Chapter 9: Chemical Bonding I: Lewis Theory Dr. Chris Kozak Memorial University of ewfoundland, Canada Lewis Theory: An verview Valence e - play a fundamental role in chemical bonding. e - transfer leads to ionic bonds. Sharing of e - leads to covalent bonds. e - are transferred or shared to give each atom a noble gas configuration the octet rule Slide 2 1
Lewis Symbols A chemical symbol represents the nucleus and the core e -. Dots around the symbol represent valence e -. P Si As Sb Bi Al Se I Ar Slide 3 Lewis Structures for Ionic Compounds Ba Ba Ba 2+ 2- Mg 2 Mg Mg 2+ - 2 Determine how many e - each atom must gain or lose. Use multiples of one or both ions to balance the number of electrons. Ionic compounds are rarely molecular, but instead merely formula representations of the charge and mass balanced ions. The ions are separated in Lewis notation using square brackets, eg: [Ba] 2+ [] 2- Slide 4 2
Covalent Bonding In true covalent bonding there is a sharing of electrons. o one atom attracts the electrons from another. Slide 5 Coordinate Covalent Bonds + - A lone pair of electrons on a Lewis Base (an electron pair donor) may donate both electrons to a Lewis acid (an electron pair acceptor). The proton ( + ion) has no valence electrons, so a coordinate bond may form that makes an ammonium cation with an octet around and a charge of +1. All - bonds are equivalent. Chloride anion ( - ) formed by the transfer of a bonding pair of electrons to chlorine resulting in an octet around and a charge of 1. Slide 6 3
Multiple Covalent Bonds C C C C Multiple bonds result from more than one bonding electron pair being formed by electron sharing. Each double bond contains 4 electrons (two per bond). Slide 7 Multiple Covalent Bonds Two nitrogen atoms each have one lone pair of electrons and three single electrons in the valence shell. The three unpaired electrons pair up with one another to generate three - bonds, a triple bond. Each has an octet (three bonding pairs and one non-bonding pair of electrons). Slide 8 4
Polar Covalent Bonds δ+ δ- Polar covalent bonds are covalent bonds which share a bonding pair of electrons unequally. Slide 9 Electronegativity Like Electron Affinity, but for atoms in molecules Electronegativity increases across a period and up a group. Difference in E between two atoms defines its covalent or ionic nature Slide 10 5
Percent Ionic Character Mostly Ionic More Covalent Slide 11 Writing Lewis Structures All the valence e - of atoms must appear. Usually, the e - are paired. Usually, each atom requires an octet. only requires 2 e -. Multiple bonds may be needed. Readily formed by C,,, S, and P. Slide 12 6
Skeletal Structure Identify central and terminal atoms. C C Slide 13 Skeletal Structure ydrogen atoms are always terminal atoms and can only accommodate two electrons. Central atoms are generally those with the lowest electronegativity, but and are common exceptions. Carbon atoms are always central atoms. Generally, structures are compact and symmetrical for inorganics. Most organic compounds, however, are not. Slide 14 7
Strategy for Writing Lewis Structures Slide 15 ormal Charge 1 C = # valence e- - # lone pair e- - # 2 bond pair e- C C: 4 valence e - - 2 lone pair e - - 3 bonds = -1 : 6 valence e - - 2 lone pair e - - 3 bonds = +1 Slide 16 8
Example Writing a Lewis structure for a polyatomic ion. Write the Lewis structure for the nitronium ion, 2+. Step 1: Total valence e - = 5 + 6 + 6 1 = 16 e - Step 2: Plausible structure: Step 3: Add e - to terminal atoms: Step 4: Determine e - left over: 16 4 12 = 0 Slide 17 Example Step 5: Use multiple bonds to satisfy octets. + == Step 6: Determine formal charges: 1 C() = 6-4 (4) = 0 2 1 C() = 5-0 (8) = +1 2 Slide 18 9
Alternative Lewis Structure + + - 1 C( ) = 6-2 (6) = +1 2 1 C() = 5-0 (8) = +1 2 1 C( ) = 6-6 (2) = -1 2 Slide 19 Alternative Lewis Structures Sum of C is the overall charge. C should be as small as possible. egative C usually on most electronegative elements. C of same sign on adjacent atoms is unlikely. + + - Slide 20 10
