Answers and Solutions to Text Problems

4 Answers and Solutions to Text Problems 4.1 Valence electrons are found in the outermost energy level in an atom. 4.2 Because the electrons in an atom s outer shell determine its chemical properties, the group number was chosen to equal the number of valence electrons present in representative elements. 4.3 a. 1s 2 2s 2 2p 3 five valence electrons b. 1s 2 2s 2 2p 4 six valence electrons c. 1s 2 2s 2 2p 6 3s 2 3p 6 eight valence electrons d. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 one valence electron e. 1s 2 2s 2 2p 6 3s 2 3p 4 six valence electrons 4.4 a. 2, 8, 1 one valence electron b. 2, 8, 3 three valence electrons c. 2, 8, 7 seven valence electrons d. 2, 8, 2 two valence electrons e. 2, 8, 5 five valence electrons 4.5 The number of dots is equal to the number of valence electrons as indicated by the group number. a. Sulfur has 6 valence electrons : S b. Nitrogen has 5 valence electrons N c. Calcium has 2 valence electrons Ca d. Sodium has 1 valence electron Na e. Potassium has 1 valence electron K 4.6 a. 4, C b. 6, : O c. 7, : F d. 1, Li e. 7, : Cl 4.7 a. M b. M 4.8 a. Nm b. : Nm 4.9 Alkali metals are members of Group 1A, and each has one valence electron. 4.10 Halogens are members of Group 7A, and each has seven valence electrons. 4.11 a. If sodium atom loses its valence electron, its second energy level has a complete octet. b. A neon atom has the same electronic arrangement as a sodium ion. c. Group 1A and Group 2A elements do not have a stable octet until each has lost one or two electrons, respectively. Electrically charged ions are formed when electrons are lost, and these positively-charged ions are attracted to negatively-charged ions resulting in the formation of compounds. Group 8A (18) elements have stable octets that remain electrically neutral, and have no tendency to form compounds.

Compounds and Their Bonds 4.12 a. When a chlorine atom gains an electron, its valence shell achieves a complete octet. b. An argon atom has the same electronic arrangement as a chloride ion. c. Group 7 elements do not have a stable octet until each has gained an electron. Electrically charged ions are formed when electrons are gained, and these negatively-charged ions are attracted to other positively-charged ions resulting in the formation of compounds. Group 8 elements have stable octets while remaining electrically neutral, and thus have no tendency to form compounds. 4.13 a. : Ne : Neon has a complete valence shell. b. : O Oxygen does not have a complete valence shell. c. Li Lithium does not have a complete valence shell. d. : Ar : Argon has a complete valence shell. 4.14 a. Mg Magnesium does not have a complete valence shell b. N Nitrogen does not have a complete valence shell c. He: Helium has a complete valence shell of two electrons. d. K Potassium does not have a complete valence shell 4.15 Atoms with 1, 2 or 3 valence electrons will lose those electrons. a. one b. two c. three d. one e. two 4.16 a. one b. two c. three d. one e. three 4.17 a. Ne b. Ne c. Ar d. Ne e. Ne 4.18 a. He b. Kr c. Ar d. Ne e. Kr 4.19 a. lose 2e b. gain 3e c. gain 1e d. lose 1e e. lose 3e 4.20 a. gain 2e b. lose 2e c. gain 1e d. lose 1e e. gain 3e 4.21 a. Li + b. F c. Mg 2+ d. Fe 3+ e. Zn 2+ 4.22 a. 8 protons, 10 electrons b. 19 protons, 18 electrons c. 35 protons, 36 electrons d. 16 protons, 18 electrons e. 38 protons, 36 electrons 4.23 a. Cl b. K + c. O 2 d. Al 3+ 4.24 a. F b. Ca 2+ c. Na + d. Li + 4.25 a. potassium b. sulfide c. calcium d. nitride 4.26 a. magnesium b. barium c. iodide d. chloride

