1. Bonding : General Concepts
Chapter 8

2. Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
participate in chemical bonding.
Group
# of valence e-
e- configuration 1A 1
ns1 2A 2
ns2 3A 3
ns2np1 4A 4
ns2np2 5A 5
ns2np3 6A 6
ns2np4 7A 7
ns2np5

3. Lewis Electron-Dot Symbols
A Lewis electron-dot symbol is a symbol in which the electrons in the valence shell of an atom or ion are represented by dots placed around the letter symbol of the element.
Note that the group number indicates the number of valence electrons. Na . Si : P S Mg Al Cl Ar
Group I
Group II
Group VII
Group VIII
Group VI
Group IV
Group V
Group III 2

4. Lewis symbol (Lewis structure) - a way to represent atoms (and their bonds) using the element symbol and valence electrons as dots.

5. Chemical Bonding
Chemical Bond - the force of attraction between any two atoms in a compound.
Interactions involving valence electrons are responsible for the chemical bond.

6. The Ionic Bond
Ionic bond: the electrostatic force that holds ions together in an ionic compound.
Li+ F - Li + F
1s2
1s22s22p6
1s22s1
1s22s22p5
[He]
[Ne]
LiF Li
Li+ + e-
e- + F - F -
Li+ +
Li+

7. Essential Features of Ionic Bonding
Atoms with low I.E. and low E.A. tend to form positive ions.
Atoms with high I.E. and high E.A. tend to form negative ions.
Ion formation takes place by electron transfer.
The ions are held together by the electrostatic force of the opposite charges.
Reactions between metals and nonmetals (representative) tend to be ionic.

8. Describing Ionic Bonds
Noble gas configurations are stable.
The atoms of metals loses electron(s) to become a cation (positive).
eg., Na →Na+
[Ne]3s1 → [Ne] + 1 e-
(Na has low I.E, easy to loose an e-)
The atom that gains the electron becomes an anion (negative), eg., Cl,
Cl + e→Cl- (Cl has high E.A)
3s23p5 + e- →3s23p6 or [Ar] 2

9. : . [ F ] Another example Mg F
The magnesium has two electrons to give, whereas the fluorines have only one “vacancy” each. : F . Mg
[ F ] - 2+
Consequently, magnesium can accommodate two fluorine atoms. 2

10. 9.1
Use Lewis dot symbols to show the formation of aluminum oxide (Al2O3).
The mineral corundum (Al2O3).

11. 9.1 Thus, the simplest neutralizing ratio of Al3+ to O2− is 2:3; two
Al3+ ions have a total charge of +6, and three O2− ions have a total charge of −6. So the empirical formula of aluminum oxide is Al2O3, and the reaction is

12. Ionic bonds
MgO – magnesium oxide
2Mg(s)+O2(g)→ 2MgO

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Isoelectronic Series
A series of ions/atoms containing the same number of electrons.
O2-, F-, Ne, Na+, Mg2+, and Al3+
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CONCEPT CHECK!
Choose an alkali metal, an alkaline earth 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.
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.
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15. Electrostatic (Lattice) Energy
Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
Q+ is the charge on the cation
E = k
Q+Q- r
Q- is the charge on the anion
r is the distance between the ions
Compound
Lattice Energy
(kJ/mol)
Lattice energy increases as Q increases and/or
as r decreases.
Q: +2,-1
Q: +2,-2
MgF2
MgO
2957
3938
E is negative for ionic bond formation, but lattice energy is the reverse, energy to break a bond, which is +ve.
LiF
LiCl
1036
853
r F- < r Cl-

16. 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]
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17. Formation of an Ionic Solid (continued)
4 Formation of nonmetal ions in the gas phase.
X(g) + e- X-(g) [exothermic]
5.Formation of the solid ionic compound.
M+(g) + X-(g) MX(s)
[quite exothermic]
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18. Born-Haber Cycle for Determining Lattice Energy
DHoverall = DH1 + DH2 + DH3 + DH4 + DH5 o

19. FromChang’s text book

20. Covalent Bond
A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.
Why should two atoms share electrons?
7e-
7e-
8e-
8e- F F + F
Noble gas config.
Lewis structure of F2
lone pairs F
single covalent bond
single covalent bond F

21. H2, N2, O2, F2, Cl2, Br2, I2 Features of Covalent Bonds
The diatomic elements have totally covalent bonds (totally equal sharing.)
H2, N2, O2, F2, Cl2, Br2, I2
Covalent compounds are usually formed from nonmetals.
Molecules - compounds characterized by covalent bonding.
not a part of a massive three dimensional crystal structure.

