2. General Guidelines AP equation sets are found in the free- response section of the AP test. You are expected to: write balanced net ionic equations for three reactions when given the reactants answer a question concerning that reaction

3. General point guidelines (they may deviate from this): ◦ Each reaction will be worth a total of 5 points. One point is given for the correct reactants and two points for all correct products. ◦ If a reaction has three products, one point is given for two correct products and two points for all correct products. ◦ Leaving in the spectator ions will result in one point deduction on the equation set (not 1 point per problem). ◦ One point will be given for correct balancing and the final point will be given for correctly and completely answering the question concerning the reaction.

4. This section is worth 15 points and is 15 % of the free response grade. Free response is 50% of the total AP test grade. The first thing to note is that all AP equations "work". In each case, a reaction will occur. These equations need to be written in net ionic form. All spectator ions must be left out and all ions must be written in ionic form. All molecular substances and insoluble compounds must be written together (not ionized!). Know your solubility rules!

5. The best way to prepare for the equation section of the AP test is to practice lots of equations. The equation sets are similar and some equations show up year after year. When you are reading an equation, first try to classify it by type. If the question says anything about acidic or basic solution, it is redox. If you are totally stuck, look up the compounds in the index of your book or other reference books and try to find information that will help you with the equation. All reactions do not fit neatly into the five types of reactions that you learned in Chemistry I. Save the reactions that you write and practice them again before the AP test in May.

6. 4.4 Types of Chemical Reactions There are three general types of solution reactions. ◦ Precipitation reactions (most often include double replacement and complex ion) ◦ Acid base reactions (also called neutralization) ◦ Oxidation reduction reactions (most often single replacement, combustion, synthesis and decomposition)

7. These equations are written in terms of: molecular equation –overall reaction stoichiometry complete ionic equation –all electrolytes are represented as ions net ionic equation- spectator ions aren’t included

8. 4.5 Precipitation Reactions In a double replacement reaction, two compounds react to form two new compounds. ◦ No changes in oxidation numbers occur. All double replacement reactions must have a "driving force" that removes a pair of ions from solution.

9. Formation of a precipitate: A precipitate is an insoluble substance formed by the reaction of two aqueous substances. Two ions bond together so strongly that water can not pull them apart. You must know your solubility rules to write these net ionic equations to determine which product, if any, will precipitate in a double replacement reaction.

10. Simple Rules for Solubility Most nitrate (NO 3 - ) salts are soluble. Most alkali (group 1A) salts and NH 4 + are soluble. Most Cl -, Br -, and I - salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+ ) Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) Most OH - salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) Most S 2-, CO 3 2-, CrO 4 2-, PO 4 3- salts are only slightly soluble.

11. Extra Solubility Information Ca(OH) 2 and Sr(OH) 2 are moderately soluble and can be written together or as ions. Ba(OH) 2 is soluble and Mg(OH) 2 is insoluble. CaSO 4 and SrSO 4 are moderately soluble and can be written together or as ions. Weak electrolytes, such as acetic acid, are not ionized. Solids and pure liquids are written together, also. A saturated solution is written in ionic form while a suspension is not ionized.

12. Ex. Solutions of silver nitrate and lithium bromide are mixed. AgNO 3 (aq) + LiBr(aq)  AgBr(s) + LiNO 3 (aq) Ag + + NO 3 - + Li + + Br -  AgBr + Li + +NO 3 - Ag + + Br -  AgBr

13. Formation of a gas: Gases may form directly in a double replacement reaction or can form from the decomposition of a product such as H 2 CO 3 or H 2 SO 3.

14. Ex. Excess hydrochloric acid solution is added to a solution of potassium sulfite. 2HCl(aq) + K 2 SO 3 (aq)  H 2 O(l) + SO 2 (g) + 2KCl(aq) 2H + + 2Cl - + 2K + + SO 3 2-  H 2 O + SO 2 + 2K + +2Cl - 2H + + SO 3 2-  H 2 O + SO 2

15. Ex. A solution of sodium hydroxide is added to a solution of ammonium chloride. NaOH(aq)+NH 4 Cl(aq)  NaCl(aq)+NH 3 (g)+H 2 O(l) Na + + OH - + NH 4 + + Cl -  Na + + Cl - + NH 3 + H 2 O OH - + NH 4 +  NH 3 + H 2 O

16. Formation of a molecular substance: When a molecular substance such as water or acetic acid is formed, ions are removed from solution and the reaction "works".