Chapter 4 Answers and Solutions 4.27 a. (Li and Cl) and c. (K and O) will form ionic compounds 4.28 b. (Mg and Cl) and d. (K and S) will form ionic compounds 4.29 a. K + Cl : K + + [: Cl :] KCl b. Ca + Cl : + Cl : Ca 2+ + 2 [: Cl :] - CaCl 2 c. Na + Na + Na + N 3 Na + + [: N :] 3 Na 3 N 4.30 a. Mg + S Mg 2+ + [: S :] --. b. Al + Cl : + Cl : + Cl : Al 3+ + [: Cl :] - + [: Cl :] - + [: Cl :] - c. 2 - Li + Li + O Li + + Li + + [: O :] 4.31 a. Na 2 O b. AlBr 3 c. BaO d. MgCl 2 e. Al 2 S 3 4.32 a. AlCl 3 b. CaS c. Li 2 S d. K 3 N e. KI 4.33 a. Na 2 S b. K 3 N c. AlI 3 d. Li 2 O 4.34 a. CaCl 2 b. BaBr 2 c. Na 3 P d. MgO 4.35 a. aluminum oxide b. calcium chloride c. sodium oxide d. magnesium nitride e. potassium iodide 4.36 a. magnesium chloride b. potassium phosphide c. lithium sulfide d. lithium bromide e. magnesium oxide 4.37 The Roman numeral is used to specify the positive charge on the transition metal in the compound. It is necessary for most transition metal compounds because many transition metals can exist as more than one cation; transition metals have variable ionic charges. 4.38 Because calcium ion only has a +2 charge, (it never varies), the name calcium is sufficient to specify the ion. However, Copper ions can exist with either a +1 or a +2 charge. Thus, the Roman numeral is used to specify which Copper ion is present in the compound. 4.39 a. iron (II); ferrous b. copper (II); cupric c. zinc d. lead (IV); plumbic 4.40 a. silver b. copper (I); cuprous c. iron (III); ferric d. tin (II); stannous

4.41 a. Tin (II) chloride; stannous chloride b. Iron (II) oxide; ferrous oxide c. Copper (I) sulfide; cuprous sulfide d. Copper (II) sulfide; cupric sulfide 4.42 a. Silver phosphide b. Lead(II) sulfide; plumbous sulfide c. Tin(IV) oxide; stannic oxide d. Gold(III) chloride; auric chloride 4.43 a. +3 b. +3 c. +4 d. +2 4.44 a. +2 b. +2 c. +3 d. +3 4.45 a. MgCl 2 b. Na 2 S c. Cu 2 O d. Zn 3 P 2 e. AuN 4.46 a. Fe 2 O 3 b. BaF 2 c. SnCl 4 d. Ag 2 S c. CuCl 2 4.47 a. : Br : Br : b. H : H c. H : F : d. : F : O : : F : : Cl : : Cl : 4.48 a. : Cl : N : Cl : b. : Cl : C : Cl : : Cl : : F : c. : Cl : Cl : d. : F : Si : F : : F : 4.49 a. H : C ::: C : H b. H : C : : O : H : O : c. : O : S : : O : 4.50 a. H : C ::: N : b. H : C :: C : H c. H : O : N :: O : H H 4.51 a. phosphorus tribromide b. carbon tetrabromide c. silicon dioxide d. hydrogen fluoride e. nitrogen triiodide Compounds and Their Bonds

