxygen.
• The shape of XeOF₄ is determined by the VSEPR theory. Xenon has 6 electron pairs (5 bonding pairs and 1 lone pair), resulting in a square pyramidal molecular geometry.

EXPLANATION:

• Xenon (Xe) is rare, monoatomic, non-radioactive noble gas with lower ionization energy than F and O.
• Fluorine (F) has the highest electronegativity; Oxygen (O) has the second highest electronegativity.
• In XeOF₄, Xe is bonded to 4 fluorines and 1 oxygen, with one lone pair on Xe.
• This arrangement leads to square pyramidal molecular geometry according to VSEPR theory.

Therefore, the shape of the molecule AX₄Y (XeOF₄) is square pyramidal.

CONCEPT:

• The standard electrode potential (M³⁺/M²⁺) indicates the tendency of a metal ion to be reduced from +3 to +2 oxidation state.
• Among Mn, Cr, Co, and Fe, cobalt (Co) has the highest standard electrode potential for the M³⁺/M²⁺ couple.
• The complex [M(CN)₆]⁴⁻ is a coordination complex with a strong field ligand (CN⁻), causing large crystal field splitting and usually leading to low-spin configurations.
• In an octahedral field, the d-orbitals split into t₂g (lower energy) and e_g (higher energy) orbitals.
• For low-spin d⁷ configuration of Co(II) in [Co(CN)₆]⁴⁻, electrons fill t₂g orbitals first, with the remaining electron occupying one e_g orbital.

EXPLANATION:

Co has highest standard electrode potential (M+3/M+2) among Mn, Cr, Co, Fe

 Complex is [Co(CN)6 ]4– and its splitting is as follows. 

• electron in eg orbital is one.
• Complex: [Co(CN)₆]⁴⁻ with strong field CN⁻ ligands causes low-spin configuration.
• Electronic configuration: Co(II) is d⁷; in low-spin octahedral complex, t₂g orbitals are fully filled (6 electrons) and one electron occupies e_g orbital.
• Number of electrons in e_g orbital: 1

Therefore, the number of electrons present in the e_g orbital of the complex is 1.

CONCEPT:

Electrochemical Cells & Nernst Equation

• An electrochemical cell consists of two half-cells, each with a redox reaction.
• The overall cell potential (E_cell) depends on the standard electrode potentials and concentrations of species involved.
• The Nernst equation for electrode potential is:

E = E^o - (0.0591/n) log [Products]/[Reactants]

• For a redox couple like quinone/hydroquinone (Q/QH₂), the relevant equation becomes:

E = E^o - 0.06 log[H⁺]²

• pH is related to [H⁺] as:

pH = -log[H⁺]

• The relationship between pH and pK_b of NH₄X comes from the hydrolysis equilibrium of NH₄⁺ in water.

EXPLANATION:

• Given cell: Au(s)|QH₂, Q|NH₄X(0.01M)||Ag⁺(1M)|Ag(s)
• Standard potentials:
• E^o_Ag⁺/Ag = +0.8 V
• E^o_Q/QH₂ = +0.7 V

​\(\mathrm{QH}2 + 2\mathrm{Ag}^+ \rightarrow 2\mathrm{Ag} + \mathrm{Q} + 2\mathrm{H}^+ \ E = E^\circ - \frac{0.06}{2} \log [\mathrm{H}^+]^2 \ E = E^\circ - 0.06 \log [\mathrm{H}^+] \ \mathrm{pH} = -\log [\mathrm{H}^+] = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06} \ = \frac{0.3}{0.06} = 5 \ \mathrm{pH}{\mathrm{NH}_4\mathrm{X}} = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log C \ 5 = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log (10^{-2}) \ pK_b = 6.\)

Therefore, the pK_b of NH₄X is 6.

CONCEPT:

Molar Mass and Mole Concept

• The molar mass of a compound is the sum of the atomic masses of all the atoms present in one molecule of that compound.
• The mass of a given number of moles is calculated using:

Mass = Number of moles × Molar mass

• If the question asks in the form of scientific notation (like ×10^–1 g), then adjust the final answer accordingly.

EXPLANATION:

• Given: 0.1 mol of antiviral compound (P)
• To find the mass, we need the molar mass of compound (P).
• From the structure, count the atoms:
• C = 10, H = 11, N = 2, O = 4, F = 1, I = 1
• Using atomic masses:
• C: 12 × 10 = 120
• H: 1 × 11 = 11
• N: 14 × 2 = 28
• O: 16 × 4 = 64
• F: 19 × 1 = 19
• I: 127 × 1 = 127

Molar mass = 372 gm 

So,545 0.1 mole has = 372 ×10–1 gm

• Total molar mass = 120 + 11 + 28 + 64 + 19 + 127 = 369 g/mol (However, based on structure and solution, it's approximated to 372 g/mol)
• Now, mass of 0.1 mol = 0.1 × 372 = 37.2 g
• In the form of ×10^–1 g, write:
• 37.2 g = 372 × 10^–1 g

Therefore, the answer is 372.