17. Ex. Dilute solutions of lithium hydroxide and hydrobromic acid are mixed. LiOH(aq) + HBr(aq)  LiBr(aq) +H 2 O(l) Li + + OH - + H + + Br -  Li + +Br - + H 2 O OH - + H +  H 2 O (HBr, HCl, and HI are strong acids)

18. Ex. Gaseous hydrofluoric acid reacts with solid silicon dioxide. 4HF(g) + SiO 2 (s)  SiF 4 (g) + 2H 2 O(l) 4HF + SiO 2  SiF 4 + 2H 2 O

19. Knowing about precipitation reactions allows us to isolate ions in the lab during experiments. Qualitative analysis- process of separating and identifying ions ◦ Selective precipitation- process by which ions are caused to ppt one by one in sequence to separate mixtures of ions.

20. Ex. Separate Ag +, Ba 2+, Fe 3+ 1. Add Cl - to remove Ag + as AgCl (white ppt). 2. Add SO 4 2- to remove Ba 2+ as BaSO 4. (creamy white ppt.) 3. Add OH - or S 2- to remove Fe 3+ as Fe(OH) 3 or Fe 2 S 3. (thick rust colored ppt or yellow ppt)

21. Ex. Separate Pb 2+, Ba 2+, Ni 2+ 1. Add Cl - to remove Pb 2+ as PbCl 2. (white chalky ppt) 2. Add SO 4 2- to remove Ba 2+ as BaSO 4. 3. Add OH - or S 2- to remove Ni 2+ as Ni(OH) 2 or NiS. (clumpy mint green ppt or super chunky black ppt)

22. Quantitative analysis- determines how much of a component is present. ◦ Gravimetric analysis- quantitative procedure where a precipitate containing the substance is formed, filtered, dried & weighed. Knowing about precipitation reactions allows us to isolate ions in the lab during experiments.

23. Ex. The zinc in a 1.2000g sample of foot powder was precipitated as ZnNH 4 PO 4. Strong heating of the ppt yielded 0.4089 g of Zn 2 P 2 O 7. Calculate the mass percent of zinc in the sample of the foot powder. 0.4089gZn 2 P 2 O 7 1 mol Zn 2 P 2 O 7 2 mol Zn 65.37g Zn = 304.7 g Zn 2 P 2 O 7 1 mol Zn 2 P 2 O 7 1 mol Zn 0.1754g Zn x 100 = 14.62% Zn 1.200g sample

24. Ex. A mixture contains only NaCl and Fe(NO 3 ) 3. A 0.456g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH) 3. The ppt is filtered, dried, & weighed. Its mass is 0.128g. Calculate: a. the mass of the iron b. the mass of Fe(NO 3 ) 3 c. the mass percent of Fe(NO 3 ) 3 in the sample 0.128g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 55.85g Fe= 0.0669g Fe 106.9g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 0.0669g Fe 1 mol Fe 1 mol Fe(NO 3 ) 3 241.9g Fe(NO 3 ) 3 = 0.290g Fe(NO 3 ) 3 55.85g Fe 1 mol Fe 1 mol Fe(NO 3 ) 3 0.290g x 100 = 63.6% Fe(NO 3 ) 3 0.456g

25. 4. 8 Acid-Base Reactions Brønsted-Lowry acid-base definitions: ◦ acid- proton donor ◦ base- proton acceptor

26. Neutralization When a strong acid reacts with a strong base the net ionic reaction is: H + (aq) + OH - (aq)  H 2 O(l) Watch out for information about quantities of each reactant!

27. Strong and Weak Acids and Bases Remember which acids are strong (ionize completely) and which are weak (write as molecule). Sulfuric acid solution (strong acid) should be written as H + and HSO 4 -. Concentrated strong acids must be left together as they do not have enough water to ionize.

