NH 3 H 2 O N 2. Why do they make chemical bonds? Molecular Orbitals

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1 N 2 NH 3 H 2 O Why do they make chemical bonds? 5 Molecular Orbitals

2 Why do they make chemical bonds? Stabilization Bond energy

3 Types of Chemical Bonds Metallic Bond Ionic Bond Covalent Bond

4 Covalent Bond A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Electron pairs in the molecule are assumed to be localized on a particular atom or in the space between two atoms Electron pairs localized on an atom : lone pairs Electron pairs found in the space between the atoms : bonding pairs Description of valence electron arrangement Prediction of geometry Description of atomic orbital types used to share electrons or hold lone pairs. Analogous to atomic orbitals for atoms, MOs are the quantum mechanical solutions to the organization of valence electrons in molecules.

5 Covalent Bond LE is great to predict bondings and structures and geometries of molecules. BUT, there are some short points. No concept of resonance. No paramagnetic properties. No information of bond energy. Fact: O 2 is paramagnetic! O O Lewis structure VSEPR Valence bond theory sp 2 hybridized lone pairs in sp 2 hybrid orbitals bonding pairs in s and p bonds All show all electrons paired. In contrast to LE, molecular orbitals describe how electrons spread over all the atoms in a molecule and bind them together, which can give correct views of concept of resonance. paramagnetic properties. bond energy.

6 Formation of Molecular Orbitals from Atomic Orbitals H 1 proton and 1 electron Schröedinger Eq => Atomic orbitals Overlap of wavefunctions (constructive Overlap) H 2 2 protons and 2 electrons Schroedinger Eq => Molecular orbitals But, no way to solve => LCAO (linear combination of atomic orbitals) => Approximate solutions of Schroedinger Eq (destructive Overlap) f A A f + f + 2 Constructive overlap enhance e - density between the nuclei attract the nuclei bonding orbital B f B f A f - 2 A B f - f B destructive overlap node(s) between the nuclei repel each other antibonding orbital

7 Formation of Molecular Orbitals from Atomic Orbitals Higher energy state Lower energy state Y = c a y a + c b y b f A A f + B f B f + 2 Constructive overlap enhance e - density between the nuclei attract the nuclei bonding orbital f A f - 2 A B f - f B destructive overlap node(s) between the nuclei repel each other antibonding orbital

8 Formation of Molecular Orbitals from Atomic Orbitals Molecular Orbitals from s Orbitals 1 Y( s ) = N[ cay (1s a) + cby (1s b)] = [ y (1s a) + y (1s b)] 2 1 Y( s*) = N[ cay (1s a) - cby (1s b)] = [ y (1s a) -y (1s b)] 2 H 2 s* 1s antibonding m.o. (higher energy than separate atoms) s 1s bonding m.o. (lower energy than separate atoms) s (C 2 symmetric about the line connecting the nuclei)

9 Formation of Molecular Orbitals from Atomic Orbitals Molecular Orbitals from p Orbitals p (C 2 antisymmetric about the line connecting the nuclei)

10 Formation of Molecular Orbitals from Atomic Orbitals Molecular Orbitals from d Orbitals No interaction

11 3 Things to Consider to Form Molecular Orbitals N atomic orbitals => N molecular orbitals Symmetry match of atomic orbitals Relative energy of atomic orbitals Forming nonbonding orbitals Remember! The more nodes, the higher energy.

