CHEMICAL BONDING Chapter 8

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1 EMIAL BNDING hapter 8 bjectives Differentiate between ionic and covalent bonds Draw Lewis Dot Structures for small molecules and ions Define resonance and use ormal harges to help determine structure Define VSEPR theory and use it to predict shapes of molecules and ions Use bond polarity and electronegativity to predict polarity of molecules 1 2 h hemical Bonding Problems and questions ow is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? an we predict the structure? ow is structure related to chemical and physical properties? 3 Structure & Bonding NN triple bond. Molecule is unreactive Phosphorus is a tetrahedron of P atoms. Very reactive! Red phosphorus, a polymer. Used in matches. 4 orms of hemical Bonds Ionic ompounds There are 2 extreme forms of connecting or bonding atoms: Ionic complete transfer of 1 or more electrons from one atom to another ovalent some valence electrons shared between atoms Most bonds are somewhere in between. Nonmetal of high EA Metal of low IE 5 2 Na(s) + l 2 (g) 2 Na l 6 1

2 oulomb s Law E = 2.31 x J*nm ( ) E = energy gained or released in Joules by ion pairing r = distance between ion centers in nm Q = numerical charge (i.e. +1 or 2) of ion A negative E (Q 1 and Q 2 opposite charges) indicates lower energy than separated ions In other words, the ions are more stable as pair then when separated A positive E (Q 1 and Q 2 like charged) indicates greater energy when ions are close together, i.e. less stable (repulsive force) ovalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. Bond is a balance of attractive and repulsive forces. 7 8 Bond ormation A bond can result from a head to head overlap of atomic orbitals on neighboring atoms. + l l Bond Length Bond length is the distance between the nuclei of two bonded atoms. verlap of (1s) and l (2p) Note that each atom has a single, unpaired electron Bond Length Bond length depends on size of bonded atoms. Bond Length Bond length depends on bond order. l Bond distances measured in Angstrom units where 1 A = 10 2 pm. I 11 Bond distances measured in Angstrom units where 1 A = 100 pm = m 12 2

3 Water Boiling point = 100 Bond Polarity Methane Boiling point = 161 Bond Polarity l is PLAR because it has a positive end and a negative end. Why do water and methane differ so much in their boiling points? + - l l has a greater share in bonding electrons than does. Why do ionic compounds dissolve in water? 13 l has slight negative charge ( ) and has slight positive charge (+ ) 14 Bond Polarity Three molecules with polar, covalent bonds. Each bond has one atom with a slight negative charge ( ) and and another with a slight positive charge (+ ) Bond Polarity This model, calc d using Ahe software for molecular calculations, shows that is + (red) and l is (yellow). alc d charge is + or l BND pure bond real bond Bond Polarity Due to the bond polarity, the l bond energy is GREATER than expected for a pure covalent bond. ENERGY 339 kj/mol calc d 432 kj/mol measured hapter Electronegativity Electronegativity is a measure of the ability of an atom in a molecule to attract electrons to itself. Difference = 92 kj. This difference is proportional to the difference in ELETRNEGATIVITY. oncept proposed by Linus Pauling

4 Electronegativity N l S P Si Electronegativity, has maximum electronegativity Atom with lowest electronegativity is the center atom in most molecules. Relative values of electronegativity determine BND PLARITY (and point of attack on a molecule) h Bond Polarity Which bond is more polar (or DIPLAR)? EN s is more polar than Molecular Polarity Molecules such as I and 2 can be PLAR (or dipolar). They have a DIPLE MMENT. The polar l molecule will turn to align with an electric field. and polarity is reversed Molecular Polarity The magnitude of the dipole is given in Debye units. Named for Peter Debye ( ). Rec d 1936 Nobel prize for work on x ray diffraction and dipole moments. Dipole Moments Property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge. Use an arrow to represent a dipole moment. Point to the negative charge center with the tail of the arrow indicating the positive center of charge

