1. Molecular Bonding
Unit 5

2. Ionic bonding
Cations (usually metals) lose electrons to anions (monoatomic and polyatomic).
Both cations and anions are stable like noble gases (octet rule)
Strong electrostatic bonds

3. Example of an ionic bond

4. Covalent Bonds Sharing pairs of electrons
Covalent bonds are the inter-atomic attraction resulting from the sharing of electrons between atoms.
They result in ‘localized overlaps’ of orbitals of different atoms.
They also are the result of the attraction of electrons for the nucleus of other atoms.
Typical of molecular substances.

5. Example of a Covalent Bond

6. Covalent Bonds Cont. Atoms bond together to form molecules
molecules are electrically neutral groups of atoms joined together by covalent bonds
strong attraction
Molecules attracted to each other weakly form molecular compounds

7. Properties of Molecular Compounds
Strong covalent bonds hold the atoms together within a molecule.
The intermolecular forces that hold one molecule to another are much weaker.
Properties vary depending on the strength of the intermolecular forces.

8. Lewis structures
A Lewis structure is a representation of the valence e- in an atom, ion or molecule.
Element symbol represents nucleus and core e-.
Dots represent valence e-.

10. Electron pairs
In covalent compounds electrons are shared between atoms creating electron pairs.
Bonding pairs: e- that are shared between 2 atoms.
Lone or unshared pairs: e- that are NOT involved in bonding.

11. Covalent bonds Hydrogen follows the duet rule: sharing 2 electrons.
Non-metals Carbon through Fluorine follow the octet rule: sharing 8 electrons.

12. Writing Lewis Structures of Molecules
1. Determine the central atom (atom in the middle)
- usually is the “single” atom
- least electronegative element
- H never in the middle; C always in the middle
2. Count the total number of valence e- (group #)
- add ion charge for “-”
- subtract ion charge for “+”

13. 3. Divide the total number of electrons by 2
- sharing involves 2 electrons
Attach the atoms together with one pair of electrons
All remaining e- = LONE PAIRS!
- lone pairs are NOT involved in bonding

14. Writing Lewis Structures Cont.
6. Place lone pairs around non-central atoms to fulfill the “octet rule” - some elements may violate this octet rule – (H=2, Be=4, B=6) 7. If more e- are still needed, create double or triple bonds around the central atom. - single = 1 pair of shared electrons (2 e-) - double = 2 pair of shared electrons (4 e-) - triple = 3 pair of shared electrons (6 e-)

15. Practice
Give the Lewis structure for:
HCl
NH3
C2H6
CO2

16. NAS
When drawing Lewis structures, remember: terminal atoms, atoms that can only make one bond, must be on the outside.
N – A = S
#e- needed for – #e- available = #e- shared
octet or duet (valence e-) (bonds)
S = the number of pairs of shared electrons
S = the number of bonds (a dash may be used to
represent a pair of shared electrons)

17. Resonance •• • O S O O S O •• •
When there is more than one Lewis structure for a molecule that differ only in the position of the electrons they are called resonance structures
Lone Pairs and Multiple Bonds in different positions
Resonance only occurs when there are double bonds and when the same atoms are attached to the central atom
The actual molecule is a combination of all the resonance forms. •• •
O S O
O S O •• •

18. Some molecules are stable when the center atom has more than an octet
Some molecules are stable when the center atom has more than an octet. i.e. SF6, PCl5
P has 10 electrons
sulfur has 12 electrons

19. Exceptions to the Octet Rule
H, Be, B (stable with 2, 4, and 6 e-, respectively)
Some molecules cannot be drawn with the Lewis structure rules, due to odd # of e-. i.e. NO & NO2 (There is no way for N to get an octet)

20. SAMPLE PROBLEM:
Writing Resonance Structures
PROBLEM:
Write resonance structures for the nitrate ion, NO3-.
PLAN:
After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms.
SOLUTION:
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
N does not have an octet; a pair of e- will move in to form a double bond.