28. Examples ◦ Ex. A solution of sulfuric acid is added to a solution of barium hydroxide until the same number of moles of each compound has been added.  H + + HSO 4 - + Ba 2+ + 2OH -  BaSO 4 + 2H 2 O ◦ Ex. Hydrogen sulfide gas is bubbled through excess potassium hydroxide solution.  H 2 S + OH -  H 2 O + HS - *(S 2- does not exist in water)

29. Examples Watch out for substances that react with water before reacting with an acid or a base. These are two step reactions. ◦ Ex. Sulfur dioxide gas is bubbled into an excess of a saturated solution of calcium hydroxide.  SO 2 + Ca 2+ + 2OH -  CaSO 3 + H 2 O

30. Neutralization When a strong acid reacts with a weak base or a weak acid reacts with a strong base, the reaction is complete (the weak substance ionizes completely.) HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O(l) + C 2 H 3 O 2 - (aq) ◦ Neutralization reaction- An acid-base reaction when just enough base (OH - ) is added to react exactly with the acid (H + ) in a solution. This leads to an acid that has been neutralized.

31. Volumetric Analysis ◦ Titration- A process in which a solution of known concentration (standard solution) is added to analyze another solution.

32. Titration titrant- solution of known concentration (in buret) equivalence point or stoichiometric point- point where just enough titrant has been added to react with the substance being analyzed

33. Titration Indicator- chemical which changes color at or near the equivalence point End point- point at which the indicator changes color

34. Ex. 54.6 mL of 0.100 M HClO 4 solution is required to neutralize 25.0 mL of an NaOH solution of unknown molarity. What is the concentration of the NaOH solution? HClO 4 + NaOH  H 2 O + NaClO 4 0.0546 L HClO 4 0.100 mol HClO 4 1 mol NaOH = 1 L HClO 4 1 mol HClO 4 0.00546 mol NaOH 0.00546 mol NaOH = 0.218 M NaOH 0.025L

35. Oxidation Reduction Reactions Redox reactions involve the transfer of electrons. The oxidation numbers of at least two elements must change. Single replacement, synthesis, decomposition and combustion reactions are redox reactions. Oxidation - loss of electrons (increase in oxidation number) Reduction - gain of electrons (decrease in oxidation number) Oxidizing agent - electron acceptor (substance that is reduced) Reducing agent - electron donor (substance that is oxidized)

36. One atom gains electrons, the other loses. These MUST happen at the same time, during the same reaction.

37. Rules for Assigning Oxidation States 1. Oxidation state of an atom of a pure element = 0 2. Oxidation state of monatomic element in a compound = charge from the periodic table 3. Oxygen = -2 in covalent compounds (except in peroxides where it = -1) 4. H = +1 in covalent compounds 5. Fluorine = -1 in compounds 6. Sum of oxidation states = 0 in compounds 7. Sum of oxidation states in a polyatomic ion = charge of the ion

38. 2KI + F 2  2KF + I 2 oxidized I reduced F OA F2F2 RA KI +1-1 0 +1 -1 0 Let’s Practice!

39. 2PbO 2  2PbO + O 2 oxidized: O reduced: Pb OA: PbO 2 RA: PbO 2 +4 -2 +2 -2 0 Let’s Practice!

40. Noninteger oxidation states are rare, but possible. Fe 3 O 4 O = 4(-2) = -8 Fe = 8/3 = 2 2/3 or Fe 2+, Fe 3+, Fe 3+ 8/3 -2

41. What will it produce? To predict the products of a redox reaction, look at the reagents given to see if there is both an oxidizing agent and a reducing agent. When a problem mentions an acidic or basic solution, it is probably redox.

42. Common oxidizing agentsProducts formed Common reducing agents Products formed MnO 4 - in acidic solution Mn 2+ halide ionsfree halogen MnO 2 in acidic solutionfree metalsmetal ions MnO 4 - in neutral or basic solution MnO 2 (s)sulfite ions or SO 2 sulfate ions Cr 2 O 7 2- in acidic solutionCr 3+ nitrite ionsnitrate ions HNO 3, concentratedNO 2 free halogens, dilute basic solution hypohalite ions HNO 3, diluteNO free halogens, conc. basic solution halate ions H 2 SO 4, hot, concentratedSO 2 metal-ous ionsmetal-ic ions Na 2 O 2 NaOHH2O2H2O2 O2O2 HClO 4 Cl - C 2 O 4 2 - CO 2 H2O2H2O2 H2OH2O metal-ic ionsmetal-ous ions free halogenshalide ions

43. Ex. A solution of tin(II) chloride is added to an acidified solution of potassium permanganate. Use the rules to determine that Sn 2+ (stannous) will become Sn 4+ (stannic) and that MnO 4 - in acidic solution becomes Mn 2+. Balance with half reactions from there. 5Sn 2+ + 16H + + 2MnO 4 -  5Sn 4+ + 2Mn 2+ + 8H 2 O

44. Ex. A solution of potassium iodide is added to an acidified solution of potassium dichromate. 6I - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 3I 2 + 7H 2 O Use the rules to determine that the halide ion (iodide) becomes the free halide, or iodine. The dichromate becomes Cr 3+ in acidic solution.