5 Dipole Moments Dipole Moments Why are some molecules polar but others are not? Molecular Polarity Molecules will be polar if a) bonds are polar AND b) the molecule is NT symmetric Polar or Nonpolar? ompare 2 and 2. Which one is polar? All above are NT polar arbon Dioxide 2 is NT polar even though the bonds are polar. 2 is symmetrical. Polar or Nonpolar? onsider AB 3 molecules: B 3, l 2, and N 3. Positive atom is reason 2 and 2 react to give

6 Molecular Polarity, B 3 Molecular Polarity, B 2 B B atom is positive and atoms are negative. B B atom is positive but & atoms are negative. B bonds in B 3 are polar. But molecule is symmetrical and NT polar B and B bonds in B 2 are polar. But molecule is NT symmetrical and is polar Is Methane, 4, Polar? Is 3 Polar? Methane is symmetrical and is NT polar. bond is very polar. Molecule is not symmetrical and so is polar l 4 Polar or Not? Substituted Ethylene nly 4 and l 4 are NT polar. These are the only two molecules that are symmetrical. bonds are MU more polar than bonds. Because both bonds are on same side of molecule, molecule is PLAR

7 Substituted Ethylene Visualizing harges and Polarity Electrostatic Potential Surfaces Electrostatic potential surfaces (EPS) can be used to visualize a) Where the charges lie in molecules b) The polarity of molecules bonds are MU more polar than bonds. Because both bonds are on opposing ends of molecule, molecule is NT PLAR. The molecule See pages Visualizing harges and Polarity Visualizing harges and Polarity The boundary surface around the molecule is made up of all the points in space where the electron density is a given value (here e - /A 3 ). The colors indicate the potential experienced by a + ion on the surface. More attraction (a negative site) is red, and repulsion (a positive site) is blue. 39 As expected, the surface near in 2 and the N is 3 -N 2 is red because that is the more electronegative atom. 2 is polar - with more negative and more positive. 3 -N 2 is also polar. 40 Visualizing harges and Polarity Visualizing harges and Polarity B 3 l 2 The EP surfaces show B 3 is not polar (but the B is slightly positively charged), whereas l 2 is polar with the more negative than l. The EP surfaces show cis- 2 2 l 2 (left) is polar whereas trans- 2 2 l 2 (right) is not polar

8 h. 8.4 Ions: Electron onfiguration and Sizes Atoms in stable compounds usually have a noble gas electron configuration. When two nonmetals react to form a covalent bond, they share electrons in a way that completes the valence electron configurations of both atoms. When a nonmetal and a representative group metal react to form a binary ionic compound, the ions form so that the valence electron configuration of the nonmetal achieves the electron configuration of the next noble gas atom. The valence orbitals of the metal are emptied. Electron onfiguration Metals look electronically like the noble gas of the previous period: Eg. Na: 1s 2 2s 2 2p 6 3p 1 1s 2 2s 2 2p 6 =Na + (+ 1 e ) and has the same # of e as neon. Nonmetals look like the noble gas at the end of their period Eg : 1s 2 2s 2 2p 4 + 2e 1s 2 2s 2 2p 6 =[ 2 ] Positive ions are smaller than the neutral atom they came from, negative ions are bigger Electron onfiguration Electron onfiguration and Size Isoelectric ions/atoms are particles that contain the same number of electrons. The particles can be positive, negative or neutral. A series of ions/atoms containing the same number of electrons. 2,, Ne, Na +, Mg 2+, and Al 3+ (All contain 10 e ) h. 8.5 Energy Effects in Binary Ionic ompuounds Energetics of Ionic Bond ormation Lattice energy: the energy of formation of 1 mole of a crystalline ionic compound from its ions in the gas phase. Example: Na + (g) + l - (g) Nal(s) Lattice energy (E l ) depends on the charges on the ions and the sizes of the ions: Q Q E l 1 2 d k is a constant, Q 1 and Q 2 are the charges on the ions, and d is the distance between ions. 47 Lattice energy Energetics of Ionic Bond ormation Lattice energy becomes more exothermic as Q Q E l 1 2 d The charges on the ions increase The distance between the ions decreases. Note: Q 1 and Q 2 are oppositely charged, so the energy is negative. The more negative, the larger the stability (release of energy) of the solid. 48 8

