1. Energy, Bonds & Chemical Structure
Chapter 7, 8 & 9

2. Determination of Atomic Radius:
Half of the distance between nuclei in covalently bonded diatomic molecule "covalent atomic radii"

3. Determination of Atomic Radius:
Periodic Trends in Atomic Radius
Radius decreases from left to right across a period
Increased effective nuclear charge due to decreased shielding
Radius increases down a group
Addition of principal quantum levels

4. Table of Atomic Radii

5. Ionic Radii Cations Positively charged ions Lose electrons
Smaller than the corresponding atom
Anions
Negatively charged ions
Gains electrons
Larger than the corresponding atom

6. Table of Ion Sizes

7. Ionization Energy
The energy required to remove one mole electrons from a mole of gaseous atoms to produce one mole of gaseous ions
M(g)  M+(g) + e-
Second ionization energy – energy change accompanying M+(g)  M2+(g) + e-
Measured in kJ/mol
Positive energy values – requires energy to remove electrons (endothermic)

8. Ionization Energy
Magnitude is determined by the attraction of the positive nucleus for the negative electrons that are being removed
Attraction is dependant on
Nuclear energy
Shielding effect of inner electrons

9. Ionization of Magnesium
Mg kJ  Mg+ + e-
Mg kJ  Mg e-
Mg kJ  Mg e-

10. Electron Affinity
The energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous ions
X(g) + e-  X-
Measured in kJ/mol

11. Electron Affinity Affinity tends to increase across a period
Affinity tends to decrease as you go down in a period
Electrons farther from the nucleus experience less nuclear attraction
Some irregularities due to repulsive forces in the relatively small p orbitals

12. Lattice Energy Atomization of Lithium +155.2 kJ/mol
Atomization of fluorine kJ/mol
Ionization of lithium kJ/mol
Electron affinity of fluorine -328 kJ/mol
Lattice Energy for LiF kJ/mol

13. Electronegativity
A measure of the ability of an atom in a chemical compound to attract electrons
Electronegativities tend to increase across a period
Electronegativities tend to decrease down a group or remain the same

14. Periodic Tables of Electronegativities

15. Example
Which bond is more polar? H—F Na—Cl C—H

16. Ionic bonds Electrons are transferred
Electronegativity differences are generally greater than 2.0
The formation of ionic bonds is always exothermic!

17. General Rule
In most chemical compounds, atoms are reacting so that they have the same electronic structure as a noble gas (they are “isoelectronic” with a noble gas).
General Rule For any set of isoelectronic atoms, the highest nuclear charge (Z) will be smallest, because that nucleus can pull on the electron clouds the hardest.
Ex Put these species in order of increasing size: Se2-, Br-, Rb+ and Sr2+

18. Covalent Bonds
Covalent bonds – sharing of electrons to form discrete molecules
Polar-Covalent bonds
Electrons are unequally shared
Electronegativity difference between 0 and 2.0
Nonpolar-Covalent bonds
Electrons are equally shared
Electronegativity difference of 0

19. Calculating Bond Energy
Values come from a table (Pg 372 in your book)
Energy (∆H) is the sum of the energy required to break old bonds plus the sum of the energy required to make new bonds
Old bonds have a positive sign (require energy to break)
New bonds have a negative sign (take in energy to make)

20. Bond Energy Bond energy is given in kJ/mol
More bonds (double or triple) contain more energy and are harder to break
If the overall value of ∆H is positive, energy must be added into the reaction (endothermic)
If the overall value of ∆H is negative, energy is released into the environment (exothermic)

21. Example
Methane is burned in oxygen gas creating water and carbon dioxide. Calculate the ∆H for the reaction.
Write the reaction.
Figure out what bonds are in each molecule.
Figure out how many of each bond there are and what the energy for each is (Use Pg. 372)
∆H = Σ(reactants) – Σ(products)

22. Lewis Structures
Shows how valence electrons are arranged among atoms in a molecule.
Reflects central idea that stability of a compound relates to noble gas electron configuration.

23. The Octet Rule
Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level.
Bonding for fluorine

24. Drawing Lewis Structures
1. Calculate the total number of valence electrons for the compound
- don’t forget to take charge into account
2. Decide which atom is the central atom
- if this is not obvious, use the least electronegative atom available
3. Use a line to indicate covalent bonds to the outer atoms
4. Arrange the rest of the electrons around the atoms to form octets
5. Form multiple bonds (double or triple bonds/lines) if necessary to complete the octets

25. Completing a Lewis Structure
Draw Lewis structures for the following
Fluorine gas
Hydrochloric acid
Carbon tetrabromide
PCl6-
NH4+

26. Comments on the Octet Rule
2nd row elements C, N, O, F observe the octet rule (HONC rule as well).
2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive.
3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.
When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.

27. Multiple Covalent Bonds: Double Bonds
Two pairs of shared electrons

28. Multiple Covalent Bonds: Triple Bonds
Three pairs of shared electrons

29. Resonance
Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule.
The actual structure is an average of the resonance structures.
Benzene, C6H6
The bond lengths in the ring are identical, and between those of single and double bonds.

30. Resonance, Bond Length and Bond Energy
Resonance bonds are shorter and stronger than single bonds.
Resonance bonds are longer and weaker than double bonds.

31. Resonance in Ozone, O3 Neither structure is correct.
Oxygen bond lengths are identical, and intermediate to single and double bonds

32. Resonance in Polyatomic Ions
Resonance in a carbonate ion:
Resonance in an acetate ion:

33. Formal Charge Formal Charges a method of determining
a. which resonance structure is most likely
b. where the electrons are most concentrated in a structure
Formal charge of an atom is
(Family # ) – ( # of lone pair e-) - (# of bonds on that atom)

34. Formal Charge
Structures that have formal charges closest to zero and any negative formal charges on the most electronegative atoms are most likely to be stable

35. Example Ex Which is the most likely Lewis structure for CO2 ?
Ex For the SCN- ion?

36. Localized Electron Model
Lewis structures are an application of the “Localized Electron Model”
L.E.M. says: Electron pairs can be thought of as “belonging” to pairs of atoms when bonding
Resonance points out a weakness in the Localized Electron Model.

