4.2. Molecular Properties from VB Theory a) Bonding and Bond distances The wavefunction that describes a bonding pair of electrons: Ψ b = a(h 1 ) + b(h 2 ) where h 1 and h 2 are HAOs on adjacent atoms pointing towards each other. (Source: Purcell + Kotz, Inorganic Chemistry, 1977) Bond strength = f(overlap integral S), i.e. the constructive overlap between h 1 and h 2. Q. How does the overlap integral change as a function of different hybridization? e.g., C-H bonds in ethane, ethylene, acetylene 95
The more s-character, the larger the overlap integral (up to a point). The more s-character the shorter the bond. In general: Shorter bonds are stronger bonds (which does not necessarily correlate with overall reactivity!!!). (Source: A. Macoll, Trans. Faraday Soc., Vol. 46, 369, (1950). C(sp n ) r CC [Å] E(C-C) [kcal/mole] H-C C-H sp 1.20 199-200 H 2 C=CH 2 sp 2 1.34 146-151 H 3 CCH 3 sp 3 1.54 83-85 Why bonds with more s-character are shorter: (Source: Purcell + Kotz, Inorganic Chemistry, 1977) Positive overlap continuously increases for s-orbitals, but goes through a maximum for p-orbitals! 96
Concept of isovalent hybdridization: BENT S RULES (cf. H.A. Bent, Chem. Rev., 61, 275, (1961)) So far we ve only considered high symmetry systems EX n or X n E-EX n. What happens in EX m Y n? The more electronegative an atom X or Y bound to a central atom E, the greater will be its demand for electron density from the central atom. From the concept of orbital electronegativity: EN (s) > EN (p) > EN (d ) the more the s-character in a hybrid orbital, the greater its electronegativity. Recall: Mulliken electronegativities of carbon: (sp) 2 π 2 10.4 ev 5.7 ev (sp 2 ) 3 π 1 8.8 ev 5.6 ev (sp 3 ) 4 8.0 ev Central atom E will direct more p (and less s) character towards the more electronegative atom (and vice versa) Lone pairs will be in orbitals with more s-character. (Think of a lone pair as a bonding pair to a very electro-positive atom ). fine tuning of VSEPR structures e.g. PCl 5 : Note that phosphorus pentachloride exists as a neutral PCl 5 molecule in the gas phase and as ionic [PCl 4 + ][PCl 6 - ] in the solid state. Q. Can you predict the structure of SbF 2 Br 3 based on isovalent hybridization? Q. Can you predict the structure of SF 4? 97
b) Molecular Dipole Moments Rationalized by VB Theory Need to find the centers of electronic (-ve) and nuclear (+ve) charge in the molecule as determined by the molecular structure and e - -distribution. ( hmm, wasn t there another concept that we used for this as well?) Consider each bond and lone-pair separately ( essence of VB theory). Def. of dipole: µ = n r where n = number of charges r= distance between charges Lone pair dipoles: µ lp = 2e<r> <r> = average value of r Lone pair dipoles are fairly large. Absolute size varies with hybrid character: sp > sp 2 > sp 3 4.4 3.7 Debye Note: Dipole moments are expressed in Debye units. 1 Debye = 10-8 pm esu Example: Two charges equal in magnitude to the charge of an electron (4.8 x 10-10 electrostatic units esu) and separated by a distance of 91.7 pm (interatomic distance of HF molecule) give a dipole of 4.4 Debye. The value 4.4 Debye represents the expected dipole moment for 100% ionic H + F -. The observed dipole moment for HF is 1.98 Debye. Therefore, the H-F bond has 1.98/4.4 = 45% ionic character. Okay, now let s look at how HAO s are useful for explaining lone pair dipole moments 98
One of the quantum mechanical postulates is that the average or expectation value of any observable property O is given by: where is the operator that corresponds to the property O (e.g. the Hamiltonian for the energy of the system, or the distance between charge centroids) and ψ is the state function (i.e. wavefunction) of the system. Consider an sp hybrid orbital: Total orbital: φ n = c s φ s + c p φ p Average (= quantum mechanical expectation) value of r for an sp hybrid orbital: The last equality makes the approximation: <r> p <r> s Furthermore because of normalization: c s 2 + c p 2 = 1 Thus: Because the s-orbital is centrosymmetric (i.e., <r s > = 0), the only dipole contribution comes from the hybrid orbital and the dipole is simply µ = 2 e <r> h = 4 e c s c p <r> sp i.e., only the interference integral that describes the hybrid contributes to the dipole. POINT: s and p orbitals are centrosymmetric. HAO s are not and therefore can be invoked in the explanation of dipoles. 99