9 Born aber ycle for Nal alculation of Lattice Energy Na + (g) + l(g) Δ ionization = +496kJ Δ EA = 349kJ Na + (g) + l (g) l(g) + Na(g) Δ sub = +107kJ l(g) + Na(s) Δ f = +121 kj Δ Lattice 1/2l 2 (g) + Na(s) Δ f = 411kJ Nal(s) alculation of Lattice Energy Δ f (Nal) = Δ f (l atoms) + Δ sub (Na) + Δ ioniz (Na) + Δ EA (l) + Δ Lattice (Nal) So Δ Lattice (Nal) = Δ f (Nal) (Δ f (l atoms) + Δ sub (Na) + Δ ioniz (Na) + Δ EA (l)) = 411 ( ) = 411 (375) = 786 kj/mol (Lattice Energy) 51 E l drops as anion gets bigger!! E l large due to +2 and -2 harges; gets smaller as ions get larger. 52 Practice Which compound in each of the following pairs should require a higher temperature to melt? Why? A. Nal or Rbl B. Ba or Mg. Nal or MgS

10 h. 8.6 Partial Ionic haracter of ovalent Bonds No bonds reach 100% ionic character even with compounds that have the maximum possible electronegativity difference. Partial Ionic haracter of ovalent Bonds The relationship between the ionic character of a covalent bond and the electronegativity difference of the bonded atoms measured dipole moment of X Y % ionic character of a bond = 100% + calculated dipole moment of X Y Partial Ionic haracter of ovalent Bonds perational Definition of Ionic ompound Any compound that conducts an electric current when melted will be classified as ionic. h. 8.7 ovalent hemical Bond: A Model Models are attempts to explain how nature operates on the microscopic level based on experiences in the macroscopic world. 57 opyright engage Learning. All rights reserved 58 undamental Properties of Models 1. A model does not equal reality. 2. Models are oversimplifications, and are therefore often wrong. 3. Models become more complicated and are modified as they age. 4. We must understand the underlying assumptions in a model so that we don t misuse it. 5. When a model is wrong, we often learn much more than when it is right. h. 8.8 ovalent Bond Energies and hemical Reactions measured by the energy req d to break a bond. See Table 8.4 opyright engage Learning. All rights reserved

11 Bond Strength measured by the energy req d to break a bond. See Table 8.4. BND Bond dissociation enthalpy (kj/mol) = N N 945 The GREATER the number of bonds (bond order) the IGER the bond strength and the SRTER the bond Bond rder Length Strength = Bond Strength pm 146 kj/mol ? Using Bond Dissociation Enthalpies Estimate the energy of the reaction (g) + l l(g) 2 l(g) Net energy = r = = energy required to break bonds energy evolved when bonds are made = 436 kj/mol l l = 242 kj/mol l = 432 kj/mol Using Bond Dissociation Enthalpies Estimate the energy of the reaction + l l 2 l Sum of + l l bond energies = 436 kj kj = +678 kj 2 mol l bond energies = 864 kj Net = r = +678 kj 864 kj = 186 kj = 436 kj/mol l l = 242 kj/mol l = 432 kj/mol Using Bond Dissociation Enthalpies Estimate the energy of the reaction 2 = + 2 Is the reaction exo or endothermic? Which is larger: A) energy req d to break bonds B) or energy evolved on making bonds?

12 Using Bond Dissociation Enthalpies 2 = + 2 Energy required to break bonds: break 4 mol of bonds = 4 (463 kj) break 2 mol bonds = 2 (146 kj) TTAL ENERGY to break bonds = 2144 kj TTAL ENERGY evolved on making = bonds and 4 bonds bonds = 2350 kj Using Bond Dissociation Enthalpies 2 = + 2 Net energy = kj kj = kj The reaction is exothermic! More energy is evolved on making bonds than is expended in breaking bonds h. 8.9/8.10 Lewis Structures bjectives are to understand: 1. valence e distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties. 69 Electron Distribution in Molecules G. N. Lewis Electron distribution is depicted with Lewis electron dot structures Valence electrons are distributed as shared or BND PAIRS and unshared or LNE PAIRS. 70 Bond and Lone Pairs Valence electrons are distributed as shared or BND PAIRS and unshared or LNE PAIRS. Valence Electrons Electrons are divided between core and valence electrons B 1s 2 2s 2 2p 1 ore = [e], valence = 2s 2 2p 1 l shared or bond pair lone pair (LP) This is called a LEWIS ELETRN DT structure. Br [Ar] 3d 10 4s 2 4p 5 ore = [Ar] 3d 10, valence = 4s 2 4p