37. VSEPR – Valence Electron Pair Repulsion
X + E
Overall Structure
Forms 2
Linear
AX2 3
Trigonal Planar
AX3 , AX2E 4
Tetrahedral
AX4 , AX3E , AX2E2 5
Trigonal bipyramidal
AX5 , AX4E , AX3E2 , AX2E3 6
Octahedral
AX6 , AX5E , AX4E2
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A

38. VSEPR - Linear
AX2
Example:
CO2

39. VSEPR – Trigonal Planar
AX3 BF3
AX2E SnCl2

40. VSEPR - Tetrahedral
AX4 CCl4
AX3E PCl3
AX2E2 Cl2O

41. VSEPR – Trigonal Bipyramidal
AX5 PCl5
AX4E SF4
AX3E2 ClF3
AX2E3 I3-

42. VSEPR - Octahedral
AX6 SF6
AX5E BrF5
AX4E2 ICl4-

43. Sigma and Pi Bonds
Sigma () bonds exist in the region directly between two bonded atoms.
Pi () bonds exist in the region above and below a line drawn between two bonded atoms.
Single Bond
1 sigma bond
Double Bond
1 sigma, 1 pi bond
Triple Bond
1 sigma, 2 pi bonds

44. Sigma and Pi Bonds
Single Bonds
1 s bond

45. Sigma and Pi Bonds
Double Bonds
1 p bond
H H
C C
H H
1 s bond

46. Sigma and Pi Bonds
Triple Bonds
1 s bond
1 p bond
1 p bond

47. The De-Localized Electron Model
Pi bonds () contribute to the delocalized model of electrons in bonding, and help explain resonance
Electron density from  bonds can be distributed symmetrically all around the ring, above and below the plane.

48. Hybridization The blending of orbitals
poodle cocker spaniel = cockapoo
S orbital p orbital = sp orbital

49. What proof exists for hybridization?
We have studied electron configuration notation and the sharing of electrons in the formation of covalent bonds.
Lets look at a molecule
of methane, CH4.
Methane is a simple natural gas. Its molecule has a carbon atom at the center with four hydrogen atoms covalently bonded around it.

50. Carbon Ground State Configuration
What is the expected orbital notation of carbon in its ground state?
Can you see a problem with this?
(Hint: How many unpaired electrons does this carbon atom have available for bonding?)

51. Carbon’s Bonding Problem
You should conclude that carbon only has TWO electrons available for bonding. That is not enough!
How does carbon overcome this problem so that it may form four bonds?

52. Carbon’s Empty Orbital
The first thought that chemists had was that carbon promotes one of its 2s electrons…

53. A Problem Arises…
However, they quickly recognized a problem with such an arrangement…
Three of the carbon-hydrogen bonds would involve an electron pair in which the carbon electron was a 2p, matched with the lone 1s electron from a hydrogen atom. 1s 1s 1s 1s 1s 2s 2p 2p 2p

54. Unequal Bond Energy
This would mean that three of the bonds in a methane molecule would be identical, because they would involve electron pairs of equal energy.
But what about the fourth bond…?

55. Unequal Bond Energy
The fourth bond is between a 2s electron from the carbon and the lone 1s hydrogen electron.
Such a bond would have slightly less energy than the other bonds in a methane molecule. 1s 1s 1s 1s 1s 2s 2p 2p 2p

56. Unequal Bond Energy
This bond would be slightly different in character than the other three bonds in methane.
This difference would be measurable to a chemist by determining the bond length and bond energy.
But is this what they observe?

57. Enter Hybridization The simple answer is, “No”.
Measurements show that all four bonds in methane are equal. Thus, we need a new explanation for the bonding in methane.
Chemists have proposed an explanation – they call it Hybridization.
Hybridization is the combining of two or more orbitals of nearly equal energy within the same atom into orbitals of equal energy.

58. In the case of methane, they call the hybridization sp3, meaning that an s orbital is combined with three p orbitals to create four equal hybrid orbitals.
1s sp3
These new orbitals have slightly MORE energy than the 2s orbital…
… and slightly LESS energy than the 2p orbitals.
sp3 Hybrid Orbitals

59. sp3 hybrid orbitals
Here is another way to look at the sp3 hybridization and energy profile…

60. sp hybrid orbitals
While sp3 is the hybridization observed in methane, there are other types of hybridization that atoms undergo.
These include sp hybridization, in which one s orbital combines with a single p orbital.
This produces two hybrid orbitals, while leaving two normal p orbitals

61. sp2 hybrid orbitals
Another hybrid is the sp2, which combines two orbitals from a p sublevel with one orbital from an s sublevel.
One p orbital remains unchanged

62. Hybridization Involving “d” Orbitals
Beginning with elements in the third row, “d” orbitals may also hybridize
dsp3 = five hybrid orbitals of equal energy
d 2sp3 = six hybrid orbitals of equal energy

63. Hybridization and Molecular Geometry
Forms
Overall Structure
Hybridization of “A”
AX2
Linear sp
AX3, AX2E
Trigonal planar
sp2
AX4, AX3E, AX2E2
Tetrahedral
sp3
AX5, AX4E, AX3E2, AX2E3
Trigonal bipyramidal
dsp3
AX6, AX5E, AX4E2
Octahedral
d2sp3
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A