Lecture B1 Lewis Dot Structures and Covalent Bonding

Transcription

1 Lecture B1 Lewis Dot Structures and Covalent Bonding

2 G.N. Lewis & Linus Pauling Two American Chemists G. N. Lewis Linus Pauling

3 The Covalent Bond 1. First proposed by G.N. Lewis in his seminal 1916 J. Am. Chem. Soc. publication (before QM was discovered!). 2. Lewis proposed that Covalent bonds consist of shared pairs of electrons. e created a powerful empirical formalism (Lewis dot structures) for understanding bonding in simple compounds. 3. Linus Pauling created a picture of covalent bonding that employed Quantum Mechanics (and won the 1954 Nobel Prize for it).

4 G.N. Lewis, circa the cubical atom.

5 G.N. Lewis, Journal of the American Chemical Society 38 (1916) 762.

6 G.N. Lewis, Journal of the American Chemical Society 38 (1916) 762. I2 molecule

7 ...we still think about covalent bonds the way G.N. Lewis did. In contrast to ionic bond, electrons are shared, not transferred......if the atoms are identical, then electrons are equally shared - an example of a perfectly nonpolar covalent bond.

8 In covalent bond formation, atoms go as far as possible toward completing their octets by sharing electron pairs. Covalent bonds are drawn as lines. Lone pairs are drawn as pairs of dots. I I I2 molecule

9 Lewis dot structures provide a simple, but extremely powerful, formalism for representing covalent bonding in molecules. a single bond, consisting of a pair of electrons.

10 Lewis dot structures provide a simple, but extremely powerful, formalism for representing covalent bonding in molecules. a nonbonding pair of electrons: a lone pair

11 Lewis dot structures provide a simple, but extremely powerful, formalism for representing covalent bonding in molecules. Note: no geometric information!

12 Lecture B1 Systematic Method for Creating Lewis Dot Structures

13 Lewis Dot Structures can be produced by following a sequence of steps. Let s produce a Lewis Dot Structure for: N4 + (the ammonium ion). Step 1: Count valence electrons: N = 5 4 x = 4 x 1 = 4 + = -1 Total = 5+4-1= 8 electrons = 4 bonds and lone pairs.

14 Lewis Dot Structures can be produced by following a sequence of steps. Let s produce a Lewis Dot Structure for: N4 + (the ammonium ion). Step 1: Count valence electrons: N = 5 4 x = 4 x 1 = 4 + = -1 Total = 5+4-1= 8 electrons = 4 bonds and lone pairs. N Step 2: Arrange the atoms (identify a central atom, if possible). +

15 Lewis Dot Structures can be produced by following a sequence of steps. Let s produce a Lewis Dot Structure for: N4 + (the ammonium ion). Step 1: Count valence electrons: + N = 5 4 x = 4 x 1 = 4 + = -1 Total = 5+4-1= 8 electrons = 4 bonds and lone pairs. Step 2: Arrange the atoms (identify a central atom, if possible). N Step 3: Place electron pairs on the atoms as either bonds or lone pairs.

16 Lewis Dot Structures can be produced by following a sequence of steps. Let s produce a Lewis Dot Structure for: N4 + (the ammonium ion). Step 1: Count valence electrons: + N = 5 4 x = 4 x 1 = 4 + = -1 Total = 5+4-1= 8 electrons = 4 bonds and lone pairs. Step 2: Arrange the atoms (identify a central atom, if possible). N Step 3: Place electron pairs on the atoms as either bonds or lone pairs.

17 Lewis Dot Structures can be produced by following a sequence of steps. Let s produce a Lewis Dot Structure for: N4 + (the ammonium ion). Step 1: Count valence electrons: + N = 5 4 x = 4 x 1 = 4 + = -1 Total = 5+4-1= 8 electrons = 4 bonds and lone pairs. Step 2: Arrange the atoms (identify a central atom, if possible). N Step 3: Place electron pairs on the atoms as either bonds or lone pairs. Step 4: Count electrons and compare with known count. If octet rule is satisfied for all atoms and the electron count is right, you re done. 8 valence electrons around N!

18 Lecture B1 Lewis Dot Structures Multiple Bonds

19 G.N. Lewis, Journal of the American Chemical Society 38 (1916) 762.

20 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 1: Count the valence electrons: 3 x C = 12 6 x = 6 x 1 = 6 O = 6 Total = = 24 electrons = 12 bonds or lone pairs.

