1. BONDING: GENERAL CONCEPTS
Chapter Eight
BONDING: GENERAL CONCEPTS

2. Questions to Consider What is meant by the term “chemical bond”?
Why do atoms bond with each other to form compounds?
How do atoms bond with each other to form compounds?
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3. A Chemical Bond No simple way to define this.
Forces that hold groups of atoms together and make them function as a unit.
A bond will form if the energy of the aggregate is lower than that of the separated atoms.
8.1
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4. The Interaction of Two Hydrogen Atoms
8.1
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5. The Interaction of Two Hydrogen Atoms
8.1
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6. Key Ideas in Bonding Ionic Bonding – electrons are transferred
Covalent Bonding – electrons are shared equally
What about intermediate cases?
8.1
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7. Polar Covalent Bond
Unequal sharing of electrons between atoms in a molecule.
Results in a charge separation in the bond (partial positive and partial negative charge).
8.1
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8. The Effect of an Electric Field on Hydrogen Fluoride Molecules
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9. Polar Molecules
8.1
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10. Concept Check What is meant by the term “chemical bond?”
Why do atoms bond with each other to form molecules?
How do atoms bond with each other to form molecules?
A chemical bond is difficult to define but in general it can be described as forces that hold groups of atoms together and make them function as a unit.
Atoms will bond with each other to form molecules if the energy of the aggregate is lower than that of the separated atoms.
Atoms bond with each other by either sharing electrons or transferring electrons to form oppositely charged ions which then attract each other.
8.1
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11. Electronegativity
The ability of an atom in a molecule to attract shared electrons to itself.
On the periodic table, electronegativity generally increases across a period and decreases down a group.
8.2
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12. The Pauling Electronegativity Values
8.2
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13. Concept Check
If lithium and fluorine react, which has more attraction for an electron? Why?
In a bond between fluorine and iodine, which has more attraction for an electron? Why?
The fluorine has more attraction for an electron than does lithium. Both have valence electrons in the same principal energy level (the 2nd), but fluorine has a greater number of protons in the nucleus. Electrons are more attracted to a larger nucleus (if the principal energy level is the same).
The fluorine also has more attraction for an electron than does iodine. In this case the nuclear charge of iodine is greater, but the valence electrons are at a much higher principal energy level (and the inner electrons shield the outer electrons).
8.2
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14. Concept Check
What is the general trend for electronegativity across rows and down columns on the periodic table?
Explain the trend.
Generally speaking, the electronegativity increases in going from left to right across a period because the number of protons increases which increases the effective nuclear charge. The electronegativity decreases going down a group because the size of the atoms increases as you go down so when other electrons approach the larger atoms, the effective nuclear charge is not as great due to shielding from the current electrons present.
8.2
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15. The Relationship Between Electronegativity and Bond Type
8.2
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16. Exercise Arrange the following bonds from most to least polar:
a) N-F O-F C-F
b) C-F N-O Si-F
c) Cl-Cl B-Cl S-Cl
The greater the electronegativity difference between the atoms, the more polar the bond.
C-F, N-F, O-F
Si-F, C-F, N-O
B-Cl, S-Cl, Cl-Cl
8.2
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17. Concept Check
Which of the following bonds would be the least polar yet still be considered polar covalent?
Mg-O C-O O-O Si-O N-O
The correct answer is N-O. To be considered polar covalent, unequal sharing of electrons must still occur. Choose the bond with the least difference in electronegativity yet there is still some unequal sharing of electrons.
8.2
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18. Concept Check
Which of the following bonds would be the most polar without being considered ionic?
Mg-O C-O O-O Si-O N-O
The correct answer is Si-O. To not be considered ionic, generally the bond needs to be between two nonmetals. The most polar bond between the nonmetals occurs with the bond that has the greatest difference in electronegativity.
8.2
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19. Dipole Moment
Property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge.
Use an arrow to represent a dipole moment.
Point to the negative charge center with the tail of the arrow indicating the positive center of charge.
8.3
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20. Dipole Moment
8.3
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21. No Net Dipole Moment (Dipoles Cancel)
8.3
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22. Stable Compounds
Atoms in stable compounds usually have a noble gas electron configuration.
8.4
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23. Isoelectronic Series
A series of ions/atoms containing the same number of electrons.
O2-, F-, Ne, Na+, Mg2+, and Al3+
8.4
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24. Concept Check
Choose an alkali metal, an alkaline metal, a noble gas, and a halogen so that they constitute an isoelectronic series when the metals and halogen are written as their most stable ions.
What is the electron configuration for each species?
Determine the number of electrons for each species.
Determine the number of protons for each species.
Rank the species according to increasing radius.
Rank the species according to increasing ionization energy.
One example could be:
K+, Ca2+, Ar, and Cl-
The electron configuration for each species is 1s22s22p63s23p6.
The number of electrons for each species is 18.
K+ has 19 protons, Ca2+ has 20 protons, Ar has 18 protons, and Cl- has 17 protons.
Increasing Radius: Ca2+, K+, Ar, Cl-
Increasing Ionization Energy: Cl-, Ar, K+, Ca2+
8.4
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25. Ionic Radii
8.4
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26. We can “read” from the periodic table:
Trends for:
Atomic size, ion radius, ionization energy,
electronegativity
Electron configurations
Formula prediction for ionic compounds
Covalent bond polarity ranking
8.4
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27. Lattice Energy
The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid.
