He 1s 2 2 Ne 1s 2 2s 2 2p 6 8 = Ar 1s 2 2s 2 2p 6 3s 2 3p 6 8 = O 1s 2 2s 2 2p 4 6 = S 1s 2 2s 2 2p 6 3s 2 3p 4 6 = 2 + 4

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1 Bonding is the joining of two atoms in a stable arrangement. CHAPTER 7: IONIC COMPOUNDS There are two different kinds of bonding: 1. Ionic bonds result from the transfer of electrons from one element to another. 2. Covalent bonds result from the sharing of electrons between two atoms. Ionic bonds form between: A metal on the left side of the periodic table. A nonmetal on the right side of the periodic table. Ionic versus Covalent Bonding Na Cl 2 NaCl Covalent bonds are formed when two nonmetals combine. A molecule is a discrete group off atoms that share electrons. The valence electrons are available to participate in bonding The valence electrons are in the outermost shell (the highest number n) of an atom. Configuration # of valence e He 1s 2 2 Ne 1s 2 2s 2 2p 6 8 = Ar 1s 2 2s 2 2p 6 3s 2 3p 6 8 = O 1s 2 2s 2 2p 4 6 = S 1s 2 2s 2 2p 6 3s 2 3p 4 6 = Mn 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 2 Lu 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 1 2 Elements in the same group have the same number of valence electrons Group # 1A 8A = # valence electrons (except He = 2) Noble Gas Configuration The noble gases are chemically very stable. They do not react readily with other substances. Number of valence electrons Ex: All elements in Group 2A have 2 valence electrons. He Ne Ar Kr Xe Rn

2 Ionic compounds consist of oppositely charged ions that have a strong electrostatic attraction for each other. Ions are charged species in which the number of protons and electrons in an atom is unequal. There are two types of ions, cations and anions. Cations are positively charged ions. A cation has fewer electrons than protons. By losing electrons, a metal atom forms a cation with a completely filled outer shell of electrons. Anions are negatively charged ions. An anion has more electrons than protons. By gaining electrons, a nonmetal atom forms an anion with a completely filled outer shell of electrons. Octet Rule = 8 Valence Electrons Atoms will gain, lose or share sufficient electrons to achieve the same number of valence electrons as a noble gas (i.e. 8 electrons) p 11 e Na 11 p 10 e isoelectronic with neon 1s 2 2s 2 2p 6 Na 7 11 p 18 e isoelectronic with argon 1s 2 2s 2 2p 6 3s 2 3p Formation of a Sodium Cation, Formation of a Magnesium Cation, Mg 2+ Cations (Positive Ions) Are Smaller Than Their Neutral Atoms Because They Have Lost Their Outermost Electrons Determining the Charge on a Metal Ion A. The number of valence electrons in aluminum is 1) 1e 2) 2e 3) 3e B. To acquire an octet of electrons in aluminum requires 1) a loss of 3e 2) a gain of 3e 3) a gain of 5e C. The ionic charge of aluminum is 1) 3 2) 5 3) 3 + D. The symbol for the aluminum ion is 1) Al 3+ 2) Al 3 3) Al + 2

3 Octet Rule = 8 Valence Electrons Atoms will gain, lose or share sufficient electrons to achieve the same number of valence electrons as a noble gas (i.e. 8 electrons). Formation of a Chloride Anion, Cl p 17 e Cl Cl p 10 e isoelectronic with neon 1s 2 2s 2 2p p 18 e isoelectronic with argon 1s 2 2s 2 2p 6 3s 2 3p 6 1 Anions (Negative Ions) Are Larger Than Their Neutral Atoms Because They Have Added Electrons Rules for Gaining or Losing Electrons 1. Metals tend to lose electrons and form cations. 2. Nonmetals tend to gain electrons and form anions. 3. The number of electrons gained or lost rarely exceeds 3. Na 7 Cl 7+ The Reason for the Formation of Ionic Compounds Lies in the Electron Configurations of the Ions Elements in the Same Group Form Ions of Similar Charge Li 1s 2s 2p The charge of a positive ion is equal to its Group number. Group 1A(1) = 1 + Group 2A(2) = 2 + Group 3A(3) = 3 + F Li + F Same configuration as [He] Same configuration as [Ne] The charge of a negative ion is obtained by subtracting 8 or 18 from its Group number. Group 6A(16) = 6 8 = 2 or 16 18= 2 3

