Chapters 8&9: Names, Formulas, and Oxidation NUmbers

Topics in this unit: Binary Ionic Compounds of Metals with Fixed Charges Formulas of Ionic Compounds: The criss-cross method Binary Compounds of Cations with Variable Charges: Stock System Polyatomics Names and Formulas of Covalent (Nonmetal) Compounds Acids Hydrates Oxidation Numbers

Binary Compounds of Metals with Fixed Charges Name the cation (metal) first, then the anion (nonmetal) Use the cation’s name directly from the periodic table (eg. Ca+2 is Calcium) Name the anion using the root of the element’s name plus the suffix “ide” (eg. S-2 would be Sulfide) Examples: CaS would be Calcium Sulfide Na2O would be Sodium Oxide Potassium Chloride would be KCl

Practice Name the Following Compounds: BaCl2 ZnF2 Ag2S AlBr3

Practice Name the Following Compounds: BaCl2 = barium chloride ZnF2 = zinc fluoride Ag2S = silver sulfide AlBr3 = aluminum bromide

Formulas of Ionic Compounds: The Criss-Cross Method Ionic compounds have to combine so that the positive charges are balanced out by the negative charges. To figure out the ratios of elements, we can use the criss-cross method Step 1: write the symbols for the ions and their charges as superscripts. Step 2: criss-cross the values of the charges, not the signs, so that they become subscripts in the formulas Step 3: make sure the subscripts are in the lowest whole number ratio

Formulas of Ionic Compounds: The Criss-Cross Method Example What is the formula for Sodium sulfide? Sodium ion = Na+1 Sulfide ion = S-2 Na+1 S-2 Final formula = Na2S1 or Na2S

Practice Write the formulas of the following ionic compounds: Magnesium chloride Lithium Sulfide Nickel Oxide Tungsten Nitride Sodium Fluoride

Practice Write the formulas of the following ionic compounds: Magnesium Chloride = MgCl2 Lithium Sulfide = Li2S Nickel Oxide = NiO Tungsten Nitride = WN2 Sodium Fluoride = NaF

Binary Compounds of Cations with Variable Charges: The Stock System Named for German chemist Alfred Stock Cations (metals) involved have multiple charges Anion has fixed charge Uses roman numerals in parentheses to show the cation’s charge Example: FeCl2 is iron(II) chloride; FeCl3 is iron (III) chloride

The Stock System Name the cation’s element first Determine the charge by multiplying the anion’s charge by it’s subscript; then divide this by the cation’s subscript Example 1: Name CuCl2 Chloride anion’s charge is -1 times it’s subscript 2 equals 2 (Ignore sign of -2) Divide by Copper’s subscript of 1 equals 2 The result is Copper(II) chloride

The Stock System Example 2: Name Fe2O3 Oxide anion’s charge is -2 times it’s subscript 3 equals -6 Divide by Iron’s subscript of 2 equals 3 The result is iron(III) oxide Example 3: Write the formula for manganese(IV) oxide The cation’s charge is given from the formula +4 Oxide’s charge is -2 Using the criss-cross method we first get Mn2O4,but this is not in the lowest ratio. So we divide each subscript by 2. The result is MnO2

Polyatomics Polyatomic ions can bond with metals that have fixed or variable charges. When more than one polyatomic ion is required, parentheses must be used around the polyatomic Example 1: Fe(NO3)2 Decide if the cation has a variable charge (iron does so you must determine the roman numeral) NO3 has a -1 charge so iron must be +2 The name is iron(II) nitrate

Polyatomics Example 2: Fe(OH)3 Example 3: Aluminum Phosphate Determine the charge of iron: Hydroxide is -1 so iron must be +3 The compound is iron(III) hydroxide Example 3: Aluminum Phosphate Aluminum is a fixed-charge of +3 Phosphate ion is PO4-3 The formula is AlPO4 Example 4: Ammonium Sulfate Ammonium is a polyatomic cation NH4+ Sulfate is SO4-2 The formula is (NH4)2SO4

Names and Formulas of Covalent (Nonmetal) Compounds Name the first element: if there are two or more, add a prefix to match the subscript Name the second element also using a prefix to match the subscript (including “mono-” for one). Change the ending to “ide” Number Prefix 1 Mono 6 Hexa 2 Di 7 Hepta 3 Tri 8 Octa 4 Tetra 9 Nona 5 Penta 10 Deca