Example Using the ormal Charge Concept in Writing Lewis Structures. Write the most plausible Lewis structure of nitrosyl chloride,, one of the oxidizing agents present in aqua regia. 2+ 2- - 2+ - - + Slide 21 Resonance zone, 3, has two possibilities, but neither is the real structure Localized Double Bond + - - + both structures exist simultaneously Localized Single Bond -½ + -½ The real structure is the average of the Lewis structures, the Resonance Structure has 6 delocalized electrons Electrons resonate between two bonds: Each bond has a Bond rder of 1.5 Slide 22 11
Resonance Structures 1. Write resonance structures and show the formal charges for the following molecules: 3 - P 4 3- P 3 3-2 - C - S 4 2-3 x = 18 1 x = 5 1 x e- = 1 24 ve - 2. Determine the average bond order for the above molecules. 4 bonds 3 - contacts = 1.33 Slide 23 Bond rder and Bond Length Bond rder Single bond, order = 1 Double bond, order = 2 Can be non-integer values due to resonance! Bond Length Distance between two nuclei igher bond order Shorter bond Stronger bond (E = hν, increased vibrational frequency) Slide 24 12
Exceptions to the ctet Rule dd e - species. = itrosyl radical Methyl radical C ydroxyl radical Radical species, those with electrons left over and are highly reactive! They are paramagnetic. Lewis theory doesn t tell us where to put the extra electron (the radical). Slide 25 Exceptions to the ctet Rule Incomplete octets. B + - B + B - Some elements may be electron deficient. B is a classic case. (electron deficient = less than the expected octet in the valence) B 3 is a very strong Lewis acid (accepts electron pairs where possible) Slide 26 13
Exceptions to the ctet Rule Expanded octets. P 3+ P P 5+ Atoms in Period 3 and below may have more than eight electrons (electron rich) Where do the extra electrons go? ne theory states that the 3d orbitals of period 3 main group elements are low enough in energy to accept the extra electrons. Some people disagree, saying they are really just ionic, [P 4 ] + [] - This is actually what Lewis would have preferred: Maintain the octet and any Slide 27 other bonds are just ionic interactions stabilized by resonance P S 6+ S Expanded Valence Shell Structure with the octet is preferred, but the shorter than expected S- bonds is because we have ionic bonding on top of covalent bonds. Therefore the average bond order is less than 2 but greater than on In reality, all S- bond lengths are equal due to resonance and this achieves highest symmetry. Slide 28 14
The Shapes of Molecules Lewis Structures don t actually tell us anything about the shape of molecules, only where the electrons are, and a simple picture of the bonding: 2 Water is T linear! It is actually BET Why? VSEPR Theory! (Pronounced vesper ) Valence Shell Electron Pair Repulsion Slide 29 Terminology Bond length distance between nuclei Bond angle angle between adjacent bonds VSEPR Theory R.J Gillespie (Canada, McMaster U.) and R. S. yholm (Britain, UCL) Electron pairs repel each other whether they are in chemical bonds (bond pairs) or unshared (lone pairs). Electron pairs assume orientations about an atom to minimize repulsions Electron group geometry distribution of e - pairs (bonding or non-bonding) Molecular geometry distribution of nuclei Slide 30 15
VSEPR Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement, but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. A - central atom X -surrounding atom E -nonbonding lone pairs of electrons AX m E n integers Slide 31 5 basic molecular shapes linear trigonal planar tetrahedral trigonal bipyramidal octahedral Slide 32 16
Methane, Ammonia and Water Tetrahedral Pyramidal Bent A is the central atom, X are the terminal atoms, E are the electron lone pairs These are all based on a triangular based pyramid, a tetrahedron Slide 33 Table 11.1 Molecular Geometry as a unction of Electron Group Geometry Slide 34 17
AX 3 and AX 2 E Geometries Examples: S 3, B 3, 3-, C 3 2- Examples: S 2, 3, Pb 2, SnBr 2 Slide 35 Tetrahedral type arrangements Examples: C 4, Si 4, S 4 2-, 4-3 P 3 3 3 + 2 2 S 2 Slide 36 18
