What Do Molecules Look Like?

Transcription

1 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.

2 Electron pairs (both bonding and lone) are distributed around a central atom such that electron-electron repulsions are minimized.

3 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

4 Arranging Electron Pairs Must consider both bonding and lone pairs when minimizing repulsion. Example: CH 4 (bonding pairs only) H H C H H Lewis Structure VSEPR Structure

5 Arranging Electron Pairs (cont.) Example: NH 3 (both bonding and lone pairs). H H N H Lewis Structure VSEPR Structure Note: electron pair geometry vs. molecular shape

6 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: CH 4 : AX 4 NH 3 : AX 3 E H 2 O: AX 2 E 2 BF 3 : AX 3

7 VSEPR: 2 electron pairs Linear (AX 2 ): angle between bonds is 180 Experiments show that molecules with multiple bonds can also be linear. Example: BeF 2 F Be F Multiple bonds are treated as a single effective electron group. F Be F F Be F 180 More than one central atom? Determine shape around each.

8 VSEPR: 3 electron pairs Trigonal Planar (AX 3 ): angle between bonds is 120 Example: BF 3 Multiple bond is treated as a single effective electron group. F F B F F B 120 F F

9 VSEPR: 4 electron pairs (cont.) Tetrahedral (AX 4 ): angle between bonds is ~109.5 Example: CH 4 H H C H H tetrahedral e- pair geometry AND tetrahedral molecular shape

10 Bonding vs. Lone pairs Bond angle in a tetrahedral arrangement of electron pairs may vary from 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).

11 VSEPR: 4 electron pairs Trigonal pyramidal (AX 3 E): Bond angles are <109.5, and structure is nonplanar due to repulsion of lone pair. Example: NH 3 H H N H 107 tetrahedral e- pair geometry; trigonal pyramidal molecular shape

12 VSEPR: 4 electron pairs (cont.) Classic example of tetrahedral angle shift from is water (AX 2 E 2 ): o bent tetrahedral e- pair geometry; bent molecular shape

13 VSEPR: 4 electron pairs (cont.) Comparison of CH 4 (AX 4 ), NH 3 (AX 3 E), and H 2 O (AX 2 E 2 ):

14 AX 2 E AX 3 E AX 2 E 2 1. Refer to central atom as A 2. Attached atoms are referred to as X 3. Lone pair are referred to as E

15 Molecular vs. Electron-Pair Geometry H H H O C N F H Central Atom Compound Electron-Pair Geometry Molecular Shape Carbon, C CH 4 tetrahedral tetrahedral Nitrogen, N NH 3 tetrahedral trigonal pyramidal Oxygen, O H 2 O tetrahedral bent Fluorine, F HF tetrahedral linear

16 What is the electron-pair geometry and the molecular shape for HCFS? a) trigonal planar, bent H C S b) trigonal planar, trigonal planar F c) tetrahedral, trigonal planar d) tetrahedral, tetrahedral

17 VSEPR: Beyond the Octet Systems with expanded valence shells will have five or six electron pairs around a central atom. Cl Cl P Cl Cl Cl 90 F F F F S F F F F F S F F 90 F

18 VSEPR: 5 electron pairs Consider the structure of SF 4 (34 e -, AX 4 E) What is the optimum arrangement of electron pairs around S? F F S F F?? Compare e pair angles lone-pair / bond-pair: bond-pair / bond-pair: F F S F F two at 90 o, two at 120 o four at 90 o, one at 120 o F F S F F three at 90 o three at 90 o, three at 120 o Repulsive forces (strongest to weakest): lone-pair/lone-pair > lone-pair/bond-pair > bond-pair/bond-pair

19 VSEPR: 5 electron pairs The optimum structure maximizes the angular separation of the lone pairs. I 3 - (AX 2 E 3 ):

20 5-electron-pair geometries AX 4 E our previous example AX 3 E 2 AX 2 E 3

21 VSEPR: 6 electron pairs Which of these is the more likely structure? See-saw Square Planar

22 6-electron-pair geometries AX 5 E AX 4 E 2 our previous example

23 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.

24 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.

