1. Molecular Geometry And Bonding Theories
Chapter 9
Molecular Geometry
And Bonding Theories

2. Molecular Shape
A bond angle is the angle defined by lines joining the centers of two atoms to a third atom to which they are covalently bonded
The molecular geometry or shape is defined by the lowest energy arrangement of its atoms in three-dimensional space.

3. Valence-Shell Electron-Pair Repulsion Theory (VSEPR)
The geometric arrangement of atoms bonded to a given atom is determined principally by minimizing electron pair repulsions between the bonds to, and lone-pairs on the atom.

4. Determining Molecular Shape
Steric number (SN) = (# of atoms bonded to central atom) + (# of lone pairs on the central atom)

5. Each Molecular Shape has characteristic bond angles

6. Geometric Forms

7. Predicting a VSEPR Structure
Draw Lewis structure.
Determine the steric number of the central atom.
Use the SN to determine the geometry around the central atom.
The name for molecular structure is determined by the number of lone pairs and bonding pairs of electrons.

8. Examples What is the molecular geometry of BF3?
What is the molecular geometry of CH4?

9. Central Atoms with Lone Pairs
Electron-pair geometry describes the arrangement of atoms and lone pairs of electrons about a central atom.
The electron-pair geometry will always be one of the five geometries presented previously.
The molecular geometry in these molecules describes the shape of the atoms present (it excludes the lone pairs).

10. Lone Pairs
Lone pairs of electrons occupy more space around a central atom than do bonding electrons.
Lone pair-lone pair repulsion is the largest.
Lone pair-bonding pair repulsion is the next largest.
Bonding pair-bonding pair repulsion is the smallest.
In structures with lone pairs on the central atom, the bond angles are a little smaller than predicted based on the electron-pair geometry.

11. SN = 4, Electron-pair Geometry = Tetrahedral
Electron “cloud” geometry versus Molecular Geometry
SN = 4, Electron-pair Geometry = Tetrahedral
No. of Bonded Atoms
No. of Lone Pairs
Molecular Geometry
Bond Angles 4
Tetrahedral
109.5o 3 1
Trigonal Pyramidal
<109.5o 2
Bent

12. Lone Pairs of Electrons Affect the Bond Angles

13. SN = 5, Electron-pair Geometry = Trigonal Bipyramidal
No. of Bonded Atoms
No. of Lone Pairs
Molecular Geometry
Bond Angles 5
Trigonal Bipyramidal
120o & 90o 4 1
Seesaw
<120o & 90o 3 2
T-shaped
Linear
180o
The lone pairs of electrons are always found in the trigonal
planar part of the structure to minimize repulsion.

14. SN = 6, Electron-pair Geometry = Octahedral
No. of Bonded Atoms
No. of Lone Pairs
Molecular Geometry
Bond Angles 6
Octahedral
90o 5 1
Square Pyramidal
<90o 4 2
Square Planar 3
Although these arrangements are possible, we will not generally encounter any molecules with these arrangements.

15. Practice
What are the molecular geometries of the ions: SCN- and NO2- ?

16. Polar Bonds and Polar Molecules
Two covalently bonded atoms with different electronegativities have partial electric charges of opposite sign creating a bond dipole.
A molecule is called a polar molecule when it
has polar bonds and a shape where the bond
dipole vector do no cancel each other. Non-polar molecules (with polar bonds) are typically high symmetric, e.g. BF3, CS2, CCl4

17. Fig. 7.30: Polar bonds can lead to polar molecules (but not necessarily). A molecule will have a permanent dipole moment ( = q۰r) when it has an asymmetric orientation of polar bonds.
1D = 3.336E-30 C-m (dipole moments are measured in Debye units)

18. Measuring Polarity
The permanent dipole moment () is a measured value that defines the extent of separation of positive and negative charge centers in a covalently bonded molecule
 = q۰r

19. Dipole Moment (m)
The dipole moment in CCl4 is zero
The calculated dipole moment of CH2Cl2 is 2.05 Debye.
-0.33
+0.33

20. Dipole Moment (m)
m = Qr where Q is the partial charge and r is the distance between partial charges q+ and q-.
The dipole moment of H2O is 1.85 Debye.
-0.33
+0.33

21. Dipole Moment (m) Problem
The dipole moment on HBr has been measured to be 0.78 D. If the H-Br bond distance is 141 pm, calculate the partial charges on the H and Br atoms relative to the charge on a single electron. H Br
141 pm

22. Atomic Orbitals and Bonds
A tetrahedral electron-pair geometry (SN=4) requires that four orbitals of the central atom must overlap with an orbital of an outer atom to form a bond.
The central atom would use its s orbital and its three p orbitals, but these orbitals would not yield the 109° bond angles observed in the tetrahedral molecule.

