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CHAPTER 5 Ionic and Covalent Compounds

CHAPTER 5 Ionic and Covalent Compounds. Chemical Compound A compound is a pure chemical substance composed of two or more elements. Properties of compounds include the following:

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CHAPTER 5 Ionic and Covalent Compounds

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  1. CHAPTER 5 Ionic and Covalent Compounds

  2. Chemical Compound A compound is a pure chemical substance composed of two or more elements. Properties of compounds include the following: 1) The relative amounts of each element in a compound is the same for all samples of the compound (Law of Definite Proportion). 2) Compounds are held together by chemical bonds. 3) Compounds have specific chemical and physical properties that distinguish them from other compounds.

  3. Types of Chemical Bonds There are three general types of chemical bonds that substances form, two of which we discuss in this chapter. Ionic bonding - Electrons are transferred; bonding is due to electrostatic attraction between cations (positive ions) and anions (negative ions). Covalent bonding - Pairs of electrons are shared between atoms. Metallic bonding - Electrons are pooled (spread out) among positively charged metal cations.

  4. Lewis Dot Structure In ionic and covalent bonding it is the valence electrons that are usually involved in bond formation. One method used to focus on changes that occur when bonds formed is to represent atoms and ions in terms of dot structures. The method was developed by the American chemist G. N. Lewis, and so is often called the Lewis dot structure method. In the dot structure method valence electrons are represented as dots around the symbol for the element. For simple cases a maximum of 8 valence electrons can occur (2 for hydrogen and helium). The electrons are placed around the atom or ion, in pairs when necessary. Dot structures are most commonly used for main group elements. Example: What are the Lewis structures for H, N, and F?

  5. Example: What are the Lewis structures for H, N, and F? H 1s1 N 1s22s22p3 F 1s22s22p5

  6. Dot Structure For Main Group Elements The dot structures for atoms of main group elements are given below.

  7. Dot Structure For Atomic Ions In addition to dot structures for atoms, we can also write dot structures for ions formed from atoms by adding or removing the appropriate number of electrons. Example: Give the dot structures for O, O2-, and O2+.

  8. Example: Give the dot structures for O, O2-, and O2+. O 1s2 2s2 2p4 O2- 1s2 2s2 2p6 O2+ 1s2 2s2 2p2 O O2- O2+

  9. Octet Rule In studying the behavior of chemical compounds, bonding is often observed to occur by one of three processes: 1) Atoms lose electrons to completely empty their valence shell. The atom therefore forms a cation. This most often occurs for metal atoms. 2) Atoms gain electrons to completely fill their valence shell. The atom therefore forms an anion. This most often occurs for nonmetal atoms. 3) Atoms share electrons with other electrons so that when the shared electrons are counted the atom completely fills the valence shell. This again most often occurs with nonmetal atoms. Since a filled valence shell for a main group elements has eight electrons (except for H and He) the above behavior is called the octet rule. It is often, but not always, obeyed in chemical compounds.

  10. Ionic Bonding Ionic bonding refers to the bonding that occurs between cations and anions due to the attractive force that acts between particles of opposite charge. We may say the following: 1) Ionic bonding usually occurs between a metal cation and a nonmetal anion (or anion group). 2) The attractive forces in ionic bonding are isotropic (the same in all directions). 3) Ionic compounds are usually solids at room temperature. 4) Ionic compounds usually have high melting points and high boiling points relative to those observed for other substances. 5) Ionic solids are usually hard and brittle, and easily cleaved.

  11. Formation of Ionic Compounds To form a binary ionic compound one or more electrons are transferred between a metal atom and a nonmetal atom to form ions. NaCl Na 1s2 2s2 2p6 3s1 Na+ 1s2 2s2 2p6 Cl 1s2 2s2 2p6 3s2 3p5 Cl- 1s2 2s2 2p6 3s2 3p6 MgCl2 Mg 1s2 2s2 2p6 3s2 Mg2+ 1s2 2s2 2p6 Cl 1s2 2s2 2p6 3s2 3p5 Cl- 1s2 2s2 2p6 3s2 3p6 Cl 1s2 2s2 2p6 3s2 3p5 Cl- 1s2 2s2 2p6 3s2 3p6 The general tendency is for atoms to gain or lose electrons to either completely fill or completely empty the ns np valence orbitals. Note that the formation of ions is similar to the processes of ionization (for formation of cations) and electron affinity (for the formation of anions).