25 Molecular Dipoles (cont.) symmetric asymmetric symmetric

26 Molecular Dipole Example Write the Lewis dot and VESPR structures for CF 2 Cl 2. Does it have a dipole moment? F F 32 e - Cl C Cl F Cl Cl F Tetrahedral

27 Advanced VSEPR Application Molecules with more than one central atom methanol (CH 3 OH) H H C O H H tetrahedral e- pairs tetrahedral shape tetrahedral e- pairs bent shape

28 The VSEPR Table # e - pairs e - Geom. Molec. Geom. 2 AX 2 BeF 2 linear linear 3 AX 3 BF 3 trigonal planar trigonal planar AX 2 E O 3 trigonal planar bent 4 AX 4 CH 4 tetrahedral tetrahedral AX 3 E NH 3 tetrahedral pyramidal AX 2 E 2 H 2 O tetrahedral bent

29 The VSEPR Table # e - pairs e - Geom. Molec. Geom. 5 AX 5 PF 5 trigonal bipyramidal AX 4 E SF 4 trigonal bipyramidal AX 3 E 2 ClF 3 trigonal bipyramidal AX 2 E 3 I - 3 trigonal bipyramidal trigonal bipyramidal see saw T-shaped linear 6 AX 6 SF 6 octahedral octahedral AX 4 E 2 XeF 4 octahedral square planar

30 What is the expected shape of ICl 2+? + 20 e- Cl I Cl AX 2 E 2 A. linear C. tetrahedral B. bent D. square planar

31 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.

32 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.

33 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

34 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.

35 Example: Methane 4 equivalent C-H covalent bonds VSEPR predicts a tetrahedral geometry

36 The Valence Orbitals of a Carbon Atom Carbon: 2s 2 2p 2 How do we explain formation of 4 equivalent C-H bonds?

37 Hybridization: Mixing of Atomic Orbitals to form New Orbitals for Bonding

38 Other Representations of Hybridization: y 1 = 1/2[(2s) + (2p x ) + (2p y ) + (2p z )] y 2 = 1/2[(2s) + (2p x ) - (2p y ) - (2p z )] y 3 = 1/2[(2s) - (2p x ) + (2p y ) - (2p z )] y 4 = 1/2[(2s) - (2p x ) - (2p y ) + (2p z )]

39 Hybridization is related to the number of valence electron pairs determined from VSEPR: Methane (CH 4 ) VSEPR: AB 4 tetrahedral sp 3 hybridized Electron pair geometry determines hybridization, not vice versa!! º

40 Hybridization is related to the number of valence electron pairs determined from VSEPR: Ammonia (NH 3 ) VSEPR: AB 3 E tetrahedral sp 3 hybridized H º N H H

41 Hybridization is related to the number of valence electron pairs determined from VSEPR: Water (H 2 O) VSEPR: AB 2 E 2 tetrahedral sp 3 hybridized º

42 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

43 Problem: Describe the hybridization and bonding of the carbon orbitals in ethylene (C 2 H 4 ) VSEPR: AB 3 trigonal planar sp 2 hybridized orbitals for s bonding sp 2 hybridized orbitals used for s bonding remaining p orbital used for p bonding

44 Bonding in ethylene (C 2 H 4 )

45 Problem: Describe the hybridization and bonding of the carbon orbitals in Carbon Dioxide (CO 2 ) VSEPR: AB 2 linear sp hybridized orbitals for s bonding

46 Bonding in Carbon Dioxide (CO 2 )

47 Atoms of the same kind can have different hybridizations CH 3 C N : Acetonitrile (important solvent and industrial chemical) H H C 2 C 1 N H s Bonds C 2 : AB 4 C 1 : AB 2 2s 2 2p x 2p y sp sp p p s p p sp 3 N: ABE 2s 2 2p x 2p y 2p sp p p z sp lone pair sp

48 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

49 sp 3 Hybridization (CH 4 ) Hybridization This is the sum of one s and three p orbitals on the carbon atom We use just the valence orbitals to make bonds sp 3 hybridization gives rise to the tetrahedral nature of the carbon atom

50 Hybridization sp 2 Hybridization (H 2 C=CH 2 ) 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)

51 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 CO 2 ), or with each other as in HC CH

52 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.

53 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: CH 3 -CH 2 -CHO Step 1: Draw the Lewis Dot Structure

54 What is important to know about hybridization? Step 2: What is the electron pair geometry and molecular shape? AXE 2 Trigonal Planar AX 4 Tetrahedral AX 4 Tetrahedral AX 3 Trigonal Planar

55 What is important to know about hybridization? Step 3: Use the molecular shape to determine the hybridization. AXE 2 Trigonal Planar sp 2 AX 4 Tetrahedral sp 3 AX 4 Tetrahedral sp 3 sp 2 AX 3 Trigonal Planar

56 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: O Lewis dot structure O=O.... All electrons are paired Contradicts experiment!

