1. 1 Chemical Bonding Compounds Ionicmolecular consist of ionssharing of electrons Ionic bondcovalent bond III/

2. 2 Energy: Ability to do work. Work=force x distance Potential energy:  energy possessed by an object due to its presence in a force field. i.e. under the effect of external force.  Object attracted/repelled by external force.  stored energy! force causes the object to move  Gravitational force causes the water to fall.  can generate electricity E pot =mgh Attraction causes the ball to fall, h smaller, E pot smaller. Attraction causes the potential energy to decrease. Repulsion causes the potential energy to increase. III/

3. 3 Covalent Bond sharing of electrons (non-metals +non-metals) H ● + ● H → H ● ● HH + H → H - H H HHH ● ● ● ● High probability of finding the electrons III/

4. 4 + + e-e- e-e- attraction repulsion attraction dominates repulsion dominates III/

5. 5 Covalent bond: sharing electrons (no loss, no gain) trying to reach octet (for H, duet, 2, like He) bonding electron pair Non-bonding Electron pairs, Lone electron pairs Lewis structure: shows how the atoms are bound together in a molecule. O = C = OH — O — H III/

6. 6 ElectronegativityCl ● ● ● ● ● ●● ● ● ● ●● ●● Cl ● ● ● ● ● ●● ● ● ● ●● ●● Equally shared Bonding electrons pure covalent bond Na Cl ● ● ● ● ● ●● ● ● ● ● ● ● ●● ● ● ● ● ● ● ●● ● + - Bonding electrons completely attracted to Cl pure ionic bond H Cl ● ● ● ● ● ●● ● ● ● ● ● ● ●● ● shared electrons attracted more to one atom polar covalent bond ● ● rHrH r Cl expected BL=r H +r Cl ● ● actual BL < r H +r Cl ++ -- Dipole Attraction between the two poles (distance between the two nuclei become shorter). III/

7. 7 Dipole moment  : -- ++ vector ℓ ++ A covalent bond is polar when it has a dipole moment!!! A dipole moment exists if there is an electronegativity difference between the two bonded atoms. Electronegativity  Ability of an atom in a particular molecule to attract the bonding electrons to itself.  ↑ III/30

8. 8 Arrange the following molecules in order of decreasing polarity: H-FH-ClHBrHI  F >  Cl >  Br >  I >  H H→FH→ClH→BrH→I  +→  -  +→  -  (H-F) >  (H-Cl) >  (H-Br) >  (H-I)  (H-F) >  (H-Cl) >  (H-Br) >  (H-I) polarity HF > HCl > HBr > HI  ↑  ↑ polarity ↑ III/

9. 9 Polarity of molecules A molecule is polar if it has a net dipole moment. Sum of all bonds dipole moments ≠ zero. CO 2 O = C = O -- -- ++ 11 22  total =  1 +  2 = 0 nonpolar H2OH2O O H -- ++ ++ 22 11  total =  1 +  2 ≠ 0  total polar III/

10. 10 Ionic compounds consist of ions (cations + anions): - cations come from metals. - anions come from nonmetals. Examples: LiF (Li +, F - ) ; CaO (Ca 2+, O 2- ) ; AlCl 3 (Al 3+, Cl - ) ; Mg 3 N 2 (Mg 2+, N 3- ) Charge of cations = charge of anions 1 Al 3+ = 3 Cl - 3 Mg 2+ = 2 N 3- Strong electrostatic forces dominate - Attraction between cations-anions - Repulsion between anions-anions ; cations-cations III/

11. 11 Sodium chloride lattice Lattice energy:  Energy needed to separate the crystal into separate ions (i.e. no interaction between them):  NaCl (s) → Na + (g) + Cl - (g)  The lattice energy is the crucial factor responsible for the stability of ionic compounds. Lattice: Arrangement in a periodic manner giving ordered structure. III/

12. 12 Why metals form cations: Small I.E. (easy to lose electrons) Only slightly negative E.A. (addition of electrons not favorable) By losing electrons, metals come easily to the stable noble gas configuration: Li (1s 2 2s 1 ) → Li + (1s 2, He) + 1e - Li (1s 2 2s 1 ) +7e - → Li 7- (1s 2 2s 2 2p 6, Ne) Why nonmetals form anions: Largre I.E. (difficult to lose electrons) Largely negative E.A. (addition of electrons favorable) By adding electrons, metals come easily to the stable noble gas configuration: F (1s 2 2s 2 2p 5 ) + 1e - → F - (1s 2 2s 2 2p 6, Ne) F (1s 2 2s 2 2p 5 ) → F 7+ (1s 2, He) + 7e- Octet Rule: When metals or nonmetals of the A group react, they often tend to gain or lose electrons until there are 8 electrons (no. of electrons in the outer shell of noble gases, except He) in the outer main shell. III/

