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What Do Molecules Look Like? The Lewis Dot Structure approach provides some insight into molecular structure in terms of bonding, but what about 3D geometry? Recall that we have two types of electron pairs: bonding and lone. Valence-Shell Electron-Pair Repulsion (VSEPR). 3D structure is determined by minimizing repulsion of electron pairs. Electron pairs (both bonding and lone) are distributed around a central atom such that electron-electron repulsions are minimized. Electron pairs (both bonding and lone) are distributed around a central atom such that electron-electron repulsions are minimized. 2 electron pairs 3 electron pairs 4 electron pairs Period 1, 2 5 electron pairs 6 electron pairs Period 3 & beyond Arranging Electron Pairs • Must consider both bonding and lone pairs when minimizing repulsion. • Example: CH4 (bonding pairs only) H H C H H Lewis Structure VSEPR Structure Arranging Electron Pairs (cont.) Example: NH3 (both bonding and lone pairs). H H N H Lewis Structure VSEPR Structure Note: “electron pair geometry” vs. “molecular shape” VSEPR Structure Guidelines The previous examples illustrate the strategy for applying VSEPR to predict molecular structure: 1. Construct the Lewis Dot Structure 2. Arrange bonding/lone electron pairs in space such that repulsions are minimized (electron pair geometry). 3. Name the molecular shape from the position of the atoms. VSEPR Shorthand: 1. Refer to central atom as “A” 2. Attached atoms are referred to as “X” 3. Lone pair are referred to as “E” Examples: CH4: AX4 NH3: AX3E H2O: AX2E2 BF3: AX3 VSEPR: 2 electron pairs Linear (AX2): angle between bonds is 180° Example: BeF2 F Be F Experiments show that molecules with multiple bonds can also be linear. Multiple bonds are treated as a single effective electron group. F F Be F Be F 180° More than one central atom? Determine shape around each. VSEPR: 3 electron pairs Trigonal Planar (AX3): angle between bonds is 120° Multiple bond is treated as a single effective electron group. Example: BF3 F F F B 120° F B F F VSEPR: 4 electron pairs (cont.) Tetrahedral (AX4): angle between bonds is ~109.5° Example: CH4 H 109.5° H C H H tetrahedral e- pair geometry AND tetrahedral molecular shape Bonding vs. Lone pairs Bond angle in a tetrahedral arrangement of electron pairs may vary from 109.5° due to size differences between bonding and lone pair electron densities. bonding pair is constrained by two nuclear potentials; more localized in space. lone pair is constrained by only one nuclear potential; less localized (needs more room). VSEPR: 4 electron pairs Trigonal pyramidal (AX3E): Bond angles are <109.5°, and structure is nonplanar due to repulsion of lone pair. Example: NH3 H H N H 107° tetrahedral e- pair geometry; trigonal pyramidal molecular shape VSEPR: 4 electron pairs (cont.) Classic example of tetrahedral angle shift from 109.5° is water (AX2E2): 104.5o “bent” tetrahedral e- pair geometry; bent molecular shape VSEPR: 4 electron pairs (cont.) Comparison of CH4 (AX4), NH3 (AX3E), and H2O (AX2E2): AX2E AX3E AX2E2 1. Refer to central atom as “A” 2. Attached atoms are referred to as “X” 3. Lone pair are referred to as “E” Molecular vs. Electron-Pair Geometry H O N C F H H H Central Atom Compound Electron-Pair Geometry Molecular Shape Carbon, C CH4 tetrahedral tetrahedral Nitrogen, N NH3 tetrahedral trigonal pyramidal Oxygen, O H2O tetrahedral bent Fluorine, F HF tetrahedral linear What is the electron-pair geometry and the molecular shape for HCFS? S a) trigonal planar, bent H b) trigonal planar, trigonal planar c) tetrahedral, trigonal planar d) tetrahedral, tetrahedral C F VSEPR: Beyond the Octet Systems with expanded valence shells will have five or six electron pairs around a central atom. F Cl Cl Cl F P Cl S F F Cl F 90° 120° F 90° F F F S F F F 90° VSEPR: 5 electron pairs • Consider the structure of SF4 (34 e-, AX4E) • What is the optimum arrangement of electron pairs around S? F F F F F ?? S F S F F F S F F F Compare e– pair angles lone-pair / bond-pair: two at 90o, two at 120o bond-pair / bond-pair: four at 90o, one at 120o three at 90o three at 90o, three at 120o Repulsive forces (strongest to weakest): lone-pair/lone-pair > lone-pair/bond-pair > bond-pair/bond-pair VSEPR: 5 electron pairs The optimum structure maximizes the angular separation of the lone pairs. I3- (AX2E3): 5-electron-pair geometries AX4E AX3E2 AX2E3 our previous example VSEPR: 6 electron pairs Which of these is the more likely structure? See-saw Square Planar 6-electron-pair geometries AX5E AX4E2 our previous example Molecular Dipole Moments We can use VSEPR to determine the polarity of a whole molecule. 