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Chapter #10 Chemical Bonding. CHAPTER 12 Forces Between Particles  Noble Gas Configurations  Ionic Bonding  Covalent Bonding  VSEPR Theory and Molecular.

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Presentation on theme: "Chapter #10 Chemical Bonding. CHAPTER 12 Forces Between Particles  Noble Gas Configurations  Ionic Bonding  Covalent Bonding  VSEPR Theory and Molecular."— Presentation transcript:

1 Chapter #10 Chemical Bonding

2 CHAPTER 12 Forces Between Particles  Noble Gas Configurations  Ionic Bonding  Covalent Bonding  VSEPR Theory and Molecular Geometry  Electronegativity  Polar Bonds and Molecules

3 It has been recognized for a long time that the noble gases have great chemical stability. With few exceptions they are unreactive or inert. The noble gases have 8 valence electrons with the exception of He which has 2. He1s 2 Ne1s 2 2s 2 2p 6 Ar1s 2 2s 2 2p 6 3s 2 3p 6 Kr1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 Xe1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 Atomic Stability

4 The electronic configuration of the noble gases is described as being energetically stable. We can draw a Lewis diagram to illustrate the number of valence electrons an atom has. In a Lewis diagram valence electrons are represented by dots placed above, below and to the left and right of the atoms symbol. E e.g. element with 4 valence electrons Lewis Diagrams

5 There are two simple rules to keep in mind when drawing Lewis diagrams: Place one dot in each of the four locations before doubling up. There can be only a maximum of 2 dots in any one location. EE  EE  Lewis Diagrams

6 What is the Lewis diagram for H? H 1.First write the electron configuration: 1s 1 2.Identify the number of valence electrons. 1 valence electron. For a representative element it is easy to identify the number of valence electrons as this is equal to the group number. Lewis Diagrams

7 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

8 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

9 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

10 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

11 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

12 What is the Lewis diagram for S? 1.First write the electron configuration: [Ne]3s 2 3p 4 2.Identify the number of valence electrons. 6 valence electrons S Alternatively you can recognize that S is in group VIA so has six valence electrons Lewis Diagrams

13 LEWIS STRUCTURES OF THE ELEMENTS 1 2 13 14 15 16 17 18 H Li Na Be Mg B Al C Si P N O F Cl S He Ne Ar

14 LEWIS STRUCTURES OF IONS (AFTER REMOVAL OR ADDITION OF ELECTRONS) 1 2 13 14 15 16 17 18 2+ 4- 3- 2- 1- H Li Na Be Mg B Al C Si P N O F Cl S He Ne Ar 1+ 3+

15 The octet rule states that: “Atoms interact in order to obtain a stable octet of eight valence electrons” The octet rule works extremely well at describing the interactions of the representative elements. Lewis Diagrams

16 One way in which atoms can interact to satisfy the octet rule is by transferring electrons between each other. Transferring of electrons results in the atoms acquiring net positive and negative charges. When an atom loses or gains electrons a simple ion is formed. Cations have more protons than electrons and are positive. Anions have more electrons than protons and are negative. Lewis Diagrams

17 Na [Ne]3s 1 Na + +1e - Cl +1e - [Ne]3s 2 3p 5 Cl Consider a Na atom what happens if it loses one electron? Consider a Cl atom would you expect it to lose or gain electrons? [Ne] [Ne]3s 2 3p 6 11 P and 11 e - 11 P and 10 e - I.E. E.A. 17 P and 17 e- 17 P and 18 e- Ion Formation -

18 Metals tend to lose electrons forming positively charged ions called cations. A representative metal will lose its group number of electrons to obtain a stable octet. Na→Na + +1e - ( Isoelectronic with Ne) Mg→Mg 2+ +2e - (isoelectronic with Ne) What would the charge be of the ion formed by a Li atom? And which Noble gas is it isoelectronic with? +1 The ion formed would be Li + Isoelectronic with He Ion Formation

19 Non-metals tend to gain electrons forming negatively charged ions called anions. A representative non-metal will gain (8 - group number) electrons to obtain a stable octet. O+2e - →O 2- (isoelectronic with Ne) S+2e - →S 2- (isoelectronic with Ar) What would the charge be of the ion formed by a I atom? Which Noble gas is it isoelectronic with? -1 The ion formed would be I - Isoelectronic with Xe Ion Formation

20 Lewis Structure of NaCl Na + Cl - Na + Cl - Na + Cl - Cl - Na + Cl - Na + Cl - Na + Forces between oppositely charged ions are called Ionic bonds. Each ion is surrounded by an octet of Electrons, thus making the ions stable.