23. Valence-Bond Theory
Valence-bond theory assumes that covalent bonds form when atomic orbitals on different atoms overlap or occupy the same region of space.
Each pair of electrons (SN) requires an orbital. e.g. for SN=4, four orbitals are needed.
A sigma () bond is a covalent bond in which the greatest electron density lies between the two atoms forming the bond.

24. Sigma (s) bond examples

25. Valence Bond Theory
Hybridization is the mixing of atomic orbitals (AOs) to generate new sets of orbitals that are then available to overlap and form covalent bonds (with the assumed geometry & based upon SN) with other atoms or lone pair electrons.
A hybrid atomic orbital is one of a set of equivalent orbitals about an atom created when specific atomic orbitals are “mixed”. The # of AOs need is equal to the steric number (SN).

26. Tetrahedral Geometry: sp3 Hybrid Orbitals
A tetrahedral orientation of valence electrons is achieved by forming four sp3 hybrid orbitals form one s and three p atomic orbitals (SN=4…required 4 hybrid orbitals).

27. Other sp3 Hybrid Examples (SN=4)

28. sp2 Hybridization (SN=3)
In a covalent pi () bond, electron density is greatest above and below the bonding axis.

29. sp Hybridization
Pi bonds will not exist between two atoms unless a sigma bond forms first.

30. The Bonding in Carbon Dioxide (SN=2)
The carbon atom is sp hybridized and these orbitals form the two sigma bonds. The  bonds are form from pure-p-type orbital on the carbon and are thus rotated 90° from one another.

31. dsp3 Hybridization (SN=5)

32. d2sp3 Hybridization (SN=6)

34. Practice
What are the hybridizations of the central atoms of the ions: XeF4
and Cl-I-Cl ?

35. Practice
What are the hybridizations of the central atoms of the molecules: XeF4 and HXeSH ?

36. Problems with Bonding Theories
Lewis structure and valence bond theory help us understand the bonding capacities of elements.
VSEPR and valence bond theories account for the observed molecular geometries.
None of these models enables us to explain other properties of molecules, e.g why O2 is attracted to a magnetic field while N2 is repelled slightly.

37. Molecular oxygen (but not N2) is attracted to the poles of a magnet.

38. Molecular Orbital (MO) Theory
The wave functions() of atomic orbitals (AOs) of atoms are combined to create molecular orbitals (MOs) in molecules.
Each MO is associated with an entire molecule, not just a single atom. MOs are spread out, or delocalized, over all the atoms in a molecule.

39. MOs for H2
The two wavefunctions (1s orbitals) may be added or subtracted to yield two MOs, a s1s bonding orbital and a s*1s anti-bonding MO.

40. Bonding and Anti-bonding MOs
Electrons in bonding orbitals serve to hold atoms together in molecules by increasing the electron density between nuclear centers.
A sigma (2s) orbital

41. Bonding and Anti-bonding MOs
Electrons in anti-bonding orbitals decrease electron density between the atoms
a sigma(2s)
antibonding

42. Bond Types
A sigma () bond is a covalent bond in which the highest electron density lies along the bond axis.
A pi () bond is formed by the mixing of atomic orbitals that are not oriented along the bonding axis in a molecule.

43. MO Guidelines
The total number of MO formed equals the number of atomic orbitals used in the mixing process.
Orbitals with similar energy and shape mix more effectively than do those that are different.
Orbitals of different principal quantum numbers have different sizes and energies, resulting in less effective mixing.
A MO can accommodate two electrons with opposite spin.
Electrons are placed in MO diagrams according to Hund’s rule.

44. Bond Order
Bond Order = 1/2 (# bonding electrons - # antibonding electrons)
The bond order is zero in He2 and the molecule is not stable.

45. Combinations of p-type Atomic Orbitals to form MOs for N2 and O2

46. p-type AOs can also combine to form sigma type MOs that can be bonding or anti-bonding
s(2pz)

47. p-type AOs can also combine to form p-type MOs that can be bonding or anti-bonding
p(2py)

48. p*(2py)

49. s*(2pz)

50. MO Diagrams for N2 and O2

51. MO Scheme for N2
Figure 9.27
Electron configuration for N2: 2s22s*22p22p4
Bond order = 1/2 (8 - 2) = 3
N2 has three bonds
N2 has no unpaired electrons

52. MO Scheme for O2
Electron configuration for O2: 2s22s*22p22p4 2p*2
Bond order = 1/2 (8 - 4) = 2
O2 has two bonds
O2 has two unpaired electrons in 2p*

53. Relative Energies of Bonding and Antibonding MO’s in Second Row Diatomics
1. The bonding and antibonding 2s orbitals are more stable than any of the six molecular orbitals derived from the 2p orbitals. This is because the 2s orbitals that give rise to 2s and *2s are more stable than the 2p atomic orbitals.
2. The two 2p bonding orbitals have identical energies, because the atomic orbitals from which they are constructed have identical energies. Likewise, the two *2p orbitals have identical energies.
3. The antibonding orbitals formed from the atomic 2p orbitals are the least stable molecular orbitals, with the *2p orbital less stable than the * orbitals.
4. The relative energies of the 2p and 2p molecular orbitals varies depending on several factors including electron-electron repulsion.