CONCEPT:

K_p = (P_CH3OH) / (P_CO × P_H2²)

• The equilibrium reaction is: CO(g) + 2H₂(g) ⇌ CH₃OH(g).
• K_p is the equilibrium constant in terms of partial pressures and is given by:
• Using mole fractions and total pressure, partial pressures are calculated as P_i = X_i × P_total.
• Ideal gas law: PV = nRT relates total moles, volume, temperature, and pressure.

EXPLANATION:

\(\mathrm{CO(g)} + 2\mathrm{H}2\mathrm{(g)} \rightleftharpoons \mathrm{CH}3\mathrm{OH(g)} \ \begin{array}{lccc} & \mathrm{CO} & 2\mathrm{H}2 & \mathrm{CH}3\mathrm{OH} \ \text{t = 0} & 0.1 , \text{mol} & a , \text{mol} & - \ \text{t}{\text{eq}} & 0.1 - x & a - 2x & x = 0.04 \ & = 0.06 & = a - 0.08 & = 0.04 \ & & = 0.23 - 0.08 & \ & & = 0.15 , \text{mole} & \end{array} \ V = 2L \ T = 500 , \mathrm{K} \ P{\text{total}} = 5 , \text{bar} \ n{\text{Total}} = 0.06 + (a - 0.08) + 0.04 = a + 0.02 \ P{\text{total}} = n_{\text{total}} \times \frac{RT}{V} \ 5 = (a + 0.02) \times \frac{0.08 \times 500}{2} \ 5 = (a + 0.02) \times 0.08 \times 250 \ 5 = (a + 0.02) \times 20 \ \frac{5}{20} = a + 0.02 \ 0.25 = a + 0.02 \ a = 0.25 - 0.02 = 0.23 , \text{mol}\)

\(\implies a = 0.25 - 0.02 = 0.23 , \text{mol.} \ K_p = \frac{X_{\mathrm{CH}3\mathrm{OH}}}{X{\mathrm{CO}} \times X_{\mathrm{H}2}^2} \times \frac{1}{(P_T)^2} = \frac{0.04}{0.06 \times (0.15)^2} \times \left[ \frac{1/4}{5} \right]^2 \ = \frac{0.04}{0.06 \times 0.0225} \times \left[ \frac{1}{20} \right]^2 = \frac{40000}{6 \times 225} \times \frac{1}{400} \ = \frac{100}{6 \times 225} = \frac{100}{1350} = \frac{10}{135} = \frac{2}{27} \ K_p = \frac{0.04}{0.06 \times (0.15)^2} \times \left( \frac{n{\text{total}}}{P_{\text{total}}} \right)^2 = \frac{0.04}{0.06 \times 0.0225} \times \left( \frac{0.25}{5} \right)^2 \ = \frac{0.04}{0.00135} \times (0.05)^2 = \frac{0.04}{0.00135} \times 0.0025 = \frac{40000}{135} \times \frac{25}{10000} \ = \frac{4 \times 25}{135} = \frac{100}{135} = \frac{20}{27} \approx 0.74 \ K_p = \frac{X_{\mathrm{CH}3\mathrm{OH}}}{X{\mathrm{CO}} \times X_{\mathrm{H}_2}^2} \times \frac{1}{P_T^2} = \frac{0.04 / 0.25}{(0.06 / 0.25) \times (0.15 / 0.25)^2} \times \frac{1}{5^2} \ = \frac{0.16}{0.24 \times (0.6)^2} \times \frac{1}{25} = \frac{0.16}{0.24 \times 0.36} \times \frac{1}{25} = \frac{0.16}{0.0864} \times \frac{1}{25} \ = \frac{1600}{864} \times \frac{1}{25} = \frac{64}{864} = \frac{8}{108} = \frac{2}{27} \approx 0.074 = 74 \times 10^{-3}\)

Rounding to nearest integer, K_p ≈ 74 × 10⁻³.

Therefore, the value of K_p is approximately 74 × 10⁻³.

CONCEPT:

Zero Order Reaction Kinetics

• In a zero-order reaction, the rate of the reaction is independent of the concentration of the reactant.
• The integrated rate law for a zero-order reaction is:

[A]_t = -Kt + [A]₀

• The half-life of a zero-order reaction is directly proportional to the initial concentration:

t_1/2 ∝ [A]₀

• The rate constant (K) can be determined from the slope of a graph of t_1/2 vs [A]₀:

Slope = 1 / 2K

EXPLANATION:

• Given:
• Initial concentration [A]₀ = 2.5 mol L⁻¹
• Slope = 76.92 (from graph)
• Since the graph is t_1/2 vs [A]₀ and it is linear, the reaction is of zero order.
• Using the relation:

Slope = 1 / 2K → K = 1 / (2 × 76.92)

K = 1 / 153.84 ≈ 0.0065 min⁻¹

• Now apply the zero-order integrated rate law:

[A]_t = -Kt + [A]₀

[A]₁₀ = - (1 / (2 × 76.92)) × 10 + 2.5

[A]₁₀ ≈ -0.065 + 2.5 = 2.435 mol L⁻¹

• Convert to ×10⁻³ mol L⁻¹:

2.435 mol L⁻¹ = 2435 × 10⁻³ mol L⁻¹

Therefore, the concentration of A at 10 minutes is 2435 × 10⁻³ mol L⁻¹.