Bond pair dipoles: more difficult to estimate Need to identify the charge centers along a bond: +ve: In 1 st approximation simply the halfway point of the bond -ve: Need to know the electron density distribution along the bond as f(r) Consider bond H-E, where E = any element except hydrogen: Expectation value of r (as before): ; φ Eh = sp-hybrid orbital on E Physical meaning of all the terms: a 2 <r> H b 2 <r> E 2ab<r> HE electron density contribution of the hydrogen 1s AO to center of ve charge electron density contribution of the E sp-hybrid orbital to center of ve charge electron density contribution of the overlap density to center of ve charge 2+ = positive charge midpoint (because we are dealing with electron pairs, we must count up the positive charge in pairs as well). R = distance between nuclei 100
From figure: - The electron density center for hydrogen is at <r> H = -(R/2) - The electron density center for the hybrid orbital is at (using the result for r sp from the derivatization for the lone pair): For the complete electron pair: <r> E = <r> s + 2c s c p <r> sp = R/2 + 2 c s c p <r> sp where the substitution i = b 2 - a 2 is a measure of the ionic character of the bond: a 2 = b 2 i = 0 purely covalent bond a 2 or b 2 = 1 i = 1 purely ionic bond (also: a 2 + b 2 = 1) In summary: i(r/2) 2ab<r> HE 2b 2 c s c p <r> sp orientation of the bond-dipole to the more electronegative atom contribution from the overlap density of both atoms in the bond to the center of negative charge along the bond contribution from the hybrid orbital itself on E to the charge center; orientation depends on the radial distribution of electron density in φ E. (non-existent for φ 1s because of radial symmetry) The orientation and size of a bond-dipole moment cannot solely be predicted on the basis of the difference in EN values of the two atoms involved!!! Have to consider the charge distribution contributions from the orbitals involved, which can lead to a dipole that is reversed from the one expected based on ΔEN!!! 101
In general: LP dipole moments >> BP dipole moments In bonds of the type E 1 -E 2 (E 1 E 2 ) with hybrid orbitals from both atoms forming the bond, the hybrid contributions to the dipole moment tend to cancel. Examples: NH 3 vs. NF 3 µ = 1.5 D µ = 0.2 D Χ N > Χ H Χ N < Χ F no hybrid dipole moment on H Bent s rule requires more s-character on N LP overall weak BP dipole moment dominated by LP dipole moment F atoms have LP aligned with bond overall weak N LP dipole moment dominated by BP NF and LP F overall dipole moment orientation?? Carbon monoxide: Lone pairs measured dipole moment µ = 0.1 D, i.e. very small Bond pairs From Bent s rule: both sp lone pairs have more than 50 % s-character with carbon more so than oxygen dipole towards oxygen but: C(sp) more diffuse (i.e. spreads further out) dipole towards carbon net effect µ 0 Dominated by ΔEN: More compact oxygen orbitals dipole towards oxygen but: +ve charge center closer to the oxygen (higher nuclear charge) net effect µ 0 and now you understand why CO is a gas!!! 102
c) Bond Energies and VB theory VB Theory microscopic picture/molecular scale isolated, localized electron pairs form bonds Thermodynamics macroscopic picture/bulk scale bond energies within a molecule are additive Diatomic molecules: Experimentally, bond energy (D) is simply the internal energy change required to separate the gas phase molecule into its two constituent atoms (also in the gas phase.) A-B(g) A(g) + B(g) D AB = ΔE (Note: D is discrete bond energy, E is average) Polyatomic molecules: 1) Removal of one terminal atom to leave a fragment AB n-1 alters the bonding between A and the remaining B atoms (changes hybridization / multiple bonding). 