CH 101 1st Edition Lecture 10 Outline of Last Lecture I Electron configuration of ions II Oxidation states A The Six Rules of Oxidation III Predicting Oxidation Formulas IV Possible Oxidation States for Unknown Elements X Y and Z Outline of Current Lecture I Types of Bonds II Naming Compounds III Finding the Chemical Formula IV Polyatomic Ions A Oxyanions or Oxoanions Current Lecture Side Note Last lecture notes I included the statement To find the electron configuration of ions you take the configuration of the element and either subtracting in the case of a cation electrons from the orbitals in this order p s d f etc I would like to amend that by saying that you take out of the p orbital first UNLESS it is filled if it is filled then you take from the s orbital first I Types of Bonds In CH 101 you are going to be working with ionic covalent and metallic bonds Ionic bonds are between a metal and a non metal where electrons are transferred from the low electronegative element metal to the element with the higher electronegativity non metal This bond occurs when the difference in electronegativity between the two elements is greater than or equal to 1 8 The bond between rubidium and sulfur is ionic Covalent bonds are between two non metals that share electrons without creating an ion This bond occurs when the difference in electronegativity between the two elements is less than or equal to 0 4 The bond between silicon and phosphorus is covalent Polar covalent bonds are created when there is an uneven sharing of electrons this bond occurs when the difference in electronegativity is greater than 0 4 but less than 1 8 The bond between Carbon and Fluorine is polar covalent Metallic bonds are between two metals and are held together by the sea of electrons that surround and atom Use this chart to calculate the difference in electronegativity or use the periodic table to estimate the type of bond II Naming Compounds You will learn that when naming compounds you must look at the placement of the element on the periodic table and the amount of atoms of each element in the compound Endings ide ite ate Monoatomic anion cation An anion cation that consist of only one element whereas a polyatomic ion is an ion that consists of more than one element Monoatomic O 2 F 1 As 3 Polyatomic SO42 With cation molecules Group 1 and Group two the compound is just called the element name with the word ion after it K1 Potassium ion Na1 Sodium ion Ra2 Radium ion With compounds that include transition metals you use roman numerals to represent the oxidation state of that transition metal The ending of the word will be ide if it s a monoatomic ion or ate or ite depending on the polyatomic ion NiSO3 Nickel II Sulfite TiBr3 Thallium III Bromide There are a few elements near the transition metals who will never use roman numerals because they only have one oxidation state Al3 Ag1 Zn2 and Cd2 review how to find the oxidation state of an ionic compound from last lectures notes because a bond between a transition metal and a non metal is an ionic bond ide Monoatomic anions changes to ide N3 Nitride S2 Sulfide O2 Oxide This also occurs when there is an ionic bond a bond between a metal and a non metal The anion in the compound changes its ending to ide CO2 Carbon Oxide LiBr Lithium Bromide Covalent bonds between two non metals will have the suffix ide Also the nonmetals will take on different prefix depending on how many atoms of that element are in the molecule Here is a chart of the prefix s that correspond with the amount of atoms of each element note that if the non metal that comes first when naming the compound is the element that is most metal like which would be furthest to the bottom left of the periodic table and if the first element only has one atom do not include the prefix mono Amount of atoms Prefix 1 Mono 2 Di 3 Tri 4 Tetra 5 Penta 6 Hexa 7 Hepta 8 Octa 9 Nona 10 Deca Examples of covalent compounds N2O5 dinitrogen pentoxide CO carbon monoxide PH3 phosphorus trihydride ate A compound will end in ate if the oxidation number of the cation is equal to it s main group number Sulfur s group number is 6 in the polyatomic ion SO42 the oxidation state of Sulfur is 6 and therefore we call this compound sulfate If you are given a compound with the ending of ite when you add one more oxygen to this compound you will get the compound with the ending ate Sulfite SO32 and Sulfate SO4 2 If you add another oxygen to any compound with the ending ate you must add the prefix per along with the ending ate this is the highest oxidation state of the compound For example Chlorate ClO31 and Perchlorate ClO41 ite If the cation s oxidation number is two less than the group number it ends in ite The group number for nitrogen is 5 the oxidation number for nitrogen in NO2 is 3 making it Nitrite Example SO32 Sulfite ClO21 Chlorite If an oxygen is taken away from the compound decreasing the oxidation level you add the prefix hypo to the name of the compound ClO1 Hypochlorite III Finding the Chemical Formula Last lecture we talked about predicting the chemical formulas of ions Now lets try to predict the chemical formula when given the name of the compound Ex What is the chemical equation of Thallium III Bromide Tl and Br First define the elements and notice that the use of roman numerals tells you that Thallium is a transition metal and that this is an ionic bond Tl3 Br1 The roman numeral III tells you that Thallium s oxidation state is 3 and since Bromide ends in ide you know that the oxidation state is the most common ion in its group on the periodic table 1 TlBr3 Now criss cross the oxidation numbers to get the chemical formula for Thallium III Bromide Ex What is the chemical equation of Nitrate N and O First identify the elements we know oxygen is present because nitrate is a polyatomic ion if it was just nitrogen it would be called nitride N5 and O Here is where it gets more difficult because this is a polyatomic ion The suffix is ate so the oxidation number of nitrogen equals its group number 5 You will have to remember that the polyatomic ions that contain nitrogen either gain 1 electron or lose 1 electron here it gains one electron giving it a charge of 1 5 6 1 Find the oxidation state of oxygen so that you can find the amount of oxygen atoms there are in nitrate 6 2 3 Divide the oxidation state of oxygen in Nitrate by the oxidation state of one oxygen atom which is 2 This tells you there are 3 oxygen atoms in one