Chapter 4 Answers and Solutions 4.52 a. carbon disulfide b. diphosphorus pentoxide c. dichlorine oxide d. phosphorus trichloride e. dinitrogen tetroxide 4.53 a. dinitrogen trioxide b. nitrogen trichloride c. silicon tetrabromide d. phosphorus pentachloride e. sulfur trioxide 4.54 a. silicon tetrafluoride b. iodine tribromide c. carbon dioxide d. sulfur dioxide e. dinitrogen oxide 4.55 a. CCl 4 b. CO c. PCl 3 d. N 2 O 4 4.56 a. SO 2 b. SiCl 4 c. IF 5 d. N 2 O 4.57 a. OF 2 b. BF 3 c. N 2 O 3 d. SF 6 4.58 a. SBr 2 b. CS 2 c. P 4 O 6 d. N 2 O 5 4.59 a. nonpolar covalent b. none c. nonpolar covalent d. ionic e. polar covalent 4.60 a. ionic b. polar covalent c. ionic d. nonpolar covalent e. polar covalent 4.61 a. polar covalent b. ionic c. polar covalent d. nonpolar covalent e. polar covalent 4.62 a. ionic b. nonpolar covalent c. polar covalent d. polar covalent e. polar covalent 4.63 a. F b. F c. Cl d. Br e. Cl 4.64 a. Li b. Cl c. O d. F e. Br 4.65 In a polar covalent bond, the more electronegative atom is assigned δ. δ + δ δ + δ δ + δ δ + δ a. H F b. C Cl c. N O d. N F δ δ + δ δ + δ + δ δ + δ 4.66 a. O H b. O S c. P Cl d. S Cl 4.67 a. HCO 3 4.68 a. NO 2 b. NH 4 + b. SO 3 2 c. PO 4 3 d. HSO 4 c. OH d. PO 3 3 4.69 a. sulfate b. carbonate c. phosphate d. nitrate 4.70 a. hydroxide b. hydrogen sulfite (or bisulfite) c. cyanide d. nitrite 4.71 OH NO 2 CO 3 2 HSO 4 PO 4 3 Li + LiOH LiNO 2 Li 2 CO 3 LiHSO 4 Li 3 PO 4 Cu 2 + Cu(OH) 2 Cu(NO 2 ) 2 CuCO 3 Cu(HSO 4 ) 2 Cu 3 (PO 4 ) 2 Ba 2 + Ba(OH) 2 Ba(NO 2 ) 2 BaCO 3 Ba(HSO 4 ) 2 Ba 3 (PO 4 ) 2

Compounds and Their Bonds 4.72 OH NO 3 HCO 3 SO 3 2 PO 4 3 NH 4 + NH 4 OH NH 4 NO 3 NH 4 HCO 3 (NH 4 ) 2 SO 3 (NH 4 ) 3 PO 4 Al 3 + Al(OH) 3 Al(NO 3 ) 3 Al(HCO 3 ) 3 Al 2 (SO 3 ) 3 AlPO 4 Pb 4 + Pb(OH) 4 Pb(NO 3 ) 4 Pb(HCO 3 ) 4 Pb(SO 3 ) 2 Pb 3 (PO 4 ) 4 4.73 a. CO 2 3, sodium carbonate b. NH + 4, ammonium chloride c. PO 3 4, lithium phosphate d. NO 2, copper (II) or cupric nitrite e. SO 2 3, iron (II) or ferrous sulfite 4.74 a. K OH Potassium Hydroxide b. Na NO 3 Sodium Nitrate c. Cu CO 3 Copper(II) or Cupric Carbonate d. NaHCO 3 Sodium Bicarbonate e. Ba SO 4 Barium Sulfate 4.75 a. Ba(OH) 2 b. Na 2 SO 4 c. Fe(NO 3 ) 2 d. Zn 3 (PO 4 ) 2 e. Fe 2 (CO 3 ) 3 4.76 a. AlCl 3 b. (NH 4 ) 2 O c. Mg(HCO 3 ) 2 d. NaNO 2 e. Cu 2 SO 4 4.77 a. This is an ionic compound with Al 3+ ion and the sulfate SO 4 2 polyatomic ion. The correct name is aluminum sulfate b. This is an ionic compound with Ca 2+ ion and the carbonate CO 3 2 polyatomic ion. The correct name is calcium carbonate c. This is a covalent compound because it contains two nonmetals. Using prefixes, it is named dinitrogen monoxide. d. This is an ionic compound with sodium ion Na + and the PO 4 3 polyatomic ion. The correct name is sodium phosphate e. This ionic compound contains two polyatomic ions ammonium NH 4 + and sulfate SO 4 2. It is named ammonium sulfate f. This is an ionic compound containing the variable metal ion Fe 3+ and oxide ion O 2. It is named using the Roman numeral as iron(iii) or ferric oxide 4.78 a. nitrogen b. magnesium phosphate c. iron(ii) or ferrous sulfate d. magnesium sulfate e. copper(i) or cuprous oxide f. tin(ii) or stannous fluoride 4.79 a. Linear. Two groups of electrons bonded to a central atom with no lone pairs gives a linear shape to the molecule. b. Pyramidal. Four electron pairs are arranged in a tetrahedron. The shape is pyramidal because there are three atoms and one lone pair bonded to the central atom. 4.80 a. Four groups of electrons bonded to a central atom with no lone pairs give a tetrahedral shape. b. Four electron pairs are arranged as a tetrahedron with 109 angles. When only two pairs are bonded to atoms, it has a bent shape. 4.81 The four electron groups in PCl 3 have a tetrahedral arrangement, but three bonded atoms around a central atom give a pyramidal shape.

