Worked solutions to student book questions Chapter 7 Covalent molecules, networks and layers

Transcription

1 E1. a Give the electronic configuration for an atom of beryllium. b How many electrons are in the outer shell of an atom of beryllium in the molecule BeH 2? AE1. a 1s 2 2s 2 b 4 E2. The noble gases helium and neon do not form any compounds. The noble gas xenon, however, does form a covalent molecular compound with fluorine. Suggest a reason for this difference. AE2. The two valence electrons of helium fill helium s outer shell (shell number 1 can contain a maximum of 2 electrons). Similarly, the outer shell of neon, shell number 2, is filled with neon s 8 valence electrons. So, helium and neon have filled outer shells and do not form compounds by sharing electrons with other non metals. Xenon also has 8 electrons in its outer shell. However, the outer shell of xenon is shell number 5, which can hold a maximum of 50 electrons. So, under appropriate circumstances, xenon can form covalent compounds in which xenon shares electrons with another non-metal, such as fluorine. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

2 Q1. Draw electron dot and valence structures for each of the following molecules: fluorine (F 2 ), hydrogen fluoride (HF), water (H 2 O), tetrachloromethane (CCl 4 ), phosphine (PH 3 ), butane (C 4 H 10 ), carbon dioxide (CO 2 ). A1. fluorine (F 2 ) hydrogen fluoride (HF) tetrachloromethane (CC1 4 ) water (H 2 O) butane (C 4 H 10 ) phosphine (PH 3 ) carbon dioxide (CO 2 ) Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 2

3 Q2. What is the maximum number of covalent bonds an atom of each of the following elements can form? a F b O c N d C e H f Ne A2. a 1 b 2 c 3 d 4 e 1 f 0 Q3. A knowledge of electronic configurations would lead you to predict that H 2 O is the empirical formula for water. Why? A3. Oxygen has six outer-shell electrons and needs two more electrons to complete its outer shell. It gains access to two more electrons by forming two covalent bonds with two different hydrogen atoms. Each of the hydrogen atoms gains an electron and also completes its outer shell. Q4. When oxygen forms covalent molecular compounds with other non-metals, the valence structures that represent the molecules of these compounds all show each oxygen atom with two lone non-bonding electron pairs. Why are there always two lone pairs? A4. To complete its outer shell, the oxygen uses two of its outer-shell electrons to form two single bonds or a double bond with suitable non-metal atoms. The remaining four electrons in the outer shell are not required for bonding, as the outer shell is now complete, and they arrange themselves as two lone pairs around the oxygen atom. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 3

4 Q5. Suggest the most likely formula of the compound formed between the following pairs of elements: a C, Cl b N, Br c Si, O d H, F e P, F A5. a CCl 4 b NBr 3 c SiO 2 d HF e PF 3 Q6. How many lone pairs would you expect atoms of the following elements to have when they form covalent bonds with other non-metal atoms? a H b F c C d N A6. a 0 b 3 c 0 d 1 Q7. Draw the valence structure for each of the following molecules: a H 2 S b HI c CCl 4 d PH 3 e CS 2 A7. a b Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 4

5 c d e Q8. Describe the shape of each of the molecules given in Question 7. A8. a V-shaped b linear c tetrahedral d pyramidal e linear Q9. Covalent bonds can form between the following pairs of elements in a variety of compounds. Use the electronegativity values given in Table 7.4 (page 123) to identify the atom in each pair that would have the larger share of bonding electrons: a S and O b C and H c C and N d N and H e F and O f P and F A9. a O b C c N d N e F f F Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 5

6 Q10. The greater the differences in electronegativity between two atoms, the more polar is the bond formed between them. a Which of the examples in Question 9 would be the most polar bond? b Which of the examples in Question 9 would be the least polar bond? A10. a P F b C H Q11. Figure 7.30 (page 126) represents models for a number of molecules. Examine the models and identify the polar molecules. A11. CH 3 OH; CH 3 F Q12. Which of the molecules in Question 11 would be capable of forming hydrogen bonds? A12. CH 3 OH. Hydrogen bonding is often called FON bonding because fluorine, oxygen and nitrogen are almost always the other elements involved. Q13. Consider each of the following substances. In which ones are the molecules held to one another by: i dipole dipole attraction? ii hydrogen bonds? a NH 3 b CHCl 3 c CH 3 Cl d F 2 O e HBr f H 2 S g HF h H 2 O i H 2 A13. i a to h ii a, g, h Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 6