12 Homonuclear Diatomic Molecules 1 st Row Diatomic Molecules H 2 s g2 (1s) b.o. = 1 b.l = 74.1 pm BE = 436 kj/mol H 2 + s g1 (1s) b.o. = 0.5 b.l = pm BE = 258 kj/mol Y( s*) = N[ cay (1s a) -cby (1s b)] Y( s ) = N[ cay (1s a) + cby (1s b)] He 2 s g2 s u * 2 (1s) b.o. = 0 *detected at very low T and P *BE = 0.01 J/mol He 2 + s g2 s u * 1 (1s) b.o. = 0.5 b.l. =108 pm BE = 233 kj/mol bond order = (no. of e in bonding m.o.s) - (no. of e in antibonding m.o.s) 2 b.o. > 0 (i.e., lower energy than separate atoms)

13 Homonuclear Diatomic Molecules 2 st Row Diatomic Molecules No mixing (O 2, F 2, Ne 2 ) Mixing (Li 2 - N 2 ) Not big Y = N[ cay (2sa) + cby (2sb) + ccy (2pa) + ccy (2pb )]

14 Homonuclear Diatomic Molecules 2 st Row Diatomic Molecules Li 2 s g2 (2s) b.o. = 1 * found in gas phase Be 2 s g2 s u * 2 (2s) b.o. = 0 B 2 s g2 s u * 2 (2s)p u1 p u1 (2p) b.o. = 1 * Paramagnetic (gas phase) * B 12 (solid phase) C 2 s g2 s u * 2 (2s)p u2 p u2 (2p) b.o. = 2 (p) * Found in gas phase * rare * C 2-2 is more common C 2 d(c-c) pm (b.o. = 2) CaC 2 d(c-c) pm (b.o. = 3) C 2 H 2 d(c-c) pm (b.o. = 3) N 2 s g2 s u * 2 (2s) p u2 p u2 s 2 g (2p) b.o. = 3 (2p+s) b.l. = pm BE = 942 kj/mol

15 Homonuclear Diatomic Molecules 2 st Row Diatomic Molecules O 2 s g2 s u * 2 (2s)s g2 p u2 p u2 p *1 g p *1 g (2p) b.o. = 2 b.l. = pm * Paramagnetic Other forms of O n 2 O + 2 b.o = 2.5, b.l = pm O - 2 b.o = 1.5, b.l = 135 pm O 2-2 b.o = 1, b.l = 149 pm F 2 s g2 s u * 2 (2s)s g2 p u2 p u2 p *2 g p *2 g (2p) b.o. = 1 Ne 2 s g2 s u * 2 (2s)s g2 p u2 p u2 p *2 g p *2 g s u * 2 (2p) b.o. = 0

16 Homonuclear Diatomic Molecules 2 st Row Diatomic Molecules HOMO (highest occupied molecular orbital) LUMO (lowest unoccupied molecular orbital) Big triumph of MO theory SOMO (singly occupied molecular orbital) Frontier orbitals

17 Homonuclear Diatomic Molecules Bond Lengths Bond length Covalent radius 2 H-X H-B H-C H-N H-O H-F Length (pm) Any trend found? OK with electronegativity difference 2. Any trend found? Don t be fooled by the text book. Covalent radii are defined in X-X single bond (Table 2-8).

18 Homonuclear Diatomic Molecules How to measure the energy levels of MOs? (Photoelectron spectroscopy) hn = UV UPS : outer electrons hn = X-ray XPS : inner electrons photoelectron v IE A + hn A + + e - Ionization energy = hn - ½ mv 2

19 Homonuclear Diatomic Molecules How to measure the energy levels of MOs? (Photoelectron spectroscopy) N 2 O 2 Why fine structure? Ionization energy = hn - ½ mv 2

20 Homonuclear Diatomic Molecules How to measure the energy levels of MOs? (Photoelectron spectroscopy) Franck-Condon Principle: Classically, the Franck Condon principle is the approximation that an electronic transition is most likely to occur without changes in the positions of the nuclei in the molecular entity and its environment. The resulting state is called a Franck Condon state, and the transition involved, a vertical transition. M nucleus >> M electron e - : faster motion Ionization energy = hn - ½ mv 2 E + vib

21 Homonuclear Diatomic Molecules How to measure the energy levels of MOs? (Photoelectron spectroscopy) N 2 O 2 stronger bonding involved less bonding involved Ionization energy = hn - ½ mv 2 E vib +