21 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 2: Arrange the atoms: Identify a central atom. If you have C or O to choose from, go with C. O C C C

22 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 3: Place 12 electron pairs on the atoms. O C C C

23 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 3: Place 12 electron pairs on the atoms. O C C C Nine Bonds, Three Lone Pairs

24 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 4: Check for the octet rule. oops -- only 6 e! O C C C Nine Bonds, Three Lone Pairs

25 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 4: Check for the octet rule. Better! O C C C Ten Bonds (one double bond), Two Lone Pairs Done!

26 ...another example: Let s produce a Lewis Dot structure for: acetone: (C3)2CO. Step 4: Check for the octet rule. O C C C This takes practice, but then becomes tedious. Getting the polyatomic arrangement can be problematic.

27 ow about Double 07 (Nitrogen)? Step 1: 10 valence electrons/five bonds and lone pairs

28 ow about Double 07 (Nitrogen)? Step 1: 10 valence electrons/five bonds and lone pairs Step 2: Arrange atoms. N N

29 ow about Double 07 (Nitrogen)? Step 1: 10 valence electrons/five bonds and lone pairs Step 2: Arrange atoms. Step 3: Fill in electron pairs and bonds. N N

30 ow about Double 07 (Nitrogen)? Step 1: 10 valence electrons/five bonds and lone pairs Step 2: Arrange atoms. Step 3: Fill in electron pairs and bonds. N N

31 ow about Double 07 (Nitrogen)? Step 1: 10 valence electrons/five bonds and lone pairs Step 2: Arrange atoms. Step 3: Fill in electron pairs and bonds. Step 4: Octet rule obeyed! N N A Triple Bond!

32 multiple bonds have higher bond dissociation energies...

33 A quantitative comparison of CC bonds N2 Bond Dissociation Energy: 941 kj mol -1

34 bond lengths are inversely correlated with bond energies - shorter bonds are stronger... Double 07 is very strong!

35 Lecture B1 Lewis Dot Structures Formal Charge

36 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs.

37 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S

38 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S Step 3: Fill in the electron pairs.

39 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S Step 3: Fill in the electron pairs.

40 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S Step 3: Fill in the electron pairs. Step 4: Check for the octet rule.

41 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S Step 3: Fill in the electron pairs. Step 4: Check for the octet rule. Good!

42 Let s produce a Lewis Dot structure for carbon disulfide, CS2. Step 1: Count valence electrons: C = 4 2 x S = 6 x 2 = 12 Total = = 16 e/8 electron pairs. Step 2: Arrange the atoms. Go with Carbon in the center. S C S Step 3: Fill in the electron pairs. Step 4: Check for the octet rule. BUT wait, what about?? C S S

43 To decide between these two candidate structures, we need to calculate the formal charges on the atoms.

44 Less formal charge is better. Thus, SCS wins! SCS wins!

45 Carbon dioxide, CO2, is isoelectronic with carbon disulfide, CS2.

46 Nitrous oxide, N2O, is also isoelectronic with carbon disulfide, CS2.

47 Lecture B1 Lewis Dot Structures Resonance

48 When multiple Lewis Dot Structures exist that are differentiated only by the positions of the electrons (the positions of the atoms are identical), the ACTUAL structure is a weighted average of these resonance structures. We predict that the three N-O bonds in the nitrate ion are identical in length.

49 Another Example: We predict that the two C-O bonds in the acetate ion are identical in length.

50 And Another Example: Benzene

51 Two resonance structures for benzene... We predict that all of the C-C bonds in benzene have the same length.

52 So normally, Chemists represent benzene like this:

53 Lecture B1 Lewis Dot Structures Exceptions to the Octet Rule

54 In the benzene, the actual structure is the average of the possible resonance structures, but this is often not the case when resonance structures with a range of formal charge values are possible. In this case, the actual structure will be a weighted average of the available resonance structures. A what? an example: Grades are often computed using a weighted average. Suppose that homework counts 10%, quizzes 20%, and tests 70%. If Pat has a homework grade of 92, a quiz grade of 68, and a test grade of 81, then Pat's overall grade = (0.10)(92) + (0.20)(68) + (0.70)(81) = 79.5 ow do we weight resonance structures?

55 An example: Which is the better Lewis Dot representation for the phosphate ion, PO4 3-?

56 An example: Which is the better Lewis Dot representation for the phosphate ion, PO4 3-? We have to look at the formal charge arguments here. The structure shown at right has fewer formal charges, but... P has 10 electrons - is this ok?

