George Mason University General Chemistry 211 Chapter 10 The Shapes (Geometry) of Molecules

Transcription

1 Acknowledgements George Mason University General Chemistry 211 Chapter 10 The Shapes (Geometry) of Molecules Course Text Chemistry the Molecular Nature of Matter and Change, 7 th edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material. Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor. 1/13/2015 1

2 Lewis Electron-Dot Symbols A Lewis electron-dot symbol is a symbol in which the electrons in the valence shell of an atom or ion are represented by dots placed around the letter symbol of the element Note that the group (column) number indicates the number of valence electrons Group I Group II Group III Group IV Group V Group VI Group VII Group VIII Na..... Mg Al... Ṣi.. P.. Ṣ. Cl. Ar 3s 1 3s 2 3s 2 3p 1 3s 2 3p 2 3s 2 3p 3 3s 2 3p 4 3s 2 3p5 3s 2 3p6 1/13/2015 2

3 Lewis Electron-Dot Formulas A Lewis electron-dot formula is an illustration used to represent the transfer of electrons during the formation of an ionic bond The Magnesium has two electrons to give, whereas the Fluorines have only one vacancy each F... Mg F. [ F ] - 2+ [ Mg F ] Consequently, Magnesium can accommodate two Fluorine atoms 1/13/

4 Lewis Structures The tendency of atoms in a molecule to have eight electrons (ns 2 np 6 ) in their outer shell (two for hydrogen) is called the octet rule You can represent the formation of the covalent bond in H 2 as follows H + H H H This uses the Lewis dot symbols for the Hydrogen atom and represents the covalent bond by a pair of dots 1/13/2015 4

5 The Electron Probability Distribution for the H 2 Molecule 1/13/2015 5

6 Lewis Structures The shared electrons in H 2 spend part of the time in the region around each atom H H In this sense, each atom in H 2 has a helium (1s 2 ) configuration 1/13/2015 6

7 Lewis Structures The formation of a bond between H and Cl to give an HCl molecule can be represented in a similar way H. +. Cl H Cl Thus, Hydrogen has two valence electrons about it (as in He) and Cl has eight valence electrons about it (as in Ar) 1/13/2015 7

8 Lewis Structures Formulas such as these are referred to as Lewis electron-dot formulas or Lewis structures H Cl An electron pair is either a bonding pair (shared between two atoms) lone pair (electron pair that is not shared) Hydrogen has no unbonded pairs Chlorine has 3 unbonded pairs 1/13/2015 8

9 Lewis Structures Rules for obtaining Lewis electron dot formulas Calculate the number of valence electrons for the molecule from group # for each atom (1-8) add the charge of Anion subtract the charge of a Cation Put atom with the lowest group number and lowest electronegativity as the central atom Arrange the other elements (ligands) around the central atom 1/13/2015 9

10 Lewis Structures Rules for Lewis Dot Formulas Distribute electrons to atoms surrounding the central atom to satisfy the octet rule for each atom Distribute the remaining electrons as pairs to the central atom If the Central atom is deficient in electrons to complete the octet; move electron pairs from surrounding atoms to complete central atom valence electron needs, that is form one or more double bonds (possibly triple bonds) around the central atom 1/13/

11 Practice Problem Write a lewis structure for CCl 2 F 2 Step 1 Arrange Atoms (Carbon is Central Atom because is has the lowest group number and lowest electronegativity Step 2 Determine total number of valence electrons 1 x C(4) + 2 x Cl(7) + 2 x F(7) = 32 Step 3 Draw single bonds to central atom and subtract 2 e- for each single bond (4 x 2 = 8) 32 8 = 24 remaining Step 4 Distribute the 24 remaining electrons in pairs around surrounding atoms (3 electron pairs around each Fluoride atom) 1/13/

12 Writing Lewis Dot Formulas The Lewis electron-dot formula of a covalent compound is a simple two-dimensional representation of the positions of electrons in a molecule Bonding electron pairs are indicated by either two dots or a dash In addition, these formulas show the positions of lone pairs of electrons 1/13/