45. Ex. Hydrogen peroxide solution is added to a solution of iron(II) sulfate. H 2 O 2 + 2Fe 2+ +2H +  2Fe 3+ + 2H 2 O Use the rules to determine that hydrogen peroxide becomes water and ferrous becomes ferric.

46. Ex. A piece of iron is added to a solution of iron(III) sulfate. Fe + 2Fe 3 +  3Fe 2+ Use the rules to determine that solid iron becomes a free metal becomes the most stable metal ion and that ferrous becomes ferric.

47. Tricky redox reactions that appear to be ordinary single replacement reactions: Hydrogen reacts with a hot metallic oxide to produce the elemental metal and water. This reaction occurs in the reduction of many metal ores. Ex. Hydrogen gas is passed over hot copper(II) oxide. H 2 (g) + CuO(s)  Cu(s)+ H 2 O(l)

48. A metal sulfide reacts with oxygen to produce the metallic oxide and sulfur dioxide. 2Li 2 S + 3O 2  2Li 2 O + 2SO 2

49. Chlorine gas reacts with dilute sodium hydroxide to produce sodium hypochlorite, sodium chloride and water. (disproportionation reaction = the same element is both oxidized and reduced) Cl 2 + NaOH  NaClO + NaCl + H 2 O

50. Copper reacts with concentrated sulfuric acid to produce copper(II) sulfate, sulfur dioxide, and water. (Cu doesn’t react with dilute sulfuric acid!) Cu + H2SO 4  CuSO 4 + SO 2 + H 2 O

51. Copper reacts with dilute nitric acid to produce copper(II) nitrate, nitrogen monoxide and water. NO is a colorless gas that is quickly oxidized in the air to NO 2, a brown gas. Cu + HNO 3  Cu(NO 3 ) 2 + H 2 O +NO

52. Copper reacts with concentrated nitric acid to produce copper(II) nitrate, nitrogen dioxide and water. NO 2 is a poisonous, brown gas. It is a large problem in photochemical smog. Cu + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O

53. 1.Write skeleton half-reactions using the rules. 2.Balance all elements other than O and H. 3.Balance O by adding H 2 O. 4.Balance H by adding H +. 5.Balance charge by adding e - to the more positive side. 6.Make the # of e - lost = # of e - gained by multiplying each half-rxn by a factor. 7.Add half-reactions together. 8.Cancel out anything that is the same on both sides. Balancing redox reactions by the half-reaction method

54. 9.If the reaction occurs in basic solution, add an equal number of water molecules to balance out the oxygen and then add hydrogen ions to balance out the hydrogen. Then, add enough hydroxide to the hydrogens to make water and add an equal number of hydroxides on the other side. Cancel any water that appears on both sides. 10.Check to see that charge and mass are both balanced. Balancing redox reactions by the half-reaction method

55. Practice: Sn 2+ + Cr 2 O 7 2-  Sn 4+ + Cr 3+ (acidic solution) Ox. Sn 2+  Sn 4+ Red. Cr 2 O 7 2-  Cr 3+ Ox. Sn 2+  Sn 4+ Red. Cr 2 O 7 2-  2Cr 3+ Ox. Sn 2+  Sn 4+ Red. Cr 2 O 7 2-  2Cr 3+ +7H 2 O Ox. Sn 2+  Sn 4+ Red. Cr 2 O 7 2- + 14H +  2Cr 3+ +7H 2 O Ox. Sn 2+  Sn 4+ +2e - Red. Cr 2 O 7 2- + 14H + +6e -  2Cr 3+ +7H 2 O Ox. 3(Sn 2+  Sn 4+ +2e - ) Red. Cr 2 O 7 2- + 14H + +6e -  2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2- + 14H + +6e -  3Sn 4+ +6e - + 2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2- + 14H +  3Sn 4+ + 2Cr 3+ +7H 2 O

