CHAPTER 11 CHEMICAL BONDS: THE FORMATION OF COMPOUNDS FROM ATOMS SOLUTIONS TO REVIEW QUESTIONS

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1 EINS v4.qxd 12/30/06 4:11 PM Page 129 APTER 11 EMIAL BNDS: TE FRMATIN F MPUNDS FRM ATMS 1. smallest l, Mg, Na, K, Rb largest. SLUTINS T REVIEW QUESTINS 2. More energy is required for neon because it has a very stable outer electron structure consisting of an octet of electrons in filled orbitals (noble gas electron structure). Sodium, an alkali metal, has a relatively unstable outer electron structure with a single electron in an unfilled orbital. The sodium electron is also farther away from the nucleus and is shielded by more inner electrons than are neon outer electrons. 3. When a third electron is removed from beryllium, it must come from a very stable electron structure corresponding to that of the noble gas, helium. In addition, the third electron must be removed from a +2 beryllium ion, which increases the difficulty of removing it. 4. The first ionization energy decreases from top to bottom because in the successive alkali metals, the outermost electron is farther away from the nucleus and is more shielded from the positive nucleus by additional electron energy levels. 5. The first ionization energy decreases from top to bottom because the outermost electrons in the successive noble gases are farther away from the nucleus and are more shielded by additional inner electron energy levels. 6. Barium and beryllium are in the same family. The electron to be removed from barium is, however, located in an energy level farther away from the nucleus than is the energy level holding the electron in beryllium. ence, it requires less energy to remove the electron from barium than to remove the electron from beryllium. Barium, therefore, has a lower ionization energy than beryllium. 7. The first electron removed from a sodium atom is the one valence electron, which is shielded from most of its nuclear charge by the electrons of lower levels. To remove a second electron from the sodium ion requires breaking into the noble gas structure. This requires much more energy than that necessary to remove the first electron, because the Na + is already positive. 8. (a) K 7 Na (d) (b) Na 7 Mg (e) (c) 7 F I 7 Br Zr 7 Ti

2 EINS v4.qxd 12/30/06 4:11 PM Page hapter The first element in each group has the smallest radius. 10. Atomic size increases down a column since each successive element has an additional energy level which contains electrons located farther from the nucleus. 11. A Lewis structure is a representation of the bonding in a compound. Valence electrons are responsible for bonding, therefore they are the only electrons that need to be shown in a Lewis structure. 12. Metals are less electronegative than nonmetals. Therefore, metals lose electrons more easily than nonmetals. So, metals will transfer electrons to nonmetals leaving the metals with a positive charge and the nonmetals with a negative charge. 13. The noble gases are the most stable of all the elements because they have a complete octet (8 electrons) in their valence level. When the elements in Groups 1A, 2A, 6A and 7A form ions, they want to establish a stable electron structure. (a) The elements in Group 1A lose an electron (obtain a positive charge) in order to achieve a noble gas configuration. (b) The elements in Group 2A lose electrons (obtain a positive charge) in order to achieve a noble gas configuration. (c) The elements in Group 6A gain electrons (obtain a negative charge) in order to achieve a noble gas configuration. (d) The elements in Group 7A gain an electron (obtain a negative charge) in order to achieve a noble gas configuration. 14. A polar bond is a bond between two atoms with very different electronegativities. A polar ecule is a dipole due to unequal electrical charge between bonded atoms resulting from unequal sharing of electrons. 15. The electron pair arrangement is the arrangement of both bonding and nonbonding electrons in a Lewis structure. The ecular shape of a ecule is the three dimensional arrangement of its atoms in space. 16. A Lewis structure is a visual representation of the arrangement of atoms and electrons in a ecule or an ion. It shows how the atoms in a ecule are bonded together. 17. The more electronegative atom in the bond between two atoms will more strongly attract electrons so it will have a partial negative charge ( -). The less electronegative atom will have a partial positive charge ( +) because the bonding pair of electrons has been pulled away by the more electronegative atom. 18. The dots in a Lewis structure represent nonbonding pairs of electrons. The lines represent bonding pairs of electrons

3 EINS v4.qxd 12/30/06 4:11 PM Page hapter Group 1A 2A 3A 4A 5A 6A 7A E E Eº Eº Eªº Eª E 20. Lewis structures: s Ba Tl ª Pb ª Po ª At Rn Each of these is a representative element and has the same number of electrons in its outer energy level as its periodic group. 21. (a) Elements with the highest electronegativities are found in the upper right hand corner of the periodic table. (b) Elements with the lowest electronegativities are found in the lower left of the periodic table. 22. Valence electrons are the electrons found in the outermost energy level of an atom. 23. By losing one electron, a potassium atom acquires a noble gas structure and becomes a K + ion. To become a K 2+ ion requires the loss of a second electron and breaking into the noble gas structure of the ion. This requires too much energy to generally occur. K An aluminum ion has a +3 charge because it has lost 3 electrons in acquiring a noble gas electron structure