13 Writing Lewis Dot Formulas The following rules allow you to write electrondot formulas even when the central atom does not follow the octet rule To illustrate, draw the structure of Phosphorus Trichloride PCl 3 Con t 1/13/

14 Writing Lewis Dot Formulas Step 1 Total all valence electrons in the molecular formula. That is, total the group numbers of all the atoms in the formula P 3s 2 3p 3 Cl 3s 2 3p 5 PCl 3 5 e - (7 e - ) x 3 (2+3) + 3x(2+5) = total electrons For a polyatomic anion, add the number of negative charges to this total For a polyatomic cation, subtract the number of positive charges from this total Con t 1/13/

15 Step 2 Writing Lewis Dot Formulas Arrange the atoms radially, with the least electronegative atom in the center Place one pair of electrons between the central atom and each peripheral atom Cl Cl P 26 6 = 20 remaining Cl Con t 1/13/

16 Writing Lewis Dot Formulas Step 3 Distribute the remaining electrons to the peripheral atoms to satisfy the octet rule Cl Cl P Cl 26 (3 x 6 + 6) = 2 remaining Con t 1/13/

17 Writing Lewis Dot Formulas Step 4 Distribute any remaining electrons (2) to the central atom. If the number of electrons on the central atom is less than the number of electrons required to complete the octet for that atom, use one or more electrons pairs from other atoms to form double or triple bonds Cl Cl P Phosphorus has an octet of electrons No double bonds required Cl 1/13/

18 Exceptions to the Octet Rule Although many molecules obey the octet rule, there are exceptions where the central atom has more than eight electrons Generally, if a nonmetal is in the third period or greater it can accommodate as many as twelve electrons, if it is the central atom These elements have unfilled d subshells that can be used for bonding 1/13/

19 Exceptions to the Octet Rule For example, the bonding in phosphorus pentafluoride, PF 5, shows ten electrons surrounding the phosphorus Total valence electrons 5 x 7 (F) + 5 (P) = 40 F Distribute electrons to F atoms 5 x 6 = 30 F F Establish bonding pairs P 5 x 2 = 10 Remaining electrons F F = 0 Phosphorus has 0 non-bonding pairs Since Phosphorus is in Period 3, PF 5 is a hypervalent molecule The Phosphorus utilizes electrons from vacant 3d orbitals to create a valence shell with more than 8 electrons 1/13/

20 Exceptions to the Octet Rule In Xenon Tetrafluoride, XeF 4, the Xenon atom must accommodate two extra lone pairs F F Xe F F Total valence electrons 4 x = 36 Distribute electrons to F atoms 4 x 6 = 24 Establish bonding pairs 4 x 2 = 8 Remaining electrons = 4 Add 2 non-bonding pairs to Xe Xe violates octet rule XeF 4 is a hypervalent molecule and utilizes vacant d orbitals to create a valence shell with more than 8 electrons 1/13/

21 Delocalized Bonding Resonance The structure of Ozone, O 3, can be represented by two different Lewis electron-dot formulas Ozone (O 3 ) O O O or O O O Experiments show, however, that both bonds are identical 1/13/

22 Delocalized Bonding Resonance According to Resonance Theory, these two equal bonds are represented as one pair of bonding electrons spread over the region of all three atoms Ozone (O 3 ) O O O This is called delocalized bonding, in which a bonding pair of electrons is spread over a number of atoms 1/13/

23 Resonance & Bond Order Recall (Chap 9) Bond Order The number of electron pairs being shared by any pair of Bonded Atoms or The number of electron pairs divided by the number of bonded-atom pairs Ex. Ozone Bond Order = Electron Pairs Bonded - Atom Pairs Electron Pairs 3 Bond Order = = = 1.5 Bonded - Atom Pairs 2 1/13/