4 Cation Charge = Group Number Anion Charge = Group number 8 group 1A: M M + + e 1 valence e group 6A: X + 2e X 2 6 valence e charge = 6 8 = 2 group 2A: group 3A: M M e 2 valence e M M e 3 valence e group 7A: X + e X 7 valence e charge = 7 8 = 1 Remembering the Charges on Ions +1, +2, +3, Skip, -3, -2, -1, Zip Identify the element X in the following ions: a. X 2+, a cation with 36 electrons The +2 charge tells us it has lost 2 electrons. The ion must have 38 protons. So this element is strontium. b. X 3, an anion with 18 electrons The 3 charge tells us it has gained 3 electrons. The ion must have 15 protons. So this element is phosphorus. Common Ions in the Human Body Write symbols for the following ions: First, figure out how many electrons have been gained or lost. a.the ion formed from a nitrogen atom gained 3 e b.the ion formed from an oxygen atom gained 2 e c.the ion formed from a lithium atom lost 1 e N 3- O 2 Li + 4

5 Ionic compounds consist of oppositely charged ions that have a strong electrostatic attraction for each other. The attraction between oppositely charged ions constitutes an ionic bond. The Sum of the Charges in an Ionic Compound Must be Zero Overall In ionic compounds, the + s have to balance the s. The formula unit is the smallest whole-number repeating ratio of ions present in an ionic compound that results in charge neutrality. Ex: NaCl There really isn t such a thing as a specific ionic bond between specific pairs of ions. There are many ions attracted by ionic bonds to their nearest neighbors. Binary Ionic Compounds Compounds like NaCl, MgCl 2, LiF, KCl, CaF 2, LiBr, SrF 2, BaCl 2, NaI, etc. are called Binary Ionic Compounds Binary means two. Binary ionic compounds contain only 2 types of elements; one element is a metal and the other is a nonmetal. How to write a formula for an ionic compound How to write a formula for an ionic compound Step [1] Identify which element is the cation and which is the anion. Step [2] Determine how many of each ion type is needed for an overall charge of zero. Metals form cations and nonmetals form anions. Use the group number of a main group element to determine the charge. K + Cl Ca 2+ O 2 metal nonmetal metal nonmetal group 1A group 7A group 2A group 6A When the cation and anion have the same charge, only one of each is needed. K + + Cl KCl Ca 2+ + O 2 CaO zero charge zero charge One of each ion is needed to balance charge. 5

6 How to write a formula for an ionic compound Step [2] Determine how many of each ion type is needed for an overall charge of zero. When the cation and anion have different charges, Use the ion charges to determine the number of ions of each needed. Ca 2+ A +2 charge means 2 Cl anions are needed. Cl A -1 charge means 1 Ca 2+ cation is needed. How to write a formula for an ionic compound Step [3] To write the formula, place the cation first and then the anion, and omit charges. Example: KCl CaO CaCl 2 Use subscripts to show the number of each ion needed to have a zero overall charge. When no subscript is written, it is assumed to be 1. Ca 2+ + Cl CaCl 2 2 Cl for each Ca 2+ Writing Formulas of Ionic Compounds Case 1: the number of electrons lost by the metal atom is exactly the same number gained by the nonmetal NaCl Ca 2+ O CaO In MgCl 2, Charge Balance in MgCl 2 A Mg atom loses two valence electrons Two Cl atoms each gain one electron Susbscripts indicate the number of ions needed to give charge balance Case 2: the number of electrons lost by the metal atom is not the same number gained by the nonmetal. Mg MgCl 2 K + K + K + +3 K 3 N N 3 3 In Na 2 S, Charge Balance in Na 2 S Two Na atoms lose one valence electron each One S atom gains two electrons Susbscripts shoe the number of ions needed to give charge balance Summary: Writing Ionic Formulas An ionic formula/compound consists of positively and negatively charged ions is neutral has charge balance total positive charge = total negative charge The symbol of the metal is written first, followed by the symbol of the nonmetal written second. 6