Remember they are 3-dimensional! Slide 37 Shapes of the 5 electron-group arrangement P 5 As 5 S 4 S 4 Xe 2 2 I 4 + I 2 2-3 Br 3 Xe 2 I 3 - I 2 - Slide 38 19
Shapes of the octahedral electron-group arrangement S 6 I 5 Br 5 Te 5 - Xe 4 Xe 4 I 4 - Slide 39 Slide 40 20
actors affecting bond angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. Effect of Double Bonds ideal 120 0 Effect of onbonding(lone) Pairs C 120 0 greater electron density larger E 116 0 122 0 C real Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. Sn 95 0 Slide 41 Applying VSEPR Theory Draw a plausible Lewis structure. Determine the number of e - groups and identify them as bond or lone pairs. Establish the e - group geometry. Determine the molecular geometry. Multiple bonds count as one group of electrons. More than one central atom can be handled individually. Slide 42 21
Predicting Molecular Shapes with Two, Three, or our Electron Groups PRBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) P 3 and (b) C 2. SLUTI: (a) or P 3 - there are 26 valence electrons, 1 nonbonding pair P The shape is based upon the tetrahedral arrangement. P <109.5 0 The -P- bond angles should be <109.5 0 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. The type of shape is AX 3 E Slide 43 Predicting Molecular Shapes with Two, Three, or our Electron Groups (b) or C 2, C has the lowest E and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. C C does not have an octet; a pair of nonbonding electrons will move in from the to make a double bond. C The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The -C- bond angle will be less than 120 0 due to the electron density of the C=. 124.5 0 C 111 0 Type AX 3 Slide 44 22
Predicting Molecular Shapes with ive or Six Electron Groups PRBLEM: SLUTI: Sb Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) Sb 5 and (b) Br 5. (a) Sb 5-40 valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5 - trigonal bipyramidal. Sb (b) Br 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal. Br Slide 45 Predicting Molecular Shapes with More Than ne Central Atom PRBLEM: Determine the shape around each of the central atoms in acetone, (C 3 ) 2 C=. PLA: ind the shape of one atom at a time after writing the Lewis structure. SLUTI: tetrahedral tetrahedral C C C trigonal planar C C C >120 0 <120 0 Slide 46 23
The tetrahedral centers of ethane and ethanol. ethane C 3 C 3 ethanol C 3 C 2 Slide 47 Dipole Moments A pair of electrodes separated by a medium that does not conduct electricity (a non-electrolyte) When the field is off the molecules orient randomly. When the filed is on the molecules align with the field. The alignment can be detected Slide 48 24
Dipole Moments Use the special arrows to show direction of dipole moments Dipole moments result from differences in Electronegativity If a net difference is observed in the directions of the arrows, the molecule has a dipole moment (a polar molecule) Slide 49 Predicting the Polarity of Molecules PRBLEM: rom electronegativity (E) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: (a) Ammonia, 3 (b) Boron trifluoride, B 3 (c) Carbonyl sulfide, CS (atom sequence SC) PLA: Draw the shape, find the E values and combine the concepts to determine the polarity. SLUTI: (a) 3 E = 3.0 E = 2.1 bond dipoles molecular dipole The dipoles reinforce each other, so the overall molecule is definitely polar. Slide 50 25
Bond Length 10.2 Slide 51 Bond Energies Bond lengths and their energies are related. The shorter the bond length, the stronger the bond. Breaking all - bonds requires the same Energy In 2, more E is required to break the first - bond than the second one The second - bond broken is that of the product, an radical The - bond energy in water is the average of the two processes Slide 52 26
Bond Energies 10.3 Slide 53 Bond Energies and Enthalpy of Reaction Example: Use bond energies in exactly the same way you use enthalpies of formation Bond breaking: endothermic Bond forming: exothermic Δ rxn = Σ Δ(product bonds) + ΣΔ(reactant bonds) = Σ Δ bonds formed + Σ Δ bonds broken = -770 kj/mol + 657 kj/mol = -114 kj/mol Slide 54 27