24 Practice Problem In the following compounds, the Carbon atoms form a single ring. Draw a Lewis structure for each compound, identify cases for which resonance exists, and determine the C-C bond order(s). C 3 H 4 Electron Pairs Bond Order = Bonded - Atom Pairs 1 = = 1 1 Electron Pairs 2 Bond Order = = = 2 Bonded - Atom Pairs 1 C 3 H 6 Electron Pairs Bond Order = Bonded - Atom Pairs 1 = = 1 1 1/13/

25 Practice Problem C 4 H 6 Electron Pairs 1 Bond Order = = = 1 Bonded - Atom Pairs 1 Electron Pairs 2 Bond Order = = = 2 Bonded - Atom Pairs 1 C 4 H 4 Electron Pairs 2 Electron Pairs 1 Bond Order = = = 2 Bond Order = = = 1 1/13/2015 Bonded - Atom Pairs 1 Bonded - Atom Pairs 1 25

26 Practice Problem C 6 H 6 Electron Pairs 9 Bond Order = = = 1.5 Bonded - Atom Pairs 6 Note There are 6 equivalent C - C bonds in the aromatic Benzene molecule 1/13/

27 Formal Charge & Lewis Structures In certain instances, more than one feasible Lewis structure can be illustrated for a molecule For example, H, C and N H C N or H N C The concept of formal charge can help discern which structure is the most likely Formal Charge An atom s formal charge is Total number of valence electrons Minus all unshared electrons Minus ½ of its shared electrons Formal Charges must sum to actual charge of species Zero Charge for a Molecule Ionic Charge for an Ion 1/13/

28 domain electrons Formal Charge & Lewis Structures When you can write several Lewis structures, choose the one having the least formal charge Form I 1 e - 4 e - 5 e - 1 e - 5 e - 4e - H C N or H N C I IV V I V IV FC H [1-0 - ½(2)] = 0 FC C [4-0 - ½(8)] = 0 FC N [5-2 - ½(6)] = 0 Note HCN is a neutral molecule Form II FC Total Valence e - unshared e - ½ shared e - group number 1/13/ FC H [1-0 - ½(2)] = 0 FC C [4-2 - ½(6)] = -1 FC N [5-0 - ½(8)] = +1 Preferred Form - Form I (Least Formal Charge) Sum of Formal Charges in the preferred form (0) equals molecular charge (0)

29 Formal Charge & Lewis Structures Ozone FC OA [6-4 - ½(4)] = 0 FC OB [6-2 - ½(6)] = +1 FC OC [6-6 - ½(2)] = -1 FC OA [6-6 - ½(2)] = -1 FC OB [6-2 - ½(6)] = +1 FC OC [6-4 - ½(4)] = 0 Both Resonance forms have the same formal charge and thus, are identical Note Ozone (O 3 ) is a neutral molecule Sum of Formal Charges (0) equals molecular charge (0) 1/13/

30 Formal Charge & Lewis Structures F B F F FC B = 3 0 -(1/2 * 6) = 0 FC F = 7-6 (1/2 * 2) = 0 Even though B violates Octet Rule, this is the preferred form because it has less formal charge O S O Boron Trifluoride BF 3 Sulfur Dioxide SO 2 F B F F FC B = 3 0 -(1/2 * 8) = -1 FC F = (1/2 * 4) = +1 O S O FC S = 6 2 (1/2 * 6) = 1 FC S = 6 2 (1/2 * 8) = 0 Preferred Form (Less Formal Charge) 1/13/

31 Resonance/Formal Charge Nitrate Ion Total Valence electrons - 3 x 6 (O) + 1 x 5 (N) + 1 (ion charge) = 24 Add 1 pair electrons between central atom and each other atom 3 x 2 = 6 Add electrons to oxygen atoms to complete octet 3 x 6 = 18 Nitrogen still missing 2 electrons to complete octet = 0 Borrow 2 electrons from one oxygen to form double bond to complete N octet Formal Charge Nitrogen 5 (0 + ½*8) = 5 4 = +1 Formal Charge Single bonded Oxygen 6 (6 + ½*2) = 6 7 = -1 x 2 = -2 Formal Charge Double bonded Oxygen 6 (4 + ½*4) = 6 6 = 0 Net Charge of the ION is +1 +(-2) + 0 = -1 1/13/