  12. Ionic Bonding With Dot Structures Formation of ionic bonds can be pictured in terms of dot structure as the transfer of electrons from metal atoms to nonmetal atoms. Example: Na + Cl  NaCl Note that we place ions within brackets, with the charge of the ion indicated outside of the bracket. This dot structure notation is not particularly useful for discussing ionic bonding, but is extremely useful in discussing covalent bonding.

  13. Lattice Energy Lattice energy (Hlattice) is defined as the energy required to convert exactly one mole of an ionic compound into gas phase ions. NaCl(s)  Na+(g) + Cl-(g) Elattice(NaCl) = 787. kJ/mol Lattice energy is expected to be positive when defined in this way, as we must add energy to break apart an ionic solid

  14. Lattice Energy and Coulomb’s Law We may use Coulomb’s law to predict general trends in lattice energy. F~ Q+ Q- Q+ = charge of cation Q- = charge of anion d2 d = distance between ion centers d = r+ + r- r+ = radius of cation r- = radius of anion The larger in magnitude the attractive force between ions the larger the value for lattice energy.

  15. Trends in Lattice Energy We may use Coulomb’s law to predict general trends in lattice energy. F~ Q+ Q- d2 The following follows from the above relationship. 1) As Q+ increases in magnitude the size of the lattice energy increases. 2) As Q- increases in magnitude the size of the lattice energy increases. 3) As the sizes of the ions decreases the size of the lattice energy increases (since smaller ion size means a smaller value for d). So the lattice energy increases as the magnitude of the charges of the ions increases and as the size of the ions decreases.

  16. Examples of Trends (1) 1) Cations in same group forming ionic compounds with the same anion. Lattice energy decreases in size in moving from top to bottom within a group. Ionic compound Elattice(kJ/mol) LiCl 834. NaCl 787. KCl 701. CsCl 657.

  17. Examples of Trends (2) 2) Anions in the same group forming ionic compounds with the same cation. Lattice energy decreases in size in moving from top to bottom within a group. Ionic compound Elattice(kJ/mol) LiF 1017. LiCl 860. LiBr 787. LiI 632.

  18. Examples of Trends (3) 3) As |Q+Q-| increases, the size of the lattice energy increases. Elattice= 910. kJ/mol Elattice= 3414. kJ/mol

  19. Example Arrange the following ionic compounds in order from largest to smallest size of lattice energy. 1) KF, KCl, KBr, KI 2) MgO, CaO, SrO, BaO 3) KF, K2O, K3N

  20. Example: Arrange the following ionic compounds in order from largest to smallest size for lattice energy. 1) KF, KCl, KBr, KI KF > KCl > KBr > KI Cation is the same, anion is larger as we go from F- to I-. 2) MgO, CaO, SrO, BaO MgO > CaO > SrO > BaO Anion is the same, cation is larger as we go from Mg2+ to Ba2+. 3) KF, K2O, K3N K3N > K2O > KF Cation is the same, anion has a larger charge as we go from F- to N3-.

  21. “Ionic Bonding” in Nonmetals Given the success of ionic bonding in explaining substances composed of metals + nonmetals, it is reasonable to attempt to apply the model for bonding between nonmetal atoms. Cl2 Cl [Ne] 3s2 3p5 Cl- [Ne] 3s2 3p6 Cl [Ne] 3s2 3p5 Cl+ [Ne] 3s2 3p4 Problems: 1) While the Cl- anion now satisfies the octet rule, the Cl+ cation is worse off than before. 2) Experimentally, both Cl atoms in Cl2 are the same. 3) Cl2 exists as molecules and is a gas at room temperature. We would expect ionic substances to exist as crystalline solids.

  22. Covalent Bonding We can get around the problems associated with transferring electrons by sharing one or more pairs of electrons. The shared electrons can be counted by both atoms in satisfying the octet rule. Bonding by sharing of one or more electron pairs is called covalent bonding. Notation 1) Bonding electron pairs are indicated by lines. 2) Nonbonding electrons, called lone pair electrons, are indicated by dots.