7 Write the Formula for the Compound Formed from Na and N Charge balance is used to write the formula for sodium nitride, a compound containing and N N 3 = Na 3 N 3(1+ ) + 1(3 ) = 0 Writing Formulas of Ionic Compounds Select the correct formula for each of the following ionic compounds: A. Mg 2+ and O 2 check: Mg 2+ + O 2 = 1(2+) + 1(2 ) = 0 1) Mg 2 O 2 2) MgO 3) MgO 2 B. Al 3+ and 1) AlCl 3 2) AlCl 3) Al 3 Cl check: Al = (3+) + 3(1 ) = 0 C. Mg 2+ and N 3 If is combined with N 3, you will need 3 ions for every 1 N 3 ion to balance the charges; therefore, the formula must be Na 3 N. 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 check: 3Mg 2+ +2N 3 = 3(2+) + 2(3 ) = 0 Writing Formulas for Ionic Compounds Using the Crisscross Method Writing Formulas for Ionic Compounds Using the Crisscross Method 1. Write the symbol for the metal cation and its charge. 2. Write the symbol for the nonmetal anion and its charge. 3. Charge (without sign) becomes subscript for other ion. 4. Reduce subscripts to smallest whole number ratio. 5. Check that the total charge of the cations cancels the total charge of the anions. 1. Ba +2 column 2A 2. Cl- column 7A 3. Ba +2 Cl - 4. BaCl 2 5. Ba = (1) (+2) = +2 Cl = (2) ( 1) = 2 1. Write the symbol for the metal cation and its charge. 2. Write the symbol for the nonmetal anion and its charge. 3. Charge (without sign) becomes subscript for other ion. 4. Reduce subscripts to smallest whole number ratio. 5. Check that the total charge of the cations cancels the total charge of the anions. 1. Al 3+ column 3A 2. O 2- column 6A 3. Al 3+ O 2-4. Al 2 O 3 5. Al = (2) (+3) = +6 O = (3) ( 2) = 6 Writing Formulas for Ionic Compounds Using the Crisscross Method 1. Write the symbol for the metal cation and its charge. 2. Write the symbol for the nonmetal anion and its charge. 3. Charge (without sign) becomes subscript for other ion. 4. Reduce subscripts to smallest whole number ratio. 5. Check that the total charge of the cations cancels the total charge of the anions. 1. Ca 2+ column 2A 2. S 2- column 6A 3. Ca 2+ S 2-4. Ca 2 S 2 CaS 5. Ca = (1) (+2) = +2 S = (1) ( 2) = 2 Practice Writing Formulas of Binary Ionic Compounds Potassium iodide Gallium bromide Strontium fluoride Aluminum selenide Potassium nitride Magnesium phosphide Zinc chloride Calcium nitride 7

8 Naming Cations Main group cations (in groups 1 (IA), 2 (IIA), and 13 (IIIA)) are named by identifying the metal, followed by the word ion. Examples: Li Lithium Li + Lithium ion Ca Calcium Ca 2+ Calcium ion Al Aluminum Al 3+ Aluminum ion Naming Anions Nonmetal anions are named by adding the suffix -ide to the stem of the nonmetal, followed by the word ion. Element Stem Formula of ion Name of ion Bromine brom- Br bromide ion Chlorine chlor- chloride ion Fluorine fluor- F fluoride ion Iodine iod- I iodide ion Nitrogen nitr- N 3 nitride ion Oxygen ox- O 2 oxide ion Phosphorus phos- P 3 phosphide ion Sulfur sulf- S 2 sulfide ion Naming Binary Ionic Compounds Name the cation and then the anion. Do not specify the charge on the ion. Do not specify how many ions of each type are needed to balance charge. Metal sodium Mg 2+ magnesium + Nonmetal Stem+ide + F fluoride Cl chloride NaF sodium fluoride MgCl 2 magnesium chloride Practice Naming Binary Ionic Compounds NaCl KI CaS Li 3 N CsBr MgO CsF AlCl 3 LiH Sodium chloride Potassium iodide Calcium sulfide Lithium nitride Cesium bromide Magnesium oxide Cesium fluoride Aluminum chloride Lithium hydride Some Metals Form More than One Type of Cation Some transition metals, inner-transition metals, and a few other metals form more than one type of charged ion. Examples: Fe 2+ and Fe 3+, Cu + and Cu 2+, Au + and Au 3+, Mn 2+ and Mn 7+, etc. To name binary ionic compounds that contain these elements, we must indicate which ion is present. There are two ways to do this: a modern system, and an older system that is still in use. Modern (Systematic) Naming System Indicate the ionic charge of the metal by putting a roman numeral in parentheses following the name of the metal. Ionic charge of metal = roman numeral Examples: Fe FeCl 2 Iron(II) chloride Fe FeCl 3 Iron(III) chloride 8

