Book1Part04.docx

1、Book1Part04Part 4 Bonding and Structure (Book 1, p.320 p.325)29. Ne and Ar: simple molecular structure with atoms attracted together by very weak instantaneous dipole-induced dipole interactions. Hence, both Ne and Ar have very low melting points. The strength of intermolecular forces increases with

2、 the number of electrons. Ar has a larger atomic size and has more electrons than Ne. Ar has a higher melting point than Ne. H2O: polar molecule. Strong hydrogen bonding exists between H2O molecules. Ice: simple molecular structure with each H2O molecule hydrogen-bonded to four other H2O molecules t

3、o give an open network structure. Ice or H2O(s) has a low melting point compared with ionic solids or giant covalent compounds. However, because of the presence of strong hydrogen bonding, its melting point is higher than other simple molecular substances with similar molecular masses. NaF: giant io

4、nic structure. Na+ and F are held in lattice sites by very strong ionic bonds. A large amount of energy is required to separate the ions for melting to occur very high melting point. Diamond: giant covalent structure with C atoms covalently bonded to each other. Each C atom is strongly bonded to fou

5、r other C atoms to give a giant network of C atoms. Therefore, infinite number of strong covalent bonds are to be broken in melting very high melting point.30. The larger the molecular size due to the greater number of electrons in the molecule, the stronger the intermolecular forces. More energy is

6、 required to separate the molecules during boiling and hence a higher boiling point is expected. B has a smaller molecular size than A. Hence, B is expected to have a boiling point lower than 17 C. C, having a larger molecular size (since it has more Cl atoms), has a higher boiling point than A.31.

7、(a) (i) BE = bond energy IE1 = first ionization energy IE2 = second ionization energy EA1 = first electron affinity LE = lattice energy (ii) Hf (CaF2(s) = Hat(Ca) + BE(F F) + IE1(Ca) + IE2(Ca) + 2 EA1(F) + LE(CaF2) 1220 = +178 + 158 + 590 + 1150 + 2(328) + LE(CaF2) LE(CaF2) = 2640 kJ mol1 (b) (i) Ca

8、F2(s) + H2SO4(aq) 2HF(g) + CaSO4(s) (ii) Hr = Hf (CaF2) Hf (H2SO4) + 2 Hf (HF) + Hf (CaSO4) = (1220) (814) + 2(271) + (1434) kJ mol1 = +58 kJ mol1 (iii) Since the forward reaction is endothermic, heating the system helps to drive the reaction forwards according to Le Chateliers principle.32.(a) Ther

9、e are 3 regions of electron clouds around B. To minimize electronic repulsion, they are directed in a trigonal planar manner. Since all are bonds and there is no lone electron pair, BF3 is trigonal planar. In both CF4 and NF3, there are 4 regions of electron clouds around C and N respectively. To mi

10、nimize electronic repulsion, they are directed tetrahedrally. In CF4, there is no lone electron pair. Hence, CF4 is tetrahedral. In NF3, there is one lone electron pair. Hence, NF3 is trigonal pyramidal. There are 6 regions of electron clouds around S. To minimize electronic repulsion, they are dire

11、cted octahedrally. Since there is no lone electron pair, SF6 is octahedral. (b) The bond enthalpy of C Cl is 340 kJ mol1, which is weaker than C F bond. Unlike C Cl bond, C F is not easily broken down to give F atoms. Hence, replacing CF2Cl2 by C2H2F4 prevents the formation of reactive halogen atoms

12、 and therefore helps to reduce ozone depletion.33. (a) (i) A: CH4, non-polar molecule Discrete molecules are held together by weak instantaneous dipole-induced dipole interactions very low boiling point. D: SiH4 similar to A Weak intermolecular forces (instantaneous dipole-induced dipole interaction

13、s) very low boiling point. Since SiH4 has more electrons than CH4, the strength of intermolecular forces is stronger. Therefore, the boiling point of SiH4 is higher than that of CH4. B: NH3 is a polar molecule with highly electronegative atom N. Since strong hydrogen bonding exists between NH3 molec

14、ules, the boiling point of B is higher than that of A and D. C: H2O is a polar molecule has highly electronegative atom O. The hydrogen bonding between H2O molecules is stronger than that between NH3 molecules because O is more electronegative than N. Therefore, the boiling point of H2O is higher th

15、an that of NH3. (ii) Boiling point of H2O Boiling point of HF Boiling point of NH3 (forms more hydrogen (F more bonds than HF) electronegative than N) Boiling point of GeH4 Boiling point of SiH4 GeH4 has more electrons, therefore stronger van der Waals forces are formed between the molecules.(b)(i)

16、(ii) Hydrogen bonds 34. (a) CH4(g) C(g) + 4H(g) H = 4 BE(C H) It is the average energy required to break 1 mole of C H bonds of methane. (b) (i) CH4(g) + H2O(g) CO(g) + 3H2(g) H1 = +206 kJ mol1 H1 = 4 BE(C H) + 2 BE (O H) BE(C O) 3 BE (H H) 260 = 4 410 + 2 460 BE(C O) 3 436 BE(C O) = 1046 kJ mol1 (i

