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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.

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Presentation on theme: "1 CHEMICAL BONDING Cocaine. 2 Chemical Bonding Problems and questions — How is a molecule or polyatomic ion held together? Why are atoms distributed at."— Presentation transcript:

1 1 CHEMICAL BONDING Cocaine

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

4 4 Ionic Bonds Essentially complete electron transfer from an element of low IE (metal) to an element of high affinity for electrons (nonmetal) 2 Na(s) + Cl 2 (g) ---> 2 Na + + 2 Cl - 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 5 Covalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. (Screen 9.5) Bond is a balance of attractive and repulsive forces.

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

7 7 Electron Distribution in Molecules Electron distribution is depicted with Lewis electron dot structuresElectron distribution is depicted with Lewis electron dot structures Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. G. N. Lewis 1875 - 1946

8 8 Bond and Lone Pairs Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. HCl lone pair (LP) shared or bond pair This is called a LEWIS ELECTRON DOT structure.

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

10 10 Valence Electrons Electrons are divided between core and valence electrons B 1s 2 2s 2 2p 1 Core = [He], valence = 2s 2 2p 1 Br [Ar] 3d 10 4s 2 4p 5 Core = [Ar] 3d 10, valence = 4s 2 4p 5

11 11 Rules of the Game No. of valence electrons of a main group 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. For Groups 5A (15)-7A (17), BP’s = 8 - Grp. No.

12 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.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 13 Building a Dot Structure Ammonia, NH 3 1. Decide on the central atom (the atom with lowest electron affinity); never H. Hydrogen atoms are always terminal. Hydrogen atoms are always terminal. Therefore, N is central Therefore, N is central 2. Count valence electrons H = 1 and N = 5 H = 1 and N = 5 Total = (3 x 1) + 5 Total = (3 x 1) + 5 = 8 electrons; 4 pairs = 8 electrons; 4 pairs

14 14 3.Form a single bond between the central atom and each surrounding atom H H H N Building a Dot Structure H H H N 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 15 10 pairs of electrons are now left. 10 pairs of electrons are now left. Sulfite ion, SO 3 2- Step 1. Central atom = S Step 2. Count valence electrons S = 6 3 x O = 3 x 6 = 18 3 x O = 3 x 6 = 18 Negative charge = 2 Negative charge = 2 TOTAL = 26 e- or 13 pairs Step 3. Form bonds

16 16 Sulfite ion, SO 3 2- Remaining pairs become lone pairs, first on outside atoms and then on central atom. OO O S Each atom is surrounded by an octet of electrons.

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

18 18 Carbon Dioxide, CO 2 4. Place lone pairs on outer atoms. The second bonding pair forms a pi (π) bond. 5. So that C has an octet, we shall form DOUBLE BONDS between C and O.

19 19 Double and even triple bonds are commonly observed for C, N, P, O, and S H 2 CO SO 3 C2F4C2F4C2F4C2F4

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

21 21 Sulfur Dioxide, SO 2 This leads to the following structures. These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two. RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two.

22 22 Urea, (NH 2 ) 2 CO

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

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

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

26 26 Boron Trifluoride Central atom = _____________Central atom = _____________ Valence electrons = __________ or electron pairs = __________Valence electrons = __________ or electron pairs = __________ Assemble dot structureAssemble 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. Violations of the Octet Rule

27 27 Sulfur Tetrafluoride, SF 4 Central atom =Central atom = Valence electrons = ___ or ___ pairs.Valence electrons = ___ or ___ pairs. Form sigma bonds and distribute electron pairs.Form sigma bonds and distribute electron pairs. 5 pairs around the S atom. A common occurrence outside the 2nd period. Violations of the Octet Rule

28 28 Odd # of electrons, NO 2 Central atom =Central atom = Valence electrons = ___ or ___ pairs.Valence electrons = ___ or ___ pairs. Form sigma bonds and distribute electron pairs.Form sigma bonds and distribute electron pairs. O O N Violations of the Octet Rule O O N

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

30 30 OOC 4 - (1/2)(8) - 0 = 0 6 - (1/2)(4) - 4 = 0 Carbon Dioxide, CO 2

31 31 Calculated Partial Charges in CO 2 Yellow = negative & red = positive Relative size = relative charge Yellow = negative & red = positive Relative size = relative charge

32 32 Thiocyanate Ion, SCN - 6 - (1/2)(2) - 6 = -15 - (1/2)(6) - 2 = 0 4 - (1/2)(8) - 0 = 0 SNC

33 33 Thiocyanate Ion, SCN - SNC SNC SNC Which is the most important resonance form?

34 34 Calculated Partial Charges in SCN - All atoms negative, but most on the S SNC

35 35 Boron Trifluoride, BF 3 What if we form a B—F double bond to satisfy the B atom octet? F F F B ++ --

36 36 MOLECULAR GEOMETRY

37 37 VSEPR VSEPR V alence S hell E lectron P air R epulsion theory.V alence S hell E lectron P air R epulsion theory. Most important factor in determining geometry is relative repulsion between electron pairs.Most important factor in determining geometry is relative repulsion between electron pairs. Molecule adopts the shape that minimizes the electron pair repulsions. MOLECULAR GEOMETRY

38 38 Electron Pair Geometries Figure 9.12

39 39

40 40 Electron Pair Geometries Figure 9.12

41 41 Structure Determination by VSEPR Ammonia, NH 3 There are 4 electron pairs at the corners of a tetrahedron. The ELECTRON PAIR GEOMETRY is tetrahedral.