4 EINS v4.qxd 12/30/06 4:11 PM Page 132 APTER 11 SLUTINS T EXERISES 1. (a) A calcium atom is larger because it has electrons in the 4 th shell, while a calcium ion does not. In addition, a 2+ ion has 20 protons and 18 electrons, creating a charge imbalance and drawing the electrons in towards the nucleus. (b) A chloride ion is larger because it has one more electron than a chlorine atom in its outer shell. Also, the ion has 17 protons and 18 electrons, creating a charge imbalance, resulting in a lessening of the attraction of the electrons towards the nucleus. (c) A magnesium ion is larger than an aluminum ion. Both ions will have 10 electrons in their valence shells, but the aluminium ion will have a greater charge imbalance since it has 13 protons and the magnesium ion has 12 protons. The charge imbalance draws the electrons in more closely to the nucleus. (d) A sodium atom is larger than a silicon atom. Sodium and silicon are both in period 3. Going across a group, the radii of atoms decrease. (e) A bromide ion is larger than a potassium ion. The bromide ion has 35 protons and 36 electrons, creating a charge imbalance that results in a lessening of the attraction of the electrons towards the nucleus. The potassium ion has 19 protons and 18 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus. 2. (a) Fe 2+ has one electron more than Fe 3+, so will have a larger radius. (b) A potassium atom is larger than a potassium ion. They both have 19 protons. The potassium atom has 19 electrons. The potassium ion has 18 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus. (c) A chloride ion is larger than a sodium ion. The chloride ion has 17 protons and 18 electrons, creating a charge imbalance that results in a lessening of the attraction of the electrons towards the nucleus. The sodium ion has 11 protons and 10 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus. (d) A strontium atom is larger than an iodine atom. Strontium and iodine are both in period 5. Going across a group, the radii of atoms decrease. (e) A rubidium ion is larger than a strontium ion. Both ions will have 36 electrons in their valence shells, but the strontium ion will have a greater charge imbalance since it has 38 protons and the rubidium ion has 37 protons. The charge imbalance draws the electrons in more closely to the nucleus

5 EINS v4.qxd 12/30/06 4:11 PM Page hapter (a) (d) Pb S (b) Na F (e) N (c) N (f) (a) l (d) I Br (b) Li (e) Mg (c) l (f) F 5. (a) ionic (b) covalent (c) covalent (d) ionic 6. (a) covalent (b) ionic (c) covalent (d) ionic 7. Magnesium has an electron structure 1s 2 2s 2 2p 6 3s 2, while the structure for chlorine is 1s 2 2s 2 2p 6 3s 2 3p 5. When these two elements react with each other, each magnesium atom loses its 3s 2 electrons, one to each of two chlorine atoms. The resulting structures for both magnesium and chlorine are noble gas configurations. 8. (a) F + 1 e - F - (b) a a e - 9. (a) (b) Mg + F + F MgF 2 K + K + K (a) (b) a + Na + Br a NaBr 11. Valence electrons: (1) K(1) Mg(2) e(2) Al(3) 12. Valence electrons: Si(4) N(5) P(5) (6) l(7) 13. Noble gas structures: (a) alcium atom, loses (b) Sulfur atom, gains 2 e - (c) elium, none 2 e

6 EINS v4.qxd 12/30/06 4:11 PM Page hapter (a) hloride ion, none (b) Nitrogen atom, gain 3 e - or lose (c) Potassium atom, lose 1 e - 5 e 15. (a) ionic, Nal (c) ionic, MgBr 2 (e) ecular, 2 (b) ecular, 4 (d) ecular, Br (a) Two oxygen atoms will form a nonpolar covalent compound. The formula is 2. (b) ydrogen and bromine will form a polar covalent compound. The formula is Br. (c) xygen and two hydrogen atoms will form a polar covalent compound. The formula is 2. (d) Two iodine atoms will form a nonpolar covalent compound. the formula is I (a) Na, Na 2 (c) Al 3, Al 2 3 (b) a 2, a (d) Sn 4, Sn (a) Sb 3, Sb 2 3 (c) l, l 2 7 (b) 2 Se, Se 3 (d) 4, 2 K S 19. Li 2 S 4 lithium sulfate potassium sulfate Rb 2 S 4 rubidium sulfate 2 4 s 2 S 4 cesium sulfate Fr 2 S 4 francium sulfate BaBr 20. BeBr 2 beryllium bromide barium bromide MgBr 2 magnesium bromide 2 RaBr 2 radium bromide SrBr 2 strontium bromide 21. Lewis structures: (a) Na (b) Br (c) (a) Ga (b) Ga 3+ (c) a (a) covalent (c) ionic (b) ionic (d) covalent 24. (a) covalent (c) covalent (b) ionic (d) covalent 25. (a) ionic (b) covalent (c) covalent 26. (a) covalent (b) covalent (c) ionic

