1. Chapter 9: Molecular Geometry and Bonding Theories
SF6
Prof. Dr. Nizam M. El-Ashgar
Chemistry Department, IUG

2. Molecular Shapes
Lewis do not indicate the shapes of molecules; they simply show the number and types of bonds between atoms.
However, we use Lewis structures to help us determine shapes.
The overall shape of a molecule is determined by its bond angles, the angles made by the lines joining the nuclei of the atoms in the molecule.
PF NH HClO4

3. Molecular Shapes

4. VSEPR Theory
Electron pairs, whether they be bonding or nonbonding (valence electrons) repel each other.
By assuming the electron pairs are placed as far as possible from each other (to minimize repulsions), we can predict the shape of the molecule (3D).
We call this process:(VSEPR) theory
Valence Shell Electron Pair Repulsion Theory

5. “The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.”
(The balloon analogy in the figure to the left demonstrates the maximum distances that minimize repulsions.)

6. Electron Domains
Electron domains: They referred to the electron pairs as.
Each pair of electrons count as an electron domain, whether they are in:
A lone pair.
A single bond
A double bond
A triple bond.
This molecule has four electron domains.

7. Electron domain geometry  Molecular geometry
Electron domain geometry: When considering the geometry about the central atom, we consider all electrons (lone pairs and bonding pairs).
When naming the molecular geometry (shape) we focus only on the positions of the atoms (nonbonding electrons are dropped).
Note:
Electron domain geometry  Molecular geometry

8. Molecular Shapes
ABn molecule A = the central atom. B = the bonded atoms n = the number of bonded atoms. The shape of any particular ABn can usually be derived from one of five basic geometric structures:

9. The 5 Basic Geometries

10. To determine the electron pair (electron domain) geometry:
Draw the Lewis structure.
Count the total number of electron pairs around the central atom.
Arrange the electron pairs in one of the above geometries to minimize e-- e- repulsion, and count multiple bonds as one bonding pair.
But then we have to account for the shape of the molecule (based on the bonded atoms only).

11. An AB: one electron domain (Two atoms). The molecule has a linear shape. Example: HCl, O2, N2
NOTE: If there are only two atoms in the molecule, the molecule will be linear no matter what the electron domain is.

12. BeH2 CO2 Molecular Shapes
1) Two electron domains; 3 atoms (AB2): The molecule has a linear shape with 180o bond angles.
Examples:
BeH CO2

13. 2) Three electron domains; 4 atoms (AB3): The molecule has a trigonal planar shape with 120o bond angles.
Examples:
BCl H2CO x

14. AB3 Derivative Geometries: (one derivative). AB2E: Elect. Dom
AB3 Derivative Geometries: (one derivative). AB2E: Elect. Dom. Geometry Molecular Shape Trigonal planar bent; V-shape; nonlinear; angular Example: NO2-

15. 3) Four electron domains; 5 atoms (AB4): The molecule has a tetrahedral shape with 109.5o bond angles.
Example: Methane

16. AB4 Derivative Geometries: (two derivatives). I) AB3E: Elect. Dom
AB4 Derivative Geometries: (two derivatives). I) AB3E: Elect. Dom. Geometry Molecular Shape Tetrahedral Trigonal Pyramidal Example: NH3

17. II) AB2E2: Elect. Dom. Geometry Molecular Shape Tetrahedral Bent Example: H2O

18. 4) Five electron domains; 6 atoms (AB5): The molecule has a trigonal bipyramidal shape with 90o and 120o bond angles.
Example:
PCl5

19. The axial position: is at a 90o angle to its neighbors.
There are two different positions of the electron domains on this molecule: axial and equatorial.
The axial position: is at a 90o angle to its neighbors.
The equatorial position: is at 120o or 90o angles to its neighbors.
Lone pairs occupy equatorial positions.

