(a) The gas-phase reaction between methane (CH₄) and diatomic sulphur (S₂) is given by the following reaction: CH₄ (g) + 2S₂ (g)…

(a)

The temperature dependence of the rate constant is given by the Arrhenius equation. In its two-temperature form, ln(k₂/k₁) = Eₐ/R (1/T₁ − 1/T₂) so Eₐ = [R T₁ T₂/(T₂ − T₁)] ln(k₂/k₁).

Given: k₁ = 1.1 × 10⁻³ m³ mol⁻¹ s⁻¹ at T₁ = 823 K k₂ = 6.4 × 10⁻³ m³ mol⁻¹ s⁻¹ at T₂ = 898 K R = 8.3145 J K⁻¹ mol⁻¹

Thus, k₂/k₁ = (6.4 × 10⁻³)/(1.1 × 10⁻³) = 6.4/1.1 = 64/11 = 5.818 ln(k₂/k₁) = ln(5.818) = 1.761

Also, T₁T₂ = 823 × 898 = 739054 K² T₂ − T₁ = 898 − 823 = 75 K

Therefore, Eₐ = 8.3145 × 739054/75 × 1.761 = 8.3145 × 9854.053 × 1.761 = 1.4428 × 10⁵ J mol⁻¹ = 144.3 kJ mol⁻¹

Final answer: Eₐ ≈ 1.44 × 10⁵ J mol⁻¹ = 144.3 kJ mol⁻¹.

(b)

By the Beer–Lambert law, A = log₁₀(I₀/I) = ε c l where A is absorbance, I₀ is incident intensity, I is transmitted intensity, c is concentration, and l is path length.

For the original solution, c = 1 g per 100 cc and l = 1 cm. It transmits 60%, so I/I₀ = 0.60 A₁ = −log₁₀(0.60) = log₁₀(5/3) = 0.2218

(i) For c = 2 g per 100 cc in the same cell, the concentration is doubled. Since absorbance is proportional to concentration, A₂ = 2A₁ = 2 × 0.2218 = 0.4436 Hence transmittance is T₂ = 10⁻ᴬ² = 10⁻⁰·⁴⁴³⁶ = 0.36 So percentage transmitted = 36%. Therefore percentage absorbed = 100 − 36 = 64%.

Final answer: 64% of the light is absorbed.

(ii) For 90% absorption, transmitted light is 10%, so T = 0.10. Required absorbance: A = −log₁₀(0.10) = 1

For the original solution, absorbance per cm is A₁ = 0.2218. Since A ∝ l, l = A/A₁ = 1/0.2218 = 4.51 cm

Exact value: l = 1/log₁₀(5/3) cm = 4.5076 cm

Final answer: required cell thickness ≈ 4.51 cm.

This assumes monochromatic blue light, no scattering, and validity of the Beer–Lambert linear region.

(c)

The volume of N₂ required for monolayer formation is 155.5 cm³ g⁻¹ of charcoal. At 0°C and 1 atm, molar volume is 22.414 dm³ mol⁻¹ = 22414 cm³ mol⁻¹.

Number of moles of N₂ in monolayer per gram: n = 155.5/22414 = 6.938 × 10⁻³ mol

Number of N₂ molecules per gram: N = n × Nₐ = 6.938 × 10⁻³ × 6.022 × 10²³ N = 4.178 × 10²¹ molecules g⁻¹

Area of cross-section of one N₂ molecule: 0.160 nm² = 0.160 × 10⁻¹⁸ m² = 1.60 × 10⁻¹⁹ m²

Therefore surface area per gram: A = N × area per molecule = 4.178 × 10²¹ × 1.60 × 10⁻¹⁹ m² g⁻¹ = 668.5 m² g⁻¹

In cm² g⁻¹: A = 668.5 × 10⁴ = 6.685 × 10⁶ cm² g⁻¹

Final answer: surface area ≈ 668.5 m² g⁻¹ = 6.685 × 10⁶ cm² g⁻¹.

This assumes monolayer adsorption and a close-packed arrangement of N₂ molecules.

(d)

For the decomposition of ammonium nitrite, NH₄NO₂(aq) → N₂(g) + 2H₂O(l)

The volume of N₂ liberated is proportional to the amount of ammonium nitrite decomposed. Let V∞ = 35.05 cc be the final volume at infinite time. For a first-order reaction, k = (2.303/t) log₁₀[V∞/(V∞ − Vₜ)]

For t = 10 min: V∞ − Vₜ = 35.05 − 6.25 = 28.80 cc k = (2.303/10) log₁₀(35.05/28.80) = 0.2303 × log₁₀(1.2170) = 0.2303 × 0.0853 = 1.96 × 10⁻² min⁻¹

For t = 15 min: V∞ − Vₜ = 35.05 − 9.00 = 26.05 cc k = (2.303/15) log₁₀(35.05/26.05) = 0.1535 × log₁₀(1.3455) = 0.1535 × 0.1289 = 1.98 × 10⁻² min⁻¹

For t = 20 min: V∞ − Vₜ = 35.05 − 11.40 = 23.65 cc k = (2.303/20) log₁₀(35.05/23.65) = 0.11515 × log₁₀(1.4820) = 0.11515 × 0.1709 = 1.97 × 10⁻² min⁻¹

For t = 25 min: V∞ − Vₜ = 35.05 − 13.65 = 21.40 cc k = (2.303/25) log₁₀(35.05/21.40) = 0.09212 × log₁₀(1.6379) = 0.09212 × 0.2143 = 1.97 × 10⁻² min⁻¹

The rate constant is nearly constant: k ≈ 1.96 × 10⁻², 1.98 × 10⁻², 1.97 × 10⁻², 1.97 × 10⁻² min⁻¹.