  23. Molecules A molecule is a neutral combination of two or more atoms held together by covalent chemical bonds. Molecules represent the smallest particle of covalently bonded chemical substances. Because of this, these substances are often called molecular compounds. Note that ionic compounds do not exist as individual molecules in the solid phase, but as collections of cations and anions held together by electrostatic forces. water (molecular) sodium chloride (ionic)

  24. General Properties of Chemical Systems As scientists began studying chemical systems they discovered several general properties of such systems. 1) Conservation of Mass. The total mass in a closed system remains constant, even if chemical reactions occur. 2) Law of Definite Proportions. All samples of a particular pure chemical substance contain the same relative amounts of each element making up the substance. Examples: methane 74.9 % C, 25.1 % H water 88.8 % O, 11.2 % H copper (II) sulfate 39.8 % Cu, 20.1 % S, 40.1 % O

  25. 3) Law of Multiple Proportions. When two elements can combine to form several different chemical compounds, the ratio of the amount of the second element combining with a fixed amount of the first element will be the ratio of small whole numbers. Example: There are two common compounds of carbon and oxygen carbon monoxide 1.000 g of C reacts with 1.332 g of O carbon dioxide 1.000 g of C reacts with 2.664 g of O g O in carbon dioxide = 2.664 g = 2.000 2 g O in carbon monoxide 1.332 g 1 It is easier to calculate the ratios with the larger number on top, but that is not required.

  26. Example: The following pure chemical substances can be formed out of the elements nitrogen and oxygen nitrogen monoxide 1.000 g of N reacts with 1.142 g of O nitrogen dioxide 1.000 g of N reacts with 2.285 g of O nitrous oxide 1.000g of N reacts with 0.5711 g of O Do these substances demonstrate the law of multiple proportions?

  27. nitrogen monoxide 1.000 g of N reacts with 1.142 g of O nitrogen dioxide 1.000 g of N reacts with 2.285 g of O nitrous oxide 1.000g of N reacts with 0.5711 g of O g O in nitrogen monoxide = 1.142 g = 2.000  2 g O in nitrous oxide 0.5711g 1 g O in nitrogen dioxide = 2.285 g = 4.001 4 g O in nitrous oxide 0.5711 g 1 g O in nitrogen dioxide = 2.285 g = 2.001 2 g O in nitrogen monoxide 1.142 g 1 So yes, these data are consistent with the law of multiple proportions.

  28. Connection With Atomic Theory Recall the second statement in Dalton’s Atomic Theory 2) Chemical substances (compounds) are composed of atoms of more than one element. a) In any particular pure chemical compound the same kinds of atoms are present in the same relative numbers. This statement accounts for both the Law of Definite Proportions and the Law of Multiple Proportions. For example, consider the nitrogen compounds we just discussed. nitrogen monoxide NO nitrogen dioxide NO2 nitrous oxide N2O

  29. Chemical Formula The chemical formula for a substance provides information concerning the composition of the substance. We can divide substances into two general types. 1) Substances that exist as collections of molecules. In this case the chemical formula indicates the number of atoms of each elements present per molecule. water phosphorus pentachloride nitrous acid (H2O) (PCl5) (HNO2)

  30. For organic molecules, the chemical formula is often given in a way that indicates how the molecule is put together. ethyl alcohol dimethyl ether acetone CH3CH2OH = C2H6O CH3OCH3 = C2H6O CH3COCH3 = C3H6O Notice that this longer notation makes it possible to distinguish among different forms (isomers) of organic molecules.