21 Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites. Ions combine in the ratio that results in zero charge to form ionic compounds. Which ions are the smaller ones? Crystal Lattice of NaCl

22 Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites. Ions combine in the ratio that results in zero charge to form ionic compounds. Which ions are the smaller ones? Sodium Crystal Lattice of NaCl

23 In our early lectures we defined a molecule as “as a compound made of nonmetals.” Molecules exist as particles containing the number of atoms specified by their formula. e.g. a water molecule is a particle containing 2 hydrogen atoms and one oxygen atom and has the formula H 2 O. Molecular Compounds

24 Non-metals may also complete their octets by sharing electrons. This may occur between non-metal atoms of the same type: e.g. H 2, O 2, N 2, Cl 2, F 2, I 2, etc Or between different types of non-metal atoms: e.g. CO 2, H 2 O, CH 4, etc Molecular Compounds

25 Consider two hydrogen atoms separated by a large distance. Each has 1 electron in a 1s atomic orbital. + - + - Now lets bring the two atoms together so there orbitals overlap. Why does the electron stay around the nucleus? Molecular Compounds

26 + - + - The atomic orbitals overlap to form a new molecular orbital. This is a stable configuration as each H atom can have a full 1s subshell (like He) where the electrons spend most of their time shared between the atoms. In this arrangement each nucleus feels an inwards attraction to the two electrons. This is called covalent bonding. Molecular Compounds

27 This new arrangement of protons and electrons is more stable than separate hydrogen atoms since the attraction of a proton to two electrons is a stronger attraction compared to one proton to one electron of a hydrogen atom. + - + - Molecular Compounds

28 We can draw Lewis diagrams showing the arrangement of valence electrons in covalent compounds. In these diagrams we represent each pair of electrons between atoms as a line. So for the H 2 molecule discussed previously the Lewis diagram would be: H – H All other electrons are represented by dots as described previously. Molecular Compounds

29 Lewis Structures Draw Lewis Structures of the following molecular compounds H a. H 2 O O H H H O Note each element has a Noble gas structure by electron sharing b. NH 3 N H H H N H H H Covalent bonding e’s Nonbonding electons

30 Simplified Lewis Structures O H H Straight lines are used to indicate a shared pair, or a covalent bond. Nonbonding electrons Covalent bond

31 Lewis Structure Construction Step 1 Connect each element with a single line Step 2 Use the “P” formula to determine extra bonds Step 3 Insert the extra bonds, to make double or triple bonds. Step 4 Give each atom an octet of electrons, except hydrogen Step 5 Determine the formal charge of each element Examples: Give Lewis Structures for the following P = 8(n-q) +2q - 2(n-1) - v N = number of atoms in molecule Q = number of hydrogen atoms V = total number of valence electrons CO 2 H 2 CO 3 SO 3 NO 2 +

32 Lewis Structure of Carbon Dioxide CO O First, connect atoms with lines

33 Lewis Structure of Carbon Dioxide CO O First, connect atoms with lines Second, use “p” formula to determine the number of extra bonds. P = 8(n-q) + 2q – 2(n-1) - v P = 8(3-0) + 2(0) – 2(3-1) - 16 P = 24 + 0 – 4 - 16 P = 4 4 extra bonding electrons 2 extra bonds 2 extra lines

34 Lewis Structure of Carbon Dioxide C O O Third, add extra lines (Three possible locations) O C O O C O

35 Lewis Structure of Carbon Dioxide C O O Third, add extra lines (three possible locations) Fourth, give each atom an octet of electrons O C O O C O

36 Lewis Structure of Carbon Dioxide C O O Third, add extra lines (three possible locations) Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C

37 Lewis Structure of Carbon Dioxide C O O Third, add extra lines (three possible locations) Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge

38 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative

39 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative O O OO O O C C C

40 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative O O OO O O C C C -

41 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative O O OO O O C C C + -

42 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative O O OO O O C C C + - -

43 Lewis Structure of Carbon Dioxide C O O Third, add extra lines Fourth, give each atom an octet of electrons O C O O C O O CO OOO O C C Fifth, give each an atom a formal charge If the element owns less than its valence, then it is positive If the element has more than its valence, then it is negative O O OO O O C C C + + - -

44 There are actually three possible Lewis structures for SO 3. S O O O S O O O S O O O Each of these three structures is equivalent. We say they are in “resonance” or that they are “resonance structures”. Resonance demonstrates how the loosely held pi electrons are free to move about. Or another way to look at it is that the pi electrons are spread over the entire molecule, making it more stable - - - - - - 2+