7 Q14. In ice, each water molecule is surrounded, at equal distances, by four other water molecules. In each case, there is an attraction between the positive hydrogen atom on one water molecule and a lone pair associated with the oxygen atom of another water molecule. Draw a diagram to show the arrangement of four water molecules around another water molecule. A14. Because of hydrogen bonding, ice is less dense than liquid water, and so ice floats on water. (For most liquids, the solid is denser than the liquid.) This is good news for fish, but not good news for travellers on the Titanic! A15. Cloudy ammonia is often used as a cleaning solution in bathrooms. This solution contains ammonia dissolved in water. Draw a diagram to represent hydrogen bonding between a water molecule and an ammonia molecule. A15. E3. The molecular formula of ethane and propane are C 2 H 6 and C 3 H 8, respectively. For each, give the: a empirical formula b structural formula c semistructural formula Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 7

8 AE3. a CH 3 ; C 3 H 8 b c CH 3 CH 3 ; CH 3 CH 2 CH 3 Q16. Explain the following properties of: i diamond, and ii graphite in terms of their respective structures: a high melting temperature b hardness or softness c ability or inability to conduct electricity A16. a i A lot of energy is required to disrupt the strong covalent bonding in three dimensions in diamond. ii A large amount of energy is required to disrupt the strong covalent bonding in two dimensions in graphite. Therefore, diamond and graphite both have high melting temperatures. b i Diamond is hard because it has strong covalent bonds throughout the lattice, with all atoms being held in fixed positions. ii Graphite is soft because there are weak dispersion forces between the layers in graphite, so layers can be made to slide over each other easily. c i Diamond is a non-conductor of electricity because all of its electrons are localised in covalent bonds and are not free to move. ii Graphite is able to conduct electricity because it has delocalised electrons between its layers of carbon atoms. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 8

9 Q17. Explain the following uses in terms of the structures of graphite and diamond: a Graphite is used as a lubricant. b Diamond is often used as an edge on saws and a tip on drills. A17. a Graphite is used as a lubricant because it is greasy. The dispersion forces between the layers in graphite enable the layers to slide over each other easily. b The strong covalent bonding throughout the lattice means that the carbon atoms are fixed in place. This makes the diamond very hard and suitable as a material for cutting other less hard materials. Chapter review Q18. Draw electron dot formulas for each of the following molecules and identify the number of bonding and non-bonding electrons in each molecule: a HBr b H 2 O 2 c CF 4 d C 2 H 6 e PF 3 f Cl 2 O g CH 4 h H 2 S A18. a 2 bonding electrons, 6 non-bonding electrons b 6 bonding electrons, 8 non-bonding electrons c 8 bonding electrons, 24 non-bonding electrons d 14 bonding electrons, 0 non-bonding electrons Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 9

10 e 6 bonding electrons, 20 non-bonding electrons f 4 bonding electrons, 16 non-bonding electrons g 8 bonding electrons h 4 bonding electrons, 4 non-bonding electrons Q19. Identify the number of bonding and non-bonding electrons in the following molecules: a N 2 b CHCl 3 c O 2 A19. a 3 bonding pairs, 2 non-bonding pairs b 4 bonding pairs, 9 non-bonding pairs c 2 bonding pairs, 4 non-bonding pairs Q20. Draw valence structures for the following molecules. Underneath each structure indicate whether the molecules are: i symmetrical or non-symmetrical ii polar or non-polar a CO 2 b PH 3 c N 2 A20. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 10

