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Chemical Bonding

Chemical Bonding. Compounds Ionic molecular consist of ions sharing of electrons Ionic bond covalent bond. III/. Energy: Ability to do work. Work=force x distance. force causes the object to move. Gravitational force causes the water to fall. can generate electricity.

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Chemical Bonding

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

  2. Energy: Ability to do work. Work=force x distance force causes the object to move • Gravitational force causes the water to fall. • can generate electricity • 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! Epot=mgh Attraction causes the ball to fall, h smaller, Epot smaller. Attraction causes the potential energy to decrease. Repulsion causes the potential energy to increase. III/

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

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

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

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

  7. ℓ d- d+ Dipole moment m: vector d+ 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 ( c ) :Ability of an atom in a particular molecule to attract the bonding electrons to itself. c ↑ c ↑ III/30

  8. Dc↑ m↑ polarity ↑ • Arrange the following molecules in order of decreasing polarity: • H-F H-Cl HBr HI cF > cCl > cBr > cI > cH • H→F H→Cl H→Br H→I d+→d- d+→d- d+→d- d+→d- Dc(H-F) > Dc(H-Cl) > Dc(H-Br) > Dc(H-I) m(H-F) > m(H-Cl) > m(H-Br) > m(H-I) polarity HF > HCl > HBr > HI III/

  9. mtotal Polarity of molecules • A molecule is polar if it has a net dipole moment. • Sum of all bonds dipole moments ≠ zero. H2O CO2 d- d- d+ d- O H H O = C = O m1 m2 d+ d+ m2 m1 mtotal = m1 + m2 ≠ 0 mtotal = m1 + m2 = 0 polar nonpolar III/

  10. Ionic compounds • consist of ions (cations + anions): -cations come from metals. - anions come from nonmetals. • Examples: LiF (Li+ , F-) ; CaO (Ca2+, O2-) ; AlCl3 (Al3+, Cl-) ; Mg3N2 (Mg2+ , N3-) • Charge of cations = charge of anions 1Al3+ = 3 Cl- 3 Mg2+ = 2 N3- • Strong electrostatic forces dominate - Attraction between cations-anions - Repulsion between anions-anions ; cations-cations III/

  11. Sodium chloride lattice Lattice: Arrangement in a periodic manner giving ordered structure. • 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. III/

  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 (1s2 2s1) → Li+ (1s2, He) + 1e- • Li (1s2 2s1) +7e- → Li7- (1s2 2s2 2p6, 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 (1s2 2s2 2p5) + 1e- → F- (1s2 2s2 2p6, Ne) • F (1s2 2s2 2p5) → F7+(1s2, 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. Failure of the octet rule • A-groups: • Sn: [Kr] 5s2 4d10 5p2 Sn2+: [Kr] 5s2 4d10 Sn4+: [Kr] 4d10 • Pb: similar to Sn → Pb2+ and Pb4+ • B-groups: • Fe: [Ar] 4s2 3d6 Fe2+: [Ar] 3d6 Fe3+: [Ar] 3d5 • Zn: [Ar] 4s2 3d10 Zn2+: [Ar] 3d10 * [Ne] 3s2 3p6 3d10 • Cu: [Ar] 4s1 3d10 Cu+: [Ar] 3d10 * Cu2+: [Ar] 3d9 *Pseudo-noble gas configuration: all subshells in outermost shell filled. III/

  14. II/

  15. II/

  16. II/

  17. II/

  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. 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. H2O 2 H + O 2x1 + 6 = 8 Ve- 1. Count the valence electrons of all atoms: 8 - 4 O H H 2. Determine the central atom. 4 - 4 3. Make single bonds. 0 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 6. If the central atom has less than octet. III/

  21. CCl4 4 Cl + C 4x7 + 4 = 32 Ve- 1. Count the valence electrons of all atoms: Cl 32 - 8 C Cl Cl 2. Determine the central atom. 24 - 24 Cl 3. Make single bonds. 0 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 6. If the central atom has less than octet. III/

  22. C O O C O O CO2 2 O + C 2x6 + 4 = 16 Ve- 1. Count the valence electrons of all atoms: 16 - 4 C O O 2. Determine the central atom. 12 - 12 3. Make single bonds. 0 4. Complete the octet for each of the surrounding atoms. 5. Place any leftover electrons on the central atom. 6. If the central atom has less than octet, make multiple bonds. III/

  23. H C N H C N N N N N H + C + N 1 + 4 + 5 = 10 Ve- HCN 10 - 4 6 H C N - 6 0 10 N2 2 N 2x5 = 10 - 2 8 N N - 6 2 - 2 0 III/

  24. F F B F F 16 - 4 • Be does not obey the octet rule. • satisfied with 4 electrons in outer shell. • may have octet, not allowed to exceed octet. BeCl2 Be + 2 Cl 2 + 2x7 = 16 Ve- 12 -12 Cl Be Cl 0 24 - 6 BF3 B + 3 F 3 + 3x7 = 24 Ve- • B does not obey the octet rule. • satisfied with 6 electrons in outer shell. • may have octet, not allowed to exceed octet. 18 F F B F -18 0 BF4- B + 4 F + 1 3 + 4x7 = 32 Ve- III/

  25. 40 -10 PCl5 P + 5 Cl 5 + 5x7 = 40 Ve- • P does not obey the octet rule. • allowed to exceed octet (usually, 10). • But never less than octet. 30 Cl Cl P Cl -30 0 Cl Cl SF6 48 -12 S + 6 F 6 + 6x7 = 48 Ve- • S does not obey the octet rule. • allowed to exceed octet (usually, 12). • But never less than octet. 36 F F F S F F F -36 0 III/

  26. Octet Rule • 2nd 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. 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 H H – C – C – H H – C = C – H H – C ≡ C – H H H H H ethane ethene ethyne 1 2 3 154 pm 137 pm 120 pm 348 607 833 BO BL BE (kJ/mol) III/

  28. O N O O N O O N O ↔ O N O Resonance: 18 - 4 N + 2 O + 1 5 + 2x6 + 1 = 18 Ve- NO2- 14 -12 2 O N O - 2 0 BO 2 1 BLshort long BO 1 2 BLlong short • Experimentally: • the two bonds are equally long!!! • bond order greater than one, smaller than 2. • both structures are wrong The real structure is an average of the two structures: III/

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

  30. O C O O O C O O O C O O 24 - 6 CO32- C + 3 O + 2 4 + 3x6 + 2 = 24 Ve- 18 O C O O -18 0 ↔ ↔ BO = 4/3 = 1.33 BO = 1 + 1/3 = 1.33 III/

  31. O S O O O Formal Charge: Draw the Lewis structure of SO42-: 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. 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 CH4: FC = 4 – 4 =0 For H in CH4: FC = 1 – 1 =0 III/

  33. O S O O O O S O O O FC = no. of group – no. of bonds to that atom – no. electrons in lone pairs - FC (S) = 6 – 4 – 0 = +2 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. FC (S) = 6 – 6 – 0 = 0 FC (O) = 6 – 2 – 4 = 0 FC (O) = 6 – 1 – 6 = -1 - Resonance! All bonds are Equally long. 2 4 BO (S-O)= 1 + = 1.5 - Remember: S can exceed the octet!! III/

  34. - + Which structure is the right one for POCl3? O O Cl Cl Cl P Cl P Cl Cl 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. In which one of the following species is the Cl-O bond the shortest? ClO- ClO2- ClO3- ClO4- • 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. Coordinative Covalent BondDative Bond • The two bond electrons come from one atom and no electrons from the other. F F B F F F B F H N H H H N H H III/

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