  31. 2) Substances that exist as collections of atoms or ions in the form of a crystal structure, a regular arrangement of the particles making up the substance. For these substances, the formula that is given is usually the empirical formula. An empirical formula gives the relative number of atoms of each element making up the compound, reduced to the smallest set of whole number coefficients. Ca2+ F- sodium chloride polonium calcium fluoride NaCl Po CaF2

  32. Empirical Formula for Molecular Compounds The empirical formula for a substance gives the relative number of atoms of each element making up a pure chemical substance, reduced to the smallest set of integer values. For substances that exist as molecules, the molecular formula must be an integer multiple of the empirical formula. SubstanceChemical formulaEmpirical formula water H2O H2O hydrogen peroxide H2O2 HO benzene C6H6 CH dichloroethane C2H4Cl2 CH2Cl acetic acid CH3COOH CH2O

  33. Molecular Compound A molecular compound is a compound composed of individual particles called molecules. The bonding between atoms is in such compounds is due to the sharing of one or more pairs of electrons. Molecular compounds are usually made up of one or more nonmetallic elements. Note that these compounds are sometimes called covalent compounds since the molecules are held together by covalent bonding. Examples: HBr hydrogen bromide SF6 sulfur hexafluoride CS2 carbon disulfide N2O4 dinitrogen tetroxide CH3Cl chloromethane SO2 sulfur dioxide Some elemental substances, such as O2 and N2, exist as indi-vidual molecules, but are not compounds, since they are composed of atoms of a single element.

  34. Ionic Compound An ionic compound is a compound formed from positive ions (cations) and negative ions (anions) held together by electrostatic attraction. For these compounds we usually give the empirical formula, (sometimes called the formula unit) or smallest electrically neutral collection of ions making up the compound. Binary ionic compounds (compounds formed from ions of two different elements) are usually a combination of a metal cation and a nonmetal anion. Examples: NaCl sodium chloride (Na+ and Cl-) Fe2O3 iron (III) oxide (Fe3+ and O2-) Na2O sodium oxide (Na+ and O2-) CuS copper (II) sulfide (Cu2+ and S2-)

  35. Chemical Formulas For Main Group Ionic Compounds Because main group elements form ions with a particular charge (depending on which group the element is from) we can predict the chemical formula for a main group ionic compound. We do this by first finding the charges of the ions formed, and then combining them to get an overall neutral compound using the smallest set of whole number coefficients. Examples: What is the chemical formula for the ionic compound formed from magnesium and chlorine, from sodium and sulfur, and from calcium and oxygen?

  36. Examples: What is the chemical formula for the ionic compound formed from magnesium and chlorine, from sodium and sulfur, and from calcium and oxygen? Mg and Cl Ions are Mg2+ and Cl-, so formula is MgCl2, magnesium chloride. Na and S Ions are Na+ and S2-, so formula is Na2S, aluminum sulfide. Ca and O Ions are Ca2+ and O2-, so formula is CaO, calcium oxide. magnesium chloride sodium sulfide calcium oxide

  37. Transition Metal Cations Transition metals can usually form ions with several different charges (this is also true for a few main group metallic elements like tin (Sn) and lead (Pb)). While we cannot easily predict which compounds will form between a transition metal and a main group nonmetal, we can usually figure out the charge of the transition metal cation if we know the chemical formula for the compound. This is indicated in the name of the compound. Examples: CuCl CuCl2 TiO2 MnS2

  38. Examples: CuCl Cl- ion, so Cu+ ion copper (I) chloride CuCl2 Cl- ion, so Cu2+ ion copper (II) chloride TiO2 O2- ion, so Ti4+ ion titanium (IV) oxide MnS2 S2- ion, so Mn4+ ion manganese (IV) sulfide CuCl CuCl2 TiO2

  39. Polyatomic Ions A polyatomic ion is a group of atoms which collectively has a charge and acts as an ion in an ionic compound. ion name example of ionic compound NO3- nitrate ion NaNO3, Ca(NO3)2, Ni(NO3)2 SO42- sulfate ion CuSO4, K2SO4, Al2(SO4)3 Note that when more than one polyatomic ion is present in a chemical compound the ion is placed in parentheses and the number of ions per formula unit of compound is given as a subscript outside the parentheses.

  40. Hydrates Some ionic compounds can exist in forms where there is a specific number of water molecules associated with every formula unit of the ionic compound. Such substances are called hydrates. Cobalt (II) chloride hexahydrate Cobalt (II) chloride

  41. Molecular Mass and Formula Mass The molecular mass (M) of a substance is the average mass of one molecule of the substance. For substances that do not exist as molecules, such as ionic compounds, we use the formula mass, which represents the average mass of one formula unit of the compound. Example: Find the molecular mass for nitrogen dioxide (NO2) and the formula mass for potassium chloride (KCl).