57 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.

58 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 Y 2. The electron probability for the Molecular Orbital is given by Y Orbitals are conserved - in bringing together 2 atomic orbitals, we have to end up with 2 molecular orbitals! How does this work?

59 Molecular Orbitals are simply Linear Combinations of Atomic Orbitals Example: H 2 s anti-bonding (s*) + - Molecular Orbitals have phases (+ or -) + s bonding Next Question: Why does this work?

60 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.

61 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

62 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

63 s* M.O. is raised in energy s M.O. is lowered in energy H atom: (1s) 1 electron configuration H 2 molecule: (s 1s ) 2 electron configuration

64 Same as previous description of bonding s* s

65 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

66 Filling Molecular Orbitals with Electrons 1) Orbitals are filled in order of increasing Energy (Aufbau principle) H 2

67 Filling Molecular Orbitals with Electrons He 2 2) An orbital has a maximum capacity of two electrons with opposite spins (Pauli exclusion principle)

68 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)

69 Bond Order Bond Order = # bonding #anti-bonding electrons electrons 2 The bond order is an indication of bond strength: Greater bond order Greater bond strength (Shorter bond length)

70 Bond Order: Examples Bond order = (2-0)/2 = 1 Single bond H 2 Stable molecule (436 kj/mol bond) Bond order = (2-2)/2 = 0 He 2 No bond! Unstable molecule (0 kj/mol bond)

71 He 2 + 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! H 2 + Bond order = (1-0)/2 = 1/2 Half of a single bond Can be made, but its not very stable (255 kj/mol bond)

72 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

73 Anti-bonding Orbitals For every bonding orbital we form, we also form an antibonding orbital:

74 MO Theory in Bonding Homonuclear atoms (H 2, O 2, F 2, N 2 ) H 2 (Only 1s orbitals available for bonding)

75 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)

76 Covalent Bonding in Homonuclear Diatomics Region of shared e - density + +

77 Valence configurations of the 2nd row atoms: Li Be B C N O F 2s 1 2s 2 2s 2 2p 1 2s 2 2p 2 2s 2 2p 3 2s 2 2p 4 2s 2 2p 5 So far we have focused on bonding involving the s orbitals. What happens when we have to consider the p orbitals?

78 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

79 ( ) destructive mixing (+) constructive mixing

80 Major limitations of the LE model: 2) Doesn t easily deal with unpaired electrons (incorrectly predicts physical properties in some cases) Example: O Lewis dot structure O=O All electrons are paired Contradicts experiment! Experiments show O 2 is paramagnetic

81 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

82 Example: the O 2 Diatomic Oxygen atom has a 2s 2 2p 4 valence configuration Bond Order = (8-4)/2 = 2 O 2 is stable O atom 2p M.O. O 2 s 2p * p 2p * O atom 2p (498 kj/mol bond strength) Energy p 2p s 2p 2s (s 2s ) 2 (s 2s *) 2 (s 2p ) 2 (p 2p ) 4 (p 2p* ) 2 s 2s * s 2s 2s Both have degenerate orbitals

83 A prediction from the M.O. diagram of O O=O.. The Lewis dot structure predicts O 2 should be diamagnetic-all electrons are paired. The unpaired electrons predicted by the M.O. diagram should behave as small magnets- O 2 should be magnetic!

84 N 2 Video O 2 Video

85 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)

86 MO Diagram for H 2 vs. N 2 N 2 s p * p 2p * H 2 s 2p p 2p Atomic orbital overlap sometimes forms both s and p bonds. Examples: N 2, O 2, F 2 s 2s * s 2s

87 Electron energy (kj mol -1 ) M.O. Diagram for N 2 s*(2p) p* p* -1,155 s(2p) -1,240-1,240 p p Valence Core -1,479-2,965 s*(2s) s(2s) Valence Core -37,875-37,871 1s(N) + 1s(N) 1s(N) 1s(N)

88 A Complication M.O. Diagram for B 2 (similar for C 2 and N 2 ) M.O. Diagram for O 2 (similar for F 2 and Xe 2 ) O O 2 O

