1. Chapter #10 - The Shapes of Molecules
Depicting Molecules and Ions with Lewis Structures
Using Lewis Structures and Bond Energies to Calculate
Heats of Reaction
Valence-Shell Electron-Pair Repulsion (VSEPR)
Theory( VB ) Theory and Molecular Shape
Molecular Shape and Molecular Polarity

2. Lewis Structures
1) Only the valence electrons appear in a Lewis structure.
2) The line joining two atoms represents a pair of electrons shared between two atoms.
single bond - two shared electrons, one line
double bond - four shared electrons, two lines
triple bond - six shared electrons, three lines
3) Dots placed next to an atom represent nonbonding electrons.

3. Lewis Structures of the Elements by Group in the Periodic TableII
III IV V VI . . . .
. . . . H . . . B
. .
. . C . . O Be . N
. . . . Li
. . . Mg . .
. . . . Na Al . . . .
. . . Si P
. . S . .
. . . Ca
VIII
VII
. .
. .
. .
. . . .
. .
. .
. .
. .
. .
. .
. . F Cl He Ne Ar
. .
. .
. .
. .

4. Fig 10.1 (P 362)

5. Writing Lewis Structures for molecules with one central atom.
Problem: Write a Lewis structure for the molecule CHCl3, Chloroform,
a molecules that has been used to put people to sleep!
Solution: Step 1: Place the atoms next to each other with carbon in the
center, since it is the lowest element in a group with more that one
electron. Place the others around the carbon in the four locations.
Step 2: Count valence electrons.
Step 3: Draw single bonds between the atoms, and subtract 2 electrons
per bond. 26 electrons - 8 electrons = 18 electrons.
Step 4: Distribute the remaining electron in pairs beginning with the
surrounding atoms.
Check:
[1xC(4e-)] [1xH(1e-)] [3xCl(7e-)] = 26 electrons 1. C Cl H C Cl H C Cl H 3. 4. .. .. .. .. .. .. .. .. ..

6. Lewis Structures of Simple Molecules - I
. . .
. .
. . .
. .
H Cl .
. .
. .
. .
. . .
F F
H H
. .
. .
Hydrogen Chloride
Molecular Fluorine
. .
H H
. .
. .
C C H .
. .
. .
F F
. .
. .
H H
. .
H F
. .
H H
Hydrogen Fluoride
Ethane
Molecular
Hydrogen - -
. .
. . -
. .
. .
. .
. .
. . . Cl
Mg+2 Cl
. .
. . .
. .
Na+ Cl
. .
Magnesium
Chloride
Sodium Chloride

7. .. .. .. .. Writing lewis Structures for Molecules with more than one
Central atom!
Problem: Write the Lewis structure for Hydrogen peroxide(molecular
formul, H2O2 ) an important household bleech.
Solution: Step 1. Place the atoms in the best geometry, with the
hydrogen atoms having only one bond, they are on the ends or outside,
and oxygen can have up to two bonds so put them in the middle.
H O O H
Step 2. Find the sum of electrons: [2 x H(1e-)] + [2 x O(6e-)] = 14e-
Step 3. Add single bonds and subtract 2e- for each bond:
H - O - O - H
14e- - 6e- = 8e-
Step 4. Add the remaning electron in pairs around the oxygen atoms as
Hydrogen can only have two! .. ..
H - O - O - H .. ..
Check: oxygen has an octet of 8e-
and hydrogen ha sit’s two electrons.

8. Writing Lewis Structures for Molecules with Multiple Bonds.
Problem: Write lewis structures for Oxygen and Acetylene(C2H2):
Plan: We begin with the first 4 steps we have done: placing atoms,
counting electrons, placing single bonds, and completing octets, and if
needed we finishas follows by placing multiple bonds in the molecules.
Solution: .. .. ..
a) For oxygen: O O - O .. ..
Change one of the lone pairs to a bonding pair. The oxygen on the
right has an octet of electrons, while the oxygen aton on the left
only has six electrons, so we convert the lone pair to another bonding
pair between the two oxygen atoms. .. .. ..
O O .. .. ..
b) For Acetylene: C2H2
H - C - C - H
Neither of the carbon atoms has an octet, or if they are placed around
one atom, the other has only 4! so no octet! Therefore place both pairs
into forming multiple bonds, a tripple bond!
H - C C - H

9. Lewis Structures of Simple Molecules - II.. .. H C H O .. .. .. H C H ..
Cl Cl H
Cl2 Chlorine
C2H4O2 Acetic Acid
CH4 Methane .. .. .. .. .. O C O .. .. . .. .. Cl ..
CO2 Carbon Dioxide .. .. C O . .. .. .. .. Cl C Cl .. .. ..
CO Carbon Monoxide O Cl H H
H2O Water
Hydrogen Oxide
CCl4 Carbon Tetrachloride

10. Lewis Structures of Simple Molecules - III
Multiple Bonds
. .
N N
. .
Nitrogen N2
. .
H C N
H C C H
Hydrocyanic acid
Hydrogen Cyanide : HCN
Acetylene : C2H2
Molecular Oxygen : O2
Ethylene : C2H4
. .
O O H H
C C
. .
O O
. . H H
. .