32 Resonance/Formal Charge Cyanate Ion FC N = 5 (6 + ½*2) = -2 FC C = 4 (0 + ½*8) = 0 FC O = 6 (2 + ½*6) = +1 Preferred Form Eliminate I FC N = 5 (4 + ½*4) = -1 FC C = 4 (0 + ½*8) = 0 FC O = 6 (4 + ½*4) = 0 FC N = 5 (2 + ½*6) = 0 FC C = 4 (0 + ½*8) = 0 FC O = 6 (6 + ½*2) = -1 Higher formal charge on Nitrogen than Carbon & Oxygen Positive formal charge on Oxygen, which is more electronegative than Nitrogen(?) Not Likely!! Eliminate II Forms II & III have the same magnitude of formal charges, but form III has a -1 charge on the more electronegative Oxygen atom Forms II & III both contribute to the resonant hybrid of the Cyanate Ion, but form III is the more important Note Net formal charge in form III is same as ionic charge (-1) Total valence e- in NCO - 5(N) + 4(C) + 6(O) + 1 (ion charge) = 16 After forming e- pairs between C-O & C-N and assigning 6 unshared e- pairs on Oxygen & Nitrogen to complete octets we have = 0 Thus; no unpaired e- needed for carbon, but carbon needs to borrow e- pairs from nitrogen to complete octet; thus forming a triple bond with nitrogen 1/13/

33 Formal Charge vs Oxidation No Formal Charge is used to examine resonance hybrid structures, whereas Oxidation Number is used to monitor REDOX reactions Formal Charge - Bonding electrons are assigned equally to the atoms as if the bonding were Nonpolar covalent, i.e., each atom has half the electrons making up the bond Formal Charge = valence e - (unbonded e - + ½ bonding e - ) Oxidation Number - Bonding electrons are transferred completely to the more electronegative atom, as if the bonding were Ionic Ox No. = valence e - (unbonded e - + bonding e - ) 11/9/2016 1/13/

34 Formal Charge vs Oxidation No Formal Charge valence e - (unbonded e - + ½ bonding e - ) N 5 (6 + ½ (2)) = -2 C 4 (0 + ½ (8)) = 0 O 6 (2 + ½ (6)) = +1 N 5 (4 + ½ (4)) = -1 C 4 (0 + ½ (8)) = 0 O 6 (4 + ½ (4)) = 0 Oxidation Number valence e - (unbonded e - + bonding e - ) N 5 (6 + 2) = -3 C 4 (0 + 0) = +4 O 6 (2 + 6) = -2 N 5 (4 + 4) = -3 C 4 (0 + 0) = +4 O 6 (4 + 4) = -2 (-3) (+4) (-2) (-3) (+4) (-2) (-3) (+4) (-2) N 5 (2 + ½ (6)) = 0 C 4 (0 + ½ (8)) = 0 O 6 (6 + ½ (2)) = -1 N 5 (2 + 6) = -3 C 4 (0 + 0) = +4 O 6 (6 + 2) = -2 Note Both Nitrogen (N) & Oxygen (O) are more electronegative than Carbon (C); thus, in the computation of Oxidation Number all the electrons are transferred to the N & O leaving C with no lone pairs and no bonded pairs Note Oxidation Nos do not change from one resonance form to another (electronegativities remain same) 1/13/

35 The Valence-Shell Electron Pair Repulsion Model (VSEPR) The Valence-Shell Electron Pair Repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence shell electron pairs are arranged as far from one another as possible Molecular geometry The shape of a molecule is determined by the positions of atomic nuclei relative to each other, i.e., angular arrangement Central Atom Place atom with Lower Group Number in center (N in NF 3 needs more electrons to complete octet) If atoms have same group number (SO 3 or ClF 3 ), place the atom with the Higher Period Number in the center (Sulfur & Chlorine) 1/13/