13. 13 Failure of the octet rule A-groups: Sn: [Kr] 5s 2 4d 10 5p 2 Sn 2+ : [Kr] 5s 2 4d 10 Sn 4+ : [Kr] 4d 10 Pb: similar to Sn → Pb 2+ and Pb 4+ B-groups: Fe: [Ar] 4s 2 3d 6 Fe 2+ : [Ar] 3d 6 Fe 3+ : [Ar] 3d 5 Zn: [Ar] 4s 2 3d 10 Zn 2+ : [Ar] 3d 10 * [Ne] 3s 2 3p 6 3d 10 Cu: [Ar] 4s 1 3d 10 Cu + : [Ar] 3d 10 * Cu 2+ : [Ar] 3d 9 * Pseudo-noble gas configuration: all subshells in outermost shell filled. III/

14. 14 II/

15. 15 II/

16. 16 II/

17. 17 II/

18. 18 Lewis Symbols  A Lewis Symbol consists of the element symbol surrounded by "dots" to represent the number of electrons in the outer energy level (valence electrons).  The number of electrons in the outer energy level is correlated by simply reading the Group number. III/

19. 19 Rules for drawing Lewis structures 1. Count the valence electrons of all atoms (add one for each negative charge, subtract one for each positive charge). 2. Determine the central atom (that supposed to make the largest number of bonds, usually the atom with smallest number). Place the other atoms around the central atom. 3.Make single bonds between each surrounding atom and the central atom. Subtract two electrons for each bond. 4.Complete the octet for each of the surrounding atoms (for H only two electrons). 5.Place any leftover electrons on the central atom as pairs. 6.If the central atom has less than octet, use lone electron pairs on the surrounding atom to make double or triple bonds. III/

20. 20 H2OH2O 1. Count the valence electrons of all atoms: 2 H + O 2x1 + 6 = 8 Ve - 2. Determine the central atom. O H H 3. Make single bonds. 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 8 - 4 4 6. If the central atom has less than octet. - 4 0 III/

21. 21 CCl 4 1. Count the valence electrons of all atoms: 4 Cl + C 4x7 + 4 = 32 Ve - 2. Determine the central atom. C Cl 3. Make single bonds. 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 32 - 8 24 6. If the central atom has less than octet. - 24 0 Cl III/

22. 22 CO 2 1. Count the valence electrons of all atoms: 2 O + C 2x6 + 4 = 16 Ve - 2. Determine the central atom. C O O 3. Make single bonds. 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 16 - 4 12 6. If the central atom has less than octet, make multiple bonds. - 12 0 C O O C O O III/

23. 23 HCN H + C + N 1 + 4 + 5 = 10 Ve - HCNHCN 10 - 4 6 - 6 0 HCNHCN HCNHCN N2N2 2 N 2x5 = 10 N 10 - 2 8 - 6 2 - 2 0N N III/

24. 24 BeCl 2 Be + 2 Cl 2 + 2x7 = 16 Ve - ClBeCl 16 - 4 12 -12 0 BF 3 B + 3 F 3 + 3x7 = 24 Ve - F F B F 24 - 6 18 -18 0 BF 4 - B + 4 F + 1 3 + 4x7 = 32 Ve - F F B F F  Be does not obey the octet rule.  satisfied with 4 electrons in outer shell.  may have octet, not allowed to exceed octet.  B does not obey the octet rule.  satisfied with 6 electrons in outer shell.  may have octet, not allowed to exceed octet. III/

25. 25 PCl 5 P + 5 Cl 5 + 5x7 = 40 Ve - Cl Cl P Cl 40 -10 30 -30 0 Cl  P does not obey the octet rule.  allowed to exceed octet (usually, 10).  But never less than octet. SF 6 S + 6 F 6 + 6x7 = 48 Ve - F FFFSFFFFFFSFFFF 48 -12 36 -36 0  S does not obey the octet rule.  allowed to exceed octet (usually, 12).  But never less than octet. III/