54. Paramagnetism and Diamagnetism
Paramagnetism - atoms or molecules having unpaired electrons are attracted to magnetic fields
Diamagnetism - atoms or molecules having all paired electrons are repelled by magnetic fields

55. Other Diatomic Molecules
The 2s and 2p interactions are strong in Li2 through N2 but weaker in O2 through Ne2.

56. List each species that is paramagnetic.
Use MO theory predict the bond orders for each of the following molecular ions: N2+, N2-, He2+, Br2+
List each species that is paramagnetic.
Answer: A

57. Other Diatomic Molecules
The MO diagram illustrates how the effective nuclear charge alters the diagram.
The odd electron is more likely to be found on nitrogen since it is in an orbital closer in energy to the atomic orbitals of the nitrogen atom.

58. Delocalization of Electrons
The electrons in the -system with alternating single and double bonds can be delocalized over several atoms or even an entire molecule.

59. Comparison of Theories
MO theory may provide the most complete picture of covalent bonding, but it is also the most difficult to apply to large molecules. Using Valence Bond theory, the general shape of larger molecules can be quickly estimated.
MOs in CH4

60. Assigning Hybridization (VBT)
Morphine

61. "I think I can safely say that nobody understands quantum mechanics.“
Richard Feyman

62. ChemTour: Partial Charges and Bond Dipoles
Click to launch animation
PC | Mac
Students learn that covalent bonds often include unequal distribution of electrons leading to partial charges on atoms, bond dipole moments, and molecule polarity. Interactive Practice Exercises ask students to calculate dipole moments of polar molecule.

63. ChemTour: Greenhouse Effect
Click to launch animation
PC | Mac
This unit explores how excess carbon dioxide and CFCs in the atmosphere contribute to global warming.

64. ChemTour: Vibrational Modes
Click to launch animation
PC | Mac
This tutorial illustrates the three vibrational modes: bending, symmetric stretching, and asymmetric stretching. Students learn that molecules can absorb specific wavelengths of infrared radiation by converting this energy into molecular vibrations.

65. ChemTour: Hybridization
Click to launch animation
PC | Mac
This tutorial animates the formation of hybrid orbitals from individual s and p orbitals, shows examples of their geometry, and describes how they can produce single, double, and triple bonds. Includes Practice Exercises.

66. ChemTour: Chemistry of the Upper Atmosphere
Click to launch animation
PC | Mac
This ChemTour examines how particles of the upper atmosphere absorb and emit electromagnetic radiation.

67. ChemTour: Molecular Orbitals
Click to launch animation
PC | Mac
This animated tutorial offers a patient explanation of molecular orbital theory, an alternative to the bonding theory depicted by Lewis dot structures. Includes Practice Exercises.

68. A) H2C=C=CH2 B) H2C=C=C=CH2 C) Neither
Ethylene, which has the molecular formula C2H4, is a rigid molecule in which all 6 atoms lie in a plane. Which of the following molecules also has a rigid planar structure?
A) H2C=C=CH2 B) H2C=C=C=CH2 C) Neither
© 2008 W. W. Norton & Company Inc. All rights reserved.
Planar Hydrocarbons

69. Consider the following arguments for each answer and vote again:
A combination of 3 carbons and 4 hydrogens can form the rigid planar molecule H2C=C=CH2.
The orientations of the π bonds in H2C=C=C=CH2 alternate in such a way as to create a planar structure.
The hybridization of the atomic orbitals on the carbons prevents the retention of a planar structure in molecules longer than C2H4.
Answer: B
Planar Hydrocarbons

70. Molecular Structures of Pentane
Which of the following depicts a molecule that is different from the one shown to the left? A) B) C)
© 2008 W. W. Norton & Company Inc. All rights reserved.
Molecular Structures of Pentane

71. Molecular Structures of Pentane
Please consider the following arguments for each answer and vote again:
This molecule, known as isopentane, has a unique T-shaped arrangement.
In this molecule, only two carbon atoms are bonded to the second carbon.
This molecule is unbranched, whereas the one pictured in the question is not.
Answer: C
Molecular Structures of Pentane

72. What is the bond order of the N-O bond in nitrate, NO3-?
A) B) 11/ C) 2
© 2008 W. W. Norton & Company Inc. All rights reserved.
Bond Order of Nitrate