13 Rules of the Game No. of valence electrons of a main group atom = Group number or Groups 1A 4A, no. of bond pairs = group number. or Groups 5A 7A, BP s = 8 Grp. No. Rules of the Game No. of valence electrons of an atom = Group number or Groups 1A 4A, no. of bond pairs = group number or Groups 5A 7A, BP s = 8 Grp. No. Except for (and sometimes atoms of 3rd and higher periods), BP s + LP s = 4 This observation is called the TET RULE Building a Dot Structure Ammonia, N 3 1. Decide on the central atom; never. entral atom is atom of lowest affinity for electrons. Therefore, N is central 2. ount valence electrons = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs Building a Dot Structure 3. orm a single bond between the central atom and each surrounding atom 4. Remaining electrons form LNE PAIRS to complete octet as needed. 3 BND PAIRS and 1 LNE PAIR. Note that N has a share in 4 pairs (8 electrons), while shares 1 pair. N N Sulfite ion, S 2-3 Step 1. entral atom = S Step 2. ount valence electrons S = 6 3 x = 3 x 6 = 18 Negative charge = 2 TTAL = 26 e or 13 pairs Step 3. orm bonds 10 pairs of electrons are now left. S Sulfite ion, S 2-3 Remaining pairs become lone pairs, first on outside atoms and then on central atom. S Each atom is surrounded by an octet of electrons

14 arbon Dioxide, 2 1. entral atom = 2. Valence electrons = 16e - or 8 pairs 3. orm bonds. arbon Dioxide, 2 4. Place lone pairs on outer atoms. This leaves 6 pairs. 5. So that has an octet, we shall form DUBLE BNDS between and. 4. Place lone pairs on outer atoms. The second bonding pair forms api (π) bond Double and even triple bonds are commonly observed for, N, P,, and S S 3 Practice Drawing Lewis Structures Draw the Lewis Structures of the following compounds and ions: 4 2 N 4 + S (There are two correct answers!) h Exceptions to the ctet Rule Usually occurs with B and elements of higher periods. Boron Trifluoride entral atom = Valence electrons = or electron pairs = Assemble dot structure The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient. B 3 S 4 B

15 Boron Trifluoride, B 3 Is There a B= Double Bond in B 3 B What if we form a B double bond to satisfy the B atom octet? B + - alc d partial charges in B 3 is negative and B is positive Sulfur Tetrafluoride, S 4 entral atom = Valence electrons = or pairs. orm sigma bonds and distribute electron pairs. h Resonance Sometimes, more than one valid Lewis structure can be drawn for a molecule. N 3 = 24e S 5 pairs around the S atom. A common occurrence outside the 2nd period. N N N Resonance Actual structure is an average of the resonance structures. Electrons are really delocalized they can move around the entire molecule. Resonance Practice Draw the Lewis structures of 3 2 ion. N N N

16 ormal harge Atoms in molecules often bear a charge (+ or ). The predominant resonance structure of a molecule is the one with charges as close to 0 as possible. Defined as charge an atom would have if all the atoms shared the electrons equally. ormal charge = # of valence electrons 1/2 (# of bonding electrons) - (# of LP electrons) ~ r ~ ormal charge = # Val e #bonds # LP e arbon Dioxide, (1/ 2)(4) - 4 = (1/2)(8) - 0 = alculated Partial harges in 2 Thiocyanate Ion, SN (1/2)(2) - 6 = (1/2)(6) - 2 = 0 Yellow = negative & red = positive Relative size = relative charge S N 4 - (1/2)(8) - 0 = Thiocyanate Ion, SN - alculated Partial harges in SN - S N S N S N Which is the most important resonance form? 95 N is most electronegative, so this is most predominant structure S N 96 16

