1 structure prediction, chemical bonding 2 Lewis structures Atoms listed in order of increasing EN, no connectivity implied CSPF (PNF 2 ) 4
3 after the Lewis Structure determine the steric number number of bonds (single and multiple) number of unshared pairs or sometimes single unpaired electrons such as in NO 2 then assign geometry using 4 the following guidelines SN geometry of electron groups 2 linear (180 ) 3 trigonal planar (120 ) 4 tetrahedral (109.5 ) 5 trigonal pyramidal 6 octahedral (90 ) 7 pentagonal bipyramid
5 6 Cartoon representations
7 8 will need to refine bond angles considering the different sizes of single bonds vs. multiple bonds lone pairs vs. single bonds
9 today in dj One chemical fad is/was the isolobal analogy. Draw a chemical species or fragment that is isolobal with the methylene group, C 2, derived from methane by loss of two (2) atoms. 10 Repulsions between electron groups are in the order LP-LP > LP-BP > BP-BP when lone pairs are present the bond angles are smaller than those predicted for the idealized geometry in rule 1. lone pairs choose the site with the most volume available to them, or the least sterically hindered site. if all sites are equivalent, the lone pairs will be trans to one another. multiple bonds occupy more space than single bonds. bonding pairs to electronegative substituents occupy less space than those to less electronegative substituents
11 If the central atom is in third row or below in the periodic chart there are two possibilities: if the substituents are oxygen atoms or halogens, the above rules hold. if the substituents are less electronegative than halogens, the lone pairs will occupy a non-bonding s orbital and the bonding to the substituents will be primarily through p orbintals with resulting bond angles of about 90. Describe Structures by the arrangement of atoms. Unshared electrons (lone pairs) influence the structure (arrangement of atoms about central atom) but are not considered when describing the molecular structure. We see atoms positions, but not the position of the unshared electrons. 12 some examples SeBr 5 - ICl 3 CoF 6 3- OF 2 COCl 2
13 Today s stuff Next time, 63 thru 72 Look at 76-82 as well Exam 1, 20 October.. Up to Matrix representations. 14 Today s q What would be the biggest bond angle in BClBrI?
15 some reality M olecule Angle( ) M olecule Angle( ) M olecule Angle( ) -X- -X- -X- 2O 104.5 N3 107.3 C4 109.5 2S 92.2 P3 93.3 Si4 109.5 2Se 91.0 As3 91.8 Ge4 109.5 2Te 89.5 Sb3 91.3 Sn4 109.5 16 more reality M olecule X-O-Y -C- 2O 104.5 C4 109.5 F2O 103.2 C3Cl 110.5 Cl2O 111 C2Cl2 112.0 (C3)2O 111 C3Br 111.2 C3O 109 C3I 111.4 C3O 109.3 3C-C3 109.3
17 any refinement? SeBr 5 - ICl 3 CoF 6 3- OF 2 COCl 2 18 experimental confirmation of the postulated hybridizations?
19 for s-p hybrids, bond angle can give % s character the relationship to right can give relative s and p character for equivalent sp x hybrids cos = s s 1 = p 1 p 20 consider O=CF 2 note that the orbitals have more p character due to either steric effects or electronegativity of F substituents Remember that p orbitals have greater spacial extension! More e - density to F cos108 = s s 1 = 0.309 s = 0.24
21 hybridization and bond angles 22 hybridization energetics orbital mixing in hybridization usually requires energy consider C hybridizing to sp 3 this promotion costs 400 kj/mole Payback? Stronger bonds, more stable product! 2p 2s sp 3 hybrids
23 factors influencing hybridization energetics of mixing how much energy needed for promotion? bond energies depends on the system will BE s return energy required for hybridization electron pair repulsions less important in larger atoms 24 consider rehybridization and repulsions for P 3 vs N 3 similar rehybrization energies are required electron pair in ns orbital must be promoted to sp 3 orbital energy (for P, E 600 kj/mole) sizes of atoms are different 75 vs 110 pm Therefore orbital mixing is not advantageous
25 Sterics vs. Electronics What is more important? Steric effects or electronic effects? 26 Lone Pairs can be insignificant Sn(C 5 Ph 5 ) 2 as two electrons in valence shell..