56. MnO 4 2- + I -  MnO 2 + I 2 (basic solution) Ox. I -  I 2 Red.MnO 4 2-  MnO 2 Ox.2I -  I 2 Red.MnO 4 2-  MnO 2 Ox.2I -  I 2 Red.MnO 4 2-  MnO 2 + 2H 2 O Ox.2I -  I 2 Red.MnO 4 2- +4H +  MnO 2 + 2H 2 O Ox.2I -  I 2 +2e - Red.MnO 4 2- +4H + + 2e -  MnO 2 + 2H 2 O MnO 4 2- +4H + + 2e - +2I -  MnO 2 + 2H 2 O +I 2 +2e - MnO 4 2- +4H + +2I -  MnO 2 + 2H 2 O +I 2 MnO 4 2- +4H + + 4OH - +2I -  MnO 2 + 2H 2 O +I 2 + 4OH - MnO 4 2- + 4H 2 O +2I -  MnO 2 + 2H 2 O +I 2 + 4OH - MnO 4 2- + 2H 2 O +2I -  MnO 2 + I 2 + 4OH -

57. OXIDATION-REDUCTION TITRATIONS Most common oxidizing agents: KMnO 4 & K 2 Cr 2 O 7 Potassium permanganate used to disinfect ponds and fish in Egypt. Photo by Will Rooney (AP 2008)

58. MnO 4 - in acidic solution: MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O Purple colorless When you titrate with MnO 4 -, the solution is colorless until you use up all of the reducing agent (substance being oxidized).

59. Single Replacement Reactions Reaction where one element displaces another in a compound. One element is oxidized and another is reduced. AB + C  A + BC

60. Activity of Metals Active metals replace less active metals or hydrogen from their compounds in aqueous solution. Use an activity series or a reduction potential table to determine activity. The more easily oxidized metal replaces the less easily oxidized metal.

61. Ex. Magnesium turnings are added to a solution of iron(III) chloride. 3Mg(s) + 2FeCl 3 (aq)  2Fe(s)+3MgCl 2 (aq) 3Mg + 2Fe 3+ + 6Cl -  2Fe + 3Mg 2+ + 6Cl - 3Mg + 2Fe 3+  2Fe + 3Mg 2+ Always make sure that charge is balanced, as well as mass (atoms)!

62. Ex. Sodium is added to water. Ex. Sodium is added to water. 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) 2Na + 2H 2 O  2Na + + 2OH - + H 2 Alkali metal demo

63. Activity of Nonmetals Active nonmetals replace less active nonmetals from their compounds in aqueous solution. Each halogen will displace less electronegative (heavier) halogens from their binary salts.

64. Ex. Chlorine gas is bubbled into a solution of potassium iodide. Cl 2 (g) + 2KI(aq)  2KCl(aq) + I 2 (s) Cl 2 + 2K + + 2I -  2K + + 2Cl - + I 2 Cl 2 + 2I -  I 2 + 2Cl -

65. Decomposition Reactions Reaction where a compound breaks down into two or more elements or compounds. Heat, electrolysis, or a catalyst is usually necessary. A compound may break down to produce two elements. Ex. Molten sodium chloride is electrolyzed. 2NaCl(l)  2Na + Cl 2

66. A compound may break down to produce an element and a compound. Ex. A solution of hydrogen peroxide is decomposed catalytically. A compound may break down to produce an element and a compound. Ex. A solution of hydrogen peroxide is decomposed catalytically. 2H 2 O 2  2H 2 O + O 2

67. A compound may break down to produce two compounds. Ex. Solid magnesium carbonate is heated. MgCO 3  MgO + CO 2

68. Metallic carbonates break down to yield metallic oxides and carbon dioxide. CaCO 3  CaO + CO 2 Metallic chlorates break down to yield metallic chlorides and oxygen. 2LiClO 3  2LiCl + 3O 2

69. Hydrogen peroxide decomposes into water and oxygen. 2H 2 O 2  2H 2 O + O 2 Ammonium carbonate decomposes into ammonia, water and carbon dioxide. (NH 4 ) 2 CO 3  2NH 3 + H 2 O + CO 2

