Predicting Ionic Charges OXIDATION NUMBER. Predicting Ionic Charges. Predicting Ionic Charges. Predicting Ionic Charges. Predicting Ionic Charges

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1 OXIDATION NUMBER Oxidation number = the charge an atom would acquire if all its bonds were treated as ionic bonds. To determine how many atoms combine with one another in a compound we must determine each element s OXIDATION NUMBER. Group 1: H + Li + Na + K + Lose 1 electron to form 1+ ions Be 2+ Mg 2+ Ca 2+ Sr 2+ Ba 2+ Group 2: Loses 2 electrons to form 2+ ions B 3+ Al 3+ Ga 3+ Group 13: Loses 3 electrons to form 3+ ions Group 14: Neither! Group 14 Lose of 4 elements rarely form electrons or gain ions. of 4 electrons? N 3- P 3- As 3- Nitride Phosphide Arsenide Group 15: Gains 3 electrons to form 3- ions 1

2 O 2- S 2- Se 2- Oxide Sulfide Selenide Group 16: Gains 2 electrons to form 2- ions F 1- Fluoride Cl 1- Chloride Br 1- Bromide Group 17: I 1- Iodide Gains 1 electron to form 1- ions Group 18: Stable Noble gases do not form ions! Groups 3-11: Many transition elements have multiple oxidation states. Iron(II) = Fe 2+ Iron(III) = Fe 3+ Groups 3-11: Some metals have only one possible oxidation state. Silver = Ag + Zinc = Zn 2+ Rules for assigning Oxidation Numbers 1. The atoms in a pure element have an oxidation number of zero. 2. Alkali metals always have an oxidation number of +1; alkaline earth metals always have an oxidation number of Fluorine always has an oxidation number of Oxygen has an oxidation number of -2 in almost all compounds. Exceptions are in compounds with a halogen, when it has an oxidation number of +2, and in peroxides (H2O2), when it has an oxidation number of Hydrogen has an oxidation number of +1 in almost all compounds except when combined with a metal when it has an oxidation number of The sum of all the oxidation numbers in a neutral compound is zero. 7. The sum of all the oxidation numbers in a polyatomic ion is equal to the charge of the ion. 2

3 ASSIGNING OXIDATION NUMBERS A compound has a total charge of ZERO so set your equation equal to ZERO. Assign the variable X to your unknown oxidation number. Given a compound, find the oxidation number of every element you know for certain. Then solve for others using algebra. KMnO 4 Do you notice a pattern with the elements in these compounds? H 2 +1 O 1 2 Mg 1 +2 F 2 1 Al 2 +3 S 3 2 Cu 3 +1 P 1 3 Ca 1 +2 Cl 2 1 Fe 2 +3 O 3 2 CaCO 3 CRISS-CROSS METHOD to determine the chemical formula 1. Write the symbols for the elements side by side. 2. Write the oxidation states of each element to the top right of the symbol. When the nonmetal is combines with a metal, the oxidation state will always be the first number (the negative one) in the list of oxidation states. 3. Criss cross the charges DOWN and use the absolute values (-2 becomes 2). 4. Check to make sure the subscripts are the lowest ratio. Practice Criss-Cross Method 1. Na and S 2. K and P 3. Al and S What about Polyatomic Ions? 4. Mg and Br Write down the only cations on Table E: (positively charges ions) 5. Al and O Write down the only polyatomic ions that end in ide. 3

4 Polyatomic Ions Chart (Table E) Find the charge of the polyatomic ion using Table E Put parenthesis around the polyatomic ion. 1. PO 4 7. HSO 4 2. CO 3 3. SO 3 4. NH 4 5. ClO 6. ClO 2 8. ClO 4 9. CN 10. OH 11. S 2 O SCN Polyatomic Ions If you see a group of atoms together with a charge it is a polyatomic ion from Table E. Put the polyatomic ion within Parenthesis. Find the charge of the polyatomic ion. Use the criss cross method to determine subscripts. Na +1 CO 3 2 How many atoms are present in the compound? Rules for writing the formula of the compound composed of ions 1. Place all Polyatomic Ions in Parenthesis (Table E) 2. Determine all oxidation numbers of elements and polyatomic ions 3. Use Criss-Cross Method 4. Reduce to Empirical Form NH 4 +1 S 2 How many atoms are present in the compound? Practice 4. Al SO 4 1. NH 4 S 5. Fe CO 3 2. Na NO 3 3. Cu Br 6. Pb PO 4 4

5 7. Ag ClO 10. Cu OH 8. Ca F 11. Ni I 12. Zn SO 4 9. NH 4 SO Pb ClO Cu CO H I 17. NH 4 O 18. Ag S 15. Fe HSO Al ClO 4 NAMING IONIC COMPOUNDS An Ionic compound can quickly be determined if a METAL is bonded to a NONMETAL. Naming Ionic compounds with metals that have only ONE oxidation state is fairly simple. Of the following metals listed below, check off all the elements that have more than one oxidation state: a. Mn b. Zn c. K d. Pb e. Mg f. Au g. Ag h. Ga i. Sn j. Li k. U l. Cr 5