57 An example: Which is the better Lewis Dot representation for the phosphate ion, PO4 3-? If 10 electrons is okay for P, then less charge is better.

58 An example: Which is the better Lewis Dot representation for the phosphate ion, PO4 3-? If 10 electrons is okay for P, then less charge is better. P can use its d electrons to expand its valence shell. More than 8 electrons = "hypervalent"

59 What if a second double bond is added? - formal charge on that O goes from -1 to 0 - formal charge on P goes from 0 to -1 - total number of formal charges is unchanged. - electron count on P goes from 10 to 12. If 12 electrons is okay for P, then is it better to charge P or O?

60 Actually, all of these resonance structures contribute to the weighted average resonance hybrid that is observed. lower weighting factor higher weighting factor lower weighting factor We need more information to determine the exact weighting factors. Pauling will introduce the concept of electronegativity to help with this.

61 Final example: Rank these Lewis dot resonance structures for cyanate (NCO - ) from best to worst:

62 Final example: Rank these Lewis dot resonance structures for cyanate (NCO - ) from best to worst: 1 2 Again, Pauling's concept of electronegativity will help with this. 3

63 Another exception to the octet rule: 1. Be, B, and Li can be electron deficient. examples: note: no formal charges here...

64 Another exception to the octet rule: 1. Be, B, and Li can be electron deficient. examples: what about this resonance structure?

65 Another exception to the octet rule: 1. Be, B, and Li can be electron deficient. examples: a smaller weighting factor for this structure

66 Another exception to the octet rule: 2. Non metals in period 3 or higher can accommodate more than 8 electrons by employing d-orbitals. examples: We talked about this one before.

67 Another exception to the octet rule: 2. Non metals in period 3 or higher can accommodate more than 8 electrons by employing d-orbitals. examples: Sulfur can do this too.

68 Another exception to the octet rule: 2. Non metals in period 3 or higher can accommodate more than 8 electrons by employing d-orbitals. examples: Even Xenon gets in on the act!

69 Lecture B1 Bond Polarity and Electronegativity

70 Pauling realized that covalent bonds between different nuclei will not be symmetric in electron density (probability): Linus Pauling

71 This will lead to a pair of net positive and negative partial charges that will result in a DIPOLE MOMENT for the molecule: or dipole moment = charge x distance Linus Pauling

72 This will lead to a pair of net positive and negative partial charges that will result in a DIPOLE MOMENT for the molecule: dipole moment = charge x distance 1 Debye = 1D = x C m Named after Peter J. W. Debye Dutch Physicist Nobel Prize in Chemistry 1936 Peter J. W. Debye

73 This will lead to a pair of net positive and negative partial charges that will result in a DIPOLE MOMENT for the molecule: Cl Dipole Moment: 1.08D Linus Pauling

74 Pauling proposed a number to quantitate the ability of an atom to polarize a bond: "Pauling Electronegativity" χ (chi) Electronegativity ranges from 0.7 to 4.0. It is unitless!

75 Three terms that are often confused: Ionization energy: the energy required to remove an electron from an atom or ion. Electron affinity: the energy change associated with the addition of one electron to an atom or ion. Electronegativity: the ability of an atom in a molecule to draw bonding electrons to itself. Important: this is a dimensionless parameter - not an energy.

76 Dipole Moments and Pauling Electronegativity Cl Dipole Moment: 1.08D χcl - χ = = 0.9 Electron density is higher on the atom with higher electronegativity. Linus Pauling

77 Dipole Moments and Pauling Electronegativity F Dipole Moment: 1.83D χf - χ = = 1.9 Cl Dipole Moment: 1.08D χcl - χ = = 0.9 Br Dipole Moment: 0.82D χbr - χ = = 0.7 Electronegativity differences correlate with dipole moment values. Linus Pauling

78 Rank these Lewis dot resonance structures for cyanate (NCO - ) from best to worst: 1 2 We can go back and use electronegativity to rank Lewis dot resonance structures. χn = 3.0 and χo = 3.5. Thus the negative charge prefers to reside on the Oxygen. 3

79 Generalizing, any bond will possess a bond dipole moment with a magnitude that is proportional to the difference in electronegativity between the bonding partners. + N For example, even though N4 + has no dipole moment, the four covalent bonds in the ammonium ion are between two elements with different electronegativities: χn - χ = = 0.9, and therefore EAC has a bond dipole moment. The fact that the four dipole moments cancel is a matter of geometric structure. That is the NEXT topic -- VSEPR Theory!