Predict whether each of the following is held together by ionic or by covalent bonds: (a) ammonia, NH 3 (b) magnesium nitride, Mg 3 N 2
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1 Example Exercise 12.1 Bond Predictions Predict whether each of the following is held together by ionic or by covalent bonds: (a) ammonia, NH 3 (b) magnesium nitride, Mg 3 N 2 A metal ion and nonmetal ion are attracted in ionic bonds; two or more nonmetal atoms are attracted in covalent bonds. (a) Ammonia contains the nonmetals nitrogen and hydrogen. It follows that NH 3 has covalent bonds. (b) Magnesium nitride contains a metal (Mg) and a nonmetal (N). It follows that Mg 3 N 2 has ionic bonds. Predict whether each of the following is held together by ionic or by covalent bonds: (a) aluminum oxide, Al 2 O 3 (b) sulfur dioxide, SO 2 Answers: (a) ionic; (b) covalent What type of chemical bond results from the attraction between a metal cation and a nonmetal anion? between two nonmetal atoms?
2 Example Exercise 12.2 Electron Configuration of Ions Which noble gas has an electron configuration identical to that of each of the following ions? (a) lithium ion (b) oxide ion (c) calcium ion (d) bromide ion Refer to the group number in the periodic table for the number of valence electrons. (a) Lithium (Group IA/1) forms an ion by losing 1 valence electron; Li + has only 2 electrons remaining and is isoelectronic with He. (b) Oxygen (Group VIA/16) forms an ion by gaining 2 electrons; thus, O 2 has 10 electrons and is isoelectronic with Ne. (c) Calcium (Group IIA/2) forms an ion by losing 2 electrons; Ca 2+ has 18 electrons remaining and is isoelectronic with Ar. (d) Bromine (Group VIIA/17) forms an ion by gaining 1 electron; Br - has 36 electrons and is isoelectronic with Kr. Which noble gas element has an electron configuration identical to that of each of the following ions? (a) potassium ion (b) nitride ion (c) strontium ion (d) iodide ion Answers: (a) K + is isoelectronic with Ar. (b) N 3 is isoelectronic with Ne. (c) Sr 2+ is isoelectronic with Kr. (d) I is isoelectronic with Xe. Which noble gas is isoelectronic with a barium ion? with an iodide ion?
3 Example Exercise 12.3 Characteristics of Ionic Bonds Which of the following statements are true regarding an ionic bond between an iron ion and a sulfide ion in an FeS formula unit? (a) The iron atom loses electrons and the sulfur atom gains electrons. (b) The iron atom has a larger radius than the iron ion. (c) Iron and sulfide ions form a bond by electrostatic attraction. All of these statements are true regarding an ionic bond. (a) An iron atom loses electrons and a sulfur atom gains electrons. (b) The radius of an iron atom is greater than its ionic radius. (c) An ionic bond is from the attraction of positive and negative ions. Which of the following statements are true regarding an ionic bond between a zinc ion and an oxide ion in a ZnO formula unit? (a) Zinc and oxygen form a bond by sharing electrons. (b) The oxygen atom is larger in radius than the oxide ion. (c) Zinc and oxide ions form a bond by the repulsion of ions. Answer: (a) False, zinc and oxygen bond by transferring electrons. (b) False, the radius of an oxygen atom is smaller than its ionic radius. (c) False, zinc and oxygen form a bond by the attraction of ions.
4 Example Exercise 12.3 Continued Characteristics of Ionic Bonds Which of the following statements are true regarding the formation of an ionic bond between a metal and a nonmetal? (a) The atomic radius of a metal atom is less than its ionic radius. (b) The atomic radius of a nonmetal atom is greater than its ionic radius. (c) Energy is released in the formation of an ionic bond.
5 Example Exercise 12.4 Characteristics of Covalent Bonds Which of the following statements are true regarding a covalent bond between a hydrogen atom and an oxygen atom in an H 2 O molecule? (a) Valence electrons are shared by hydrogen and oxygen atoms. (b) The H O bond length is less than the sum of the two atomic radii. (c) Breaking the bond between H O requires energy. All of the previous statements regarding a covalent bond in a molecule are true. (a) Valence electrons are shared by the two nonmetal atoms. (b) The bond length is less than the sum of the two atomic radii. (c) The breaking of a covalent bond requires energy. Which of the following statements are true regarding a covalent bond between a hydrogen atom and a sulfur atom in an H 2 S molecule? (a) Valence electrons are transferred from hydrogen to sulfur. (b) The H S bond length equals the sum of the two atomic radii. (c) Forming a bond between H S requires energy. Answer: (a) False, valence electrons are shared by hydrogen and sulfur. (b) False, the bond length is less than the sum of the two atomic radii. (c) False, the forming of a covalent bond releases energy.