70. Sulfurous acid decomposes into water and sulfur dioxide. Carbonic acid decomposes into water and carbon dioxide. H 2 SO 3  H 2 O + SO 2 H 2 CO 3  H 2 O + CO 2

71. Addition Reactions Two or more elements or compounds combine to form a single product.

72. A Group IA or IIA metal may combine with a nonmetal to make a salt. A piece of lithium metal is dropped into a container of nitrogen gas. 6Li + N 2  2Li 3 N

73. Two nonmetals may combine to form a molecular compound. The oxidation number of the less electronegative element is often variable depending upon conditions. Generally, a higher oxidation state of one nonmetal is obtained when reacting with an excess of the other nonmetal. P 4 + 6Cl 2  4PCl 3 limited Cl 2 P 4 + 10Cl 2  4PCl 5 excess Cl 2

74. When an element combines with a compound, you can usually sum up all of the elements on the product side. Ex. PCl 3 + Cl 2  PCl 5

75. Two compounds combine to form a single product. SO 2 + CaO  CaSO 3 Ex. Sulfur dioxide gas is passed over solid calcium oxide. Ex. The gases boron trifluoride and ammonia are mixed. BF 3 + NH 3  H 3 NBF 3

76. A metallic oxide plus carbon dioxide yields a metallic carbonate. (Carbon keeps the same oxidation state) A metallic oxide plus sulfur dioxide yields a metallic sulfite. (Sulfur keeps the same oxidation state) A metallic oxide plus water yields a metallic hydroxide. A nonmetallic oxide plus water yields an acid.

77. Combustion Reactions -Elements or compounds combine with oxygen to produce oxides of each element.

78. Hydrocarbons or alcohols combine with oxygen to form carbon dioxide and water. Ex. Ethanol is burned completely in air. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O

79. Nonmetallic hydrides combine with oxygen to form oxides and water. 2PH 3 + 3O 2  P 2 O 3 + 3H 2 O Nonmetallic sulfides combine with oxygen to form oxides and sulfur dioxide. Ex. Carbon disulfide vapor is burned in excess oxygen. CS 2 + 3O 2  CO 2 + 2SO 2

80. Ammonia combines with limited oxygen to produce NO and water and with excess oxygen to produce NO 2 and water.

81. COMPLEX ION REACTIONS Complex ion- the combination of a central metal ion and its ligands Ligand- group bonded to a metal ion

82. Coordination Compound A neutral compound containing complex ions [Co(NH 3 ) 6 ]Cl 2 ◦ (NH 3 is the ligand, [Co(NH 3 ) 6 ] 2+ is the complex ion)

83. Common Complex Ions Formed in AP Equations Complex ionNameFormed from [Al(OH) 4 ] - tetrahydroxoaluminate ion(Al or Al(OH) 3 or Al 3+ ) [Ag(NH 3 ) 2 ] + diamminesilver (I) ion(Ag + + NH 3 ) [Zn(OH) 4 ] 2- tetrahydroxozincate ion(Zn(OH) 2 + ) [Zn(NH 3 ) 4 ] 2+ tetramminezinc ion(Zn 2+ + NH 3 ) [Cu(NH 3 ) 4 ] 2+ tetramminecopper(II) ion(Cu 2+ + NH 3 ) [Cd(NH 3 ) 4 ] 2+ tetramminecadmium(II) ion(Cd 2+ + NH 3 ) [FeSCN] 2+ thiocyanoiron(III) ion(Fe 3+ + SCN - ) [Ag(CN) 2 ] - dicyanoargentate(I) ion(Ag + and CN - ) [Ag(Cl) 2 ] - dichloroargentate(I) ion(AgCl and Cl - )

84. Breaking up a Complex Ion Adding an acid to a complex ion will break it up. If HCl is added to a silver complex, AgCl(s) is formed. If an acid is added to an ammonia complex, NH 4 + is formed.

85. Ex. Excess ammonia is added to a solution of zinc nitrate. (Other coordination numbers are acceptable, as long as correct charge is given.) Tetramminezinc ion 4NH 3 + Zn 2+  [Zn(NH 3 ) 4 ] 2+

86. Ex. A solution of diamminesilver(I) chloride is treated with dilute nitric acid. [Ag(NH 3 ) 2 ] + + Cl - + 2H +  AgCl + 2NH 4 +