11. Writing Lewis Structures - IV
Step 1) Place the atoms relative to each other: For compounds of
formula ABn , place the atom with the lower group number in
the center, the one that needs more electrons to attain an octet.
In NF3 (nitrogen trifluoride), the N (Group 5A) has five
electrons so it needs three, whereas F (Group 7A) has seven so
it needs only one; thus, N goes in the center with the three
F atoms around it.
Step 2) Determine the total number of valence electrons available:
For molecules, add up the valence electrons of all atoms (the
number of valence electrons equals the A-group number). In
NF3, N has five valence electrons, and each F has seven. For
polyatomic ions, add one e - for for each negative charge, or
subtract one e - for each positive charge.

12. Writing Lewis Structures - V
Step 3) Draw a single bond from each surrounding atom to the
central atom, and subtract two valence electrons for each
bond. There must be at least a single bond between bonded
atoms.
Step 4) Distribute the remaining electrons in pairs so that each
atom obtains eight electrons (or two for H). Place lone pairs
on the surrounding (more electronegative) atoms first to give
each an octet. If electrons remain, place them around the central
atom. Then check that each atom has 8e -.

13. Lewis Structures of Simple Molecules - VIH H H ..
CH4 H C C .. O H
Methane H H
Ethyl Alcohol (Ethanol) .. .. .. .. C F .. O K+ .. .. Cl .. .. .. .. .. O .. O
CF4 ..
KClO3
Potassium Chlorate
Carbon Tetrafluoride

14. Lewis Structures of Simple Molecules - VIIH H H N
. . H H
. .
. . N N
Ammonia H H C H +
. . O
. . H N H
Urea H
Ammonium Ion

15. (P 365)

16. Resonance:
Delocalized Electron-Pair Bonding - I
Ozone : O3 .. .. .. .. .. O O .. .. .. .. .. .. O O .. O O II I
Resonance Hybrid Structure .. O .. .. .. .. O O
One pair of electron’s resonances
between the two locations!!

17. Delocalized Electron-Pair Bonding - II
Resonance:
Delocalized Electron-Pair Bonding - II H H C C C C H C C H H H H C C H C H C H H C C C H H C H C H C H H C C
Benzene
Resonance Structure H

18. Lewis Structures of Simple Molecules
Resonance Structures -III
Nitrate .. .. .. O .. N .. .. O .. .. O .. .. .. O .. .. O N .. .. N .. .. .. O .. .. O .. .. .. .. O O

19. Lewis Structures of Simple Molecules -VIII
Determine the Lewis structure of molecular Nitrogen, N2
N2 is a covalent compound.
There are 10 valence electrons.
N-N use 2 e-, leaving 8 around the 2 atoms.
Three pairs are placed around one atom, leaving 1 pair.
Provisional structure:
N N Calculate FC
Formal Charge
N = 5 valence -(1 bonding
+ 2 nonbonding) = +2
+ 6 nonbonding) = -2
Move electrons in to make a triple bond.
N N
. .
. .
. .
. .
. .
. .

20. Lewis Structures for Octet Rule Exceptions.. .. .. B Cl .. .. .. .. F .. .. .. .. F Cl .. .. F
Each Chlorine atom has
8 electrons associated.
Boron has only 6!
Each Fluorine atom has
8 electrons associated.
Chlorine has 10 electrons! . .. .. .. .. .. .. N .. .. Cl Be Cl .. .. .. O O
Each Chlorine atom has
8 electrons associated. The
Beryllium has only 4 electrons.
NO2 is an odd electron atom.
The nitrogen has 7 electrons.