11 Q21. Are the following molecules polar or non-polar? Draw valence structures or make models to help you to decide. a CS 2 b Cl 2 O c SiH 4 d CH 3 Cl e CH 3 CH 3 f CCl 4 A21. d e f Q22. Use Table 7.4 (page 123 of the student book) to determine which of the following molecules contains the most polar bond: a CO 2 b H 2 O c H 2 d H 2 S e NH 3 A22. The O H bond in water is the most polar bond. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 11

12 Q23. For each of the structures shown below, state whether: i the molecule is polar or non-polar ii the strongest intermolecular forces of attraction between molecules of each type would be dispersion forces, hydrogen bonding or dipole dipole attraction. A23. SO 3 i non-polar; ii dispersion forces SiCl 4 i non-polar; ii dispersion forces CF 4 i non-polar; ii dispersion forces NF 3 i polar; ii dipole dipole attraction CH 3 NH 2 i polar; ii hydrogen bonding Q24 Consider solid samples of the following compounds. In which cases will the only forces between molecules in the samples be dispersion forces? (You should first ascertain whether molecules of these compounds are polar or non-polar. You can do this by drawing an accurate structural formula for each one.) a tetrachloromethane CCl 4 (s) b sulfur dioxide SO 2 (s) c carbon dioxide CO 2 (s) d hydrogen sulfide H 2 S(s) A24. Compounds CCl 4 and CO 2 are both non-polar, and so intermolecular forces operating between these molecules will be dispersion forces. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 12

13 Q25. The melting temperatures of four of the halogens are given in the table below. (Refer to the periodic table on page 408 to establish where the halogens occur in the table.) Describe and explain the trend in melting temperatures of these elements. Halogen Melting temperature ( C) Fluorine (F 2 ) 220 Chlorine (Cl 2 ) 101 Bromine (Br 2 ) 7 Iodine (I 2 ) 114 A25. Melting temperatures increase down the table because the molecules increase in mass and size and there are more electrons in the molecules; therefore, the strength of the dispersion forces increases. Q26. Suppose that you had samples of the two compounds OF 2 and CF 4. Between molecules of which sample would the intermolecular forces of attraction be greater? Explain your answer. A26. CF 4 has a slightly higher boiling temperature ( 128 C) than OF 2 ( 145 C), indicating that the forces between molecules in CF 4 are stronger. OF 2 is slightly polar; CF 4 is non-polar. OF 2 molecules are held together by forces of dipole dipole attraction and dispersion forces. Although CF 4 molecules are attracted by dispersion forces only, the much larger size of CF 4 molecules makes the dispersion forces stronger than the sum of the dipole dipole forces and the dispersion forces between OF 2 molecules. Q27. The mass of a hydrogen fluoride molecule is similar to the mass of a neon atom. The boiling temperatures of these substances are very different, however. That of hydrogen fluoride is 19.5 C and that of neon is 246 C. Explain the difference in this property of the two substances. A27. Neon exists as single atoms, with the only forces of attraction being dispersion forces. As Ne atoms have very few electrons, the dispersion forces are extremely weak. Neon therefore has a very low boiling temperature. Hydrogen fluoride molecules, however, are very polar and so are held together by electrostatic attraction between permanent dipoles. Because hydrogen is bonded to the very electronegative fluorine, the forces between molecules are known as hydrogen bonds. These are relatively strong intermolecular bonds and HF, therefore, has a much higher boiling temperature than Ne. (The dispersion forces operating between HF molecules are extremely weak.) Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 13

14 Q28. At room temperature, CCl 4 is a liquid whereas CH 4 is a gas. a Which substance has the stronger intermolecular attractions? b Explain the difference in the strengths of the intermolecular attractions. A28. a CCl 4 b CH 4 and CCl 4 are both non-polar and so are held together in a lattice only by dispersion forces. CCl 4 is the larger of these two molecules and has more electrons, so the dispersion forces between CCl 4 molecules will be greater than those between CH 4 molecules. As there are stronger dispersion forces between molecules of CCl 4 than for CH 4, it takes more energy to vaporise CCl 4. Q29. What are the forces of attraction between the following molecules? a H 2 b HCl c NH 3 d CH 4 e H 2 O f C 2 H 6 A29. a dispersion forces b dispersion forces, dipole dipole attractions c dispersion forces, hydrogen bonding d dispersion forces e dispersion forces, hydrogen bonding f dispersion forces Q30. Silicon carbide is a substance that is almost as hard as diamond and is used as a commercial abrasive. It is made by heating silica and carbon to a very high temperature. Silicon carbide consists of a tetrahedral covalent network lattice, containing alternating silicon and carbon atoms. Draw a section of this lattice. A30. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 14

