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CHEMICAL BONDING

CHEMICAL BONDING

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CHEMICAL BONDING

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  1. CHEMICAL BONDING Cocaine

  2. Chemical Bonding Problems and questions — How is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? Can we predict the structure? How is structure related to chemical and physical properties?

  3. Forms of Chemical Bonds • There are 2 extreme forms of connecting or bonding atoms: • Ionic—complete transfer of 1 or more electrons from one atom to another • Covalent—some valence electrons shared between atoms • Most bonds are somewhere in between.

  4. Ionic Bonds Essentially complete electron transfer from an element oflow IE (metal)to an element ofhigh affinityfor electrons(nonmetal) 2 Na(s) + Cl2(g) ---> 2 Na+ + 2 Cl- Therefore, ionic compds. exist primarily between metals at left of periodic table (Grps 1A and 2A and transition metals) and nonmetals at right (O and halogens).

  5. Covalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons.Electron sharingresults. (Screen 9.5) Bond is a balance of attractive and repulsive forces.

  6. Chemical Bonding: Objectives Objectivesare to understand: 1. valence e- distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties.

  7. G. N. Lewis 1875 - 1946 Electron Distribution in Molecules • Electron distribution is depicted withLewis electron dot structures • Valence electrons are distributed as shared orBOND PAIRS and unshared orLONE PAIRS.

  8. •• H Cl • • •• lone pair (LP) shared or bond pair Bond and Lone Pairs • Valence electrons are distributed as shared orBOND PAIRS and unshared orLONE PAIRS. This is called a LEWIS ELECTRON DOT structure.

  9. •• •• Cl H H Cl • • + • • •• •• Bond Formation A bond can result from a “head-to-head”overlapof atomic orbitals on neighboring atoms. Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

  10. Valence Electrons Electrons are divided between core and valence electrons B 1s2 2s2 2p1 Core = [He] , valence = 2s2 2p1 Br [Ar] 3d10 4s2 4p5 Core = [Ar] 3d10 , valence = 4s2 4p5

  11. Rules of the Game •For Groups 1A-4A (14), no. of bond pairs = group number. No. of valence electrons of a main group atom = Group number • For Groups 5A (15)-7A (17), BP’s = 8 - Grp. No.

  12. Rules of the Game •No. of valence electrons of an atom = Group number •For Groups 1A-4A (14), no. of bond pairs = group number • For Groups 5A (15)-7A (17), BP’s = 8 - Grp. No. •Except for H (and sometimes atoms of 3rd and higher periods), BP’s + LP’s = 4 This observation is called the OCTET RULE

  13. Building a Dot Structure Ammonia, NH3 1. Decide on the central atom (the atom with lowest electron affinity); never H. Hydrogen atoms are always terminal. Therefore, N is central 2. Count valence electrons H = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons; 4 pairs

  14. H H N H •• H H N H Building a Dot Structure 3. Form a single bond between the central atom and each surrounding atom 4. Remaining electrons form LONE PAIRS to complete octet as needed. 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair.

  15. Sulfite ion, SO32- Step 1. Central atom = S Step 2. Count valence electrons S = 6 3 x O = 3 x 6 = 18 Negative charge = 2 TOTAL = 26 e- or 13 pairs Step 3. Form bonds 10 pairs of electrons are now left.

  16. •• O • • • • •• •• O S O • • • • •• •• Sulfite ion, SO32- Remaining pairs become lone pairs, first on outside atoms and then on central atom. •• Each atom is surrounded by an octet of electrons.

  17. Carbon Dioxide, CO2 1. Central atom = _______ 2. Valence electrons = __ or __ pairs 3. Form bonds. This leaves 6 pairs. 4. Place lone pairs on outer atoms.

  18. Carbon Dioxide, CO2 4. Place lone pairs on outer atoms. 5. So that C has an octet, we shall form DOUBLE BONDS between C and O. The second bonding pair forms api (π)bond.

  19. H2CO Double and even triple bonds are commonly observed for C, N, P, O, and S SO3 C2F4

  20. OR bring in bring in right pair left pair •• •• •• O S O • • • • •• •• Sulfur Dioxide, SO2 1. Central atom = S 2. Valence electrons = 18 or 9 pairs 3. Form double bond so that S has an octet — but note that there are two ways of doing this.