89 A Complication M.O. Diagram for B 2 (similar for C 2 and N 2 ) M.O. Diagram for O 2 (similar for F 2 and Ne 2 ) s-p mixing No s-p mixing

90 Why does s-p mixing occur? Electron repulsion!! s 2s and s 2p both have significant e - probability between the nuclei, so e - in s 2s will repel e - in s 2p Effect will decrease as you move across the Periodic Table increased nuclear charge pulls the s 2s e - closer, making the s 2s orbital smaller and decreasing the s 2s and s 2p interaction

91 Molecular Orbitals of X 2 Molecules s p orbital mixing (a little hybridization) lowers the energy of the s 2s orbitals and raises the energy of the s 2p orbitals. As a result, E(s 2p ) > E(p 2p) for B 2, C 2, and N 2. As one moves right in Row 2, 2s and 2p get further apart in energy, decreasing s p mixing E(s 2p ) < E(p 2p ) for O 2, F 2, and Ne 2. See text pages Note that s p mixing does not affect bond order or magnetism in the common diatomics (N 2, O 2, and F 2 ). Hence it is not of much practical importance.

92 s-p mixing No s-p mixing

93 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)

94 Relating the M.O. Diagrams to Physical Properties

95 Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: N 2 N + 2 O 2 O + 2 Bond energy (kj/mol) Bond length (pm) No. of valence electrons Removing an electron from N 2 decreases the bond energy of the resulting ion, whereas removing an electron from O 2 increases the bond energy of the resulting ion. Explain these facts using M.O. diagrams.

96 Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: N 2 N + 2 O 2 O + 2 Bond energy (kj/mol) Bond length (pm) No. of valence electrons Plan: We first draw the MO energy levels for the four species, recalling that they differ for N 2 and O 2. 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.

97 Sample Problem - Continued Solution: The MO energy levels are: N 2 N + 2 O 2 O 2 + s p * s p * p 2p * p 2p * s 2p p 2p p 2p s 2p s 2s * s 2s * s 2s s 2s Bond Orders: (8-2)/2 = 3 (7-2)/2 = 2.5 (8-4)/2 = 2 (8-3)/2 = 2.5

98 Sample Problem: Using MO Theory to Explain Bond Properties Problem: Consider the following data for these homonuclear diatomic species: N 2 N + 2 O 2 O + 2 Bond energy (kj/mol) Bond length (pm) No. of valence electrons Bond Order

99 What have we learned so far? 1. Molecular orbitals (MO) explain the properties of valence electrons in molecules (Example: O 2 ) 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 s 2p and p 2p molecular orbitals depends on the electron occupancy: s-p mixing

100 Bonding in Diatomic Molecules Covalent Ionic Covalent Ionic

101 Homonuclear: H 2 Heteronuclear: HF Nonpolar covalent bond (450 kj/mol bond) Electronegativity Polar covalent bond (565 kj/mol bond)

102 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 F Figure 14.45

103 M.O.s of a Polar Covalent Bond: HF s Antibonding (s*) Mostly H(1s) H F H F This approach simplifies model and only considers electrons involved in bond. s Bonding Mostly F(2p)

104 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: N 2, O 2, F 2. 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.

105 Energy MOs OF XY MOLECULES s* 2p 2s p* s p s* s 2p 2p 2s 2s Electronegativity C Atom (4e ) C δ+ O δ (10e ) O Atom (6e ) CO Bond Order = 3.0 (same as N 2 ). CO Bond Energy = 1,076 kj/mol (N 2 = 945 kj/mol). Isoelectronic to CO and N 2 : CN, NO +. NO has 1e in p* bond order = 2.5; this e is more on N than O; NO NO + easy

106 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

107 When the electronegativities of the 2 atoms are more similar, the bonding becomes less polar. 2p 2s 2p 2s Electronegativity..... N=O.. EN(N) = 3.0 EN(O) = 3.4 N NO O Bond order = 2.5, unpaired electron is in a N-like orbital

108 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

109 M.O. diagram for NO p 2p * -597 p 2p * (empty) s 2p p 2p p 2p s 2s * s 2s -3320

110 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.

111 Combining the Localized Electron and Molecular Orbital Models (into a convenient working model) Figure Only the p bonding changes between these resonance structures - The M.O. model describes this p bonding more effectively.

112 Atomic Orbitals Molecular Orbitals Figure 14.51

113 Another example: Benzene s bonding: p bonding: p atomic orbitals p molecular orbital

114 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.