36 VSEPR Model of Molecular Shapes The following rules and figures will help discern electron pair arrangements Select the Central Atom (Least Electronegative Atom) Draw the Lewis structure Determine how many bonding electron pairs are around the central atom Determine the number of non-bonding electron pairs Count a multiple bond as one pair Arrange the electron pairs as far apart as possible to minimize electron repulsions Note the number of bonding and lone pairs 1/13/

37 VSEPR Model of Molecular Shapes To predict the relative positions of atoms around a given atom using the VSEPR model, you first note the arrangement of the electron pairs around that central atom Molecular Notation A The Central Atom (Least Electronegative atom) X The Ligands (Bonding Pairs) a The Number of Bonding pairs AX E Non-Bonding Electron Pairs a E b b The Number of Non-Bonding Electron Pairs Double & Triple Bonds count as a single electron pair The Geometric arrangement is determined by sum (a + b) 1/13/

38 VSEPR Model of Molecular Shapes Molecule Lewis Structure AL x N y Notation Geometric e - Pair Arrangement BeH 2 AX 2 E 0 b = 0 a = 2 a + b = 2 BH 3 AX 3 E 0 b = 0 a = 3 a + b = 3 3-D Structure Linear Trigonal Planar 3-D View & Isomers AX 4 E 0 CH 2 Li 2 or AX 2 X 2 E 0 a = 4 b = 0 a + b = 4 Tetrahedral OH 2 AX 2 E 2 b = 2 a = 2 a + b = 4 BiF 5 AX 5 E 0 b = 0 a + b = 5 a = 5 Tetrahedral Trigonal Bipyramidal 1/13/

39 VSEPR Model of Molecular Shapes Molecule Lewis Structure AL x N y Notation Geometric e - Pair Arrangement 3-D Structure 3-D View & Selected Isomers SF 5 Cl AX 6 E 0 or AX 5 X 1 E 0 a = 6 b = 0 a + b = 6 Octahedral PCL 4 Br AX 5 N 0 or AX 4 X 1 E 0 a = 5 b = 0 a + b = 5 Trigonal Bipyramidal TeCl 3 Br AX 4 E 1 or AX 3 X 1 E 1 b = 4 a = 1 a + b = 5 Trigonal Bipyramidal AX 6 E 0 or SF 4 Cl 2 AX 4 X 2 E 0 a = 6 b = 0 a + b = 6 Octahedral XeF 2 AX 2 E 3 b = 3 a + b = 5 a = 2 Trigonal Bipyramidal 1/13/

40 Arrangement of Electron Pairs About an Atom Basic Shapes CS 2 HCN BeF 2 NO 2 + 1/13/

41 Arrangement of Electron Pairs About an Atom Basic Shapes SO 3 BF 3 NO 3 NO 2 CO3 2 SO 2 O 3 PbCl 2 SnBr 2 1/13/

42 Arrangement of Electron Pairs About an Atom Basic Shapes CH 4 SiCl 4 SO 4 2- ClO 4 - NH 3 PF 3 ClO 3 H 3 O + H 2 O OF 2 SCl 2 1/13/

43 Arrangement of Electron Pairs About an Atom Basic Shapes PF 5 AsF 5 SOF 4 SF 4, XeO 2 F 2, IF 4 +, IO 2 F 2 - ClF 3 BrF 3 XeF 2 I 3 - IF 2-1/13/

44 Arrangement of Electron Pairs About an Atom Basic Shapes SF 6 IOF 5 BrF 5 TeF 5 - XeOF 4 XeF 4 ICl 4-1/13/

45 Electron Pair Arrangement 1/13/

46 Electron Pair Arrangement 1/13/

47 Linear Geometry Two electron pairs (linear arrangement) Carbon is central atom because it has lower group number Double bonds count as a single electron pair 2 bonding pairs 0 non-bonding pairs AX a E b = a + b = = 2 (Linear) Thus, according to the VSEPR model, the bonds are arranged linearly (bond angle = 180 o ) Molecular shape of carbon dioxide is linear 1/13/