17 ormal harge Practice Draw the following structures and determine the formal charge on each atom. 1. N h MLEULAR STRUTURE: TE VSEPR MDEL 2. N 3 3. N 2 3 (There are no N N bonds and no bonds and the molecule lies in a straight line.) MLEULAR GEMETRY VSEPR Valence Shell Electron Pair Repulsion theory. Most important factor in determining geometry is relative repulsion between electron pairs. Molecule adopts the shape that minimizes the electron pair repulsions. VSEPR Model Definitions Bonding Pair a pair of electrons that occupy space between two atom Non Bonding Pair a pair of electrons that are located principally around 1 atom Electron Domain Regions about a central atom where electrons can be found. They can be nonbonding, single bonds or multiple bonds. PLAY MVIE Arrangements of Electron Pairs Around an Atom Yielding Minimum Repulsion No. of e- Pairs Around entral Atom Example 2 Be Þ B 120Þ 109Þ Geometry linear planar trigonal tetrahedral

18 VSEPR: Two Electron Pairs VSEPR: Three Electron Pairs To play movie you must be in Slide Show Mode To play movie you must be in Slide Show Mode opyright engage Learning. All rights reserved 103 opyright engage Learning. All rights reserved 104 VSEPR: our Electron Pairs Electron Pair Geometries To play movie you must be in Slide Show Mode opyright engage Learning. All rights reserved Structures of Molecules That ave our Electron Pairs Around the entral Atom Geometries for our Electron Pairs

19 Structure Determination by VSEPR Structure Determination by VSEPR Ammonia, N 3 1. Draw electron dot structure 2. ount BP s and LP s = 4 3. The 4 electron pairs are at the corners of a tetrahedron. (4 e domains) N lone pair of electrons in tetrahedral position N 109 Ammonia, N 3 There are 4 electron pairs at the corners of a tetrahedron. N N lone pair of electrons in tetrahedral positio The ELETRN PAIR GEMETRY is tetrahedral. 110 Structure Determination by VSEPR Structure Determination by VSEPR Ammonia, N 3 The electron pair geometry is tetrahedral. N lone pair of electrons in tetrahedral position TheMLEULAR GEMETRY the positions of the atoms is PYRAMIDAL. 111 Water, 2 1. Draw electron dot structure 2. ount BP s and LP s = 4 3. The 4 electron pairs are at the corners of a tetrahedron. The electron pair geometry is TETRAEDRAL. 112 Structure Determination by VSEPR Water, 2 The molecular geometry is BENT. The electron pair geometry is TETRAEDRAL 113 Structure Determination by VSEPR ormaldehyde, 2 1. Draw electron dot structure 2. ount BP s and LP s at 3. There are 3 electron domains around at the corners of a planar triangle. The electron pair geometry is TRIGNAL PLANAR with 120 o bond angles

20 Structure Determination by VSEPR Structure Determination by VSEPR ormaldehyde, 2 The molecular geometry is also trigonal planar. The electron pair geometry is TRIGNAL PLANAR. Methanol, 3 Define and bond angles = 109 o = 109 o In both cases the atom is surrounded by 4 electron pairs Structure Determination by VSEPR Acetonitrile, 3 N Define unique bond angles = 109 o N = 180 o N ne is surrounded by 4 electron domains and the other by 2 domains Phenylalanine, an amino acid N Phenylalanine Structures with entral Atoms with More Than or Less Than 4 Electron Pairs 119 ften occurs with Group 3A elements and with those of 3rd period and higher