6 Naming Ionic Compounds that contain metals with 1 oxidation state Binary Ionic Compounds = ionic compounds with only 2 different elements. Name the metal and end the nonmetal in ide. For example: CaBr 2 = Calcium Bromide Naming Binary Ionic Compounds Naming salts is very easy, because they are binary ionic compounds (made up of two elements). The cation is named by using the name of the element. The anion named by combining the name of the element with an ide ending. The name of compound is made up of both the cation and anion name Ex: NaCl = sodium chloride Ex: ZnS = zinc sulfide Ex: K 2 O = potassium oxide Ex: Mg 3 N 2 = magnesium nitride Ex: Al 2 S 3 = aluminum sulfide Name the following Binary compounds: 1. MgO= Magnesium Oxide 2. CaCl 2 = 3. AlBr 3 = 4. Ag 3 N = 5. Al 2 O 3 = 6. LiI = 7. BaF 2 = 8. Zn 2 C = 9. Ba 3 N 2 = 10.CdO = 11. Ga 2 S 3 = What about Polyatomic ions? (Table E) These are ions consisting of more than one atom. The names of polyatomic ions end in ate or ite, except for two (ammonium and cyanide) 12. K 3 N = 13. SrO = 6

7 Polyatomic ion chart (Table E) Ternary Ionic Compounds Ionic Compounds with 3 different elements. They usually contain Polyatomic Ions (Table E) Name the following Polyatomic Ions: a. NO - 3 nitrate d. SO 2-4 b. ClO 2- e. SO 3 2- c. CO 3 2- f. SCN - Naming Ternary Compounds Name the metal and then name the polyatomic ion (if it has a negative oxidation number). For example: KNO 3 = Potassium nitrate Naming Ternary Compounds Name the polyatomic ion (if it has a positive oxidation number) and then name the nonmetal. For example: NH 4 Cl = ammonium chloride METAL POLYATOMIC ION POLYATOMIC ION NONMETAL Naming Ternary Compounds If there are two polyatomic ions, name the positive polyatomic ion first and then name the negative polyatomic ion. For example: NH 4 NO 3 = ammonium nitrate POSITIVE POLYATOMIC ION NEGATIVE POLYATOMIC ION DO NOT CHANGE THE ENDINGS OF POLYATOMIC IONS! Name the following Ternary compounds: 1. NaC 2 H 3 O 2 Sodium acetate 2. AgHCO 3 Silver hydrogen carbonate 3. LiNO 2 4. Ga 2 (S 2 O 3 ) 3 5. Ca 3 (PO 4 ) 2 7

8 6. ZnSO 3 7. KClO 3 8. Al(OH) 3 9. RbSCN 10. SrCO 3 Naming Ionic Compounds with Metals with Multiple oxidation states (multiple charges): Using the STOCK SYSTEM 1. Determine the oxidation state of the metal in the compound. 2. Name the metal, put the oxidation state in ROMAN NUMERALS in parenthesis 1. I = 1, II = 2, III = 3, IV = 4, V = 5 3. End the nonmetal in ide. Review: Find the Formula Criss-Cross Find Empirical (bring down) formula Pb +4 O -2 Pb 2 O 4 Pb O 2 Cu +2 (SO 4 ) -2 Sn +2 (CO 3 ) -2 Work Backwards: Start with the Empirical Formula to determine the Oxidation state of a Metal with Multiple Oxidation States Empirical Formula Write in the oxidation number for the nonmetal or polyatomic ion you are sure of and criss-cross. Non-reduced form with Oxidation States Fe 1 O 1 Fe +2 O -2 to FeO Fe 2 O 2 reduce Fe 1 (SO 4 ) 1 Fe (SO 4 ) Cu 1 (SO 4 ) 1 Cu (SO 4 ) Work Backwards: Start with the Empirical Formula to determine the Oxidation state of a Metal with Multiple Oxidation States Empirical Formula Write in the oxidation number for the nonmetal or polyatomic ion you are sure of and criss-cross. Sn 1 (SO 3 ) 1 Sn (SO 3 3) Non-reduced form with Oxidation States Name the following compound using the Stock System: 1. Fe O Fe +2 O -2 Fe 2 O 2 Iron (II) Oxide 2. Fe Cl 2 Mn 1 (SO 4 ) 2 Mn (SO 4 ) 3. Cu SO 4 4. Pb Cl 2 Cr 1 (PO 4 ) 2 Cr (PO 4 ) 8

9 5. Pb O 2 6. Cu 3 (PO 4 ) 2 7. Cu 2 S 8. Fe 2 (CrO 4 ) Sn CO Sn F 4 Name each of the following compounds, use Roman Numerals only when necessary. Put a check next to every compound that begins with a metal with more than 1 oxidation state. Put parenthesis around all the polyatomic ions. 1. NH 4 Cl 2. Pb SO 4 3. Co Cl 3 9. Cu S 4. Ba (NO 3 ) 2 5. Co 2 (SO 3 ) 3 6. KH 7. NH 4 F 8. K 2 Cr 2 O Cu ClO Ag NO Fe Cl Cr F Na Cl 15. Fe PO Li F 17. Fe F Al (OH) Mg I Fe Cl 3 9

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