22. Four criteria for choosing the more important resonance structure:
1. Smaller formal charges (either positive or negative) are preferable to larger charges;
2. A more negative formal charge should exist on an atom with a larger EN value.
3. Get unlike charges as close together as possible
4. Avoid like charges (+ + or - - ) on adjacent atoms

23. Resonance and Formal Charge
Formal charge of atom =
# valence e- =
(# unshared electrons +
1/2 the # shared electrons)

24. Nitric acid : .. H O N ..
We will calculate the formal charge for each atom in this Lewis structure. 19 24

25. Nitric acid Formal charge of H : .. H O N ..
Hydrogen shares 2 electrons with oxygen.
Assign 1 electron to H and 1 to O.
A neutral hydrogen atom has 1 electron.
Therefore, the formal charge of H in nitric acid is 0. 19 25

26. Nitric acid : .. H O N Formal charge of O ..
Oxygen has 4 electrons in covalent bonds.
Assign 2 of these 4 electrons to O.
Oxygen has 2 unshared pairs. Assign all 4 of these electrons to O.
Therefore, the total number of electrons assigned to O is = 6. 19 26

27. Nitric acid : .. H O N Formal charge of O .. Electron count of O is 6.
A neutral oxygen has 6 electrons.
Therefore, the formal charge of oxygen is 0. 19 27

28. Nitric acid : .. H O N Formal charge of O ..
Electron count of O is 6 (4 electrons from unshared pairs + half of 4 bonded electrons).
A neutral oxygen has 6 electrons.
Therefore, the formal charge of oxygen is 0. 19 28

29. Nitric acid : .. H O N Formal charge of O ..
Electron count of O is 7 (6 electrons from unshared pairs + half of 2 bonded electrons).
A neutral oxygen has 6 electrons.
Therefore, the formal charge of oxygen is -1. 19 29

30. Nitric acid : .. H O N Formal charge of N – ..
Electron count of N is 4 (half of 8 electrons in covalent bonds).
A neutral nitrogen has 5 electrons.
Therefore, the formal charge of N is +1. 19 30

31. Nitric acid Formal charges : .. H O N + – ..
A Lewis structure is not complete unless formal charges (if any) are shown. 19 31

32. Formal Charge An arithmetic formula for calculating formal charge.
Number of
valence
electrons
number of bonds
number of unshared electrons – – 16 32

33. "Electron Counts" and Formal Charges in NH4+ and BF4-.. B F : 1 4 N H + 7 – 4 17 33

34. Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge of atom =
# valence e-
- # unshared electrons
- 1/2 # shared electrons

35. For OB
# valence e- = 6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
For OC
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
For OA
# valence e- = 6
# nonbonding e- = 4
# bonding e- = 4 X 1/2 = 2
Formal charge = 0

36. Resonance and Formal Charge
EXAMPLE: NCO- has 3 possible resonance forms - A B C

37. Now Determine Formal Charges-2 +1 -1 -1
Forms B and C have smaller formal charges; this makes them more important than form A. (rule 1)
Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid. (rule 2)

38. Coordinate Covalent Bond
A covalent bond in which one atom contributes both bonding electrons.

39. Predicting Molecular Geometry
VSEPR Theory
Valence Shell Electron Pair Repulsion
The shape around the central atom(s) can be predicted by assuming that the areas of electrons on the central atom will repel each other

40. Each Bond counts as 1 area of electrons
single, double or triple all count as 1 area
Each Lone Pair counts a 1 area of electrons
Even though lone pairs are not attached to other atoms, they do “occupy space” around the central atom
Lone pairs generally “push harder” than bonding electrons, affecting the bond angle

41. Shapes
Straight Line
molecule made up of only 2 atoms

42. Shapes Linear Trigonal Planar 180°
2 atoms on opposite sides of central atom, no lone pairs around CA
180° bond angles
Trigonal Planar
3 atoms form a triangle around the central atom, no lone pairs around CA
Planar
120° bond angles
180°
120°

43. Tetrahedral
4 surrounding atoms form a tetrahedron around the central atom, no lone pairs around the CA
109.5° bond angles
109.5°

44. Shapes Trigonal Pyramidal V-shaped or Bent
3 bonding areas and 1 lone pair around the CA
Bond angle = 1070
V-shaped or Bent
2 bonding areas and 2 lone pairs around the CA
bond angle =

48. Polar Bonds: Electronegativity
Measure of the ability of an atom to attract shared electrons
Larger electronegativity means atom attracts more strongly
Values 0.7 to 4.0
Increases across period (left to right) on Periodic Table
Decreases down group (top to bottom) on Periodic Table

49. Larger difference in electronegativities means more polar bond
negative end toward more electronegative atom
Covalent bonding between unlike atoms results in unequal sharing of the electrons
One end of the bond has larger electron density (more electronegative) than the other
Polar covalent – unequal sharing
Nonpolar covalent – equal sharing

50. Bond Polarity • H F d+ d- The result is bond polarity
The end with the larger electron density gets a partial negative charge
The end that is electron deficient gets a partial positive charge
H F • d+ d-