Names of Nonmetal Compounds: Examples Example 1: N2O Subscript for 2 is “di-” so it is dinitrogen Subscript for O is 1 so the name is dinitrogen monoxide Example 2: N2O5 Subscript for nitrogen is 2 so it is dinitrogen Subscript is 5 for oxygen or pentaoxide The name is dinitrogen pentaoxide (pentoxide is also acceptable) Example 3: What is the formula for triphosphorus hexabromide? Phosphorus is P and tri means there’s 3 P atoms Bromine is Br and hexa means there’s 6 Br atoms The formula is P3Br6 (notice we don’t reduce the subscripts)

Practice: Covalent Compounds Name the following covalent compounds: CO P2O4 PCl3 Write formulas for the following covalent compounds: Antimony tribromide Xenon hexafluoride Diphosphorus decasulfide

Practice: Covalent Compounds Name the following covalent compounds: CO = carbon monoxide N2O4 = dinitrogen tetroxide PCl3 = phosphorus trichloride Write formulas for the following covalent compounds: Antimony pentabromide= SbBr5 Xenon hexafluoride = XeF6 Diphosphorus decasulfide = P2S10

Acids Acids are a special type of ionic compound that forms when certain polar molecules dissolve in water The formulas typically have an H in front Ex: HCl, H3PO4 The cation is always H+1, so the criss-cross rule still applies for writing formulas Ex: What acid forms when hydrogen bonds with the sulfate ion? Hydrogen ion = H+1 sulfate = SO42- The formula is H2SO4 There are two types of acids, binary and ternary (oxyacids)

Binary Acids Binary acids are hydrogen with a non-metal (Examples: HCl, HF, HBr, HI) “hydro-” STEM “-ic” is used (STEM comes from the anion) The word acid is added at the end Examples: HCl is hydrochloric acid (STEM is chlor) HF is hydrofluoric acid HBr is hydrobromic acid

Oxy (Ternary) Acids Contains hydrogen and a polyatomic ion Change the “ate” ending to “ic” and the “ite” ending to “ous” and add the word acid Polyatomic ion that ends in “ate” STEM –ic acid Example HClO3 = chloric acid H2SO4 = sulfuric acid Polyatomic ion that ends in “ite” STEM –ous acid Example HClO2 = chlorous acid H3PO3 = phosphorous acid

Oxy (Ternary) Acids Polyatomic ion with prefix “hypo” Hypo- STEM –ous acid Example HClO hypochlorous acid Polyatomic ion with prefix “per” Per- STEM –ic acid Example HClO4 perchloric acid

Hydrates Ionic compounds that have water in them General Formula AB . xH2O Name the compound (AB) and identify the number of water molecules present (x equals the coefficient) with a prefix Examples: CuSO4 . 5 H2O is copper(II) sulfate pentahydrate MgSO4 . 9 H2O is magnesium sulfate nonahydrate

Oxidation Numbers: The Rules All free uncombined elements have an oxidation number of zero (In diatomic elements like F2, each fluorine’s oxidation number is 0) Hydrogen has an oxidation number of +1 Exception: when combined with metals, H’s oxidation number is -1 Oxygen has an oxidation number of -2 Exception #1: In peroxide (O22-) the oxidation number is -1 Exception # 2: When oxygen bonds with fluorine (OF2) it is +2 because fluorine is more electronegative

Oxidation Numbers: The Rules The alkali metals (group 1A) have an oxidation state of +1 The alkaline earth metals (group 2A) have an oxidation state of +2 The oxidation number of Aluminum (Al) is always +3 The halogens (group 7A) have an oxidation state of −1 (except when they are bonded to oxygen or with another halogen)

Oxidation Numbers: The Rules In a neutral molecule, the sum of the oxidation numbers of all of the atoms must be zero.  Ex.  In H2O, each hydrogen is +1 and the oxygen is -2.  So, (2 x +1) + (-2) = 0.  In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal the net charge of the ion.  Ex.  In the polyatomic ion known as hydroxide (OH-), the oxygen is -2 and the hydrogen is +1.  So, (-2) + (+1) = -1, the same as the charge on the hydroxide ion (OH-)

Oxidation Number Examples What is the oxidation number of Na in Na2O? O is -2; the two Na must be +1 each What is the oxidation number of Cl in ClO¯? The O is -2, but since a -1 must be left over, then the Cl is +1 What is the oxidation number for each element in KMnO4? K = +1; O = -2 Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn = +7.