CH 101 1st Edition Lecture 12 Outline of Last Lecture I Polarity II Molecular Bonds A Formal Charge III Lewis Structures IV Drawing out Lewis Structures Outline of Current Lecture I Equivalent Resonance Structures II Preferred Resonance Structures III Molecular Structure IV Bond Order A Finding the Oxidation State using the Electron Counting Method Current Lecture V Equivalent Resonance Structures Equivalent resonance structures occur when a molecule has the same elements surrounding it this means that if there is a double bond that needs to be placed in the Lewis structure it doesn t matter which bond the double bond is placed between This occurs in molecules like CO32 because when you look at the Lewis structure you see that there is a double bond that can be place between any of Carbon and Oxygen atoms Here are the three equivalent resonance structures of CO32 These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute VI Preferred Resonance Structures Preferred resonance structures occurs when you create different Lewis structures that AREN T the same because the central atom is bonded to different types of elements like in CHO2 Find the preferred resonance structure by looking for the structure that does not have a charge and if it does the charge is on the most electronegative element Ex What is the preferred Lewis structure for the molecule SCN ER 24 VE 16 SP 4 LP 4 Determine your Lewis structure Using the information found about the Lewis structure of SCN there are three structures that can be drawn Analyze the three structures and determine which one is preferred by finding the formal charges which has the negative charge on the most electronegative atom In A the formal charges are S1 and N2In B the formal charge is S1In C the formal charge is N1 Using the formal charge Use that fact that the electronegativity of S is greater than N help you decide the preferred Lewis structure B is the preferred Lewis structure for SCN VII Molecular Structure The molecular structure is determined by the amount of bond regions there are and if there is a lone pair on the center atom The lone pair pairs attached to the center atom are known as dipoles and they distort the angle of the bonds by pushing the angles closer together Electron Regions Lone pairs around center atom Bond Angle Name of shape 2 0 180o Linear 2 1 120o Bent 3 0 120o Trigonal Planar 3 1 109o Trigonal Pyramidal 4 2 109o Bent Angular 4 0 109o Tetrahedral VIII Visual Bond Order The bond order can be calculated by dividing the amount of bonds in the molecule by the amount of bond regions of total bonds of bond regions DON T FORGET to draw in your lone pairs because they count as bond regions if they are around the central atom When you are trying to determine the bond order of a particular part of a molecule first look at the bond then determine if the atoms bonded are bonded to the same element in another bond like O C O If she points out the bond highlighted in red and asks for the bond order you would see if the carbon in bonded to another oxygen or if the oxygen is attached to another carbon Since the carbon is attached to another oxygen the number of bonds is 2 and the number of regions is 2 therefore the bond order is 2 2 1 Ex Find the bond order of the molecule CO32 ER 32 VE 24 SP 4 LP 8 Determine the Lewis structure of CO 32 Here is the Lewis structure to find the bond order you must find the number of binds and the number of bond regions Bond 4 Bond regions 3 You can see that although there are 4 bonds there are only 3 regions because of the double bond 4 3 Plug into the equation bond bond regions to find that the bond order is 4 3 A Finding the Oxidation State using the Electron Counting Method In chapter five the counting method was introduced to find the oxidation state of an atom within a molecule The oxidation state is found by finding the valence electrons of the atom and subtracting the number of possessed electrons VEPE oxidation state When finding the possessed electrons use the electronegativity of the atoms to determine which atom takes the electrons The atom with the higher electronegativity possess both of the electrons in the bond If the atoms bonded are the same elements each atom gets one electron out of the pair Ex Find the oxidation state of the CA carbon in the molecule C2HOF2using the electron counting method Here is the Lewis structure for the molecule C2HOF2 VE for CA is 4 You can reference the Lewis structure or the periodic table to determine that the valence electrons for carbon is 4 Electronegativity of F C H Look at the periodic table to determine which atoms have the higher electronegativity in each bond this will determine the amount of possessed electrons an atom has because the more electronegative atom gets both the electrons PE for CA is 1 Fluorine take both the electrons in the bond with Carbon but carbon possesses one electron in its bond with Carbon 4 1 3 Plug into VE PE oxidation state to get 3 as the oxidation state of C A