Lattice energy = k(Q1Q2/r)
8.5
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28. Born-Haber Cycle for NaCl
8.5
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29. Formation of an Ionic Solid
1.Sublimation of the solid metal
M(s)  M(g) [endothermic]
2. Ionization of the metal atoms
M(g)  M+(g) + e [endothermic]
3. Dissociation of the nonmetal
1/2X2(g)  X(g) [endothermic]
8.5
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30. Formation of an Ionic Solid (continued)
4. Formation of X ions in the gas phase:
X(g) + e  X(g) [exothermic]
5. Formation of the solid MX
M+(g) + X(g)  MX(s)
[quite exothermic]
8.5
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31. Comparing Energy Changes
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32. Partial Ionic Character of Covalent Bonds
None of the bonds reaches 100% ionic character even with compounds that have a large electronegativity difference (when tested in the gas phase).
8.6
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33. The relationship between the ionic character of a covalent bond and the electronegativity difference of the bonded atoms
8.6
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34. Ionic Compound
Any compound that conducts an electric current when melted will be classified as ionic.
8.6
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35. Models
Models are attempts to explain how nature operates on the microscopic level based on experiences in the macroscopic world.
8.7
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36. Fundamental Properties of Models
A model does not equal reality.
Models are oversimplifications, and are therefore often wrong.
Models become more complicated and are modified as they age.
We must understand the underlying assumptions in a model so that we don’t misuse it.
8.7
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37. Bond Energies
To break bonds, energy must be added to the system (endothermic).
To form bonds, energy is released (exothermic).
8.8
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38. ΔH = ΣD(bonds broken) – ΣD(bonds formed)
Bond Energies
ΔH = ΣD(bonds broken) – ΣD(bonds formed)
D represents the bond energy per mole of bonds (always has a positive sign).
8.8
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39. Localized Electron Model
A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.
8.9
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40. Localized Electron Model
Electron pairs are assumed to be localized on a particular atom or in the space between two atoms:
Lone pairs – pairs of electrons localized on an atom
Bonding pairs – pairs of electrons found in the space between the atoms
8.9
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41. Localized Electron Model
Description of valence electron arrangement (Lewis structure).
Prediction of geometry (VSEPR model).
Description of atomic orbital types used to share electrons or hold lone pairs.
8.9
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42. Lewis Structure
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.
8.10
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43. Writing Lewis Structures
Sum the valence electrons.
Place bonding electrons between pairs of atoms.
Atoms usually have noble gas configurations. Arrange remaining electrons to satisfy the octet rule (or duet rule for hydrogen).
8.10
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44. Concept Check
Draw a Lewis structure for each of the following molecules: H2 F2 HF
8.10
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45. Concept Check
Draw a Lewis structure for each of the following molecules:
H2O
NH3
8.10
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46. Exceptions
When it is necessary to exceed the octet rule for one of several third-row (or higher) elements, place the extra electrons on the central atom.
8.11
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47. Resonance
More than one valid Lewis structure can be written for a particular molecule.
Actual structure is an average of the resonance structures.
8.12
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48. Concept Check
Draw a Lewis structure for each of the following molecules: 
CO CO2
CH3OH BF3
PCl5 NO3-
SF6
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49. Formal Charge Nonequivalent Lewis structures.
Atoms in molecules try to achieve formal charges as close to zero as possible.
Any negative formal charges are expected to reside on the most electronegative atoms.
8.12
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50. Formal Charge
Formal charge = # valence e– on free atom – # valence e– assigned to the atom in the molecule.
Assume:
Lone pair electrons belong entirely to the atom in question
Shared electrons are divided equally between the two sharing atoms
8.12
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51. VSEPR Model VSEPR: Valence Shell Electron-Pair Repulsion.
The structure around a given atom is determined principally by minimizing electron pair repulsions.
8.13
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52. Predicting a VSEPR Structure
Draw Lewis structure.
Put electron pairs as far apart as possible.
Determine positions of atoms from the way electron pairs are shared.
Determine the name of molecular structure from positions of the atoms.
8.13
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53. VSEPR
8.13
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54. VSEPR: Two Electron Pairs
8.13
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55. VSEPR: Three Electron Pairs
8.13
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56. VSEPR: Four Electron Pairs
8.13
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57. VSEPR: Iodine Pentafluoride
8.13
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58. Concept Check
Determine the shape for each of the following molecules, and include bond angles:
HCN
PH3
SF4 O3
KrF4
HCN – linear, 180o
PH3 – trigonal pyramid, 107o
SF4 – see saw, 90o, 120o
O3 – bent, 120o
KrF4 – square planar, 90o, 180o
8.13
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59. Concept Check True or false:
A molecule that has polar bonds will always be polar.
-If true, explain why.
-If false, provide a counter-example.
False, a molecule may have polar bonds (like CO2) but the individual dipoles might cancel out so that the net dipole moment is zero.
8.13
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60. Concept Check True or false: Lone pairs make a molecule polar.
-If true, explain why.
-If false, provide a counter-example.
False, lone pairs do not always make a molecule polar. They might be arranged so that they are symmetrically distributed to minimize repulsions, such as XeF4.
8.13
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