(a) Determine the electronic and molecular geometry of [BrF₅] and [ICl₂]⁻ interhalogen compounds. 10

(b) Answer the following questions based on the phase diagram given below :
(i) How many components does this phase diagram represent?
(ii) Identify

Question

(a) Determine the electronic and molecular geometry of [BrF₅] and [ICl₂]⁻ interhalogen compounds. 10

(b) Answer the following questions based on the phase diagram given below :
(i) How many components does this phase diagram represent?
(ii) Identify the points A to D with corresponding degrees of freedom.
(iii) Explain the changes expected in the paths E → B → H and E → F → G → H. 20

(c) Consider the reaction H₂(g) + ½O₂(g) → H₂O(l), which occurs in a H₂–O₂ fuel cell.
(i) Identify the elements that undergo oxidation and reduction.
(ii) Calculate the standard reaction Gibbs free energy (ΔᵣG°) at 25 °C.
(iii) Write down the two reduction half-reactions for the cell.
(iv) Calculate the E_cell.
Given that
Δ_fH°(H₂O, l) = –285.83 kJ mol⁻¹
S°_m(H₂O, l) = 69.91 J K⁻¹ mol⁻¹
S°_m(H₂, g) = 130.68 J K⁻¹ mol⁻¹
S°_m(O₂, g) = 205.14 J K⁻¹ mol⁻¹ 20

UPSC Answer Check · Accepted Answer

(a) Use VSEPR: steric number = σ bonds + lone pairs on the central atom; electron-pair geometry minimises repulsions, and molecular geometry is obtained by ignoring lone pairs.
BrF₅: Br has 7 valence e⁻. Five Br–F σ bonds use 5 e⁻, leaving 2 e⁻ as one lone pair. Steric number = 5 + 1 = 6. The six electron pairs are octahedral; one site is a lone pair, so the five F atoms form a square base and one apex. Final: BrF₅ has electronic geometry octahedral and molecular geometry square pyramidal (AX₅E).
[ICl₂]⁻: I has 7 valence e⁻; the negative charge adds 1 e⁻, and two I–Cl σ bonds account for 2 e⁻, giving (7 + 2 + 1)/2 = 5 electron pairs on I: two bonding pairs and three lone pairs. Steric number = 2 + 3 = 5. Electron-pair geometry is trigonal bipyramidal; the three lone pairs occupy equatorial positions, leaving the two Cl atoms axial. Final: [ICl₂]⁻ has electronic geometry trigonal bipyramidal and molecular geometry linear (AX₂E₃).

(b)(i) A component is an independent chemical constituent. The diagram has only one chemical species, one set of solid–liquid, liquid–vapour and sublimation curves, and one triple point; there is no composition axis. A binary system would require a third variable, usually composition. Final: the phase diagram represents one component, C = 1.

(b)(ii) Use Gibbs phase rule: F = C − φ + 2, where φ is the number of coexisting phases; valid for a non-reactive system with only T and P as intensive variables. For C = 1, F = 3 − φ.
A: solid + liquid on the melting curve, φ = 2, so F = 1. At 1 atm it is the normal melting point.
B: liquid + vapour on the liquid–vapour curve, φ = 2, so F = 1. At 1 atm it is the normal boiling point.
C: liquid + vapour on the liquid–vapour curve at higher T, φ = 2, so F = 1.
D: triple point, solid + liquid + vapour, φ = 3, so F = 0.
Final: A, B and C each have F = 1; D has F = 0.

(b)(iii) In the diagram, E, F, G and H lie in the gas region; a vertical segment is isothermal and a horizontal segment is isobaric.
Path E → B → H: E → B is isothermal compression of the gas until the liquid–vapour curve is reached at B. The gas becomes saturated; at B liquid and vapour coexist, so φ = 2 and F = 1. B → H is isothermal decompression into the gas region, so any liquid formed at B evaporates; H is a single gas phase, F = 2.
Path E → F → G → H: E → F is an isobaric temperature change, F → G is an isothermal pressure change, and G → H is an isobaric temperature change. Since the drawn route lies in the gas region, it remains a single gas phase, F = 2 throughout. No melting, condensation, or sublimation occurs.
Final: the only phase change in the first path is at B; the second route remains entirely gaseous.

(c)(i) Oxidation states: H in H₂(g) is 0 and in H₂O(l) is +1, so hydrogen loses electrons and is oxidised. O in O₂(g) is 0 and in H₂O(l) is −2, so oxygen gains electrons and is reduced. Final: H is oxidised; O is reduced.

(c)(ii) Use ΔᵣG° = ΔᵣH° − TΔᵣS° at T = 298.15 K. Standard enthalpies of formation of H₂(g) and O₂(g) in their reference states are zero; standard states are pure gases at the standard pressure and pure liquid water.
ΔᵣH° = ΔfH°(H₂O,l) − [ΔfH°(H₂,g) + ½ΔfH°(O₂,g)]
= −285.83 kJ mol⁻¹ − [0 + 0]
= −285.83 kJ mol⁻¹.
ΔᵣS° = S°ₘ(H₂O,l) − [S°ₘ(H₂,g) + ½S°ₘ(O₂,g)]
= 69.91 J K⁻¹ mol⁻¹ − [130.68 + ½(205.14)] J K⁻¹ mol⁻¹
= 69.91 J K⁻¹ mol⁻¹ − 233.25 J K⁻¹ mol⁻¹
= −163.34 J K⁻¹ mol⁻¹.
TΔᵣS° = 298.15 K × (−163.34 J K⁻¹ mol⁻¹) = −48.70 kJ mol⁻¹.
ΔᵣG° = −285.83 kJ mol⁻¹ − (−48.70 kJ mol⁻¹)
= −237.13 kJ mol⁻¹.
Final: ΔᵣG° = −237.13 kJ mol⁻¹ at 25 °C.

(c)(iii) In the acidic standard-state convention, the two reduction half-reactions are:
Hydrogen couple: 2H⁺(aq) + 2e⁻ → H₂(g)
Oxygen/water couple: ½O₂(g) + 2H⁺(aq) + 2e⁻ → H₂O(l)
In the operating H₂–O₂ fuel cell, the hydrogen couple runs in reverse: H₂(g) → 2H⁺(aq) + 2e⁻, while the oxygen/water couple runs as written. Final: the two reduction half-reactions are the hydrogen and oxygen/water couples above.

(c)(iv) Use ΔᵣG° = −nFE°cell, valid for a reversible cell under standard conditions. For the reaction as written, H₂ loses 2 e⁻, so n = 2 mol e⁻ per mol reaction; Faraday constant F = 96485 C mol⁻¹.
E°cell = −ΔᵣG°/(nF)
= −(−237.13 × 10³ J mol⁻¹)/(2 × 96485 C mol⁻¹)
= 237130 J mol⁻¹ / 192970 C mol⁻¹
= 1.229 V.
Final: E°cell = 1.229 V at 25 °C, standard states.

(a) Determine the electronic and molecular geometry of [BrF₅] and [ICl₂]⁻ interhalogen compounds. 10 (b) Answer the following…

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