22 Homonuclear Diatomic Molecules Correlation Diagram non-crossing rule

23 Heteronuclear Diatomic Molecules Polar Bonds 3 things to consider to form MOs N atomic orbitals => N molecular orbitals Symmetry match of atomic orbitals Relative energy of atomic orbitals Y = c y + c y a a b b c A = c B c A > c B c A >> c B c A = c B c A < c B c A << c B

24 Heteronuclear Diatomic Molecules Polar Bonds Average potential energy of all terms Ex) E(C 2p ) = [E( 1 D) x 5 + E( 3 P) x 9 + E( 1 S) x 1]/15

25 Heteronuclear Diatomic Molecules CO Significant mixing between O 2pz and C 2s

26 Heteronuclear Diatomic Molecules CO Significant mixing with C 2s

27 Heteronuclear Diatomic Molecules CO z C O p orbital w/o considering signs C 2v C v

28 Heteronuclear Diatomic Molecules CO M C O M O C?

29 Heteronuclear Diatomic Molecules Ionic Compounds and MOs LiF View of ionic interaction Li: 1s 2 2s 1 Li + : 1s 2 F: 1s 2 2s 2 sp 5 F - : 1s 2 2s 2 2p 6 Electrostatic interaction View of MO Transfer of Li 2s e - to F 2p orbital which is lowered F 2p character Don t forget that ionic interaction is omnidirectional and more accurate MO description requires bands.

30 Heteronuclear Diatomic Molecules Ionic Compounds and MOs LiF View of ionic interaction Li: 1s 2 2s 1 Li + : 1s 2 F: 1s 2 2s 2 sp 5 F - : 1s 2 2s 2 2p 6 Is this process really helpful? Thermodynamically unfavorable!! Lattice enthalpy is the deriving force.

31 MOs for Larger Molecules 1. Determine the point group of molecules. (D h D 2h, C v C 2v ) 2. Assign x, y, z coordinates. 3. Find reducible representations for ns orbitals on the outer atoms. Repeat for np orbitals in the same symmetry. (valence orbitals) 4. Reduce the reducible representations of step 3 to derive group orbitals or symmetry adapted linear combinations (SALCs) 5. Find the atomic orbitals of the central atoms with the same symmetries as those found in step Combine the atomic orbitals of the central atom and the SALCs of the outer atoms with the same symmetry and similar energy to form MOs.

32 MOs for Larger Molecules FHF - D h D 2h Very strong H-bonding F (2px) +F (2px) F (2py) +F (2py) F (2pz) +F (2pz) F (2s) + F (2s) B 3u + B 2g B 2u + B 3g A g + B 1u A g + B 1u

33 MOs for Larger Molecules FHF - D h D 2h A g can combine to form MOs F---F: SALCs H: 1s orbital

34 MOs for Larger Molecules FHF - D h D 2h ( ev) H 1s ( ev) ( ev) H 1s will strongly interact with F 2Pz s (A g ). Don't forget if F2s contributes, 3 MOs are formed.

35 MOs for Larger Molecules FHF - D h D 2h antibonding F H F Lewis structure non-bonding * there are slight long-range interactions. bonding MO 3-center 2-electron bond

36 MOs for Larger Molecules CO 2 D h D 2h O (2px) +O (2px) B 3u + B 2g O (2py) +O (2py) O (2pz) +O (2pz) O (2s) + O (2s) B 2u + B 3g A g + B 1u A g + B 1u

37 MOs for Larger Molecules CO 2 D h D 2h combine to form MOs O---O: SALCs C: valence orbitals

38 MOs for Larger Molecules CO 2 D h D 2h Effective to form MOs

39 MOs for Larger Molecules CO 2 D h D 2h

40 MOs for Larger Molecules CO 2 O=C=O Lewis structure 2-center 2 electron bond 16 valence e - 's non-bonding p bonding p 3-center 2 electron bond bonding s non-bonding s

