Example Exercise 12.1 Bond Predictions · PDF fileZnO formula unit? (a) Zinc and oxygen form a bond by sharing electrons. (b) ... The breaking of a covalent bond requires energy. Solution

© 2011 Pearson Education, Inc.Introductory Chemistry: Concepts and Critical Thinking, 6th EditionCharles H. Corwin

Example Exercise 12.1 Bond Predictions

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 NH3 has covalent bonds.(b) Magnesium nitride contains a metal (Mg) and a nonmetal (N). It follows that Mg3N2 has ionic bonds.

Solution

Predict whether each of the following is held together by ionic or by covalent bonds:(a) ammonia, NH3 (b) magnesium nitride, Mg3N2

Predict whether each of the following is held together by ionic or by covalent bonds:(a) aluminum oxide, Al2O3 (b) sulfur dioxide, SO2

Answers: (a) ionic; (b) covalent

Practice Exercise

What type of chemical bond results from the attraction between a metal cation and a nonmetal anion? between two nonmetal atoms?

Answer: See Appendix G.

Concept Exercise


Example Exercise 12.2 Electron Configuration of Ions

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, O2– has 10 electrons and is isoelectronic

with Ne.(c) Calcium (Group IIA/2) forms an ion by losing 2 electrons; Ca2+ 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.

Solution

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

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 ionAnswers:(a) K+ is isoelectronic with Ar. (b) N3– is isoelectronic with Ne.(c) Sr2+ is isoelectronic with Kr. (d) I– is isoelectronic with Xe.

Practice Exercise

Which noble gas is isoelectronic with a barium ion? with an iodide ion?Answer: See Appendix G.

Concept Exercise


Example Exercise 12.3 Characteristics of Ionic Bonds

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.

Solution

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.

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.

Practice Exercise


Example Exercise 12.3 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.


Concept ExerciseContinued


Example Exercise 12.4 Characteristics of Covalent Bonds

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.

Solution

Which of the following statements are true regarding a covalent bond between a hydrogen atom and an oxygen atom in an H2O 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.

Which of the following statements are true regarding a covalent bond between a hydrogen atom and a sulfur atom in an H2S 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.

Practice Exercise


Example Exercise 12.4 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.




Example Exercise 12.5 Electron Dot Formulas—Single Bonds

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.

Solution

Draw the electron dot formula and the structural formula for a chloroform molecule, CHCl3.

Chloroform, CHCl3



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:

Practice Exercise

Continued



Draw the electron dot formula for H2S. How many pairs of nonbonding electrons are in a hydrogen sulfide molecule?


Concept Exercise

Continued


Example Exercise 12.6 Electron Dot Formulas—Double Bonds

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.

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.

Solution

Draw the electron dot formula and the structural formula for a carbon dioxide molecule, CO2.

Carbon Dioxide CO2


Example Exercise 12.6 Electron Dot Formulas—Double Bonds

Draw the electron dot formula and the structural formula for a molecule of SiO2.

Answers:

Practice Exercise

Draw the electron dot formula for SO2. How many pairs of nonbonding electrons are in a sulfur dioxide molecule?


Concept Exercise

Continued


Example Exercise 12.7 Electron Dot Formulas of Polyatomic Ions

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

Solution

Draw the electron dot formula and the structural formula for the sulfate ion, SO42–.


Example Exercise 12.7 Electron Dot Formulas of Polyatomic Ions

Draw the electron dot formula for the polyatomic sulfite ion, SO32–. How many pairs of nonbonding

electrons are in a sulfite ion?


Concept Exercise

Continued

Draw the electron dot formula and the structural formula for the nitrite ion, NO2–.

Answers:

Practice Exercise


Example Exercise 12.8 Electronegativity Predictions

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.

Solution

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

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

Practice Exercise

State the general trends in the periodic table for increasing electronegativity.


Concept Exercise


Example Exercise 12.9 Delta Notation for Polar Bonds

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 3.5 – 2.5 = 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

Solution

Calculate the electronegativity difference and apply delta notation to the bond between carbon and oxygen, C—O.

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:

Practice Exercise

Figure 12.9 Electronegativity Values for the ElementsThe Pauling electronegativity value for an element is shown below the symbol. In general, electronegativity increases across a period and up a group.


Example Exercise 12.9 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.





Example Exercise 12.10 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

(3.0 – 3.0 = 0) is nonpolar.(b) The Cl—N bond

(3.0 – 3.0 = 0) is nonpolar.(c) The Cl—Br bond

(3.0 – 2.8 = 0.2) is slightly polar.(d) The Cl—H bond

(3.0 – 2.1 = 0.9) is polar.

Solution

Classify each of the following as a polar or a nonpolar bond:(a) Cl—Cl (b) Cl—N(c) Cl—Br (d) Cl—H

Classify each of the following as a polar or a nonpolar bond:(a) C—C (b) C—O(c) C—S (d) C—HAnswers: (a) nonpolar; (b) polar; (c) nonpolar; (d) slightly polar

Practice Exercise



Example Exercise 12.10 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.




Example Exercise 12.11 Coordinate Covalent Bonds

The total number of valence electrons in one molecule of SO2 is 6 e– + 2(6 e–) = 18 e–. By trial and error, we find that the electron dot formula for SO2 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:

Solution

Burning yellow sulfur powder produces sulfur dioxide, SO2. 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 SO2. Then show the formation of a coordinate covalent bond in SO3.

A nitrogen molecule can form a coordinate covalent bond with an oxygen atom to give nitrous oxide, N2O. Draw the electron dot formula for a nitrogen molecule and attach an oxygen atom.

Answers:

Practice Exercise


Example Exercise 12.11 Coordinate Covalent Bonds

Draw the electron dot formula for HClO and determine if the molecule contains a coordinate covalent bond.


Concept Exercise

Continued

Documents

Example Exercise 12.1 Bond Predictions · PDF fileZnO formula unit? (a) Zinc and oxygen form a bond by sharing electrons. (b) ... The breaking of a covalent bond requires energy. Solution