Chapter 4 Answers and Solutions 4.82 In the electron dot structure of H 2 S, the central atom S has four electrons pairs arranged as a tetrahedron. Because there are two bonded atoms and two lone pairs, H 2 S has a bent shape. The arrangement of electron pairs determine the angels between the pairs, whereas the number of bonded atom determines the shape of the molecule. 4.83 In the electron dot structure of BH 3, the central atom B has three bonded atoms and no lone pairs, which give BH 3 a trigonal planar shape with angles of 120. In the molecule NH 3, the central atom N is bonded to three atoms and one lone pair. The structure of the electron groups is tetrahedral, which gives NH 3 a pyramidal shape with angles of 109. 4.84 In CH 4 and H 2 O the central atoms of C and O have four electron pairs arranged as tetrahedrons, which have 109 angles. The shapes are different because CH 4 has four bonded atoms (tetrahedral shape), whereas H 2 O has two bonded atoms and two lone pairs (bent shape). 4.85 a. Two atoms in a molecule have a linear shape. b. The central oxygen atom has four electron pairs with two bonded to fluorine atoms. Its shape is bent with 109 angles. c. The central atom C has two electron groups bonded to two atoms; HCN is linear. d. The central atom C has four electron pairs bonded to four chlorine atoms; CCl 4 has a tetrahedral shape. 4.86 a. tetrahedral b. pyramidal c. bent d. linear 4.87 To find the total valence electrons for an ion, add the total valence electrons for each atom and add the number of electrons indicated by a negative charge. a. C (4 valence electrons ) + 3 O (3 x 6 valence electrons) + charge ( 2 electrons) = 4e - + 18e - + 2e - = 24 e - total for the electron dot structure 2- : O : : O : C : : O : 3 electron groups around C; trigonal planar shape b. S (6 valence electrons ) + 4 O (4 6 valence electrons) + charge ( 2 electrons) = 6e - + 24e - + 2e - = 32 e - total for the electron dot structure : O : 2 - : O : S : O : 4 electron pairs around S; tetrahedral shape : O : 4.88 To find the total valence electrons for an ion, add the total valence electrons for each atom and add the number of electrons indicated by a negative charge. a. N (5 valence electrons ) + 2 O (2 x 6 valence electrons) + charge ( 1 electron) = 5e - + 12e - + 1e - = 18 e - total for the electron dot structure -

