CHEM J-3 June 2014

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1 CHEM J-3 June 2014 All terpenes are derived from isoprene and many, such as myrcene, (R)-citronellal and geraniol, are used in the perfume industry. Explain the differences in boiling points of these four compounds in terms of the type and size of the intermolecular forces present. All the molecules experience dispersion forces. Dispersion forces are related to the polarisability of a molecule and increase as the number of electrons in the molecule increases (i.e. they increase with molecular size). Dispersion forces are the only intermolecular forces present in isoprene and myrcene, but are stronger for the larger myrcene, so it has the higher boiling point. Myrcene, citronellal and geraniol are all of similar size, so have similar dispersion forces.. Citronellal has a polar C=O group so can engage in dipoledipole interactions so has a higher boiling point than myrcene. Geraniol contains an OH group so can engage in hydrogen bonding, a particularly strong intermolecular force, so it has a higher boiling point than citronellal.

2 CHEM N-7 November 2014 Complete the table concerning two of the isomers of C 3 H 6 O 2. Identify the geometry around each atom marked with an asterisk and the list the major intermolecular forces present in the liquid. 5 Isomer A B Chemical structure Geometry bent tetrahedral Major intermolecular forces in liquid H-bonding, dipole-dipole and dispersion dipole-dipole and disperson The boiling point of isomer A is 141 C and that of isomer B is 60 C. Explain why the boiling point of A is higher than B? The molecules are very similar in size so dispersion forces will be of similar magnitude in each. The strong hydrogen bonding possible for A is the major reason for its higher boiling point.

3 CHEM J-3 June 2013 Intermolecular forces are responsible for the physical properties of many compounds. What are dispersion forces? 3 Transient unsymmetrical electron distribution around an atom results in an instantaneous dipole that causes an induced dipole in an adjacent molecule. Dispersion forces are the attractions between the instantaneous dipoles and the induced dipoles. The boiling points of F 2, Cl 2 and Br 2 are 85, 239 and 338 K, respectively. Where would you expect the boiling point of I 2? Give reasons. All the molecules are non-polar, so dispersion forces are the only ones relevant. Iodine is the biggest atom with the most electrons and hence its electron cloud is the most polarisable. I 2 therefore has the strongest dispersion forces and the highest boiling point. Its boiling point will be higher than 338 K.

4 CHEM N-8 November 2013 In terms of the type and size of intermolecular forces involved, explain the trend in boiling points of the following compounds. 4 Substance Stick structure Boiling Point ( C) ethane, C 2 H methylpropane, C 4 H butane, C 4 H 10 1 water, H 2 O 100 Water has hydrogen bonds, so has the highest boiling point. All the others just have dispersion forces, so the number of electrons and the surface area determine the overall magnitude of the dispersion forces. Ethane is smallest molecule, so has lowest boiling point. Butane is a longer molecule than 2-methylpropane (which has a spherical shape). Butane is better able to entangle with other butane molecules and has larger contact surface area, so its boiling point is greater than that of 2-methylpropane. THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

5 CHEM J-15 June 2012 The figure below shows the boiling points of Group 14 and Group 17 hydrides as a function of the period (row) of the periodic table. 4 Boiling Point ( C) HF HCl HBr GeH4 HI SnH CH 4 SiH Period It is apparent from this figure that: - the tetrahydrides have lower boiling points than the monohydrides, - the boiling points increase with period, with the exception of HF. Explain these features. Boiling points depend on the strength of the intermolecular forces. Higher boiling points occur when these forces are strong and lower boiling points occur when these forces are weak. Three kinds of intermolecular forces are present in these molecules: Dispersion forces. These depend on the size of the molecule and the number of electrons. Dipole-dipole interactions. These require the presence of a dipole. Hydrogen bonds. They require H to be bonded to a very electronegative element like F. The tetrahydrides are all tetrahedral. As a result, none have dipole moments so there are no dipole-dipole interactions in tetrahydrides. The monohydrides are all polar, however. The higher boiling points in the monohydrides are due to the presence of these dipole-dipole interactions. Note that the each monohydride is isoelectronic with the tetrahydride in the same period so dispersion forces are quite similar. Down a period, the number of electrons is increasing and these are held further from the nucleus. As a result, dispersion forces increase and so boiling points increase. An exception to this is HF. Because H is bonded to the very electronegative F atom, strong hydrogen bonds are present between HF molecules. Despite the weak dispersion forces in HF, it has a high boiling point as a result of the hydrogen bonds.

6 CHEM N-7 November 2012 i) N 2 O is sparingly soluble in water. What does this tell you about the strength of any hydrogen bonding that exists? Rationalise your answer in terms of the structures of the H 2 O and N 2 O molecules. 2 As N 2 O is only sparingly soluble in water, it follows that any H-bonds from H 2 O to N 2 O must be quite weak. N 2 O can only act as an H-bond acceptor, not as a donor. As nitrogen and oxygen are of similar electronegativity, the N O bond is not as polarised as the O H bonds in water. As a consequence, any H-bonds formed between water and N 2 O will be weaker than those between 2 water molecules. THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