1. Draw Lewis structures to determine 3D arrangement of atoms. 2. If one “side” of the molecule has more EN atoms than the other, the molecule has a net dipole. Shortcut: completely symmetric molecules will not have a dipole regardless of the polarity of the bonds. Molecular Dipoles The C=O bonds have dipoles of equal magnitude but opposite direction, so there is no net dipole moment. The O-H bonds have dipoles of equal magnitude that do not cancel each other, so water has a net dipole moment. Molecular Dipoles (cont.) symmetric asymmetric symmetric Molecular Dipole Example • Write the Lewis dot and VESPR structures for CF2Cl2. Does it have a dipole moment? F F 32 e- Cl C Cl F Cl F Cl Tetrahedral Advanced VSEPR Application Molecules with more than one central atom… methanol (CH3OH) H H C O H tetrahedral e- pairs tetrahedral shape tetrahedral e- pairs bent shape H The VSEPR Table # e- pairs e- Geom. Molec. Geom. 2 AX2 BeF2 linear linear 3 AX3 BF3 trigonal planar trigonal planar AX2E O3 trigonal planar bent AX4 CH4 tetrahedral tetrahedral AX3E NH3 tetrahedral pyramidal AX2E2 H2O tetrahedral bent 4 The VSEPR Table # e- pairs 5 6 e- Geom. Molec. Geom. AX5 PF5 trigonal bipyramidal trigonal bipyramidal AX4E SF4 trigonal bipyramidal see saw AX3E2 ClF3 trigonal bipyramidal T-shaped AX2E3 I3- trigonal bipyramidal linear AX6 SF6 octahedral octahedral AX4E2 XeF4 octahedral square planar What is the expected shape of ICl2+? + 20 e- Cl I Cl AX2E2 A. linear C. tetrahedral B. bent D. square planar Valence Bond Theory Basic Principle of Localized Electron Model: A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei. Rule 1: Maximum overlap. The bond strength depends on the attraction of nuclei to the shared electrons, so: The greater the orbital overlap, the stronger the bond. Valence Bond Theory Basic Principle of Localized Electron Model: A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei. Rule 2: Spins pair. The two electrons in the overlap region occupy the same space and therefore must have opposite spins. There may be no more than 2 electrons in a molecular orbital. Valence Bond Theory Basic Principle of Localized Electron Model: A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei. Rule 3: Hybridization. To explain experimental observations, Pauling proposed that the valence atomic orbitals in a molecule are different from those in the isolated atoms. We call this concept Hybridization What is hybridization? • Atoms adjust to meet the “needs” of the molecule. • In a molecule, electrons rearrange in an attempt to give each atom a noble gas configuration and to minimize electron repulsion. • Atoms in a molecule adjust their orbitals through hybridization in order for the molecule to have a structure with minimum energy. • The source of the valence electrons is not as important as where they are needed in the molecule to achieve a maximum stability. Example: Methane • 4 equivalent C-H covalent bonds • VSEPR predicts a tetrahedral geometry The Valence Orbitals of a Carbon Atom Carbon: 2s22p2 How do we explain formation of 4 equivalent C-H bonds? Hybridization: Mixing of Atomic Orbitals to form New Orbitals for Bonding + – + – + – + + – + + – – Other Representations of Hybridization: y1 = 1/2[(2s) + (2px) + (2py) + (2pz)] y2 = 1/2[(2s) + (2px) - (2py) - (2pz)] y3 = 1/2[(2s) - (2px) + (2py) - (2pz)] y4 = 1/2[(2s) - (2px) - (2py) + (2pz)] Hybridization is related to the number of valence electron pairs determined from VSEPR: Methane (CH4) VSEPR: AB4 tetrahedral sp3 hybridized Electron pair geometry determines hybridization, not vice versa!! 