20. Axial vs. Equatorial Lone Pairs
The equatorial position experiences less repulsion, so nonbonding pairs will always occupy the equatorial positions. For AB3E: The lone pair exists in the equatorial position
X 

21. AB5 Derivative Geometries: (Three derivatives).
AB4E:
Example: SF4
Elect. Dom. Geom.
Trigonalpipyramidal
Mol. Shape: Seesaw

22. II) AB3E2: Example: BrF3 Elect. Dom. Geom. Mol
II) AB3E2: Example: BrF3 Elect. Dom. Geom. Mol. Shape Trigonalpipyramidal T-Shape

23. III) AB2E3: Example: XeF2 Elect. Dom. Geom. Mol
III) AB2E3: Example: XeF2 Elect. Dom. Geom. Mol. Shape Trigonalpipyramidal Linear

24. 5) Six electron domains; 7 atoms (AB6): The molecule has an octahedral shape with 90o all bond angles.

25. I) AB5E: AB6 Derivative Geometries: (Two derivatives). Example: IF5
Elect. Dom. Geom Mol. Shape
Octahedral Square pyramidal

26. II) AB4E2 Elect. Dom. Arr. Mol. Shape Octahedral Square planar
Example: XeF4

27. Sample Exercise 9.1
Use the VSEPR model to predict the molecular geometry of
a) O3
LS : Three electron domains
AB2E
The Elec. Dom. Geom. is:
Trigonal planar
The molecular shape (geometry) is Bent

28. b) SnCl3- LS : Four electron domains AB3E The Elec. Dom. Geom. is:
Tetrahedral
The molecular shape (geometry) is Trigonal pyramidal

29. Practice Exercise
Predict the electron-domain geometry and the molecular geometry for:
SeCl2
CO32-

30. Sample Exercise 9.2
Use the VSEPR model to predict the molecular geometry of
a. SF4
LS : Five electron domains
AB4E
The Elec. Dom. Geom. is:
Trigonal bipyramidal
The molecular shape (geometry) is Seesaw

31. LS : six electron domains AB5E The Elec. Dom. Geom. Is: Octahedral
b. IF5
LS : six electron domains
AB5E
The Elec. Dom. Geom. Is:
Octahedral
The molecular shape (geometry) is Square pyramidal

32. Bonding vs. Nonbonding
A bonding pair of electrons: are located between the nuclei of the two atoms that are bonded together. A nonbonding pair (lone pair): of electrons are located on one atom. Both pairs represent an electron domain.

33. Nonbonding Pairs and Bond Angle
Nonbonding pairs are physically larger than bonding pairs.
Therefore, their repulsions are greater; this tends to decrease bond angles in a molecule.
Therefore, the bond angle decreases as the number of lone pairs increase.

34. Multiple Bonds and Bond Angles
Double and triple bonds place greater electron density on one side of the central atom than do single bonds.
Therefore, they also affect bond angles.

35. Sample Exercise
Predict the electron-domain geometry and molecular geometry of:
a. ClF Answer: (a) trigonal bipyramidal, T-shaped;
b. ICl Answer: (b) octahedral, square planar

36. Shapes of Large Molecules
Larger molecules can be broken into sections to see the shapes involved.

37. Shapes of Larger Molecules (Examples)
For larger molecules, look at the geometry about each atom rather than the molecule as a whole.

38. Sample Exercise 9.3
Eyedrops used for dry eyes usually contain the water-soluble polymer called poly(vinylalcohol), which is based on the unstable organic molecule called vinyl alcohol: Predict the approximate values for the H – O – C and the O – C – C bond angles.
<109.5 o
>120 o

39. Practice Exercise
Predict the H – C – H ad C – C – C bond angles in the following molecule, called propyne:
Answer: 109.5o , 180o

40. Molecular Shape and Molecular Polarity
General Guides: Ask yourself:
COVALENT or IONIC? If COVALENT:
Are the BONDS polar? (Do they have a bond dipole?)
NO: The molecule is NONPOLAR!
YES: Continue—Do the AVERAGE position of δ+ and δ– coincide? (Is the overall dipole moment equal to ZERO?)
YES: The molecule is NONPOLAR.
NO: The molecule is POLAR.
NOTE: Different atoms attached to the central atom have different polarity of bonds.