7 EINS v4.qxd 12/30/06 4:11 PM Page hapter (a) (b) N N (c) l l (a) (b) Br Br (c) I I 29. (a) l N l (c) l (b) (d) Na + N 30. (a) S (c) N (b) S S (d) N + l 31. (a) (d ) Ba 2+ N (b ) Al (e) (c) 3+ S (a) I (d ) l (b ) (e) S 2 N (c)

8 EINS v4.qxd 12/30/06 4:11 PM Page hapter (a) 2, polar (b) Br, polar (c) F 4, nonpolar 34. (a) F 2, nonpolar (b) 2, nonpolar (c) N 3, polar 35. (a) 4 electron pairs, tetrahedral (b) 4 electron pairs, tetrahedral (c) 3 electron pairs, trigonal planar 36. (a) 3 electron pairs, trigonal planar (b) 4 electron pairs, tetrahedral (c) 4 electron pairs, tetrahedral 37. (a) tetrahedral (b) pyramidal (c) tetrahedral 38. (a) tetrahedral (b) pyramidal (c) tetrahedral 39. (a) tetrahedral (b) pyramidal (c) bent 40. (a) tetrahedral (b) bent (c) bent 41. xygen 42. Potassium 43. N e N N N 3 16 e N = N = N 44. (a) 18 e N bent N 2 2 (b) 32 e S tetrahedral S 4 2 (c) Sl 2 26 e l S l pyramidal (d) l 2 20 e l l bent

9 EINS v4.qxd 12/30/06 4:11 PM Page hapter (a) e (b) e (c) e 46. (a) g (b) Be (c) N (d) Fr (e) Au 47. (a) Zn (b) Be (c) N 48. (1) Fluorine is missing one electron from its 2p level while neon has a full energy level. (2) Fluorine s valence electrons are closer to the nucleus. The attraction between the electrons and the positive nucleus is greater. 49. Lithium has a +1 charge after the first electron is removed. It takes more energy to overcome that charge and to remove another electron than to remove a single electron from an uncharged e atom. 50. Yes. Each of these elements have an ns 1 electron and they could lose that electron in the same way elements in Group 1A do. They would then form +1 ions and ionic compounds such as ul, Agl, and Aul. 51. SnBr 2, GeBr The bond between sodium and chlorine is ionic. An electron has been transferred from a sodium atom to a chlorine atom. The substance is composed of ions not ecules. Use of the word ecule implies covalent bonding. 53. A covalent bond results from the sharing of a pair of electrons between two atoms, while an ionic bond involves the transfer of one or more electrons from one atom to another. 54. This structure shown is incorrect since the bond is ionic. It should be represented as: Na + 2 Na The four most electronegative elements are F,, N, l. 56. highest F,, S,, Mg, s lowest

10 EINS v4.qxd 12/30/06 4:11 PM Page hapter It is possible for a ecule to be nonpolar even though it contains polar bonds. If the ecule is symmetrical, the polarities of the bonds will cancel (in a manner similar to a positive and negative number of the same size) resulting in a nonpolar ecule. An example is 2 which is linear and nonpolar. 58. Both ecules contain polar bonds. 2 is symmetrical about the atom, so the polarities cancel. In, there is only one polar bond, therefore the ecule is polar. 59. (a) 105 (b) 107 (c) (d) (a) Both use the p orbitals for bonding. B uses one s and two p orbitals while N uses one s and three p orbitals for bonding. (b) BF 3 is trigonal planar while is pyramidal. (c) BF 3 NF 3 no lone pairs one lone pair (d) BF 3 has 3 very polar covalent bonds. NF 3 has 3 polar covalent bonds 61. Fluorine s electronegativity is greater than any other element. Ionic bonds form between atoms of widely different electronegativities. Therefore, FrF, sf, RbF or KF would be ionic substances with the greatest electronegativity difference. 62. Each element in a particular column has the same number of valence electrons and therefore the same Lewis structure g 63. S = g g = g Empirical formula is S = = 3.00 S 64. We need to know the ecular formula before we can draw the Lewis structure. From the data, determine the empirical and then the ecular formula g = g 1.21 =

11 EINS v4.qxd 12/30/06 4:11 PM Page hapter g 2.41 l = 2.41 g 1.21 = l 2 is the empirical formula empirical mass = 1(12.01) + 2(35.45) = = 2.00 Therefore, the ecular formula is (l 2 ) 2 or 2 l 4 l l 65. (a) incorrect, correct structure is: (d) incorrect, correct structure is: l l (b) correct (e) incorrect, correct structure is: (c) correct (f) incorrect, correct structure is: N S N (25 g Li) a 1 kj ba520 b = 1.9 * 10 3 kj g 67. Removing the first electron from 1 e of sodium atoms requires 496 kj. To remove a second electron from 1 e of sodium atoms requires 4,565 kj. The conversion is: (15 Na) a (15 Na) a 496 KJ b = 7.4 * 10 3 kj 4565 KJ b = 6.8 * 10 4 kj (7.4 * 10 3 kj) + (6.8 * 10 4 kj) = 7.5 * 10 4 kj