48 Trigonal Planar Geometry Three electron pairs on Central atom O C Cl Cl Central Atom - Carbon 3 bonding electron pairs (double bond counts as 1 pair) 0 non-bonding electron pairs a + b = = 3 Trigonal Planar The three groups of electron pairs are arranged in a trigonal plane. Thus, the molecular shape of COCl 2 is trigonal planar. The Bond angle is 120 o 1/13/

49 Trigonal Planar Geometry Effect of Double Bonds Bond angles deviate from ideal angles when surrounding atoms and electron groups are not identical A double bond has greater electron density and repels two single bonds more strongly than they repel each other 120 o H H C Ideal 120 o O 116 o H H 122 o C Actual O 1/13/

50 Effect of Lone Pairs Trigonal Planar Geometry The molecular shape is defined only by the positions of the nuclei When one of the three electron pairs in a trigonal planar molecule is a lone (non-bonding) pair, it is held by only one nucleus It is less confined and exerts a stronger repulsive force than a bonding pair This results in a decrease in the angle between the bonding pairs The normal Trigonal Planar angle between the bonding pairs is 120 o 1/13/

51 Trigonal Planar Geometry Three electron pairs (Effect of Lone pairs) O O O Ozone has two bonding electron pairs about the central oxygen (double bond counts as 1 pair) There is one non-bonding lone pair These groups have a Trigonal Planar arrangement AX a E b (a + b) = = 3 Since one of the groups is a lone pair, the molecular geometry is described as bent or angular 1/13/ <120 o SO 3 BF 3 NO 3 - CO 3 2-

52 Four electron pairs Tetrahedral Geometry (Tetrahedral Arrangement) Cl Cl C Cl Cl H N H H H O H Four electron pairs about the central atom lead to three different molecular geometries a + b = a + b = a + b = = 4 = 4 = 4 1/13/

53 Tetrahedral Geometry Molecular Geometries produced by variable nonbonding electron pairs Cl Note impact of non-bonding electron pairs on bond angle Cl C Cl Cl H N H H O H H AX o AX 4 E 107 o AX 2 E o CH 4, SiCl 4, SO 4 2-, ClO 4 - PF 3, ClO 3 -, H3 O + OF 2, SCl 2 1/13/

54 Five electron pairs Trigonal Bipyramidal (trigonal bipyramidal arrangement) F F F P F F ASF 5 SOF 4 This structure results in both 90 o and 120 o bond angles 1/13/

55 Trigonal Bipyramidal Other molecular geometries are possible when one or more of the electron pairs is a lone pair F F F S F F Cl F F F Xe F <120 o (eq) 180 o XeO 2 F 2 IF 4+ IOF 2 - ClF 3 BrF 3 XeF 2 I 3 - IF 2-1/13/

56 Octahedral Geometry Six electron pairs F F (Octahedral arrangement) F S F F F This octahedral arrangement results in 90 o bond angles 90 o SF 6 IOF 5 1/13/

57 Six electron pairs Iodine violates octet rule Iodine is sp 3 d 2 hybridized Iodine uses d orbitals Other Geometries (octahedral arrangement) F F F I F F F F Xe F F Noble gases not always inert Xenon forms 6 electron domains square pyramidal square planar <90 o 90 o BrF 5 TeF 5 - XeOF 4 XeF 4 ICl 4-1/13/

58 Practice Problem In the ICl 4 ion, the electron pairs are arranged around the central iodine atom in the shape of a. a tetrahedron b. a trigonal bipyramid c. Octahedron(square plane) d. an octahedron Cl Cl I Cl AX 4 e. a trigonal pyramid Ans a Cl Total valence e- = 4 x (ion charge) = = 4 unaccounted for e - Add two pairs e- to Iodine atom AX a E b a + b = = 6 (AX 4 E 2 Square Planar 1/13/

59 Dipole Moment The dipole moment ( ) is a measure of the degree of charge separation (molecular polarity) in a molecule The product of the magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges = Q r Example What is the dipole moment in Debyes of a molecule with a r = Bond Length = 127 pm Q = Electron Charge (e) 1.6x10-19 C 1 Debye = 3.34 x C m m 1D μ = Qr = C (127pm) ( ) ( ) = 6.08 D -30 pm C m) 1/13/