21. Resonance Structures - IV
Expanded Valence Shells .. .. .. .. .. .. .. .. .. .. .. .. .. .. F F F F .. .. .. .. .. .. F S F .. P .. F .. .. .. .. .. .. F .. .. .. .. F F .. F
p = 10e-
S = 12e-
Sulfur hexafluoride ..
Phosphorous pentafluoride .. .. O S H .. O S H .. ..
Resonance Structures .. .. .. .. ..
Sulfuric acid
S = 12e-

22. Lewis Structures of Simple Molecules
. .
. . -2
Sulfate O
. .
. .
Resonance Structures-V
. .
. . O S O
. .
. . -2
. .
. . O
o *
o o O
o o
Plus 4 others
for a total of 6
x o o o
o o x
. .
o * O
x o S O
o o x
o o -2
. . O
. .
o o
x x
. .
. .
o o O
o o O S O
. .
. .
o o
. . O
. .
. .
x = Sulfur electrons
o = Oxygen electrons

23. Fig 10.2 (P 371)

24. Fig 10.3 (P 372)

25. Calculating H from Bond Energies - I
Problem: Using the bond energies in Table 9.2, calculate the H of
the reaction between methane and chlorine and fluorine to
give freon-12 (CCl2F2).
Plan: Look up the bond energies of the reactants and products in
table 9.2, and subtract the product bands from the reactant bonds.
Solution:
CH4 (g) +2 Cl2 (g) + 2 F2 (g) CF2Cl2 (g) + 2 HF(g) + 2 HCl (g)
Reactant bonds broken:
For Methane : mol C - H bonds
For Molecular Chlorine : 2 mol Cl - Cl bonds
For Molecular Fluorine : mol F - F bonds
Product bonds formed:
For Freon - 12 : mol C - F bonds , 2 mol C - Cl bonds
For HF : mol H - F bonds
For HCl : mol H - Cl bonds

26. Calculating H from Bond Energies - II
Solution cont.
Reactant bonds broken:
4 mol C - H bonds = 4 mol x 413 kJ/mol = 1652 kJ
2 mol Cl - Cl bonds = 2 mol x 243 kJ/mol = kJ
2 mol F - F bonds = 2 mol x 159 kJ/mol = kJ
H0bonds broken = _________
Product bonds formed:
2 mol C - F bonds = 2 mol x 453 kJ/mol = kJ
2 mol C - Cl bonds = 2 mol x 339 kJ/mol = 678 kJ
2 molH - F bonds = 2 mol x 565 kJ/mol = kJ
2 mol H - Cl bonds = 2 mol x 427 kJ/mol = 854 kJ
H0bonds formed = ________
H0rxn = H0bonds broken H0bonds formed
H0rxn = _______ - _________ = _________

27. Fig 10.4 (P 375)
Two Three Four Five Six
Number of Electron Groups

28. Fig 10.5 (P 375)

29. Fig 10.6 (P 375)

30. AX2 Geometry - Linear .. .. .. .. .. ..
Molecular Geometry =
Linear Arrangement Cl Be Cl
BeCl2
1800
GaseousBeryllium Chloride is an example of a molecule in which the
central atom - Be does not have an octet of electrons, and is electron
deficient.
Other alkaline earth elements also have the same valence electron
configuration, and the same geometry for molecules of this type.
Therefore this geometry is common to group II elements. .. .. .. .. O C O
CO2
1800
Carbon Dioxide also has the same geometry, and is a linear molecule,
but in this case, the bonds between the carbon and oxygens are double
bonds.

31. Fig 10.7 (P 376)

32. AX3 Geometry - Trigonal Planar.. .. .. .. .. ..
All of the Boron Family(IIIA)
elements have the same
geometry. Trigonal Planar ! F F
BF3 B
Boron Trifluoride
1200 .. .. .. F
AX2E SO2 .. - .. .. .. .. O S O .. ..
NO3- ..
1200 .. N .. .. .. .. O O
The AX2E molecules have a pair of
Electrons where the third atom
would appear in the space around the
central atom, in the trigonal planar
geometry.
1200
Nitrate Anion

33. (P 376)

34. (P 376)

35. (P 376)

36. (P 376)

37. (P 377)

38. Fig 10.8 (P 377)

39. Fig 10.9
(P 377)

40. The effect of lone pair replusion on bond angle.

41. (P 378)

42. (P 378)

43. Fig10.10 (P 378)

44. (P 378)

45. (P 379)

46. (P 379)

47. (P 379)

48. Fig (P 379)

49. (P 379)

50. (P 380)

51. (P 380)

52. Using VSEPR Theory to Determine Molecular Shape
1) Write the Lewis structure from the molecular formula to see the
relative placement of atoms and the number of electron groups.
2) Assign an electron-group arrangement by counting all electron
groups around the central atom, bonding plus nonbonding.
3) Predict the ideal bond angle from the electron-group arrangement
and the direction of any deviation caused by the lone pairs or double
bonds.
4) Draw and name the molecular shape by counting bonding groups
and non-bonding groups separately.