7 CHEM J-14 June 2010 Rationalise the order of the boiling points of the following liquids in terms of their intermolecular forces. 4 liquid F 2 HCl HBr HI Cl 2 HF Br 2 I 2 b.p. ( C) The boiling points in F 2, Cl 2, Br 2 and I 2 are determined by the size of the dispersion forces between molecules. The bigger the atoms, the more polarisable their electron clouds and the greater the dispersion forces. Hence boiling points are in order: I 2 > Br 2 > Cl 2 > F 2. Dispersion forces also operate in HF, HCl, HBr and HI, but here the dipole formed between the halogen atom and the hydrogen also needs to be considered. F is a very small and very electronegative atom. The H F bond is therefore highly polarised and H-bonds form in this liquid. These are much stronger than dispersion forces and so HF has an anomalously high boiling point. Cl, Br and I are not as electronegative as F: the dispersion forces in HCl, HBr and HI are more significant than the dipole-dipole forces as can be evidenced by the order of boiling points HF > HI > HBr > HCl. The values given tell us that the total of the dispersion forces in Br 2 is greater than the H-bonds in HF. Similar comparisons can be made between other members of the two series.

8 CHEM J-14 June 2009 Explain the trend in the following table in terms of the type and size of intermolecular forces. 6 Substance Boiling point ( o C) CH 3 CH 3 89 CH 3 CH 2 CH 2 CH 3 1 CH 3 CH 2 O CH 2 CH 3 35 CH 3 CH 2 OH 78 H 2 O 100 CH 3 CH 3 and CH 3 CH 2 CH 2 CH 3 have weak dispersion forces only, so have the lowest boiling points. CH 3 CH 2 CH 2 CH 3 has more atoms, so more dispersion forces and hence the higher boiling point of the two. CH 3 CH 2 OCH 2 CH 3 is similar in size to CH 3 CH 2 CH 2 CH 3, but has dipole-dipole forces as well due to the presence of polar C-O bonds. Thus, the boiling point of CH 3 CH 2 OCH 2 CH 3 is higher than that of CH 3 CH 2 CH 2 CH 3. CH 3 CH 2 OH and H 2 O have strong intermolecular H-bonds due to the presence of H atoms bonded to electronegative O atoms. Their boiling points are thus higher. Water has 2 H atoms and 2 lone pairs on O capable of H-bonding so can form on average 4 H-bonds per molecule. Ethanol just one H so can only form 1 H- bond per molecule. Water this has the higher boiling point. There are two isomers with the molecular formula C 4 H 10. CH 3H CH H 3 C C 3 CH 2 CH 2 CH 3 CH 3 butane 2-methylpropane Discuss which isomer will have the greater intermolecular forces. Both molecules have weak dispersion forces only. Butane has the greater dispersion forces as it has a greater surface area that can interact with other molecules. 2-Methylpropane is ball-like in structure with a smaller surface area will prevents close approach of the bonds.

9 CHEM J-10 June 2006 List the following five solids in order of increasing melting points. NaCl, H 2, CH 4, H 2 O, SiO 2 4 H 2 < CH 4 < H 2 O < NaCl < SiO 2 Briefly explain your ordering based on the types of forces that are involved. The interactions between the molecules of H 2 and between the molecules of CH 4 are dispersion forces, involving induced dipole-induced dipole interactions. CH 4 is a larger molecule with more electrons so has a larger polarizability than H 2 and, as a result, has stronger dispersion forces and a higher melting point. H 2 O has highly polar O-H bonds and lone pairs on the oxygen atom and hence quite strong H-bonds exist in the solid state. Hence, H 2 O has a higher melting point than CH 4. NaCl is an ionic compound with a close-packed arrangement of cations and anions which interact via strong coulombic interactions to give a solid with a high melting point. SiO 2 (s) is a covalent network with very strong Si-O bonds that need to be broken to melt the solid. Hence, SiO 2 has the highest melting point of these materials. List those that are electrical conductors when molten. Briefly explain your answers. Only NaCl conducts when molten as melting releases the Na + and Cl - ions from the lattice and these can carry electrical charge.

10 CHEM N-7 November 2006 List the following five solids in order of increasing melting points. NaCl, H 2, CH 4, H 2 O, SiO 2 4 H 2 < CH 4 < H 2 O < NaCl < SiO 2 Briefly explain your ordering based on the types of forces that are involved. The interactions between the molecules of H 2 and between the molecules of CH 4 are dispersion forces, involving induced dipole-induced dipole interactions. CH 4 is a larger molecule with more electrons so has a larger polarizability than H 2 and, as a result, has stronger dispersion forces and a higher melting point. H 2 O has highly polar O-H bonds and lone pairs on the oxygen atom and hence quite strong H-bonds exist in the solid state. Hence, H 2 O has a higher melting point than CH 4. NaCl is an ionic compound with a close-packed arrangement of cations and anions which interact via strong coulombic interactions to give a solid with a high melting point. SiO 2 (s) is a covalent network with very strong Si-O bonds that need to be broken to melt the solid. Hence, SiO 2 has the highest melting point of these materials. List those that are electrical conductors when molten. Briefly explain your answers. Only NaCl conducts when molten as melting releases the Na + and Cl - ions from the lattice and these can carry electrical charge.

No Brain Too Small CHEMISTRY AS91390 Demonstrate understanding of thermochemical principles and the properties of particles and substances

No Brain Too Small CHEMISTRY AS91390 Demonstrate understanding of thermochemical principles and the properties of particles and substances COLLATED QUESTIONS Attractive forces between atoms, ions, and molecules. These will include ionic bonds, covalent bonds, and intermolecular attractions due to temporary dipoles and permanent dipoles (including