41. Bond polarity: is a measure of how equally the electrons in a bond are shared between the two atoms of the bond.
When there is a difference in electronegativity between two atoms, then the bond between them is polar.
1) For Diatomic Molecules:
Same Atoms: non polar (H―H)
Different Atoms: Polar (H―F)

43. 2) For Polyatomic Molecules: It is possible for a molecule to contain polar bonds, but not be polar. For example: the bond dipoles in CO2 cancel each other because CO2 is linear. The dipole moment depends on: - Polarities of the individual bonds. - The geometry of the molecule. Bond dipoles and dipole moments are vector quantities – magnitude and direction.

44. The overall dipole moment is the vector sum of its bond dipoles.
In CO2: The dipoles are of equal magnitude but in opposite directions, then they cancel out making the molecule nonpolar.
In H2O: the molecule is not linear and the bond dipoles do not cancel each other (polar molecule).

45. If the dipoles do not cancel out, then the molecule is polar
If the dipoles do not cancel out, then the molecule is polar. Be sure to consider shape before deciding polarity.

46. a. BrCl (polar);  = 0.57 D b. SO2 (polar);  = 1.63 D c. SF6
Sample Exercise 9.4
Predict whether the following molecules are polar or nonpolar:
a. BrCl
(polar);  = 0.57 D
b. SO2
(polar);  = 1.63 D
c. SF6
(nonpolar);  = 0 D

47. Non-Polar: AB2E3 and AB4E2
1) AB2E3 Example: XeF2 2) AB4E2 Example: XeF4

48. Practice Exercise
Determine whether the following molecules are polar or nonpolar:
a. NF3
Answer: (a) polar because polar bonds are arranged in a trigonal-pyramidal geometry
b. BCl3
Answer: (b) nonpolar because polar bonds are arranged in a trigonal-planar geometry

49. Valence Bond Theory: Orbital Overlap
In the Lewis theory: covalent bonding occurs when atoms share electrons. Lewis structures and VSEPR do not explain why a bond forms.
In the valence-bond theory:
- Bonds form when orbitals on atoms overlap.
- The overlap of the orbitals allows two electrons of opposite spin to share the common space between the nuclei, forming a covalent bond.

50. Bond Length
Increased overlap brings the atoms together until a balance is reached between the like charge repulsions and the electron-nucleus attraction.
Atoms can’t get too close because the internuclear repulsions get too great.
The bond length of a covalent bond corresponds to the minimum of the potential energy curve.

51. Covalent Bonding and Orbital Overlap
As two nuclei approach each other their atomic orbitals overlap.
As the amount of overlap increases, the energy of the interaction decreases.
At some distance the minimum energy is reached.
The minimum energy corresponds to the bonding distance (or bond length).
As the two atoms get closer, their nuclei begin to repel and the energy increases.
At the bonding distance: the attractive forces between nuclei and electrons just balance the repulsive forces (nucleus-nucleus, electron-electron).

52. Hybrid Orbitals
To explain geometries: we assume that the atomic orbitals on an atom mix to form new orbitals called hybrid orbitals.
Hybridization: The process of mixing atomic orbitals.
The number of hybrid orbitals must equal the number of atomic orbitals.

53. SP Hybrid Orbitals BeF2 Be: Z = 4 The ground state configuration:
Promotion of an electron gives: An excited state configuration:

54. Be can now form two bonds with the two F atoms since F atom has an unpaired electron the p orbital (2p5).
However, this arrangement would not make the two Be – F bonds equal because one would involve and s orbital and one would involve a p orbital.
We can form 2 hybrid sp orbitals which have 2 lobes like a p orbital, but 1 lobe is much larger than the other
These two degenerate orbitals would align themselves 180 from each other ( linear Geometry).
s p sp

55. The hybrid orbitals and bonding in BeF2 is linear in geometry.
So BeF2 is Linear

56. Hybrid Orbitals Notes
Whenever we mix a certain number of atomic orbitals, we get the same number of hybrid orbitals.
Each of the hybrid orbitals is equivalent to the others but points in a different direction.
Thus mixing one 2s and one 2p orbital yields two equivalent sp hybrid orbitals that point in opposite directions.

57. s + 2 p  sp2 Using a similar model for Boron: B: (Z = 5)  1S22S22p3
SP2 Hybridization
Using a similar model for Boron:
B: (Z = 5)  1S22S22p3
s p  sp2

58. The three degenerate orbitals would align themselves 120 from each other (trigonal planar) geometry.
The hybrid orbitals and bonding in BF3 is trigonal planar.