109.47 º Hybridization is related to the number of valence electron pairs determined from VSEPR: Ammonia (NH3) VSEPR: AB3E tetrahedral sp3 hybridized H N 108.1 º H H Hybridization is related to the number of valence electron pairs determined from VSEPR: Water (H2O) VSEPR: AB2E2 tetrahedral sp3 hybridized 105.6 º s bonding and p bonding • Two modes of bonding are important for 1st and 2nd row elements: s bonding and p bonding • These two differ in their relationship to the internuclear axis: s bonds have electron density ALONG the axis p bonds have electron density ABOVE AND BELOW the axis Problem: Describe the hybridization and bonding of the carbon orbitals in ethylene (C2H4) VSEPR: AB3 trigonal planar sp2 hybridized orbitals for s bonding sp2 hybridized orbitals used for s bonding remaining p orbital used for p bonding Bonding in ethylene (C2H4) Problem: Describe the hybridization and bonding of the carbon orbitals in Carbon Dioxide (CO2) VSEPR: AB2 linear sp hybridized orbitals for s bonding Bonding in Carbon Dioxide (CO2) Atoms of the same kind can have different hybridizations CH3 C N: Acetonitrile (important solvent and industrial chemical) H H C2 C1 N Bonds H s C2: AB4 C1: AB2 2s2 2px2py sp3 N: sp sp p p s p sp p p p ABE 2s2 2px2py2pz sp lone pair sp What have we learned so far? • Molecular orbitals are combinations of atomic orbitals • Atomic orbitals are “hybridized” to satisfy bonding in molecules • Hybridization follows simple rules that can be deduced from the number of chemical bonds in the molecule and the VSEPR model for electron pair geometry Hybridization • sp3 Hybridization (CH4) – This is the sum of one s and three p orbitals on the carbon atom – We use just the valence orbitals to make bonds – sp3 hybridization gives rise to the tetrahedral nature of the carbon atom Hybridization • sp2 Hybridization (H2C=CH2) – This is the sum of one s and two p orbitals on the carbon atom – Leaves one p orbital uninvolved – this is free to form a p bond (the second bond in a double bond) Hybridization • sp Hybridization (O=C=O) – This is the sum of one s and one p orbital on the carbon atom – Leaves two p orbitals free to bond with other atoms (such a O in CO2), or with each other as in HC≡CH General Notes • This is a model and only goes so far, but it is especially helpful in understanding geometry and expanding Lewis dot structures. • Orbitals are waves. Hybridized orbitals are just the sums of waves – constructive and destructive interference. What is important to know about hybridization? • You should be able to give the hybridization of an atom in a molecule based on the formula given. • Example: CH3-CH2-CHO • Step 1: Draw the Lewis Dot Structure What is important to know about hybridization? • Step 2: What is the electron pair geometry and molecular shape? AXE2 Trigonal Planar AX3 AX4 Tetrahedral AX4 Tetrahedral Trigonal Planar What is important to know about hybridization? • Step 3: Use the molecular shape to determine the hybridization. AXE2 Trigonal Planar sp2 AX3 AX4 Tetrahedral sp3 AX4 Tetrahedral sp3 Trigonal Planar sp2 The Localized Electron Model is very powerful for explaining geometries and basic features of bonding in molecules, but it is just a model. Major limitations of the LE model: • • • Assumes electrons are highly localized between the nuclei (sometimes requires resonance structures) Doesn’t easily deal with unpaired electrons (incorrectly predicts physical properties in some cases) Doesn’t provide direct information about bond energies Example: O2 .. .. Lewis dot structure O=O .. .. All electrons are paired Contradicts experiment! The Molecular Orbital Model Basic premise: When atomic orbitals interact to form a bond, the result is the formation of new molecular orbitals HY = EY Important features of molecular orbitals: 1. Atomic Orbitals are solutions of the Schrödinger equation for atoms. Molecular orbitals are the solutions of the same Schrödinger equation applied to the molecule. Molecular Orbital Theory 2. Atomic Orbitals can hold 2 electrons with opposite spins. Molecular Orbitals can hold 2 electrons with opposite spins. 3. The electron probability for the Atomic Orbital is given by Y2. The electron probability for the Molecular Orbital is given by Y2. 