21 Boron ompounds onsider boron trifluoride, B 3 The B atom is surrounded by only 3 electron pairs. Bond angles are 120 o Geometry described as trigonal planar B ompounds with 5 or 6 Pairs Around the entral Atom 90Þ Trigonal bipyramid P 120Þ 5 electron pairs Trigonal Bipyramidal Structure Molecular Geometries for ive Electron Pairs All based on trigonal bipyramid Sulfur Tetrafluoride, S 4 Sulfur Tetrafluoride, S 4 Number of valence electrons = 34 entral atom = S Dot structure Electron pair geometry is trigonal bipyramid (because there are 5 pairs around the S) S 90Þ S 120Þ 125 Lone pair is in the equator because it requires more room. 90Þ S 120Þ

22 Molecular Geometries for Six Electron Pairs All are based on the 8-sided octahedron ompounds with 5 or 6 Pairs Around the entral Atom 90Þ ctahedron S 90Þ 6 electron pairs ovalent Bonding: rbitals hapter 9 hapter 9 bjectives 1. Define and apply Valence Bond (VB) theory. 2. Extend information learned in chapter 8 (on Lewis structures) and use this to draw electron domain geometry (in conjunction with the VSEPR model) 3. Take the electron domain geometry and determine molecular shape Atomic rbitals Molecules h. 9.1 ybridization and the Localized Electron Model VALENE BND TERY Linus Pauling valence electrons are localized between atoms (or are lone pairs). half filled atomic orbitals overlap to form bonds. Sigma Bond ormation by rbital verlap Two s orbitals overlap Linus Pauling,

23 Sigma Bond ormation Two s orbitals overlap Two p orbitals overlap Using VB Theory Bonding in B 3 Boron configuration B 1s 2s 2p planar triangle angle = 120 o Bonding in B 3 Bonding in B 3 ow to account for 3 bonds 120 o apart using a spherical s orbital and p orbitals that are 90 o apart? Pauling said to modify VB approach with RBITAL YBRIDIZATIN mix available orbitals to form a new set of orbitals YBRID RBITALS that will give the maximum overlap in the correct geometry for sigma bonds s hydridize orbs. three sp 2 hybrid orbitals 2p rearrange electrons unused p orbital 136 Bonding in B 3 Bonding in B 3 The three hybrid orbitals are made from 1 s orbital and 2 p orbitals 3 sp 2 hybrids. An orbital from each overlaps one of the sp 2 hybrids to form a B bond. Now we have 3, half filled YBRID orbitals that can be used to form B sigma bonds

24 B 3, Planar Trigonal Bonding in 4 ow do we account for 4 sigma bonds 109 o apart? Need to use 4 atomic orbitals s, p x, p y, and p z to form 4 new hybrid orbitals pointing in the correct direction. 109 o Bonding in a Tetrahedron ormation of ybrid Atomic rbitals 1s 2s 2p x 2p y 2p z y 4 atom orbitals hybridize to form four equivalent sp 3 hybrid atomic orbitals. x o z Methane: arbon 141 an also hybridize to include lone pairs 142 Bonding in a Tetrahedron ormation of ybrid Atomic rbitals Bonding in 4, See igure atom orbitals hybridize to form four equivalent sp 3 hybrid atomic orbitals

25 Bonding in Glycine sp 3 sp 3 N sp 2 sp rbital ybridization BNDS SAPE YBRID REMAIN 2 linear sp 2 p s 3 trigonal sp 2 1 p planar 4 tetrahedral sp 3 none h. 9.2 Molecular rbital Model MLEULAR RBITAL TERY Robert Mullikan ( ) valence electrons are delocalized valence electrons are in orbitals (called molecular orbitals) spread over entire molecule. Molecular rbital Model Regards a molecule as a collection of nuclei and electrons, where the electrons are assumed to occupy orbitals much as they do in atoms, but having the orbitals extend over the entire molecule. The electrons are assumed to be delocalized rather than always located between a given pair of atoms. 149 opyright engage Learning. All rights reserved