51. Dipole Moment
Bond polarity results in an unequal electron distribution, resulting in areas of partial positive and partial negative charge
Any molecule that has a center of positive charge and a center of negative charge in different points is said to have a dipole moment

52. If a molecule has more than one polar covalent bond, the areas of partial negative and positive charge for each bond will partially add to or cancel out each other
The end result will be a molecule with one center of positive charge and one center of negative charge
The dipole moment affects the attractive forces between molecules and therefore the physical properties of the substance

53. Charge distribution in the water molecule

54. Polarity of Molecules Molecule will be NONPOLAR if:
the bonds are nonpolar (Br-Br, F-F)
there are no lone pairs around the central atom and all the atoms attached to the central atom are the same
Molecule will be POLAR if:
the central atom has lone pairs

55. Bond Length Distance between the nuclei of two bonded atoms.
Average value
Same bond get shorter when bond doubles or triples (C-C: 154 pm; C=C: 134 pm)

56. Bond Dissociation Energy
Energy required to break a single bond
Ex  H-H: kJ  H● + H●
435 kJ of energy/mole of H2
Shorter bond = Stronger bond = higher energy
Calculate the amount of energy needed to dissociate 1 mole of C2H6.
C-C: 348 kJ/mol; C-H: 398 kJ/mol

57. Enthalphy of reaction Breaking bonds: endothermic
Formation of bonds: exothermic
Calculate the enthalpy (ΔrH) of the combustion of 1 mol CH4
Break: 4 x C-H, 2 x O=O: kJ/mol
Form: 2 x C=O, 4 x O-H: kJ/mol
ΔrH = -694 kJ/mol

58. Bondlength (pm) and bond energy (kJ/mol)
H--H 74
435
H--C
109
393
C--C
154
348
H--N
101
391
N--N
145
170
H--O 96
366
O--O
148
H--F 92
568
F--F
142
158
H--Cl
127
432
Cl-Cl
199
243
H--Br
141
Br-Br
228
193
H--I
161
298
I--I
267
151
C=C
134
614
C--N
147
308
CºC
120
839
C--O
143
360
C=O
121
736
O=O
498
C--F
135
488
C--Cl
177
330
NºN
110
945

59. Overlap of Orbitals

60. The degree of overlap is determined by the system’s potential energy
equilibrium bond distance
The point at which the potential energy is a minimum is called the equilibrium bond distance

61. Formation of sp hybrid orbitals
The combination of an s orbital and a p orbital produces 2 new orbitals called sp orbitals. 2s
These new orbitals are called hybrid orbitals
The process is called hybridization
What this means is that both the s and one p orbital are involved in bonding to the connecting atoms

62. Hybridization refers to a mixture or a blending
Biology – refers to genetic material
Chemistry – refers to blending of orbitals
Remember, orbitals can only predict an area in space where an e- may be located.
Sometimes blending orbitals can produce a lower, more stable bonding opportunity.
Orbital hybridization occurs through e- promotion in orbitals that have similar energies (i.e. same energy level).

63. Hybridization Cont.
Hybridization occurs WITHIN the atom to enhance bonding possibilities.
Do not confuse this concept with orbital overlap (bonding).
Hybridization is a concept used to explain observed phenomenon about bonding that can’t be explained by dot structures.
EXAMPLES – draw box diagrams for Be, B, and C (use noble gas core).

67. How do I know if my CA is hybridized?
If your CA is B, Be, C, Si, or Al then it is hybridized.
If your molecule has multiple bonds in it then it is hybridized.
Double bonds – sp2 hybridized
Triple bonds – sp hybridized

68. Formation of sp2 hybrid orbitals

69. Formation of sp3 hybrid orbitals

70. Hybrid orbitals can be used to explain bonding and molecular geometry

71. Multiple Bonds
Everything we have talked about so far has only dealt with what we call sigma bonds
Sigma bond (s)  A bond where the line of electron density is concentrated symmetrically along the line connecting the two atoms.

72. Pi bond (p)  A bond where the overlapping regions exist above and below the internuclear axis (with a nodal plane along the internuclear axis).

73. Example: H2C=CH2

74. Example: H2C=CH2

75. Example: HCCH

76. Delocalized p bonds
When a molecule has two or more resonance structures, the pi electrons can be delocalized over all the atoms that have pi bond overlap.

77. Example: C6H6 benzene
Benzene is an excellent example.  For benzene the p orbitals all overlap leading to a very delocalized electron system
In general delocalized p bonding is present in all molecules where we can draw resonance structures with the multiple bonds located in different places.