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

(a) Geometry of BrF₅ and [ICl₂]⁻

By VSEPR, the electron-pair (electronic) geometry follows from the steric number (bond pairs + lone pairs), and the molecular geometry from the positions of the atoms only.

BrF₅: Br has 7 valence electrons; five are used in Br–F bonds, leaving 2 electrons, i.e. one lone pair. Steric number = 5 + 1 = 6, so the electronic geometry is octahedral. With one lone pair, the molecular geometry is square pyramidal (AX₅E), with the lone pair on one octahedral position and F atoms slightly bent away from it.

[ICl₂]⁻: I has 7 valence electrons, plus 1 for the negative charge, giving 8; two are used in I–Cl bonds, leaving 6 electrons, i.e. three lone pairs. Steric number = 2 + 3 = 5, so the electronic geometry is trigonal bipyramidal. The three lone pairs occupy the equatorial positions, which minimises repulsion, and the two Cl atoms occupy the axial positions, so the molecular geometry is linear (AX₂E₃).

(b) Phase diagram: components, degrees of freedom and paths

(i) The diagram is a pressure–temperature diagram of a single substance with solid, liquid and vapour regions and no composition axis, so it represents one component (C = 1). It is water-like, because the solid–liquid line slopes backwards.

(ii) Gibbs phase rule: F = C − P + 2 = 3 − P for one component.

• A: on the solid–liquid line at 1 atm, the normal melting point. Two phases coexist, so F = 1.
• B: on the liquid–vapour line at 1 atm, the normal boiling point. Two phases coexist, so F = 1.
• C: the critical point, the end of the liquid–vapour line, where liquid and vapour become identical. T and P are both fixed, so F = 0 (invariant).
• D: the triple point, where solid, liquid and vapour coexist, so F = 3 − 3 = 0.

(iii) Path E → B → H is an isothermal compression (vertical line) at a temperature below the critical temperature. The vapour at E is compressed until it reaches the liquid–vapour line at B, where it condenses to liquid at constant pressure (two phases, F = 1). Further compression to H gives a single liquid phase (F = 2). So there is a sudden change of phase, with a discontinuity in volume and in density, at B.

Path E → F → G → H first heats the vapour at constant pressure to F (above the critical temperature), compresses it isothermally to G (above the critical pressure), and cools it at constant pressure to H. The path goes around the critical point, so the fluid crosses no phase boundary. The density changes smoothly and the vapour becomes a liquid without a sudden condensation, showing the continuity of the liquid and gaseous states. The end state H is the same liquid in both paths.

(c) H₂–O₂ fuel cell: H₂(g) + ½O₂(g) → H₂O(l)

(i) Oxidation numbers: H changes from 0 to +1, so H is oxidised; O changes from 0 to −2, so O is reduced.

(ii) ΔᵣH° = ΔfH°(H₂O,l) = −285.83 kJ mol⁻¹, since H₂ and O₂ are elements in their standard states.

ΔᵣS° = S°(H₂O,l) − [S°(H₂,g) + ½S°(O₂,g)] = 69.91 − [130.68 + 102.57] = −163.34 J K⁻¹ mol⁻¹.

ΔᵣG° = ΔᵣH° − TΔᵣS° = −285.83 kJ − (298.15 K)(−0.16334 kJ K⁻¹) = −285.83 + 48.70 = −237.13 kJ mol⁻¹.

(iii) The two reduction half-reactions (acidic medium):

• 2H⁺(aq) + 2e⁻ → H₂(g) (hydrogen electrode)
• ½O₂(g) + 2H⁺(aq) + 2e⁻ → H₂O(l) (oxygen electrode)

In the cell, the first is reversed to H₂ → 2H⁺ + 2e⁻ (anode), and the second runs as written (cathode).

(iv) n = 2, F = 96485 C mol⁻¹. ΔᵣG° = −nFE°cell, so

E°cell = 237130 J mol⁻¹ / (2 × 96485 C mol⁻¹) = 1.23 V.