chemistry - unit 1 - section 3: bonding - exam questions

The boiling point of pentan-2-ol is 119 °C The boiling point of pent-1-ene is 30 °C Explain why pentan-2-ol has a higher boiling point than pent-1-ene.

M1 idea that pentan-2-ol has stronger intermolecular forces M1 idea that hydrogen bonds are stronger than van der Waals’ forces M2 pent-1-ene has van der Waals’ forces (only) M3 pentan-2-ol (also) has hydrogen bonds

Pent-1-ene is formed by the elimination of water from pentan-2-ol. State the reagent and condition for this reaction.

reagent = conc sulfuric acid or conc phosphoric acid

condition = hot / temperature in range 150-200°(C)

The melting point of XeF4 is higher than the melting point of PF3 Explain why the melting points of these two compounds are different. In your answer you should give the shape of each molecule, explain why each molecule has that shape and how the shape influences the forces that affect the melting point.

Stage 1 electron pairs

1a XeF4 4BP and 2LP around Xe

1b PF3 3BP and 1LP around P

Stage 2 explanation of shapes

2a XeF4 is square planar

2b PF3 is pyramidal (allow tetrahedral)

2c Electron pairs repel as far as possible or Lone pair repels more than bonding pairs

Stage 3 IMF

The relative strength of the intermolecular forces in the molecules must be explained to gain maximum marks.

3a XeF4 has vdw forces and PF3 has dipole-dipole forces (and vdw)

3b Stronger/more intermolecular forces in XeF4 3c Due to larger Mr or more electrons or larger molecules or packs more closely together

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