1. 1
   Sulfur forms two fluorides, SF₄ and SF₆, and the noble gas xenon forms the fluoride XeF₄. In each molecule the central atom is bonded to every fluorine atom by a single bond. Sulfur has 6 valence electrons and xenon has 8.

   (a)

   What is the molecular geometry of SF₄?

   [1 mark]

   • ATetrahedral
   • BSquare planar
   • CSeesaw
   • DTrigonal pyramidal

   (b)

   Which gives the electron domain geometry and the molecular geometry of XeF₄?

   [1 mark]

   • AOctahedral and tetrahedral
   • BTetrahedral and tetrahedral
   • CTrigonal bipyramidal and seesaw
   • DOctahedral and square planar

   (c)

   Deduce, giving reasons, whether SF₄ and SF₆ are polar molecules.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Propenenitrile, CH₂=CH–C≡N, is the monomer used to make acrylic fibres. Its carbon atoms are numbered from the CH₂ end: C1 is the CH₂ carbon, C2 is the CH carbon and C3 is the carbon triple-bonded to nitrogen.

   (a)

   How many sigma (σ) and pi (π) bonds are present in one molecule of propenenitrile?

   [1 mark]

   • A6 σ and 3 π
   • B9 σ and 3 π
   • C6 σ and 2 π
   • D3 σ and 6 π

   (b)

   What is the hybridisation of C1, C2 and C3 respectively?

   [1 mark]

   • Asp³, sp², sp
   • Bsp², sp², sp
   • Csp², sp, sp
   • Dsp², sp², sp²

   (c)

   Deduce the approximate C1=C2–C3 and C2–C3≡N bond angles, explaining your answers using the VSEPR model.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Dinitrogen monoxide, N₂O, is a linear molecule in which the atoms are connected in the order N–N–O. Two possible Lewis formulas are proposed. In structure I, the two nitrogen atoms are joined by a triple bond and the central nitrogen is joined to oxygen by a single bond; the terminal nitrogen has one lone pair, the central nitrogen has none and the oxygen has three lone pairs. In structure II, both bonds are double bonds; the terminal nitrogen has two lone pairs, the central nitrogen has none and the oxygen has two lone pairs.

   (a)

   Calculate the formal charge on each atom in structure I.

   [3 marks]

   (b)

   Calculate the formal charges in structure II, deduce which structure is the preferred Lewis formula, and explain why structures I and II are described as resonance structures.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The carbon–carbon single bond in cyclohexane is 0.154 nm long and the carbon–carbon double bond in cyclohexene is 0.134 nm long. All six carbon–carbon bonds in benzene, C₆H₆, are 0.140 nm long. The enthalpy change for the hydrogenation of cyclohexene to cyclohexane is −120 kJ mol⁻¹; for benzene to cyclohexane it is −208 kJ mol⁻¹. Cyclohexene decolourises bromine water rapidly by addition, whereas benzene reacts with bromine only in the presence of a catalyst, forming C₆H₅Br by substitution. Only one compound with the structure 1,2-dibromobenzene exists. Benzene (M = 78.12 g mol⁻¹) boils at 80 °C and does not mix with water; methylbenzene, C₆H₅CH₃ (M = 92.15 g mol⁻¹), boils at 111 °C and does not mix with water; phenol, C₆H₅OH (M = 94.12 g mol⁻¹), boils at 182 °C and is moderately soluble in water.

   (a)

   Discuss the structure of benzene using the physical and chemical evidence.

   [6 marks]

   (b)

   Explain the differences in boiling point and solubility in water of benzene, methylbenzene and phenol, and evaluate whether the difference in molar mass alone accounts for the boiling points.

   [6 marks]

   Total for question 4: 12 marks

End of questions