4. Orbitals are conserved - in bringing together 2 atomic orbitals, we have to end up with 2 molecular orbitals! How does this work? Molecular Orbitals are simply Linear Combinations of Atomic Orbitals Example: H2 s anti-bonding (s*) - + Molecular Orbitals have phases (+ or -) + s bonding Next Question: Why does this work? Constructive and Destructive Interference Constructive interference between two overlapping orbitals leads to a bonding orbital. Destructive interference between two orbitals of opposite sign leads to an anti-bonding orbital. Bonding is driven by stabilization of electrons • Electrons are negatively charged • Nuclei are positively charged = + = nucleus The bonding combination puts electron density between the two nuclei stabilization The anti-bonding combination moves electron density away from region between the nuclei - destabilization MO Diagrams • We can depict the relative energies of molecular orbitals with a molecular orbital diagram: The new molecular orbital is lower in energy than the atomic orbitals s* M.O. is raised in energy s M.O. is lowered in energy H atom: (1s)1 electron configuration H2 molecule: (s1s)2 electron configuration Same as previous description of bonding s* s Review of Orbital Filling • Pauli Exclusion Principle: – No more than 2 e- in an orbital, spins must be paired (↑↓) • Aufbau Principle (a.k.a. “Building-Up”): – Fill the lowest energy levels with electrons first • 1s 2s 2p 3s 3p 4s 3d 4p … • Hund’s Rule: – When more than one orbital has the same energy, electrons occupy separate orbitals with parallel spins: Yes No No Filling Molecular Orbitals with Electrons 1) Orbitals are filled in order of increasing Energy (Aufbau principle) H2 Filling Molecular Orbitals with Electrons He2 2) An orbital has a maximum capacity of two electrons with opposite spins (Pauli exclusion principle) Filling Molecular Orbitals with Electrons 3) Orbitals of equal energy (degenerate orbitals) are half filled, with spins parallel, before any is filled completely (Hund’s rule) Bond Order Bond Order = # bonding electrons #anti-bonding electrons 2 The bond order is an indication of bond strength: Greater bond order Greater bond strength (Shorter bond length) Bond Order: Examples Bond order = (2-0)/2 = 1 Single bond H2 Stable molecule (436 kJ/mol bond) Bond order = (2-2)/2 = 0 He2 No bond! Unstable molecule (0 kJ/mol bond) He2+ Bond order = (2-1)/2 = 1/2 Half of a single bond Can be made, but its not very stable (250 kJ/mol bond) Fractional bond orders are okay! H2+ Bond order = (1-0)/2 = 1/2 Half of a single bond Can be made, but its not very stable (255 kJ/mol bond) Forming Bonds • A s bond can be formed a number of ways: – s, s overlap – s, p overlap – p, p overlap Only orbitals of the same phase (+, +) can form bonds Anti-bonding Orbitals • For every bonding orbital we form, we also form an antibonding orbital: MO Theory in Bonding • Homonuclear atoms (H2, O2, F2, N2) H2 (Only 1s orbitals available for bonding) Covalent Bonding in Homonuclear Diatomics • Atomic orbitals must overlap in space in order to participate in molecular orbitals • Covalent bonding is dominated by the valence orbitals (only valence orbitals are shown in the MO diagrams) Covalent Bonding in Homonuclear Diatomics Region of shared edensity – + + Valence configurations of the 2nd row atoms: Li 2s1 Be 2s2 B C 2s22p1 2s22p2 N 2s22p3 O 2s22p4 F 2s22p5 So far we have focused on bonding involving the s orbitals. What happens when we have to consider the p orbitals? For diatomic molecules containing atoms with valence electrons in the p orbitals, we must consider three possible bonding interactions: = nucleus p-type p-type s-type (–) destructive mixing (+) constructive mixing Major limitations of the LE model: 2) Doesn’t easily deal with unpaired electrons (incorrectly predicts physical properties in some cases) Example: O2 .. .. - Lewis dot structure .. O=O .. - All electrons are paired Contradicts experiment! Experiments show O2 is paramagnetic A quick note on magnetism… Paramagnetic The molecule contains unpaired electrons and is attracted to (has a positive susceptibility to) an applied