26. 26 Octet Rule 2 nd period: C, N, O, F obey strictly the octet rule. Be and B are satisfied with less than octet but never exceeds octet. Lower periods: Allowed to exceed octet (due to empty d-orbital) but never less than octet. III/

27. 27 Bond Length (BL): distance between the nuclei of atoms connected to each other by that bond. Bond energy (BE): energy needed to break a bond (separate the two bonded atoms). Bond order (BO): number of bonds connecting two atoms with each other. BO ↑BL ↓ BE ↑ H – C – C – HH – C = C – HH – C ≡ C – H H H HH H H ethaneetheneethyne 1 2 3 154 pm137 pm120 pm 348 607 833 BO BL BE ( kJ/mol ) III/

28. 28 Resonance: NO 2 - N + 2 O + 1 5 + 2x6 + 1 = 18 Ve - ONOONO 18 - 4 14 -12 2 - 2 0 ONOONO BO 1 2 BL long short Experimentally:  the two bonds are equally long!!!  bond order greater than one, smaller than 2.  both structures are wrong ONOONO BO 2 1 BL short long ONOONOONOONO ↔ The real structure is an average of the two structures: III/

29. 29 ONOONOONOONO ↔ ONOONO Bond order = Number of all bonds between the two atoms in all resonance structures The number of resonance structures BO = 3/2 = 1.5 BO = 1 + number of  bonds number surrounding atoms the  bond is moving through ONOONO The first bond is always of the  type The second/third bonds in a multiple bond are of the  type Resonance structures BO= 1 + 1212 III/

30. 30 CO 3 2- C + 3 O + 2 4 + 3x6 + 2 = 24 Ve - OCOOOCOOO 24 - 6 18 -18 0 OCOOOCOOO OCOOOCOOO OCOOOCOOO ↔↔ BO = 4/3 = 1.33 BO = 1 + 1/3 = 1.33 III/

31. 31 Formal Charge: Draw the Lewis structure of SO 4 2- : S O O O O Octet on all atoms!  However, according to above structure, the order of the S-O bond is 1.  Experimental evidence indicates that the S-O bond length is shorter than that expected for BO=1.  2 > BO > 1 Question: Since the bonding electrons are shared equally between the two atoms, then what was the number of electrons that belonged to each atom before bonding? To know, break the bonds homolytically: III/

32. 32 Formal charge (FC) = no. of valence electrons the atom actually had before bonding – no. of electrons belonging to that atom after breaking the bond homolytically. FC = Ve - in isolated atom – Ve - of atom in molecule FC = no. of group of atom – Ve - of atom in molecule For C in CH 4 : FC = 4 – 4 =0For H in CH 4 : FC = 1 – 1 =0 III/

33. 33 S O O O O FC = no. of group – no. of bonds to that atom – no. electrons in lone pairs FC (S) = 6 – 6 – 0 = 0 FC (O) = 6 – 2 – 4 = 0 FC (O) = 6 – 1 – 6 = -1 2+ - - - - Sum of all FC = charge of species. The right Lewis structure is that with the least formal charges. S O O O O - - BO (S-O)= 1 + 2424 = 1.5 Remember: S can exceed the octet!! Resonance! All bonds are Equally long. FC (S) = 6 – 4 – 0 = +2 FC (O) = 6 – 1 – 6 = -1 III/

34. 34 Which structure is the right one for POCl 3 ? P Cl O P O FC (P) = 5 – 5 – 0 = 0 FC (O) = 6 – 2 – 4 = 0 FC (Cl) = 7 – 1 – 6 = 0 FC (P) = 5 – 4 – 0 = +1 FC (O) = 6 – 1 – 6 = -1 FC (Cl) = 7 – 1 – 6 = 0 - +   III/

35. 35 In which one of the following species is the Cl-O bond the shortest? ClO - ClO 2 - ClO 3 - ClO 4 - Strategy to solve the problem:  Following the rules, draw the Lewis structure.  Check for formal charges, try to minimize by making multiple bonds.  For the “right” Lewis structure, take Resonance into consideration.  Determine the Bond Order.  Relate Bond Order to Bond Length. III/

36. 36 Coordinative Covalent Bond Dative Bond The two bond electrons come from one atom and no electrons from the other. F F B F H N H H F F B F H N H H III/