15 Q31. Look at the data for the melting temperatures of nitrogen, phosphorus, silicon and sulfur in the table below. Which do you think exist as covalent molecular substances and which are covalent network lattices? Element Melting temperature ( C) Nitrogen 210 Phosphorus 44 Silicon 1410 Sulfur 119 A31. covalent molecular: nitrogen, phosphorus and sulfur; covalent network lattice: silicon Q32. The atoms in molecules of nitrogen, oxygen and fluorine are held together by covalent bonds. How are the bonds in these molecules: a similar? b different? A32. a The bonds are similar in that they all involve the sharing of electron pairs between two atoms; that is, they are covalent bonds. b They differ in the number of electron pairs shared: one pair (fluorine), two pairs (oxygen) and three pairs (nitrogen). Q33. The elements carbon and silicon have much in common. There are four electrons in the outer shell in each of the atoms. Both form a dioxide. But, although carbon dioxide is a gas at room temperature, silicon dioxide has the very high melting temperature of 1700 C. Explain this significant difference in terms of their structure. A33. The atoms in CO 2 and SiO 2 are held together by covalent bonds. In the case of CO 2, this bonding is between atoms to form discrete molecules. The relatively low boiling temperature of CO 2 is a reflection of the weak dispersion forces between these CO 2 molecules. In SiO 2, however, the bonding is between silicon and oxygen atoms in a covalent network lattice. There are strong covalent bonds holding every atom to others in the lattice and, hence, a large amount of energy is needed to disrupt this lattice. SiO 2 has much higher melting and boiling temperatures than CO 2. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 15

16 Q34. Experimental evidence shows that the double bond between the two oxygen atoms in O 2 is much stronger than a single bond between two oxygen atoms in a compound such as hydrogen peroxide (H 2 O 2 ). a Draw electron dot diagrams and structural formulas for O 2 and H 2 O 2. b Explain why the oxygen double bond is stronger than the oxygen single bond. c Why does oxygen not form a triple bond or three single covalent bonds? A34. a b In a single bond there are two electrons that are shared between the atoms, whereas in a double bond there are four electrons. The net attraction is stronger and more energy is needed to break the double bond than the single bond. c Oxygen has six outer-shell electrons and completes its outer shell by forming two single covalent bonds or a double bond with another non-metal. There is no need for the oxygen to form a triple bond or three single covalent bonds as this exceeds its requirement of two electrons to complete its outer shell. Q35. Consider the following list of molecules: N 2, Cl 2, O 2, NH 3, HCl, CH 4, H 2 O, CO 2, CCl 4, CHCl 3. a Draw a valence structure for each of the following from the list, showing bonding and non-bonding electron pairs: i A molecule that contains one triple bond. ii A molecule that contains one double bond. iii A molecule that contains two double bonds. b From the list, which substances contain: i polar molecules? ii symmetrical molecules? iii molecules with hydrogen bonding between them? A35. a i N 2 N N ii O 2 iii CO 2 Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 16

17 b i NH 3, HCl, H 2 O, CHCl 3 ii N 2, Cl 2, O 2, CH 4, CO 2, CCl 4 iii NH 3, H 2 O Q36. Water is a polar molecule. Explain how this fact shows that water is not a linear molecule. A36. If water was a linear molecule, the polarity of the two O H bonds would cancel each other out and make the molecule non-polar. As water is polar, it cannot be a linear molecule. Q37. Draw your own bonding concept map using the terms listed as a guide: covalent bond, lattice, molecule, intermolecular bond, non-metal atom, electron, melting temperature, electron, charge, shell, stable configuration, nucleus A37. An example of a concept map using these terms is shown. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 17