Since k is constant within experimental error, the reaction is of the first order. Average k ≈ 1.97 × 10⁻² min⁻¹.

(e)

The active site of ferredoxin containing a [4Fe–4S] cluster is a distorted cubane-like Fe₄S₄ core. Four iron atoms and four inorganic μ₃-sulfide ions occupy alternate corners of a cube. Each iron atom is tetrahedrally coordinated by three inorganic μ₃-S²⁻ ions and one thiolate sulfur atom from a cysteine residue of the protein.

The core may be written as Fe₄(μ₃-S)₄(S-Cys)₄: the Fe₄ tetrahedron and the S₄ tetrahedron interpenetrate to form the distorted cube, and the four cysteine sulfur atoms point outward from the four iron corners, anchoring the cluster to the protein chain.

For the average oxidation number of iron, let the average oxidation state of Fe be x. The inorganic sulfide is S²⁻. For oxidized ferredoxin, the cluster is commonly [Fe₄S₄]²⁺: 4x + 4(−2) = +2 4x − 8 = +2 4x = +10 x = +2.5

For reduced ferredoxin, the cluster is [Fe₄S₄]⁺: 4x + 4(−2) = +1 4x − 8 = +1 4x = +9 x = +2.25

Final answer: in oxidized [4Fe–4S] ferredoxin, average Fe oxidation number = +2.5; in reduced ferredoxin, it is +2.25.

(f) Crystal field splitting energy Δ₀ depends on the oxidation state of the metal, the principal quantum number of the metal d orbitals, and the position of the ligand in the spectrochemical series.

(i) [Co(H₂O)₆]³⁺ or [Co(H₂O)₆]²⁺ Co³⁺ has a higher positive charge and smaller ionic radius than Co²⁺. Therefore it attracts H₂O ligands more strongly, decreases metal–ligand bond length, and produces larger crystal field splitting. Hence Δ₀ for [Co(H₂O)₆]³⁺ is larger.

Final: [Co(H₂O)₆]³⁺ has higher Δ₀.

(ii) [Co(NH₃)₆]³⁺ or [Rh(NH₃)₆]³⁺ Both have the same ligand and +3 oxidation state, but Rh belongs to the 4d series while Co belongs to the 3d series. The 4d orbitals are more diffuse and overlap more effectively with ligand orbitals, giving stronger metal–ligand interaction. Also, 4d and 5d metals generally have much larger Δ₀ than 3d metals.

Final: [Rh(NH₃)₆]³⁺ has higher Δ₀.

(iii) [Co(H₂O)₆]³⁺ or [Co(NH₃)₆]³⁺ Both contain Co³⁺ and have octahedral geometry. According to the spectrochemical series, NH₃ is a stronger field ligand than H₂O: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO Hence NH₃ causes greater splitting than H₂O.

Final: [Co(NH₃)₆]³⁺ has higher Δ₀.

(g) Complete reactions:

(i) S₄N₄ + SbCl₅ → S₄N₄·SbCl₅ Tetrasulfur tetranitride forms a Lewis acid–base adduct with antimony pentachloride.

(ii) XeF₆ + PtF₅ → [XeF₅]⁺[PtF₆]⁻ Xenon hexafluoride donates fluoride ion to PtF₅, giving the XeF₅⁺ cation and PtF₆⁻ anion.

(iii) B₂H₆ + 2NH₃ → [BH₂(NH₃)₂]⁺[BH₄]⁻ At −120°C, diborane reacts with ammonia to form the diammoniate of diborane.

(iv) 8KrF₂ + 2Au → 2[KrF]⁺[AuF₆]⁻ + 6Kr + F₂ Krypton difluoride, a very strong oxidative fluorinating agent, oxidises gold to Au(V) and forms the salt [KrF]⁺[AuF₆]⁻ (the usual stoichiometry is 7KrF₂ + 2Au → 2[KrF][AuF₆] + 5Kr; the eighth KrF₂ decomposes to Kr + F₂). On gentle warming the salt gives AuF₅.

(v) XeOF₄ + XeO₃ → 2XeO₂F₂ Xenon oxytetrafluoride and xenon trioxide undergo fluoride–oxide redistribution to give xenon dioxydifluoride.

(h) For d-block elements, especially 3d transition metals, the magnetic moment is largely due to spin motion of electrons because the orbital angular momentum is largely quenched. In an octahedral or tetrahedral crystal field, the d orbitals interact strongly with ligands, lifting their degeneracy. The orbital contribution is reduced or quenched by the crystal field, so the spin-only formula is usually valid: μ_eff = √[n(n + 2)] BM where n is the number of unpaired electrons.

For f-block elements, especially lanthanides, the 4f orbitals are deeply buried inside the atom and are shielded from the ligand field by the outer 5s² and 5p⁶ electrons. Therefore, the crystal field effect on 4f electrons is very small compared with spin–orbit coupling. As a result, orbital angular momentum is not quenched. The magnetic moment depends on the total angular momentum J, obtained by Russell–Saunders coupling of L and S: μ_eff = g_J √[J(J + 1)] BM where g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)]

Thus, for d-block ions, spin-only behaviour often dominates, whereas for f-block ions, both spin and orbital contributions must be considered. An exception is Gd³⁺, for which L = 0, so its magnetic moment is approximately spin-only.