27 or stereochemically active (Janiak, et al., Chem. Ber.(1988) 121 p1745) 28 consider reality M olecule Angle( ) M olecule Angle( ) M olecule Angle( ) -X- -X- -X- 2O 104.5 N3 107.3 C4 109.5 2S 92.2 P3 93.3 Si4 109.5 2Se 91.0 As3 91.8 Ge4 109.5 2Te 89.5 Sb3 91.3 Sn4 109.5
29 results smaller bond angles in P 3 and lower congeners indicate low degrees of hybridization lone pair is in s orbital Much lower basicity!! 30 more reality M olecule X-O-Y -C- 2O 104.5 C4 109.5 F2O 103.2 C3Cl 110.5 Cl2O 111 C2Cl2 112.0 (C3)2O 111 C3Br 111.2 C3O 109 C3I 111.4 C3O 109.3 3C-C3 109.3
31 bond energies, electronegativities this factor can be difficult to deconvolute since electronegativity difference leads to stronger bonding because of ionic resonance forms A-B A + -B - 32 Bent s Rule More electronegative substituents prefer hybrid orbitals of less s character and the less elctronegative substituents prefer orbitals having more s character. the p character tends to concentrate in orbitals of weaker covalency and s character concentrates in orbitals of stronger covalency covalency depends on electronegativity difference and orbital overlap
33 for example PMeCl 4 PMe 2 Cl 3 34 VSEPR is a lie that works we think of 2 O as having this structure: O
35 the lone pairs in 2 O look like this: 36 Valence Bond Theory developed by Pauling invokes overlap molecular orbitals are formed from products of 1 electron atomic orbitals = (1) (2) hybridization is required to obtain the correct geometry
37 today Introduction to the chemical bond- yet another time 38 Valence Bond Model and refinements
39 (a) = 1sA 1sB (b) delocalization = 1sA(1) 1sB(2) + 1sA(2) 1sB(1) (c) shielding by electrons (d) ionic resonance forms - +- - -- + = 1sA(1) 1sB(2) + 1sA(2) 1sB(1)+ ( 1sA(1) 1sB(1) + 1sA(2) 1sB(2)) 40 aspects of VB theory hybridization delocalization is not inherent ionic forms are not inherent
41 Linear Combinations of Atomic Orbitals a molecular orbital theory can use symmetry to make building MO s easier start with diatomics, O 2 and NO finish with several triatomics, Be 2 2 O, CO 2 and NO - 2 some these latter will be solved using group theory 42 LCAO assumptions same size/energy orbitals equal distribution of electrons, therefore each atomic orbital makes an equal contribution to the molecular orbital. This condition does not hold in heterodiatomic molecules. use the orbital approximation the wave function of the N electrons in an atom can be written as the product of the 1 electron wavefunctions, each a fcn of n, l, m l and m s. for electrons in low-lying orbitals (near the nucleus) in a molecule the wavefunction resembles that of the lowlying corresponding orbital in the atom. a reasonable first approximation to the molecular wave functioncan be obtained from linear combinations of the atomic orbitals, and specifically, the valence orbitals of the atoms involved.
43 For 2 one can form: (1) = (1) + (2) (2) = (1) (2) 44 Consider 2 : by overlap of 1s orbitals See the spreadsheets: just overlapping wavefcns VB formalism And LCAO method
45 an important foundation orbitals of different symmetry cannot mix to form molecular orbitals for diatomic molecules, the z axis contains the line joining the nuclei. the p x and p y orbitals have the same symmetry as do the p z and s orbitals. 46 combinations s orbitals
47 bonding and antibonding orbitals antibonding + + bonding + + 48 And still more Other combos Which are bonding? Antibonding?
49 Start with O 2 What are valence orbitals on O atoms? 2s and 2p orbitals What orbitals can overlap and form bonding/antibonding combinations? 2s and 2p z can form bonds 2p x and 2p y form bonds What of relative energies? 50 Cache ab initio MO s for O 2 s orbitals give these sigma bonding and antibonding orbitals in O 2 (s- )= (2s) + (2s) (s- )= (2s) - (2s)
51 p z orbitals give sigma bonding and antibonding orbitals 52 p x and p y orbitals give degenerate and * orbitals
53 MO diagram for O 2 p s orbitals give bonding and antibonding orbitals p orbitals give and bonding and antibonding orbitals *not completely true! np p np ns ns 54 expect that electron density should favor proximity to more electronegative atom (bonding MO s are located on more en atom) antibonding MO s should have more character of less en atom differences between homoand hetero- binuclear molecules np N O np ns ns
55 compare orbitals for O 2 and NO 56 compare p orbital and * orbitals on NO and O 2
57 compare p orbital and * orbitals 58 note that in most cases, the bonding orbitals that are occupied have more electron density on the more electronegative atom antibonding MO s have more character on the less en atom differences between homoand hetero- binuclear molecules np N O np ns ns
59 VSEPR is a lie that works we think of 2 O as having this structure: O 60 the lone pairs in 2 O look like this:
61 This is the orbital mixing! bonding Nonbonding 62 What is % s ard to tell Not exactly25% like sp 3, eh.
63 Isolobal analogy Orbital similarities between molecular fragments have predictive power of the types of compounds formed. 64 What is meant by isolobal? To determine whether two molecules are isolobal one must consider the frontier orbitals (the valence orbitals including the OMO and the LUMO).
65 two molecular fragments are considered isolobal if: 1) the number 2) the symmetry 3) the electron occupancy and 4) the approximate energy of the frontier orbitals are the same. 66 What s the use? Isolobal analogy: compounds which have frontier orbitals which have the same symmetry and electron occupation will tend to form analogous complexes.
67 Making fragments - -e - C 3 C 3 + 68 What is isolobal with C 3 and C 3 +? Cl B + - -e - C 3 C 3 +
69 Simple predictions 1. Since C 3 forms a compound with the :N 2 radical, so will the isolobal and Cl fragments. 70 Pictorially, C 3 + N 3 C N 2 therefore: Cl N 2 Cl + N N 2
71 however these fragments are isolobal to each other. 72 Unlike fragments, C 3 + N + exists, therefore N + B and N + probably exist