26 Molecular rbital Model The electron probability of both molecular orbitals is centered along the line passing through the two nuclei. Sigma (σ) molecular orbitals (Ms) In the molecule only the molecular orbitals are available for occupation by electrons. Molecular rbital Model ombination of ydrogen 1s Atomic rbitals to form Ms opyright engage Learning. All rights reserved 151 opyright engage Learning. All rights reserved 152 Molecular rbital Model M Energy Level Diagram for the 2 Molecule M 1 is lower in energy than the s orbitals of free atoms, while M 2 is higher in energy than the s orbitals. Bonding molecular orbital lower in energy Antibonding molecular orbital higher in energy opyright engage Learning. All rights reserved 153 opyright engage Learning. All rights reserved 154 Molecular rbital Model The molecular orbital model produces electron distributions and energies that agree with our basic ideas of bonding. The labels on molecular orbitals indicate their symmetry (shape), the parent atomic orbitals, and whether they are bonding or antibonding. Molecular rbital Model Molecular electron configurations can be written in much the same way as atomic electron configurations. Each molecular orbital can hold 2 electrons with opposite spins. The number of orbitals are conserved. opyright engage Learning. All rights reserved 155 opyright engage Learning. All rights reserved

27 Sigma Bonding and Antibonding rbitals Bond rder Larger bond order means greater bond strength. Bond order = # of bonding e # of antibonding e 2 To play movie you must be in Slide Show Mode opyright engage Learning. All rights reserved 157 opyright engage Learning. All rights reserved 158 Example: 2 Example: Bond order = = Bond order = = 2 2 opyright engage Learning. All rights reserved 160 h. 9.3 Bonding in omonuclear Diatomic Molecules omposed of 2 identical atoms. nly the valence orbitals of the atoms contribute significantly to the molecular orbitals of a particular molecule. Pi Bonding and Antibonding rbitals To play movie you must be in Slide Show Mode opyright engage Learning. All rights reserved 161 opyright engage Learning. All rights reserved

28 Induced magnetism Paramagnetism substance is attracted into the inducing magnetic field. Unpaired electrons ( 2 ) Diamagnetism substance is repelled from the inducing magnetic field. Paired electrons (N 2 ) Apparatus Used to Measure the Paramagnetism of a Sample opyright engage Learning. All rights reserved 163 opyright engage Learning. All rights reserved 164 Molecular rbital Summary of Second Row Diatomic Molecules h. 9.4 eteronuclear molecules omposed of 2 different atoms. Example: The 2p orbital of fluorine is at a lower energy than the 1s orbital of hydrogen because fluorine binds its valence electrons more tightly. Electrons prefer to be closer to the fluorine atom. Thus the 2p electron on a free fluorine atom is at a lower energy than the 1s electron on a free hydrogen atom. opyright engage Learning. All rights reserved rbital Energy Level Diagram for the Molecule eteronuclear Diatomic Molecule: The diagram predicts that the molecule should be stable because both electrons are lowered in energy relative to their energy in the free hydrogen and fluorine atoms, which is the driving force for bond formation. opyright engage Learning. All rights reserved 167 opyright engage Learning. All rights reserved

29 The Electron Probability Distribution in the Bonding Molecular rbital of the Molecule h. 9.5 ombining Localized and M Models (π bonding) onsider ethylene, Þ sp 2 opyright engage Learning. All rights reserved Sigma Bonds in Þ sp 2 π Bonding in 2 4 The unused p orbital on each atom contains an electron and this p orbital overlaps the p orbital on the neighboring atom to form the π bond. 2s 2p 3 sp 2 hybrid orbitals p orb. for š bond π Bonding in 2 4 The unused p orbital on each atom contains an electron and this p orbital overlaps the p orbital on the neighboring atom to form the π bond. Multiple Bonding in

30 and π Bonding in 2 4 and π Bonding in and π Bonding in 2 2 and π Bonding in onsequences of Multiple Bonding There is restricted rotation around = bond. onsequences of Multiple Bonding Restricted rotation around = bond

31 h. 9.6 Photoelectron Spectroscopy (PES) an be used to determine the relative energies of electrons in individual atoms and molecules. igh energy photons are directed at the sample, and the kinetic energies of the ejected electrons are measured. AP Exam Practice 2011 AP Exam #5 (a b) 2010 AP Exam #5 (a c, e) 2008 AP Exam #5 (d f) 2007 AP Exam #6 2006B AP Exam # AP Exam # AP Exam # AP Exam #8 (a,d)