6 Example Exercise 12.4 Continued Characteristics of Covalent Bonds Which of the following statements are true regarding the formation of a covalent bond between two nonmetal atoms? (a) Bonding electrons are shared between two nonmetal atoms. (b) The bond length is less than the sum of the two atomic radii. (c) The bond energy is the energy require to break a covalent bond.
7 Example Exercise 12.5 Electron Dot Formulas Single Bonds Draw the electron dot formula and the structural formula for a chloroform molecule, CHCl 3. Carbon is the central atom in the chloroform molecule and is indicated in bold. The total number of valence electrons is 4 e + 1 e + 3(7 e ) = 26 e. The number of electron pairs is 13 (26/2 = 13). We begin by placing 4 electron pairs around the carbon and adding the one H and three Cl atoms. We have 13 electron pairs minus the 4 pairs we used for the carbon. Since 9 electron pairs remain, let s place 3 pairs around each chlorine. Chloroform, CHCl 3
8 Example Exercise 12.5 Continued Electron Dot Formulas Single Bonds Each atom is surrounded by an octet of electrons, except hydrogen, which has two. This is the correct electron dot formula. In the corresponding structural formula we replace each bonding electron pair by a single dash. The structural formula is Note that the structural formula is free to rotate. We could have also written the hydrogen atom below or at the side of the carbon atom. Draw the electron dot formula and the structural formula for a molecule of SiHClBrI. Answers:
9 Example Exercise 12.5 Continued Electron Dot Formulas Single Bonds Draw the electron dot formula for H 2 S. How many pairs of nonbonding electrons are in a hydrogen sulfide molecule?
10 Example Exercise 12.6 Electron Dot Formulas Double Bonds Draw the electron dot formula and the structural formula for a carbon dioxide molecule, CO 2. Carbon is the central atom in the carbon dioxide molecule as indicated in bold. The total number of valence electrons is 4 e + 2(6 e ) = 16 e. The number of electron pairs is 8 (16/2 = 8). We begin by placing 4 electron pairs around the carbon and adding the two O atoms. The number of remaining electron pairs is 8 4 = 4 pairs. We can add 2 pairs to each oxygen. Carbon Dioxide CO 2 Each oxygen atom shares six electrons, two less than an octet. We can use the nonbonding electron pairs around carbon to complete the octet around oxygen. We move one pair to the oxygen on the left, and the other pair to the oxygen on the right to complete the octet around each oxygen. Each carbon oxygen bond shares two electron pairs. Now the octet rule is satisfied for each atom. This is the correct electron dot formula. There are two electron pairs between each oxygen and carbon. Thus, there are two double bonds in the carbon dioxide molecule. In the structural formula each double bond is indicated by a double dash.
11 Example Exercise 12.6 Continued Electron Dot Formulas Double Bonds Draw the electron dot formula and the structural formula for a molecule of SiO 2. Answers: Draw the electron dot formula for SO 2. How many pairs of nonbonding electrons are in a sulfur dioxide molecule?
12 Example Exercise 12.7 Electron Dot Formulas of Polyatomic Ions Draw the electron dot formula and the structural formula for the sulfate ion, SO 4 2. The total number of valence electrons is the sum of each atom plus 2 for the negative charge: 6 e + 4(6 e ) + 2 e = 32 e. The number of electron pairs is 16 (32/2 = 16). We begin by placing 4 electron pairs around the central sulfur atom and attaching the four oxygen atoms as follows: We have 12 remaining electron pairs, and so we place 3 electron pairs around each oxygen atom. Notice that each atom is surrounded by an octet of electrons. This is the correct electron dot formula. In the structural formula each bonding electron pair is replaced by a single dash. The structural formula is
13 Example Exercise 12.7 Continued Electron Dot Formulas of Polyatomic Ions Draw the electron dot formula and the structural formula for the nitrite ion, NO 2. Answers: Draw the electron dot formula for the polyatomic sulfite ion, SO 3 2. How many pairs of nonbonding electrons are in a sulfite ion?