60 Dipole Moment In the previous example the Dipole Moment was calculated on the assumption that each of the atoms bore a full charge of 1 electronic unit, e, (+1 & -1). That is Q = 1 e = 1.60 x Coulombs (C) Such a molecule would be nearly ionic Atoms in asymmetric covalent molecules would not exhibit full ionic charges, thus, the value of Q would be less than 1 e, depending on the relative electronegativity differences and the bond length 1/13/

61 Dipole Moment and Molecular Geometry The polarity of individual bonds within a molecule can be viewed as vector quantities Thus, molecules that are perfectly symmetric have a zero dipole moment. These molecules are considered nonpolar Consider Carbon Dioxide CO 2 d + d - d - 1/13/

62 Dipole Moment and Molecular Geometry However, molecules that exhibit any asymmetry in the arrangement of electron pairs would have a nonzero dipole moment. These molecules are considered polar d - H N H d + H NH 3 PF 3 ClO 3 H 3 O + 1/13/

63 Dipole Moment and Molecular Geometry Formula Molecular Geometry Dipole Moment AX Linear Can Be nonzero AX2E0 Linear Zero AX3E0 Trigonal Planar Zero AX2E1 Trigonal Planar Bent Can Be nonzero AX4E0 Tetrahedral Zero AX3E1 Tetrahedral Trigonal Pyramidal Can Be nonzero AX2E2 Tetrahedral Bent Can Be nonzero AX5E0 Trigonal Bipyramidal Zero AX4E1 Trigonal Bipyramidal SeeSaw Can Be nonzero AX3E2 Trigonal Bipyramidal T-Shaped Can Be nonzero AX2E3 Trigonal Bipyramidal Linear Can Be nonzero AX6E0 Octahedral Zero AX5E1 Octahedral Square Pyramidal Can Be nonzero AX4E2 Octahedral Square Planar Zero 1/13/

64 Practice Problem The Nitrogen atom would be expected to have the positive end of the dipole in the species a. NH + 4 b. Ca 3 N 2 c. HCN d. AlN e. NO + N is more Electronegative than H N is more EN than Ca N is more EN than C N is more EN than Al O is more EN than Nitrogen Ans e 1/13/

65 Practice Problem Which of the following molecules is polar? a. BF 3 b. CBr 4 c. CO 2 d. NO 2 e. SF 6 Ans d O -N- O O N O The Lewis structures for BF 3, CBr 4, CO 2, and SF 6 do not have any non-bonding electrons on the central atom The Lewis structure for NO 2 shows one double bond and a lone non-bonding electron on the Nitrogen The VSEPR Molecular Geometry for NO 2 is AX 2 E 1 (a + b = = 3) - Trigonal Planar Formal Charge on N is 5 1 ½ (6) = +1 NO 2 molecule is polar difference in oxygen and nitrogen electronegativity 1/13/ O N O -

66 Practice Problem Which of the following compounds is nonpolar? a. XeF 2 b. HCl c. SO 2 d. H 2 S e. N 2 0 Ans a HCL is ionic and very polar SO 2 has AX 2 E 1 Trigonal Planar Bent geometry with a dipole moment (polar) H 2 S has AX 2 E 2 Tetrahedral Bent geometry and with a dipole moment (polar) N 2 O has AX 2 E 0 linear with asymmetric geometry. Since oxygen is more EN than N, the molecule is polar XeF 2 has AX 2 E 3 Trigonal Bypyramidal Geometry, but linear molecular geometry (nonpolar) 1/13/

67 Equation Summary Bond Order = Electron Pairs Bonded - Atom Pairs Formal Charge (FC) = Total Valence e - (Non - Bonding e + 1 / 2 Bonding e ) Oxidation Number (ON) = Total Valence e - (Non - Bonding e + Bonding e ) VSEPR Model - AX a E b Geometric Configuration Determined by the sum (a + b) Dipole Moment = = Q x r 1/13/