Compounds and Their Bonds : O : N : : O : 3 electron groups around N; two bonded atoms and one lone pair give a bent molecular shape with 120 angles. b. P (5 valence electrons ) + 4 O (4 x 6 valence electrons) + charge ( 3 electrons) = 5e - + 24e - + 3e - = 32 e - total for the electron dot structure : O : 3- : O : P : O : 4 electron pairs around P and four bonded atoms give a tetrahedral shape : O : 4.89 Cl 2 is a nonpolar molecule because there is a nonpolar covalent bond between Cl atoms, which have identical electronegativity values. In HCl, the bond is a polar bond, which is a dipole and makes HCl a polar molecule. 4.90 CH 4 with four equal polar bonds has four equal dipoles that cancel to give a nonpolar molecule. In CH 3 Cl, the one C Cl bond is much more polar than the C H bonds. Therefore, the dipoles will not cancel, which makes CH 3 Cl a polar molecule. 4.91 Write the symbols δ + and δ over the atoms that are in polar bonds. δ + δ δ + δ a. Cl Cl b. C Cl c. N O δ + δ δ + δ δ δ + d. H O e. P F f. S H 4.92 Write the symbols δ + and δ over the atoms that are in polar bonds. δ δ + δ + δ a. F Cl b. P H c. C O δ + δ δ + δ δ δ + d. S O e. N F f. O Cl 4.93 a. two dipoles cancel; nonpolar b. dipoles do not cancel; polar c. four dipoles cancel; nonpolar d. three dipoles cancel; nonpolar 4.94 a. dipoles do not cancel; nonpolar b. dipoles do not cancel; polar c. four dipoles cancel; nonpolar d. dipoles do not cancel; polar 4.95 a. P b. Na c. Al d. Si 4.96 a. 2,8,8 b. 2,8 c. 2,8,8 d. 2,8,8 e. 2,8,8 4.97 a. 1s 2 2s 2 2p 6 b. 1s 2 2s 2 2p 6 c. 1s 2 2s 2 2p 6 d. 1s 2 2s 2 2p 6 e. 1s 2 4.98 a. Ne b. Ar c. Ar d. Ne e. Ne 4.99 a. Group 2A b. X c. Be 4.100 a. X is in Group 1; Y is in Group 6 b. ionic c. X +, Y 2 d. X 2 Y e. XCl f. YCl 2

Chapter 4 Answers and Solutions 4.101 a. Sn 4+ b. 50 protons, 46 electrons c. SnO 2 d. Sn 3 (PO 4 ) 4 4.102 Calcium, phosphate, and hydroxide ions are present in calcium hydroxyapatite. 4.103 Compounds with a metal and nonmetal are classified as ionic; two nonmetals as covalent. a. ionic b. covalent c. covalent d. covalent e. ionic f. covalent g. ionic h. ionic 4.104 With a filled energy level, helium is a stable, unreactive element. Hydrogen does not have a filled energy level, and it is a reactive element, (in fact it burns). Thus, helium is used to avoid potentially hazardous conditions. 4.105 a. iron(iii) or ferric chloride b. dichlorine heptoxide c. bromine d. calcium phosphate e. phosphorus trichloride f. aluminum carbonate g. lead(iv) or plumbic chloride h. magnesium carbonate i. nitrogen dioxide j. tin (II) or stannous sulfate k. barium nitrate l. copper(ii) or cupric sulfide 4.106 a. SnCO 3 b. Li 3 P c. SiCl 4 d. Fe 2 S 3 e. CO 2 f. CaBr 2 g. Na 2 CO 3 h. NO 2 i. Al(NO 3 ) 3 j. Cu 3 N k. K 3 PO 3 l. PbS 2 4.107 a. trigonal planar, dipoles cancel, nonpolar b. bent, dipoles do not cancel, polar c. linear, dipoles cancel, nonpolar 4.108 a. Tetrahedral; dipoles cancel; nonpolar b. Pyramid; dipoles do not cancel; polar c. Tetrahedral; dipoles do not cancel; polar 4.109 a. bent, dipoles do not cancel, polar b. pyramidal, dipoles do not cancel, polar c. trigonal planar, dipoles cancel, nonpolar 4.110 a. Linear; equal atoms, nonpolar b. Tetrahedral; dipoles cancel; nonpolar c. Linear; unlike atoms, polar 4.111 a. trigonal planar b. tetrahedral c. tetrahedral 4.112 a. trigonal planar b. pyramidal c. bent (120 )