73. Consider the following arguments for each answer and vote again:
The majority of the bonds in NO3- are single bonds, so the bond order is 1.
The N-O bond is twice as likely to be a single bond as it is to be a double bond, so the bond order should be 11/3.
The bond order is dictated by the strongest bond, which in NO3- is a double bond.
Answer: B
Bond Order of Nitrate

74. Molecular Geometry of SF2, SF3-, and SF4
According to Valence Shell Electron Pair Repulsion (VSEPR) theory, 4 objects around a central atom will have the tetrahedral arrangement shown to the left with bond angles of ~109.5º. Which of the following compounds has a bond angle of ~109.5º?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) SF2 B) SF3- C) SF4
Molecular Geometry of SF2, SF3-, and SF4

75. Molecular Geometry of SF2, SF3-, and SF4
Please consider the following arguments for each answer and vote again:
SF2 consists of a sulfur atom surrounded by 2 lone electron pairs and bonded to 2 fluorine atoms, therefore, it has an approximately tetrahedral bond angle.
The tetrahedral VSEPR arrangement of SF3- is formed by a sulfur atom surrounded by 3 fluorine atoms and by the additional electron (from the negative charge).
Sulfur tetrafluoride is the only molecule with a central atom (sulfur) surrounded by 4 additional atoms (4 fluorine atoms) and so is the only molecule with a bond angle of ~109.5º.
Answer: A
Molecular Geometry of SF2, SF3-, and SF4

76. Bond Angles of BrF2 and ICl2
Which of the following is true of the bond angle (θ1) in BrF2+ compared to the bond angle (θ2) in ICl2-?
A) θ1 = θ2 B) θ1 > θ2 C) θ1 < θ2
© 2008 W. W. Norton & Company Inc. All rights reserved.
Bond Angles of BrF2 and ICl2

77. Bond Angles of BrF2 and ICl2
Please consider the following arguments for each answer and vote again:
Both BrF2+ and ICl2- consist of a central halogen atom bonded to two halogen atoms, and therefore should have the same arrangement of atoms.
ICl2- has 1 more lone pair of electrons than BrF2+, which forces the chlorine atoms closer together.
ICl2-, with 3 lone pairs, is linear whereas BrF2+, with 2 lone pairs, is bent.
Answer: C
Bond Angles of BrF2 and ICl2

78. Reaction of Boron Trifluoride
Boron trifluoride (BF3), which has the structure shown to the left, is capable of reacting with an unknown compound to form a new compound without breaking any bonds. Which of the following could be the unknown compound?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) BF3 B) CH4 C) NH3
Reaction of Boron Trifluoride

79. Reaction of Boron Trifluoride
Please consider the following arguments for each answer and vote again:
BF3 can dimerize to BF3-BF3 by forming a boron-boron single bond.
By forming a boron-carbon bond, the carbon atom in CH4 will increase its steric number to 5, thus expanding its octet to compensate for boron's incomplete octet.
The nitrogen lone electron pair can form a nitrogen-boron bond yielding BF3-NH3, isoelectronic with CH3-CH3.
Answer: C
Reaction of Boron Trifluoride

80. Dipole Moments of Dichloroethylene
Pictured to the left is the planar molecule ethylene, C2H4, which does not have a permanent electric dipole moment.
If chlorine atoms were substituted for two hydrogen atoms, how many of the possible structures would also not possess a dipole moment?
© 2008 W. W. Norton & Company Inc. All rights reserved.
A) B) C) 2
Dipole Moments of Dichloroethylene

81. Dipole Moments of Dichloroethylene
Consider the following arguments for each answer and vote again:
Chlorine atoms always draw electron density away from carbon atoms, so all possible structures will possess a dipole moment.
Only if the chlorine atoms are diagonally opposite will the two carbon-chlorine dipole moments cancel each other.
So long as the two chlorine atoms are on different carbon atoms, no permanent dipole moment will form.
Answer: B
Dipole Moments of Dichloroethylene

82. Molecular Geometry of XF42
For which central atom "X" does the anion pictured to the left have a square planar geometry?
A) C B) S C) Xe
© 2008 W. W. Norton & Company Inc. All rights reserved.
Molecular Geometry of XF42
Molecular Geometry of XF42-

83. Molecular Geometry of XF42-
Please consider the following arguments for each answer and vote again:
CF42- forms a structure in which the 4 fluorine atoms form a square plane with one negative charge on either side of the plane.
With 2 lone electron pairs on the sulfur in SF42-, its steric number is 6.
To maximize fluorine-fluorine distances, the 4 fluorine atoms in XeF42- will lie in a plane.
Answer: B
Molecular Geometry of XF42-