42 42 Bond Properties bond order, bond length, bond energy, bond polarity Buckyball in HIV-protease

43 43 Bond Order # of bonds between a pair of atoms # of bonds between a pair of atoms Bond Order # of bonds between a pair of atoms # of bonds between a pair of atoms Double bond Single bond Triple bond AcrylonitrileAcrylonitrile

44 44 Bond Order Fractional bond orders in resonance structures. Consider NO 2 - The N—O bond order = 1.5

45 45 Bond Order Bond order is proportional to two important bond properties: (a) bond strength (b)bond length 745 kJ 414 kJ 110 pm 123 pm

46 46 Bond Length Bond length is the distance between the nuclei of two bonded atoms.

47 47 Bond length depends on bond order. Bond distances measured using CAChe software. In Angstrom units where 1 A = 10 -2 pm. Bond Length

48 48 Using Bond Energies Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl Net energy = ∆H rxn = = 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 H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol

49 49 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 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 Using Bond Energies 2 mol H-Cl bond energies = 864 kJ Net = ∆H = +678 kJ - 864 kJ = -186 kJ

50 50 Molecular Polarity Why do ionic compounds dissolve in water? Boiling point = 100 ˚C Boiling point = -161 ˚C Why do water and methane differ so much in their boiling points?

51 51 Bond Polarity HCl is POLAR because it has a positive end and a negative end. Cl has a greater share in bonding electrons than does H. Cl has slight negative charge (-  ) and H has slight positive charge (+  )

52 52 Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond. BONDENERGY “pure” bond339 kJ/mol calc’d real bond432 kJ/mol measured BONDENERGY “pure” bond339 kJ/mol calc’d real bond432 kJ/mol measured Difference = 92 kJ. This difference is proportional to the difference in ELECTRONEGATIVITY, . Bond Polarity

53 53 Electronegativity,   is a measure of the ability of an atom in a molecule to attract electrons to itself. Concept proposed by Linus Pauling 1901-1994 Concept proposed by Linus Pauling 1901-1994

54 54 Linus Pauling, 1901-1994 The only person to receive two unshared Nobel prizes (for Peace and Chemistry). Chemistry areas: bonding, electronegativity, protein structure

55 55 Electronegativity Figure 9.9

56 56 F has maximum .F has maximum . Atom with lowest  is the center atom in most molecules.Atom with lowest  is the center atom in most molecules. Relative values of  determine BOND POLARITY (and point of attack on a molecule).Relative values of  determine BOND POLARITY (and point of attack on a molecule). Electronegativity,  See Figure 9.9

57 57 Bond Polarity Which bond is more polar (or DIPOLAR)? O—HO—F O—HO—F  3.5 - 2.13.5 - 4.0  1.40.5 OH is more polar than OF OH is more polar than OF and polarity is “reversed.”

58 58 Molecular Polarity Molecules—such as HCl and H 2 O— can be POLAR (or dipolar). They have a DIPOLE MOMENT. The polar HCl molecule will turn to align with an electric field. Figure 9.15

59 59 Molecular Polarity The magnitude of the dipole is given in Debye units. Named for Peter Debye (1884 - 1966). Rec’d 1936 Nobel prize for work on x-ray diffraction and dipole moments.

60 60 Dipole Moments Why are some molecules polar but others are not?

61 61 Molecular Polarity Molecules will be polar if a)bonds are polar AND AND b)the molecule is NOT “symmetric” All above are NOT polar

62 62 Polar or Nonpolar? Compare CO 2 and H 2 O. Which one is polar?

63 63 Carbon Dioxide CO 2 is NOT polar even though the CO bonds are polar.CO 2 is NOT polar even though the CO bonds are polar. CO 2 is symmetrical.CO 2 is symmetrical. +1.5-0.75-0.75 Positive C atom is reason CO 2 + H 2 O gives H 2 CO 3

64 64 Consequences of H 2 O Polarity Microwave oven

65 65 Polar or Nonpolar? Consider AB 3 molecules: BF 3, Cl 2 CO, and NH 3.

66 66 Molecular Polarity, BF 3 B atom is positive and F atoms are negative. B—F bonds in BF 3 are polar. But molecule is symmetrical and NOT polar

67 67 Molecular Polarity, HBF 2 B atom is positive but H & F atoms are negative. B—F and B—H bonds in HBF 2 are polar. But molecule is NOT symmetrical and is polar.

68 68 Is Methane, CH 4, Polar? Methane is symmetrical and is NOT polar.

69 69 Is CH 3 F Polar? C—F bond is very polar. Molecule is not symmetrical and so is polar.

70 70 Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.C—F bonds are MUCH more polar than C—H bonds. Because both C—F bonds are on same side of molecule, molecule is POLAR.Because both C—F bonds are on same side of molecule, molecule is POLAR.

71 71 Substituted Ethylene C—F bonds are MUCH more polar than C—H bonds.C—F bonds are MUCH more polar than C—H bonds. Because both C—F bonds are on opposing ends of molecule, molecule is NOT POLAR.Because both C—F bonds are on opposing ends of molecule, molecule is NOT POLAR.


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