magnetic field Diamagnetic The molecule contains only paired electrons and is not attracted to (has a negative susceptibility to) an applied magnetic field Example: the O2 Diatomic Oxygen atom has a 2s22p4 valence configuration O atom O atom ____ s2p* ___ ___ p2p* Bond Order = (8-4)/2 = 2 O2 is stable (498 kJ/mol bond strength) M.O. O2 ___ ___ ___2p ___ ___ ___ 2p ___ ___ p2p ____ s2p Energy ____ s2s* ___ 2s ___ 2s ____ s2s (s2s)2(s2s*)2(s2p)2(p2p)4(p2p*)2 Both have degenerate orbitals A prediction from the M.O. diagram of O2 .. .. O=O .. .. The Lewis dot structure predicts O2 should be diamagnetic-all electrons are paired. The unpaired electrons predicted by the M.O. diagram should behave as small magnetsO2 should be magnetic! N2 Video O2 Video What have we learned so far? 1. Molecular orbitals (MO) are linear combinations of atomic orbitals 2. Both s and p atomic orbitals can be mixed to form MOs 3. Molecular orbitals are bonding and anti-bonding 4. Bonding and anti-bonding MOs lead to the definition of the bond order 5. Bond order is related to the bond strength (bond dissociation energy) MO Diagram for H2 vs. N2 N2 sp* H2 p2p* s2p p2p Atomic orbital overlap sometimes forms both s and p bonds. Examples: N2, O2, F2 s2s* s2s M.O. Diagram for N2 s*(2p) p* Electron energy (kJ mol-1) p* -1,155 s(2p) -1,240 -1,240 p p -1,479 s*(2s) Valence Valence -2,965 Core s(2s) Core -37,875 -37,871 1s(N) + 1s(N) 1s(N) –1s(N) A Complication… M.O. Diagram for B2 (similar for C2 and N2) M.O. Diagram for O2 (similar for F2 and Xe2) O O2 O A Complication… M.O. Diagram for B2 (similar for C2 and N2) M.O. Diagram for O2 (similar for F2 and Ne2) No s-p mixing s-p mixing Why does s-p mixing occur? Electron repulsion!! s2s and s2p both have significant e- probability between the nuclei, so e- in s2s will repel e- in s2p Effect will decrease as you move across the Periodic Table increased nuclear charge pulls the s2s e- closer, making the s2s orbital smaller and decreasing the s2s and s2p interaction Molecular Orbitals of X2 Molecules sp orbital mixing (a little hybridization) • lowers the energy of the s2s orbitals and • raises the energy of the s2p orbitals. • As a result, E(s2p) > E(p 2p) for B2, C2, and N2. • As one moves right in Row 2, 2s and 2p get further apart in energy, decreasing s–p mixing E(s2p) < E(p2p) for O2, F2, and Ne2. See text pages 680-681. • Note that s–p mixing does not affect bond order or magnetism in the common diatomics (N2, O2, and F2). Hence it is not of much practical importance. s-p mixing No s-p mixing When does s-p mixing occur? B, C, and N all have 1/2 filled 2p orbitals O, F, and Xe all have > 1/2 filled 2p orbitals • If 2 electrons are forced to be in the same orbital, their energies go up. • Electrons repel each other because they are negatively charged. • Having > 1/2 filled 2p orbitals raises the energies of these orbitals due to e- - e- repulsion s-p mixing only occurs when the s and p atomic orbitals are close in energy ( 1/2 filled 2p orbitals) Relating the M.O. Diagrams to Physical Properties Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: Bond energy (kJ/mol) Bond length (pm) No. of valence electrons N2 945 110 10 N2+ 841 112 9 O2 498 121 12 O2 + 623 112 11 Removing an electron from N2 decreases the bond energy of the resulting ion, whereas removing an electron from O2 increases the bond energy of the resulting ion. Explain these facts using M.O. diagrams. Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: Bond energy (kJ/mol) Bond length (pm) No. of valence electrons N2 945 110 10 N2+ 841 112 9 O2 498 121 12 O2 + 623 112 11 Plan: We first draw the MO energy levels for the four species, recalling that they differ for N2 and O2. Then we determine the bond orders and compare them with the data: bond order is related directly to bond energy and inversely to bond length. Sample Problem - Continued Solution: The MO energy levels are: N2 N2+ O2 O2 + sp* sp* p2p* p2p* s2p p2p p2p s2p s2s* s2s* s2s s2s Bond Orders: (8-2)/2 = 3 (7-2)/2 = 2.5 (8-4)/2 = 2 (8-3)/2 = 2.5 Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: Bond energy (kJ/mol) Bond length (pm) No. of valence electrons Bond Order