2) Ground state electron configuration of the dissociated B atom may be quite different from its valence state when bound to A. The valence state promotion energy for B may make a significant contribution to the bond energy D. Consider a simple binary compound AB n. AB n (g) AB n-1 (g) + B(g) D AB =?? D AB is a function of: Hybridization state of AB n and AB n-1 Changes in multiple bonding between AB caused by removal of one B Electrostatic configuration of B alone vs. B in AB n (in particular of the promotion energy B groundstate B hybrid ). Example: Stepwise dissociation of CO 2 (gasphase) Step 1: O=C=O C O + O D CO = 127 kcal/mol Step 2: C O C + O D CO = 256 kcal/mol In step 1: Breaking of one bond followed by configuration relaxation: O: (sp 2 ) 1 (sp 2 ) 2 (sp 2 ) 2 π 1 (= s 5/3 p 13/3 ) s 2 p 4 ( 3 P) from hybrid to ground state atomic C: (sp) 1 (sp) 1 (π) 1 (π) 1 (= s 1 p 3 ) (sp) 1 (sp) 2 π 1 (= s 1.5 p 2.5 ) from double to triple bonding In step 2: Breaking of second bond and relaxation of both atoms into g. s. atomic configuration DO THIS FOR YOURSELF AT HOME 103
A reverse example: Cl-Hg II -Cl Hg I -Cl + Cl Hg 0 + 2 Cl D HgCl = 81 kcal/mol D HgCl = 25 kcal/mol - Can you write down the electron configurations of this example? - Why is the second step energetically easier? Important conclusion: Average bond energies and individual bond energies in AB n systems are not the same!!! Average bond energies (thermodynamics): AB n (g) A(g) + n B(g) E A = 1/n ΔE Expect major consequences on stability/reactivity of molecules in general!!! Consider the reaction PCl 5 PCl 4 + Cl What are the structures of PCl 5 and PCl 4? Based on VB are all the P-Cl bonds equally strong? Thought experiment: No matter which P-Cl is broken ΔE is always the same. This is a requirement of the macroscopically observed real experimental value but how can that be?? In summary: Experimental bond energies are composed of the actual bond dissociation energy and the energy terms that account for changes in the electronic configurations of the participating atoms and/or molecular fragments. ( to be really correct would also have to consider vibrational and rotational states ) 104
More examples: Series of EH n molecules with n = 2, 3, or 4 (NOTE: These are called hydrides.) Compare CH 4 vs. SiH 4 : Average <E EH > [kcal/mole] Promotion energy [kcal/mole] s 2 p n (n = 2,3,4) 4 sp 3 CH 4 99 150 SiH 4 76 115 NH 3 93 228 PH 3 78 133 OH 2 111 194 SH 2 88 109 4 <E CH > = 4 99 = 4 E * CH P C 396 = 4 E * CH 150 E * CH = 137 kcal/mole 4 <E SiH > = 4 76 = 4 E * SiH P Si 304 = 4 E * Si 115 E * Si = 105 kcal/mole The C-H bond is truly stronger than the Si-H bond. Also true for N/P and O/S pairs. Why? Because of greater diffuseness of the atomic orbitals of heavier elements see below! Consequence of more diffuse orbitals on heavier elements resulting in smaller overlap integrals and thus in weaker bonds: (Source: Purcell + Kotz, Inorganic Chemistry, 1977) 105
4.3 Concept of Isolobality (Nobel laureate Roald Hoffman) See: Angew. Chem. Int. Ed. 1982, 21, 711. Contributes to the understanding of parallels between organic and inorganic chemistry. Structurally analogous fragments of molecules are described as isolobal. Studying isolobal molecular fragments can help suggest patterns of bonding / reactivity. Concept of hybridization can be used to identify isolobality. Def. Two molecular fragments are said to isolobal if the number, symmetry properties, approximated energy and shape of the frontier orbitals and the number of electrons in them are similar (NOT identical, but similar). Example: The following orbitals have σ-type symmetry, are similar in energy, and are occupied by one electron therefore we can expect analogous bonding properties. We can also make the isolobal analogy with the following two and three orbitals examples: NOTE: Although the isolobal analogy can give clues about what kind of compounds might be expected to exist, it is only a useful tool and not a hard and fast rule. HOMEWORK: Exercise 2.25 106
4.4. Limitations of VB Theory Valence shell expansion is often required to draw satisfactory Lewis diagrams, e.g. for PCl 5 and SF 6, even though the incorporation of d-orbitals does energetically not make sense the process: s 2 p n d 0 s 2 p n-m d m Costs too much hybridization energy. So how can it be that e.g. sulfur can bind six fluorine atoms??? Why is O 2(g) paramagnetic (i.e., has unpaired electrons)? How can we describe the bonding in B 2 H 6 using only H 1s orbitals? How can we explain these bond angles through hybridization? No hybridization can explain bond angles < 90!! We need a new theory! 107