1. Lewis Structure A molecule is composed of atoms that are bound together by sharing electron pairs. A bonding pair is of which two electrons are localized between two adjacent nuclei. A lone pair is that the electron pair is localized in an atom. Localized Electron Bonding Model

2. Octet Rule 1electron dot (bar) formula 2 valence eletrons & net charge 3 multiple bond 4 dative bond (arrow to the acceptor) 5formal charge What is the difference between f. c. & ox. no 6 generally good thru the 3rd period

3. What is the difference between f. c. & ox. no Formal charge is for the comparison of the elctron change between the bonded and the non-bonded atomic molds. Oxidation number is for the comparison of the electron change between the oxidized and the reduced molds. Formal charge and oxidation numbers are the same for the ionic compounds.

4. Disadvantage of Lewis Structure 1 Octet rule is no good for d orbital or d electrons involved species – use 18 electron rule 2 no good for electron delocalization – use resonance 3 Octet rule is no good for odd electron species 4 no information for geometry- use VSEPR 5 It does not explain the bonding character – use VBT or MOT

5. Valence Bond Theory (VBT) Atomic orbitals that involve in chemical bonding are the major concern. ψ i ψ j ≠ 0, wherein ψ i and ψ j represent the orbital wave functions of different atoms.

7. VBT for H 2 Consideration of simple orbital coulomb overlap: Ψ I = ψ 1sa (1) ψ 1sb (2) ― curve a

8. Ψ I = ψ 1sa (1) ψ 1sb (2) experimental

9. Consideration of indistinguishability of the electrons: Heitler –London functions: Attractive Ψ II + = ψ 1sa (1) ψ 1sb (2) + ψ 1sa (2) ψ 1sb (1) ― curve b Repulsive Ψ II − = ψ 1sa (1) ψ 1sb (2) − ψ 1sa (2) ψ 1sb (1) ― curve g

10. Ψ I = ψ 1sa (1) ψ 1sb (2) experimental Ψ II + = ψ 1sa (1) ψ 1sb (2) + ψ ψ 1sa (2) ψ 1sb (1) Ψ II – = ψ 1sa (1) ψ 1sb (2) – ψ 1sa (2) ψ 1sb (1)

11. Consideration of effective nulear charge: Ψ 1s = N(z/a 0 ) 3/2 exp(-zr/a 0 ), z = 1.17 ― curve c The z-modification gives good bond length, but energy is still substantially off.

12. Ψ I = ψ 1sa (1) ψ 1sb (2) experimental Ψ II + = ψ 1sa (1) ψ 1sb (2) + ψ 1sa (2) ψ 1sb (1) Ψ II – = ψ 1sa (1) ψ 1sb (2) – ψ 1sa (2) ψ 1sb (1) z = 1.17

13. Consideration of polarizability or hybridization: ψ a = N (1s a +  2p za ) ψ b = N (1s b +  2p zb ), assuming z is the nuclear axis.

14. Normalization condition gives N = 1/(1 +  2 ) 1/2 Contribution of 1s a to ψ a (or 1s b to ψ b ) is 1/(1 +  2 ) Contribution of 2p za to ψ a (or 2p zb to ψ b ) is  2 /(1 +  2 ) If  is set as 0.1, which gives 99% (1/1.01) of 1s and 1% (0.01/1.01) of 2p z. Note that 1% of contribution of pz accounts for about 5% energy stabilization, mainly due to the enhancement of orbital overlap efficiency.

15. Ψ I = ψ 1sa (1) ψ 1sb (2) experimental Ψ II + = ψ 1sa (1) ψ 1sb (2) + ψ 1sa (2) ψ 1sb (1) Ψ II – = ψ 1sa (1) ψ 1sb (2) – ψ 1sa (2) ψ 1sb (1) z = 1.17 ψ a = N (1s a +  2p za ) ψ b = N (1s b +  2p zb )  = 0.1

16. Combine indistinguishability and hybridization: Attractive Heitler –London functions: Ψ III = ψ a (1) ψ b (2) + ψ a (2) ψ b (1) Consideration of other overlaps 1s a –2p b /2p a –1s b and 2p a –2p b

17. H―H ↔ H a − ―H b + ↔ H a + ―H b − Ψ IV = ψ a (1) ψ b (2) + ψ a (2) ψ b (1) + ψ a (1)ψ a (2) +  ψ b (1)ψ b (2) = 0.25 would minimize the total energy. The value of 2 ~ 0.06, indicating that about 6% ionic contribution will only improve the energy to 2~3%. Consideration of ionization:

18. Ψ I = ψ 1sa (1) ψ 1sb (2) experimental Ψ II + = ψ 1sa (1) ψ 1sb (2) + ψ 1sa (2) ψ 1sb (1) Ψ II – = ψ 1sa (1) ψ 1sb (2) – ψ 1sa (2) ψ 1sb (1) z = 1.17 ψ a = N (1s a +  2p za ) ψ b = N (1s b +  2p zb )  = 0.1 Ψ IV = ψ a (1) ψ b (2) + ψ a (2) ψ b (1) + ψ a (1) ψ a (2) +  ψ b (1) ψ b (2)

19. Valence Bond of HF Consideration of indistinguishable orbital overlap Ψ I = 1s (H) (1) 2p z(F) (2) + 1s (H) (2) 2p z(F) (1)

20. Consideration of hybridization: Hybridization for A.O. of F would give the configuration of 2s 1 2p 6 which allows the use of 2s for bonding. ψ F = N (2p z(F) +  2s (F) ),  < 0.5 and  2 < 0.25, the contribution of 2s (i.e.  2 /(1+  2 ) in F is less than 20%. Consideration of ionic model: Ψ = ψ 1s(H) (1) ψ F (2) + ψ 1s(H) (2) ψ F (1) + ψ F (1)ψ F (2) Pauling estimated that 2 ~ 50%.

21. 1It helps to have the electron density concentrated between the nuclei, thus leading to the enhancement of localized bonding character. Advantages from hybridization : 2It helps to decrease the electron-electron repulsion. ψ lp(F) = N (2s (F) −  2pz (F) )