  21. Sulfur Dioxide, SO2 This leads to the following structures. These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is aHYBRIDof the two.

  22. Urea, (NH2)2CO

  23. Urea, (NH2)2CO 1. Number of valence electrons = 24 e- 2. Draw sigma bonds.

  24. Urea, (NH2)2CO 3. Place remaining electron pairs in the molecule.

  25. Urea, (NH2)2CO 4. Complete C atom octet with double bond.

  26. Violations of the Octet Rule Boron Trifluoride • Central atom = _____________ • Valence electrons = __________ or electron pairs = __________ • Assemble dot structure The B atom has a share in only 6 pairs of electrons (or 3 pairs). B atom in many molecules is electron deficient.

  27. Violations of the Octet Rule Sulfur Tetrafluoride, SF4 • Central atom = • Valence electrons = ___ or ___ pairs. • Form sigma bonds and distribute electron pairs. 5 pairs around the S atom. A common occurrence outside the 2nd period.

  28. Violations of the Octet Rule Odd # of electrons, NO2 • Central atom = • Valence electrons = ___ or ___ pairs. • Form sigma bonds and distribute electron pairs. • • •• •• N O N O • • •• •• •• O O •• • • •• ••

  29. Formal Atomic Charges Definition of Formal Charge: • Formal charge= Group no. – 1/2 BEs - LPEs

  30. 6 - ( 1 / 2 ) ( 4 ) - 4 = 0 • • • • O C O • • • • 4 - ( 1 / 2 ) ( 8 ) - 0 = 0 Carbon Dioxide, CO2

  31. Calculated Partial Charges in CO2 Yellow = negative&red = positive Relative size = relative charge

  32. • S C N • • • • • • Thiocyanate Ion, SCN- 6 - (1/2)(2) - 6 = -1 5 - (1/2)(6) - 2 = 0 4 - (1/2)(8) - 0 = 0

  33. • S C N • • • • • • Calculated Partial Charges in SCN- All atoms negative, but most on the S

  34. • • • • • S C N S C N • • • • • • • • • • • • S C N • • • • • • Thiocyanate Ion, SCN- Which is the most important resonance form?

  35. F • • • • •• F B • • •• F • • • • •• Boron Trifluoride, BF3 +1 -1 What if we form a B—F double bond to satisfy the B atom octet?

  36. MOLECULAR GEOMETRY

  37. MOLECULAR GEOMETRY Molecule adopts the shape that minimizes the electron pair repulsions. VSEPR • Valence Shell Electron Pair Repulsion theory. • Most important factor in determining geometry is relative repulsion between electron pairs.

  38. Electron Pair GeometriesFigure 9.12

  39. Electron Pair GeometriesFigure 9.12

  40. Structure Determination by VSEPR Ammonia, NH3 There are 4 electron pairs at the corners of a tetrahedron. The ELECTRON PAIR GEOMETRY is tetrahedral.

  41. Bond Propertiesbond order, bond length, bond energy, bond polarity Buckyball in HIV-protease

  42. Double bond Single bond Triple bond Bond Order # of bonds between a pair of atoms Acrylonitrile

  43. Bond Order Fractional bond ordersin resonance structures. Consider NO2- The N—O bond order = 1.5

  44. 414 kJ 123 pm 745 kJ 110 pm Bond Order Bond order is proportional to two important bond properties: (a) bond strength (b) bond length

  45. Bond Length • Bond length is the distance between the nuclei of two bonded atoms.

  46. Bond Length Bond length depends on bond order. Bond distances measured using CAChe software. In Angstrom units where 1 A = 10-2 pm.

  47. Using Bond Energies Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl Net energy = ∆Hrxn = = energy required to break bonds - energy evolved when bonds are made H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol

  48. Using Bond Energies Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol Sum of H-H + Cl-Cl bond energies = 436 kJ + 242 kJ = +678 kJ 2 mol H-Cl bond energies = 864 kJ Net = ∆H = +678 kJ - 864 kJ = -186 kJ

  49. Molecular Polarity Boiling point = 100 ˚C Why do ionic compounds dissolve in water? Boiling point = -161 ˚C Why do water and methane differ so much in their boiling points?