59. SP3 Hybridization s + 3 p  sp3 Using the model for Carbon:
C: (Z = 6)  1S22S22p4
s p  sp3

61. The four degenerate orbitals would align themselves 109
The four degenerate orbitals would align themselves  from each other (tetrahedral) geometry.
The hybrid orbitals and bonding in CH4 is tetrahedral.

62. What Happens with Water?
We started this discussion with H2O and the angle question: Why is it 104.5° instead of 90°?
Oxygen has two bonds and two lone pairs—four electron domains.
The result is sp3 hybridization!

63. Octahedral electron domain geometries require sp3d2 hybridization.
Hybridization Involving d Orbitals
Since there are only three p-orbitals, trigonal bipyramidal and octahedral electron domain geometries must involve d-orbitals.
Trigonal bipyramidal electron domain geometries require sp3d hybridization.
Octahedral electron domain geometries require sp3d2 hybridization.

64. Hypervalent Molecules
The elements that have more than an octet.
Valence-bond model would use d orbitals to make more than four bonds.
This view works for period 3 and below.
Theoretical studies suggest that the energy needed would be too great for this.
A more detailed bonding view is needed than we will use in this course.

65. sp3d Hybridization:
For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids.

66. sp3d2 Hybridization:
s + 3p + 3d  sp3d2
Octahedral

67. Hybridization Summary
Draw the Lewis structure.
Determine the electron domain geometry with VSEPR.
Specify the hybrid orbitals required for the electron pairs based on the electron domain geometry.
4pairs = sp3

68. Electron pairs: electron domains (bonding and nonbonding)

69. Hybrid Orbital Chart

71. Sample Exercise 9.5
Indicate the hybridization of orbitals employed by the central atom in NH2-.
The Lewis structure of NH2– is:
There are four electron domains around N.
The electron-domain geometry is tetrahedral.
The hybridization that gives a tetrahedral electron-domain geometry is sp3

72. Practice Exercise
Predict the electron-domain geometry and the hybridization of the central atom in
(a) SO32-
Answer: (a) tetrahedral, sp3
(b) SF6
b) octahedral, sp3d2

73. Examples – Determine the hybridization on the central atom of each:
NCl3 sp3
CO2 sp
H2O sp3
SF4 sp3d
BF3 sp2
XeF4 sp3d2

74. Examples – Determine the hybridization on EACH atom:
sp2 sp
sp3
sp2

75. Multiple Bonds
-Bonds: electron density lies on the axis between the nuclei.
All single bonds are -bonds.
-Bonds: electron density lies above and below the plane of the nuclei.
A double bond consists of one -bond and one -bond.
A triple bond has one -bond and two -bonds.
Often, the p-orbitals involved in -bonding come from unhybridized orbitals.

76. Sigma () Bonds Sigma bonds are characterized by Head-to-head overlap.
Cylindrical symmetry of electron density about the internuclear axis.

77. Multiple Bonds A pi (p): characterized by
Side-to-side overlap: Overlap of the orbitals is perpendicular to the internuclear axis.
Electron density above and below the internuclear axis.

78. Pi vs. Sigma
Unlike a sigma bond, in a pi bond there is no probability of finding the electron on the internuclear axis.
Since the p orbitals in a pi bond overlap sideways rather than directly facing each other, the total overlap in a pi bond tends to be less than in a sigma bond.
This means that pi bonds tend to be weaker than sigma bonds.

79. Type of Bond Single bonds are sigma bonds.
A double bond consists of 1 sigma bond and 1 pi bond.

80. Type of Bond A triple bond consists of 1 sigma bond and 2 pi bonds.
Since pi bonds occur using unhybridized p orbitals, it can only occur with sp and sp2 hybridization.

81. Sample Exercise 9.6 Formaldehyde has the Lewis structure:
Describe how the bonds in formaldehyde are formed in terms of overlaps of appropriate hybridized and unhybridized orbitals.