9 Only metals which can form more than one ion need the Roman numeral. Metals with a fixed ionic charge do NOT need the Roman numeral. Ag +, Zn 2+, and Cd 2+ form only one ion and do not require a Roman numeral Most Transition Metals and Group 4(14) Metals Form 2 or More Positive Ions How to Name an Ionic Compound That Contains a Metal with Variable Charge Example Step [1] Give the name for CuCl 2 Determine the charge on the cation. CuCl 2 2 Cl anions = 2 total negative charge Cu cation must have a +2 charge to make the overall charge zero How to Name an Ionic Compound That Contains a Metal with Variable Charge How to Name an Ionic Compound That Contains a Metal with Variable Charge Step [2] Name the cation and the anion. Step [3] Write the name of the cation first, then the anion. The cation is named one of two possible ways: Systematic Common Cu 2+ copper(ii) cupric The anion changes ending of element name to ide Answer = copper(ii) chloride or cupric chloride Chloride How to Name an Ionic Compound That Contains a Metal with Variable Charge 1. Determine the charge of the cation from the anion. 2. Name the cation by the element name, and use a Roman numeral to show its charge. Ionic charge of metal = roman numeral Cu? F +1 1 CuF Cu? +2 F F 2 CuF 2 Practice Naming Ionic Compounds Containing a Transition Metal Cr 2 O 3 Fe 2 S 3 CuO Copper(I) fluoride Copper(II) fluoride 9

10 Naming Binary Ionic Compounds Using the Older Common Name System The suffix -ous is added to the root metal name with the ion of lower charge. The suffix -ic is added to the root metal name with the ion of high er charge. Compound Modern Name Older Name FeCl 2 FeCl 3 CuF CuF 2 CoBr 2 CoBr 3 iron(ii) chloride iron(iii) chloride copper(i) fluoride copper(ii) fluoride cobalt(ii) bromide cobalt(iii) bromide ferrous chloride ferric chloride cuprous fluoride cupric fluoride cobaltous bromide cobaltic bromide Polyatomic Ions A Polyatomic ion is a cation or anion that contains more than one atom. Both the formul as and the charges of the polyatomic ion must be memorized. NH 4 + OH CO 2 3 SO 2 4 CN NO 3 PO 3 4 ammonium ion hydroxide ion carbonate ion sulfate ion cyanide ion nitrate ion phosphate ion Rules For Naming Polyatomic Ions The names of common polyatomic anions End in ate ClO 3 - chlorate With one less oxygen end in it e ClO - 2 chlorit e With two less oxygen end in it e and have prefix hypo ClO - hypochlorit e With one more oxygen end in ate with prefix per ClO - 4 perchlorate With hydrogen attached use prefix hydrogen (or bi) HCO 3- hydrogen carbonate (bicarbonate) Ionic Compounds Containing Polyatomic Ions The + s have to balance the s. In the formula, parentheses are used around the polyatomic ion if more than one is used. NaOH K 2 HPO 4 Ca(MnO 4 ) 2 NH 4 NO 3 contains OH contains K + HPO 2 4 contains Ca 2+ MnO 4 contains NH 4+ NO 3 10

11 Naming Compounds Containing Polyatomic Ions Flowchart for Naming Ionic Compounds Polyatomic ion present Positive ion is present Negative ion is present Both ions are polyatomic 1. Polyatomic ion name 2. Stem of nonmetal name 3. Suffix ide 1. Full metal name 2. Possible Roman numeral 3. Polyatomic ion name 1. Positive polyatomic ion name 2. Negative polyatomic ion name NH 4 Cl Ammonium chloride Ca(MnO 4 ) 2 Calcium permanganate NH 4 NO 3 Ammonium nitrate Practice Naming Ionic Compounds MgS MgSO 3 MgSO 4 Ca(ClO 3 ) 2 CaCl 2 Ca(ClO 2 ) 2 Writing Formulas for Compounds Containing Polyatomic Ions Write the formula of each of the following substances: Determine the charge on the given metal ion Determine the charge on the given polyatomic ion In ionic compounds, the + s have to balance the s. Mg(NO 3 ) 2 PbO 2 Fe 2 (SO 4 ) 3 a. Sodium bicarbonate, HCO 3 - NaHCO 3 b. Aluminum hydroxide Al 3+, OH - Al(OH) 3 c. Lithium carbonate Li +, CO 3 2- Li 2 CO 3 Ba 3 (PO 4 ) 2 Writing Formulas for Compounds Containing Polyatomic Ions Select the correct formula for each: a. Aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 b. Copper(II) nitrate 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) c. Iron(III) hydroxide 1) FeOH 2) FeOH 3 3) Fe(OH) 3 d. Tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) Writing Formulas for Ionic Compounds Potassium bromate Calcium carbonate Sodium phosphate Iron(III) oxide Iron(II) nitrate Iron(II) hydroxide Copper(II) bromide Lithium phosphate 11

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