Prof Gilbert LECTURE 22 CHEM 1211 11 1 10 Last time Chapter 8 Chemical Bonding Three types ionic metallic and covalent The Octet Rule Lewis symbols unpaired dots bonding capacity Lewis structures and the S N A rule This time Chapter 8 Chemical Bonding continued Polar bonds Sharing electrons does not necessarily mean equal sharing Different elements have different attraction for the shared pairs In other words different elements have different electronegativities Electronegativity is expressed using a relative scale that goes from 0 8 Cs to 4 0 F Most metals have values of 1 0 1 9 metalloids are 2 0 2 4 and non metals are 2 5 4 0 Note the periodic trends in electronegativities the overall height of the bars of the first 20 elements less the noble gases in this figure and how they map onto the trends in ionization energies the heights of the darker color portions different electronegativities such as H and Cl share a pair of electrons the pair spends more time closer to the more electro negative atom Cl giving that end of the bond a negative polarity and the other H end a positive polarity Here are some ways to represent unequal sharing and bond polarity When two atoms with Ions may be considered extreme cases of unequally shared electrons Bonds are considered ionic rather than compounds form when the difference in electronegativity between the two atoms is 2 0 NOTE the difference between a polar covalent bond and an ionic bond is a matter of the degree of inequality in electron sharing and not clearly defined Resonance Sometimes a single Lewis structure does not adequately describe the bonding in a molecules of molecular ion We have already seen some examples HNO3 NO3 CO3 2 Here is one more ozone O3 has two identical bonds but its Lewis Structure requires one single and one double bond To rationalize the difference we assume that the single double bond structures resonate Resonance can happen when bonding and lone pairs are free to move without breaking any single bonds Choosing Between Lewis Structures Formal Charges You may be able to draw more than one Lewis structure for a molecule in which the octet rule is obeyed by all the atoms To determine which structure is best calculate the formal charge on each atom and determine which structure minimizes these charges that one is preferred How to calculate formula charges Formal charge of valence electrons on the free atom of valence electrons assigned to the atom in the structure Inquiry the molecular structure of N2O How do we choose between NOTE When the of bonds to an atom in a Lewis structure matches the atom s bonding capacity its formal charge FC is 0 If there is one fewer bonds then FC 1 if one more bond then FC 1 Exceptions to the Octet Rule 1 Odd number of electrons Examples NO and NO2 Applying S N A to NO we get an N of 5 or 2 5 bonds Trying 2 or 3 Which atom gets the odd electron Let formal charges decide Applying S N A to NO2 we get an N of 7 or 3 5 bonds Sticking with 3 1 2 3 4 Structure 3 looks like the winner except the more electronegative element O is the one with an incomplete octet BAD FORM Never do that If any element is short changed electrons be sure it is the less electronegative one nitrogen So focus on 1 and 2 2 Less than an octet If we apply S N A to BeCl2 However FC s indicate the structure on the right is preferred even though the octet on Be is not complete Similarly for BF3 and for AlCl3 no it s not ionic 3 More than an octet Many compounds and polyatomic ions with atoms from the 3rd period onwards can accommodate more than an octet because the d orbitals are low enough in energy to participate in bonding Consider SF6 SO4 2 PO4 3 Bond length bond strength and bond order Bond Order the of shared pairs of electrons between atoms For example a double bond has a bond order of 2 Bond order is related to bond length which we can measure Consider these O O bonds Note inverse relationship between bond order and bond length This trend applies to bonds between all the elements Bond Strength average bond energy the average change in enthalpy H that is required to break a mole of bonds of a given type in the gas phase Bond energies vary depending on their molecular environment e g the bond energy of C O bonds is 743 kJ mol but to break the C O bonds in CO2 requires 799 kJ mol The trends in bond strength and bond length are illustrated for C C and C N bonds in the table Calculation Estimating Hrxn from bond energies BE Add up the energy required to breaks the bonds that hold together the molecules of the reactants with the energy released when the bonds in the products form from free atoms NOTE The signs of the H values for breaking bonds are but the signs of the H values for forming bonds are i e H Consider the combustion of methane Hrxn 4 H C H 2 H O O 2 H C O 4 H O H 808 kJ