82. Explanation:
Structure: Trigonalplanar geometry with bond angles of about 120. It is sp2 hybrid orbitals on C .
These hybrids are used to make the two C—H and one C—O σ bonds to C. There remains an unhybridized 2p orbital on carbon, perpendicular to the plane of the three sp2 hybrids.
The O atom also has three electron domains around it : sp2 hybridization as well.
One of these hybrids participates in the C—O σ bond, while the other two hybrids hold the two nonbonding electron pairs of the O atom. The remain unhybridized 2p orbital form  bond with the unhybridized 2p orbital with carbon

83. Practice Exercise Consider the acetonitrile molecule:
a) Predict the bond angles around each carbon atom.
Answer: (a) approximately 109around the left C and 180on the right C

84. b. Describe the hybridization at each carbon atom.
Answer: (b) sp3, sp
c. Determine the total number of s and p bonds in the molecule.
Answer: ( (c) five σbonds and two πbonds

85. Delocalized Electrons: Resonance
When writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion.

86. Localized and Delocalized Electrons
Localized electrons: belong to the two atoms that form the bond.
When resonance structures exist for a molecule, the electrons move around and are called delocalized.

87. Delocalized p Bonding in Benzene

88. Practice Exercise
Which of the following molecules or ions will exhibit delocalized bonding: SO3, SO32-, H2CO, O3, NH4+?
Answer: SO3 and O3, as indicated by the presence of two or more resonance structures involving π bonding for each of these molecules

89. Sample Integrative Exercise
Elemental sulfur is a yellow solid that consists of S8 molecules. The structure of the S8 molecule is an eight- membered ring. Heating elemental sulfur to high temperatures produces gaseous S2 molecules:
S8(s)  4S2(g)
a) With respect to electronic structure, which element in the second row of the periodic table is most similar to sulfur?
Sulfur: [Ne]3s23p4 electron configuration.
It is expected to be most similar electronically to oxygen (electron configuration, [He]2s22p4).

90. b. Use the VSEPR model to predict the S – S – S bond angles in S8, and the hybridization at S in S8.
sp3hybridization
d. Use average bond enthalpies (Table 8.4) to estimate the enthalpy change for the reaction just described. Is the reaction exothermic or endothermic?
The reaction is endothermic

91. HW Exercises:

92. {وَعَسَى أَنْ تَكْرَهُوا شَيْئًا وَهُوَ خَيْرٌ لَكُمْ وَعَسَى أَنْ تُحِبُّوا شَيْئًا وَهُوَ شَرٌّ لَكُمْ وَاللَّهُ يَعْلَمُ وَأَنْتُمْ لَا تَعْلَمُونَ} [البقرة: 216]}.
* لما مات زوج أم سلمة أبو سلمة رضي الله عنهم جميعاً، تقول أم سلمة رضي الله عنها: سمعت رسول الله صلى الله عليه وسلم عليه وسلم يقول: «ما من مسلم تصيبه مصيبة فيقول ما أمره الله إنا لله وإنا إليه راجعون اللهم أجرني في مصيبتي واخلف لي خيرا منها. إلا أخلف الله له خيرا منها». قالت: فلما مات أبو سلمة، قلت: أي المسلمين خير من أبى سلمة؟ أول بيت هاجر إلى رسول الله صلى الله عليه وسلم عليه؟ ثم إني قلتها، فأخلف الله لي رسول الله صلى الله عليه وسلم عليه! * ذكر الطنطاوي رحمه الله عن صاحب له: أن رجلاً قدم إلى المطار، وكان حريصا على رحلته، وهو مجهد بعض الشيء، فأخذته نومةٌ ترتب عليها أن أقلعت الطائرة، وفيها ركاب كثيرون يزيدون على ثلاث مائة راكب، فلما أفاق إذا بالطائرة قد أقلعت قبل قليل، وفاتته الرحلة، فضاق صدره، وندم ندماً شديداً، ولم تمض دقائق على هذه الحال التي هو عليها حتى أعلن عن سقوط الطائرة، واحتراق من فيها بالكامل! والسؤال ألم يكن فوات الرحلة خيراً لهذاالرجل؟! وفي هذا عبرة وعظة. إن المؤمن عليه: أن يسعى إلى الخير جهده *** وليس عليه أن تتم المقاصد