53. Fig (P 380)

54. Predicting Molecular Shapes
Problem: Determine the molecular shape and ideal bond angles for:
a) NCl3 b) COCl2
Solution:

55. Predicting Molecular Shapes
Problem: Determine the molecular shape and ideal bond angles for:
a) NCl3 b) COCl2
Solution: a) for NCl3
1) Write the Lewis structure: .. .. .. .. N Cl ..
2) Assign the electron arrangement:
Four electron groups around N,
( three bonding, and one non-bonding), so we have the tetrahedral
arrangement.
3) For the tetrahedral arrangement, the ideal angle is Since
there is one lone pair, the actual bond angle should be less than
4) Draw and name the molecular shape: .. .. N
NCl3 Has a trigonal pyramidal shape .. .. .. Cl .. .. Cl .. .. .. Cl

56. .. ..
Solution:
b) for COCl2
1) Write the Lewis structure: O .. .. .. .. Cl .. C .. Cl
2) Assign the electron-group arrangement: three electron groups around
carbon ( two single, and one double bond ) which gives the trigonal
planar arrangement.
3) Predict the bond angles: the ideal angle is 1200, but the double bond
between the Carbon and the oxygen should compress the Cl - C - Cl
bond angle by repelling the chlorine atoms, and the bonds between
them and the carbon atom.
4) Draw and name the molecular shape: .. .. O
124.50 .. .. C .. .. .. Cl .. Cl
1110

57. Predicting Molecular Shapes with five or six
Electron Groups
Problem: Determine the molecular shape and predict the bond angles
( relative to the ideal angles ) of (a) AsI5 (b) BrF5
Solution: .. .. .. .. .. I
(a) 1)Lewis structure for AsI5: .. .. .. .. I I
2) Electron group
arrangement with five
groups, this is the
trigonal bipyramidal
arrangement. .. As .. .. .. .. .. .. .. .. I .. I I .. .. I
900 .. .. ..
1200 As .. I ..
3) Bond angles: since all the groups and
surrounding atoms are indentical, the
bond angles are ideal: 1200 between equatorial
groups and 900 between axial and equatorial groups. .. I .. .. .. I
4) Molecular arrangement: Trigonal bipyramidal

58. .. .. .. .. ..
b) BrF5
1) Lewis structure for BrF5:
2) Electron Group arrangement
6 electron groups- octhedral!
3) Bond angles: Lone pair should
make all angles less than 900.
4) Molecular shape: one lone pair,
and five bonding pairs give square pyrimidal: F .. .. .. F .. .. F .. .. .. .. .. F Br .. F .. .. .. .. .. F .. .. .. F .. F .. .. .. .. .. .. F Br .. F

59. Fig (P 382)

60. (P 383)

61. (P 383)

62. (P 383)

63. (P 383)

64. Fig (P 384)

65. (P )

66. (P 386)

67. .. .. .. Predicting The Polarity of Molecules -I
Problem: From electronegativity, and their periodic trends, predict
whether each of the following molecules is polar and show the direction
of bond dipoles and the overall molecular dipole.
(a) Phosphine, PH3
(b) Carbon Disulfide, CS2 (atom sequence SCS)
(c) Auminum Chloride, AlCl3
Plan: First we draw and name the molecular shape. Then using relative
EN values, we decide on the direction of each bond polarity. Finally we
decide upon the polarity of the molecule based upon the geometry.
Solution: .. .. ..
(a) P P P H H H H H H H H H
Molecular Dipole
Bond dipoles
Molecular shape

68. (II)
(b) Carbon Disulfide, CS2 .. .. .. .. .. .. S C .. S C .. S C ..
Molecular shape
Bond dipoles
No -Molecular dipole
(c) Aluminum Chloride, AlCl3 .. .. .. .. .. .. .. .. .. Cl Cl Cl Al Al Al .. .. .. .. .. .. .. .. .. .. .. .. .. Cl Cl .. .. Cl .. Cl .. Cl .. Cl
Molecular shape
Bond dipoles
No - Molecular dipole

69. Seven Primary Odors or Olfactory Receptor Sites
1. Camphor - like
2. Musky
3. Floral
4. Pepperminty
5. Etheral
6. Pungent
7. Putrid

70. Fig 10A (P 387)

71. Fig 10B (P 387)

72. Fig 10C (P 388)