N2 945 110 10 N2+ 841 112 9 3 O2 498 121 12 2.5 2 O2 + 623 112 11 2.5 What have we learned so far? 1. Molecular orbitals (MO) explain the properties of valence electrons in molecules (Example: O2) 2. s and p atomic orbitals can be mixed to form s, s*, p, and p* molecular orbitals 3. Electrons in p or p* molecular orbitals can have the same energies: Degenerate orbitals 4. The ordering of s2p and p2p molecular orbitals depends on the electron occupancy: s-p mixing Bonding in Diatomic Molecules Covalent Ionic Ionic Covalent Homonuclear: H2 Heteronuclear: HF Electronegativity Nonpolar covalent bond (450 kJ/mol bond) Polar covalent bond (565 kJ/mol bond) Electrons are not equally shared in heteronuclear bonds HF Electronegativity Because F (EN = 4.0) is more electronegative than H (EN = 2.2), the electrons move closer to F. This gives rise to a polar bond: H Figure 14.45 F M.O.s of a Polar Covalent Bond: HF s Antibonding (s*) Mostly H(1s) This approach simplifies model and only considers electrons involved in bond. H F H F s Bonding Mostly F(2p) MOs OF XY MOLECULES Equal or unequal e sharing between 2 atoms is reflected in the composition of the MOs: When 2 atoms X and Y have the same electronegativity (purely covalent bond), their overlapping AOs have the same energy, and the bonding and antibonding MOs are each half X and half Y AO. All electrons spend equal time near X and Y. Examples: N2, O2, F2. If EN(Y) > EN(X) (polar covalent X+Y), the Y AO has lower energy than the X AO. The bonding MO is more like the Y AO and the anti-bonding MO more like the X AO. Bonding e spend more time near Y than X; vice versa for anti-bonding e. Example: CO. MOs OF XY____ MOLECULES s* ___ ___ p* ↑ Energy ___ ___ ___ 2p ____ s ___ ___ p ___ ___ ___ 2p ____ s* ___ 2s ___ 2s ____ s C Atom (4e–) Electronegativity ___ ___ ___2p ___ 2s Cδ+Oδ– (10e–) O Atom (6e–) CO Bond Order = 3.0 (same as N2). CO Bond Energy = 1,076 kJ/mol (N2 = 945 kJ/mol). Isoelectronic to CO and N2: CN–, NO+. NO has 1e– in p* bond order = 2.5; this e– is more on N than O; NO NO+ easy… Bonding in NO • Two possible Lewis dot structures for NO • The simplest structure minimizes formal charges and places the lone (unpaired) electron on the nitrogen. • The Lewis structure predicts a bond order of 2, but experimental evidence suggests a bond order between 2 and 3. • How does MO theory help us understand bonding in NO? . .. N=O .. .. .. . N=O .. .. -1 +1 When the electronegativities of the 2 atoms are more similar, the bonding becomes less polar. 2p 2p 2s NO Electronegativity 2s N . .. N=O .. .. EN(N) = 3.0 EN(O) = 3.4 O Bond order = 2.5, unpaired electron is in a N-like orbital NO is easily oxidized to form NO+. Why? What changes can we predict in the bonding and magnetism of the molecule? NO NO+ oxidation Bond Order = (8-3)/2 = 2.5 Paramagnetic Bond Order = (8-2)/2 = 3 Diamagnetic M.O. diagram for NO -597 p2p * p2p * (empty) s2p -1307 -1444 p2p p2p s2s* s2s -1835 -3320 -1374 Key Points of MO Theory – Heteronuclear Molecules • The more electronegative atom has orbitals lower in energy than the more positive atom. • Electrons in bonding orbitals are closer to the more electronegative atom, anti-bonding electrons are closer to the more positive atom. • For most diatomic molecules, s-p mixing changes the orbital energy levels, but since these orbitals are almost always fully occupied, their order is less important to us. Combining the Localized Electron and Molecular Orbital Models (into a convenient working model) Figure 14.47 Only the p bonding changes between these resonance structures - The M.O. model describes this p bonding more effectively. Atomic Orbitals Molecular Orbitals Figure 14.51 Another example: Benzene s bonding: p bonding: p atomic orbitals p molecular orbital MO Theory Expectations • You should be able to: – predict which atomic orbitals are higher or lower in energy (based on electronegativity differences). – correctly fill a molecular orbital diagram. – correctly calculate bond order. – predict molecular magnetic properties based on orbital occupation. – understand how molecular properties change upon ionization (oxidation or reduction) of molecules.