  42. Example: Find the molecular mass for nitrogen dioxide (NO2) and the formula mass for potassium chloride (KCl). NO2 N 1 . 14.0067 amu = 14.0067 amu O 2 . 15.9994 amu = 31.9988 amu M = 46.0055 amu  46.01 amu KCl K 1 . 39.0983 amu = 39.0983 amu Cl 1 . 35.453 amu = 35.453 amu M = 74.551  74.55 amu We often round off molecular mass or formula mass to two places to the right of the decimal point. Also recall that we can interpret the molecular mass or formula mass as the mass in grams per mole of substance. For example, for NO2, M= 46.01 g/mol.

  43. Percent By Mass The percent by mass represents the percentage of each element present in a sample of a pure chemical substance. The percent by mass of a particular element is simply the mass of that element found in one molecule (formula unit) divided by the total mass of the molecule (or formula unit), converted to a percentage by multiplying by 100%. percent X by mass = mass of X • 100% formula mass Note that the sum of the percent by mass of each element present in a substance should be equal to 100% (to within roundoff error). Example: What is the molecular mass, and the percent by mass of carbon, hydrogen, and oxygen, in propionaldehyde (C3H6O)?

  44. Example: What is the molecular mass, and the percent by mass of carbon, hydrogen, and oxygen, in propionaldehyde (C3H6O)? First step: Find the average mass of each element per molecule and the average mass per molecule (or formula unit). C3H6O C 3 . 12.0107 amu = 36.0321 amu H 6 . 1.00794 amu = 6.04764 amu O 1 . 15.9994 amu = 15.9994 amu M = 58.0791 amu  58.08 amu

  45. Second step: Find the percent composition. C3H6O C 3 . 12.0107 amu = 36.0321 amu H 6 . 1.00794 amu = 6.04764 amu O 1 . 15.9994 amu = 15.9994 amu M = 58.0791 amu  58.08 amu % C by mass = 36.0321 amu • 100% = 62.04 % C by mass 58.0791 amu % H by mass = 6.04764 amu • 100% = 10.41 % H by mass 58.0791 amu % O by mass = 15.9994 amu • 100% = 27.55 % O by mass 58.0791 amu Note 62.04% + 10.41% + 27.55% = 100.00%, as expected.

  46. Use of Percent Composition as a Conversion Factor The percent composition of a chemical substance can be used as a conversion factor in calculations. Example: Titanium metal is often extracted from rutile, an ore that is 59.9 % titanium by mass. What mass of rutile would be needed to produce 100.0 kg of titanium metal?

  47. Example: Titanium metal is often extracted from rutile, an ore that is 59.9 % titanium by mass. What mass of rutile would be needed to produce 100.0 kg of titanium metal? The percent composition tells us that for every 100.0 kg of rutile we will have 59.9 kg of titanium. Therefore kg rutile = 100.0 kg Ti 100.0 kg rutile = 167. kg rutile 59.9 kg Ti

  48. Use of the Mole Concept (Example) A chemist has a 14.38 g sample of carbon tetrachloride (CCl4). a) How many moles of CCl4 does she have? b) How many atoms of C and Cl are present in the sample?

  49. Use of the Mole Concept (Example) A chemist has a 14.38 g sample of carbon tetrachloride (CCl4). a) How many moles of CCl4 does she have? CCl4 C 1 . 12.0107 amu = 12.0107 amu Cl 4 . 35.453 amu = 141.81 amu M = 153.82 amu = 153.82 g/mol Moles CCl4 = 14.38 g 1 mol = 0.09349 mol CCl4 153.82 g = 9.349 x 10-2 mol CCl4

  50. Use of the Mole Concept (Example) b) How many atoms of C and Cl are present in the sample? # C atoms = 9.349 x 10-2 mol CCl41 mol C 6.022 x 1023 atoms C 1 mol CCl4 mol = 5.630 x 1022 C atoms # Cl atoms = 9.349 x 10-2 mol CCl44 mol Cl 6.022 x 1023 atoms C 1 mol CCl4 mol = 2.252 x 1023 Cl atoms

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