17、i) BE(C O) = 360 kJ mol1 BE(C = O) = 740 kJ mol1 In carbon monoxide, the C O bond is a triple bond. It is stronger than a C = O double bond and a C O single bond.35. (a) (i) A: Enthalpy change of atomization of sodium B: First ionization energy of sodium C: Enthalpy change of neutralization (ii) D:

18、Enthalpy change of hydration of gaseous sodium ion and electron affinity of H+(aq) (b) (i) B: H = +494 kJ mol1 (ii) 58 494 107 + HE = 850 HE = 191 kJ mol136. (a) In NH3, polar N H bonds are present. In HCl, a polar H Cl bond is present. In both cases, the bonds are formed through the head-on overlap

19、 between orbitals. In CO2, C = O bonds are present. The bond is formed through a head-on overlap ( bond) and a side-way overlap ( bond) of orbitals. Between NH3 molecules, relatively strong hydrogen bonds exist. Hence, its boiling point is the highest among the three. Between HCl molecules, permanen

20、t dipole-permanent dipole interactions exist as a result of the polarity of HCl molecules. These interactions are not as strong as the hydrogen bonds in NH3. Hence, its boiling point is lower. In CO2, instantaneous dipole-induced dipole interactions exist between the non-polar molecules. Due to its

21、larger molecular size than HCl, such interactions are slightly stronger than those in HCl. Hence, it has a slightly higher boiling point than HCl.(b)37.(a)Aluminium chlorideAmmonia (b) A dative bond is formed between N and Al.(c)Aluminium chloride (trigonal planar)Ammonia (trigonal pyramidal)The pro

22、duct (all the bond angles are about 109.5)(d)(i)(ii)38. (a) Standard enthalpy change of formation of a compound is the energy change when 1 mole of that compound in the specified state is formed from its constituent elements at standard state at 298 K and 1 atm.(b) (c) 610 = +338 + 4(+122) 4 BE(Si C

23、l) BE(Si Cl) = +359 kJ mol1 (d) (i) SiCl3H + H2 Si + 3HCl (ii) BE(Si H) = 320 kJ mol1 At a high temperature, the heat energy supplied is sufficient to break the Si H bond which does not have an extremely high bond energy. (iii) Pure silicon is used in making semi-conductors.39. (a) (i) Na+(g) + Cl(g

24、) NaCl(s) H = lattice energy Lattice energy is the energy released when 1 mole of solid crystalline NaCl is formed from the combination of 1 mole of gaseous Na+ and 1 mole of gaseous Cl. (ii) Li+ has a smaller cationic size () than Na+. Hence, LiCl has a more exothermic lattice energy.(b) (i) LiCl :

25、 = + = 0.074 + 0.181 = 0.255 nm (ii) NaCl : = + = 0.095 + 0.181 = 0.276 nm40. As the molecular size and the number of electrons per molecule increase, the strength of van der Waals forces between the molecules also increase. More energy is required to separate the molecules during boiling. Hence, th

26、e boiling point increases from CH4 to C20H42. With increasing masses, the density is expected to increase since density is defined as mass per unit volume. However, due to the larger volume of the molecules from CH4 to C20H42, the increase in density is not as drastic as the increase in mass.41. (a)

27、 BF3 : trigonal planar CF4 : tetrahedral NF3 : trigonal pyramidal (b) BF3 has a vacant p orbital for occupation by the lone pair of electron of oxygen in water; whilst NF3 does not have such vacant orbital. BF3 + 3H2O B(OH)3 + 3HF (c) (i) Mole ratio of K : Sn : Cl : O : H =: = 0.56 : 0.28 : 0.85 : 0

28、.85 : 0.85 = 2 : 1 : 3 : 3 : 3 Hence, the empirical formula is K2SnCl3O3H3. (ii) SnCl3(OH)32; octahedral (iii) SnCl4 + 3KOH K2SnCl3(OH)3 + KCl42. (a) (i) The hydrogen bonds between the O H groups of glucose molecules are broken. Hydrogen bonds between the O H groups of glucose molecules and water mo

29、lecules are formed. (ii) (b) CO2(aq) + H2O(l) H2CO3(aq) Covalent bond43. In pentane, weak van der Waals forces exist between the molecules while strong hydrogen bonds are present in the other three substances. Hence, the least amount of energy is required to separate the molecules apart and pentane

30、has the lowest boiling point. The molecules are relatively far apart and it has the lowest density. Since it cannot form hydrogen bonds with water molecules, it is insoluble in water. On the other hand, the other three are soluble in water since they are able to form hydrogen bonds with water molecu

31、les. Propanoic acid forms the most extensive hydrogen bonds due to the presence of group and it has the shortest hydrocarbon chain. Hence, it has the highest boiling point and its molecules are relatively closer to each other compared to the alcohol and amine. Hence, it has the highest density. Oxygen is more electronegative than nitrogen. Hence, the hydrogen bonds formed