45 So far we have been considering how electrons are distributed between atoms in molecules and polyatomic ions. An important question is: How can we predict the shape of molecules and polyatomic ions? Molecular Geometry

46 The simplest polyatomic ion or molecule is made of two atoms: What is the shape of this type of molecule or ion? The only one way to join two atoms is with a line. All diatomic molecules and ions have a linear geometry. Molecular shape is the geometry defined by the atoms making up the molecule. Molecular Shape

47 Valence shell electron pair repulsion theory (VSEPR theory) allows us to predict the 3 dimensional shape of molecules and polyatomic ions with >2 atoms. VSEPR theory states that electrons in lone pairs and bonds move as far away from one another as possible to minimize repulsive interactions. Predicting Molecular Shape

48 Repulsion Angles

49 For a central atom with three electron regions there are two possibilities. Predicting Molecular Shape

50 For a central atom with four electron regions there are three possibilities. Predicting Molecular Shape

51 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. Predicting Molecular Shape

52 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. Predicting Molecular Shape

53 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. Predicting Molecular Shape

54 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. Predicting Molecular Shape

55 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. 3 regions Predicting Molecular Shape

56 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. 3 regions Predicting Molecular Shape

57 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. 3 regions4 regions Predicting Molecular Shape

58 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. 3 regions4 regions Predicting Molecular Shape

59 To determine the correct shape of a molecule we must first begin with the correct Lewis structure. We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs. 3 regions4 regions4 regions Predicting Molecular Shape

60 The arrangement of outer atoms is determined by how many nonbonding pairs there are around the central atom. If there are 2 or less electron regions then the arrangement will always be linear. The situation becomes more complex if there are more than 2 electron regions around the central atom. Predicting Molecular Shape

61 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 Predicting Molecular Shape

62 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. Predicting Molecular Shape

63 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. Predicting Molecular Shape N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) -10 P = 4, 2 extra lines, right?

64 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) -10 P = 4, 2 extra lines, right?

65 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) - 10 P = 4, 2 extra lines, right? N

66 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) -10 P = 4, 2 extra lines, right? N Adding extra bonds N Adding nonbonding pairs

67 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH First we will start with the Lewis structure of nitrogen. N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) -10 P = 4, 2 extra lines, right? N Adding extra bonds N Adding nonbonding pairs Consider either nitrogen to be the central atom. Notice there are two volumes of space for electrons

68 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 First we will start with the Lewis structure of nitrogen. N P = 8(n-q) + 2q -2(n-1) – v P = 8(2-0) + 2(0) -2(3-1) -10 P = 4, 2 extra lines, right? N Adding extra bonds N Adding nonbonding pairs Consider either nitrogen to be the central atom. Notice there are two volumes of space for electrons Two volumes of electrons repel to 180°, thus linear shape

69 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 Now consider the Lewis Structure of water. H H O

70 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 Now consider the Lewis Structure of water. H In water there are two bonding and two nonbonding electron pairs O H Molecular Shape of Water

71 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 H In water there are two bonding and two nonbonding electron pairs. H H Nonbonding electrons H O Molecular Shape of Water

72 Use VSEPR theory to determine as much as possible about the structure of N 2, H 2 O, SO 3, CH 4, NH 4 + and NH 3 H In water there are two bonding and nonbonding electron pairs H Nonbonding electrons Bent or v-shape, bond angles less than 109.5 ° O H H Molecular Shape of Water

73 Electronegativity Electronegativity is an atoms affinity for electrons. Affinity is a Greek word for loving. Since electrons are attracted to the nucleus, then it makes sense that small atoms with more protons should attract electrons stronger. Therefore, we should predict electronegativity should increase from left to right and bottom to top on the Periodic Chart.

74 Electronegative Chart

75 Molecule Polarity A molecule will be polar if –it has polar bonds, and –its centers of partial positive and partial negative charges lie at different places within the molecule Carbon dioxide, CO 2, has two polar bonds but, because of its geometry, is a nonpolar molecule

76 Water Water, H 2 O, has two polar bonds and, because of its geometry, is a polar molecule

77 Ammonia Ammonia, NH 3, has three polar bonds and, because of its geometry, is a polar molecule

78 Dichloromethane Both dichloromethane, CH 2 Cl 2, and formaldehyde, CH 2 O, have polar bonds and are polar molecules

79 CHAPTER #10 The End


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