41 MOs for Larger Molecules H 2 O H (1s) +H (1s) A 1 + B 1 A 1 B 1 H H H H Y a1 = (1/ 2){f a (H 1s )+f b (H 1s )} Y b1 = (1/ 2){f a (H 1s )-f b (H 1s )} 2p y B 2 2p x B 1 2p z A 1 2s A 1 H------H: SALCs O: valence orbitals

42 MOs for Larger Molecules H 2 O B 1 2p x B 1 H H 1b 1 bonding 2b 1 antibonding A 1 2s A 1 H H 2a 1 bonding 3a 1 nearly non-bonding 4a 1 antibonding 2p z A 1 O H------H: SALCs H 1b 2 non-bonding H 2 O MOs 2p y B 2 O: valence orbitals

43 MOs for Larger Molecules H 2 O Y 2 Y 6 Y 1 Y 3 Y 5 Y 4 Very weak contribution of H group orbital (1s) weak contribution of O(2p z )

44 MOs for Larger Molecules H 2 O Lewis structure MO lone pairs Two lone pairs are equivalent (2 x sp 3 ). Two O-H bonds are equvalent [2 x (sp 3 + H(1s))]. O-H bonds

45 MOs for Larger Molecules H 2 O Other approach O A 1 2s 2p z sp sp b 1 b 2 B 2 B 1 H----H SALCs b 1 A 1 B 1

46 MOs for Larger Molecules H 2 O Why o in MO theory? Walsh Diagram - a diagram showing the variation of orbital energy with molecular geometry

47 MOs for Larger Molecules H 2 O Walsh Diagram - a diagram showing the variation of orbital energy with molecular geometry * When x-, y- axes are defined differently, B 1 and B 2 symmetries are exchanged.

48 MOs for Larger Molecules NH 3 3H (1s) A 1 + E N 3H SALCs A 1 A 1 A 1 s p z Y = 1 [ y ( Ha) + y ( Hb) + y ( H 3 c )] E E p x p y 1 1 Y = [2y ( Ha) -y ( Hb) -y ( Hc)] Y = [ y ( Hb) -y ( H c )] 6 2

49 MOs for Larger Molecules NH 3 A 1 s A 1 A 1 p z E mostly p z character p x p y N 2s E 3H SALCs

50 MOs for Larger Molecules BF 3 F F B F 3F (2s) A 1 ' + E' 3F (2px) A 2 ' + E' 3F (2py) A 1 ' + E' 3F (2pz) A 2 '' + E''

51 MOs for Larger Molecules BF 3 F 2s F B F 2p x, 2p y A 1 ' 3F (2s) A 1 ' + E' 3F SALCs 3F (2py) A 1 ' + E' E' 3F (2px) A 1 ' A 2 ' + E' E' A 1 ' 3F (2pz) E' A 2 '' + E'' 2p z A 2 '' B orbitals A 2 ' E' A 2 '' E''

52 MOs for Larger Molecules BF 3 A 2 '' E'' 3F (2pz) 2p y 2p x E' A 2 ' E' 3F (2px) 2p z A 2 '' A 1 ' E' 3F (2py) 2s A 1 ' B orbitals A 1 ' E' 3F (2s) 3F SALCs

53 MOs for Larger Molecules BF 3 antibonding Lewis? VBT?? s and? p non-bonding bonding (a 2 :slightly bonding) almost non-bonding same for SO 3, NO 3-, CO 3 2-

54 MOs for Larger Molecules SALCs

55 MOs for Larger Molecules SALCs

56 MOs for Larger Molecules SALCs

57 MOs for Larger Molecules Hybrid Orbitals One of the ways to predict the hybrid orbitals is.. (Which atomic orbitals form hybrides?)

58 MOs for Larger Molecules Hybrid Orbitals Ex) CH 4 Ex) BF 3 F 4 C-H bonds A 1 + T 2 s, (p x, p y, p z ) sp 3 why not d? B F F 3B-F bonds = 3F (2s) A 1 ' + E' s, (p x, p y ) sp 2