14 Example Exercise 12.8 Electronegativity Predictions Predict which element in each of the following pairs is more electronegative according to the general electronegativity trends in the periodic table: (a) N or O (b) Br or Se (c) F or Cl (d) Si or C According to the trends in the periodic table, elements that lie to the right in a series or at the top of a group are more electronegative. (a) O is more electronegative than N. (b) Br is more electronegative than Se. (c) F is more electronegative than Cl. (d) C is more electronegative than Si. Predict which element in each of the following pairs is more electronegative according to the general trends in the periodic table: (a) H or Cl (b) Br or I (c) P or S (d) As or Sb Answers: (a) Cl (b) Br (c) S (d) As State the general trends in the periodic table for increasing electronegativity.
15 Example Exercise 12.9 Delta Notation for Polar Bonds Calculate the electronegativity difference and apply delta notation to the bond between carbon and oxygen, C O. From Figure 12.9, we find that the electronegativity value of C is 2.5 and of O is 3.5. By convention, we always subtract the lesser value from the greater. The difference between the two elements is = 1.0. The result shows that the C O bond is slightly polarized. Since O is the more electronegative element, the bonding electron pair is drawn away from the C atom and toward the O atom. The O atom thus becomes slightly negatively charged, whereas the C atom becomes slightly positively charged. Applying the delta convention, we have From the trends in the periodic table, apply delta notation and label each atom in the following polar covalent bonds: (a) N O (b) Br F Answers: Figure 12.9 Electronegativity Values for the Elements The Pauling electronegativity value for an element is shown below the symbol. In general, electronegativity increases across a period and up a group.
16 Example Exercise 12.9 Continued Delta Notation for Polar Bonds Refer to the electronegativity values in Figure 12.9 and predict which of the following bonds is most polar: H N, H O, or H F. Figure 12.9 Electronegativity Values for the Elements The Pauling electronegativity value for an element is shown below the symbol. In general, electronegativity increases across a period and up a group.
17 Example Exercise Classify each of the following as a polar or a nonpolar bond: (a) Cl Cl (b) Cl N (c) Cl Br (d) Cl H Polar versus Nonpolar Bonds From Figure 12.9 we find the following electronegativity values: Cl = 3.0, N = 3.0, Br = 2.8, and H = 2.1. (a) The Cl Cl bond ( = 0) is nonpolar. (b) The Cl N bond ( = 0) is nonpolar. (c) The Cl Br bond ( = 0.2) is slightly polar. (d) The Cl H bond ( = 0.9) is polar. Classify each of the following as a polar or a nonpolar bond: (a) C C (b) C O (c) C S (d) C H Answers: (a) nonpolar; (b) polar; (c) nonpolar; (d) slightly polar Figure 12.9 Electronegativity Values for the Elements The Pauling electronegativity value for an element is shown below the symbol. In general, electronegativity increases across a period and up a group.
18 Example Exercise Continued Polar versus Nonpolar Bonds Refer to the electronegativity values in Figure 12.9 and predict which of the following bonds is most nonpolar: C N, C O, or C I.
19 Example Exercise Coordinate Covalent Bonds Burning yellow sulfur powder produces sulfur dioxide, SO 2. Sulfur dioxide is a colorless gas with a suffocating odor. It is used to kill insect larvae and produces the odor released when a match is ignited. Draw the electron dot formula for SO 2. Then show the formation of a coordinate covalent bond in SO 3. The total number of valence electrons in one molecule of SO 2 is 6 e + 2(6 e ) = 18 e. By trial and error, we find that the electron dot formula for SO 2 has a double bond. Since the sulfur atom has a nonbonding electron pair, it can bond to an additional oxygen atom. A diagram of the formation of the coordinate covalent bond is as follows: A nitrogen molecule can form a coordinate covalent bond with an oxygen atom to give nitrous oxide, N 2 O. Draw the electron dot formula for a nitrogen molecule and attach an oxygen atom. Answers:
20 Example Exercise Continued Coordinate Covalent Bonds Draw the electron dot formula for HClO and determine if the molecule contains a coordinate covalent bond.
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