22. Hydrogen molecule, diatomic
Lewis structure of water
single covalent bonds
2e-
8e-
2e- H + O + H O H or
Hydrogen molecule, diatomic

23. Carbon dioxide and nitrogen molecule
Double bond – two atoms share two pairs of electrons
8e-
8e-
8e- O C or O C
double bonds
Triple bond – two atoms share three pairs of electrons
8e- N
8e- or N
triple bond

24. Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond

25. From Chang’s text book

26. Polar Covalent Bonds
A polar covalent bond is one in which the bonding electrons spend more time near one of the two atoms involved.
When the atoms are alike, as in the H-H bond of H2 , the bonding electrons are shared equally (a nonpolar covalent bond).
When the two atoms are of different elements, the bonding electrons need not be shared equally, resulting in a “polar” bond. 2

27. Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms H F
electron rich
region
electron poor
region
e- poor
e- rich F H d+ d-

28. Electron Affinity - measurable, Cl is highest
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e X-(g)
Electronegativity - relative, F is highest

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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.
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30. Dipole Moment

31. No Net Dipole Moment (Dipoles Cancel)
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32. Polar Covalent Bonds
Electronegativity is a measure of the ability of an atom in a molecule to draw bonding electrons to itself.
In general, electronegativity increases from the lower-left corner to the upper-right corner of the periodic table.
The current electronegativity scale, developed by Linus Pauling, assigns a value of 4.0 to fluorine and a value of 0.7 to cesium. 2

33. The Electronegativities of Common Elements

34. Variation of Electronegativity with Atomic Number

35. Classification of bonds by difference in electronegativity
Bond Type
Covalent
 2
Ionic
0 < and <2
Polar Covalent
Increasing difference in electronegativity
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-

36. Which would be more polar, a H-F bond or a H-Cl bond?
The greater the difference in electronegativity between two atoms, the greater the polarity of a bond.
Which would be more polar, a H-F bond or a H-Cl bond?
H-F … = 1.9
H-Cl… = 0.9
HF bond is more polar than the HCl bond.

37. Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent

38. Chemical bonding (cont.)
Classify the following bonds as ionic, polar covalent, or covalent:
E.N. diff
the bond in HCl 3.0 – 2.1 = 0.9 polar covalent
The bond in KF 4.0 – 0.8 = 3.2 Ionic
the CC bond in H3CCH – 2.5 = 0 covalent

39. Writing Lewis Structures
Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.
Count total number of valence e-.
Complete an octet for all atoms except hydrogen.
If structure contains too many electrons, form double and triple bonds on central atom as needed.

40. 9.3
Write the Lewis structure for nitrogen trifluoride (NF3) in which all three F atoms are bonded to the N atom.
NF3 is a colorless, odorless, unreactive gas.

41. 9.3
Solution We follow the preceding procedure for writing Lewis structures.
Step 1: The N atom is less electronegative than F, so the skeletal structure of NF3 is
Step 2: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, respectively. Thus, there are 5 + (3 × 7), or 26, valence electrons to account for in NF3.

42. 9.3
Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N:
lone pair of electrons
bonding pair
Because this structure satisfies the octet rule for all the atoms, step 4 is not required.
Check Count the valence electrons in NF3 (in bonds and in lone pairs). The result is 26, the same as the total number of valence electrons on three F atoms (3 × 7 = 21) and one N atom (5).
How about NH3? Similar to NF3, Difference for H ?

43. Multiple Bonds
In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared.
It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms. or C : H 2

44. Multiple Bonds
Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms. C or H :
::: 2

45. Formal Charge of an atom
An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.
formal charge on an atom in a Lewis structure =
total number of valence electrons in the free atom -
total number of nonbonding electrons - 1 2
total number of bonding electrons ( )
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

46. ( ) - Formaldehyde, CH2O H C O = 1 2 = 4 - 0 - ½ x 8 = 0 = 6 - 4 -H C O
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
formal charge on an atom in a Lewis structure = 1 2
total number of bonding electrons ( )
total number of valence electrons in the free atom -
total number of nonbonding electrons
formal charge on C
= 4 -
0 -
½ x 8 = 0
formal charge on O
= 6 -
4 -
½ x 4 = 0

47. ( ) - The Lewis structure for the carbonate ion is, 1 = 2
Lewis structure and formal charge of carbonate ion
The Lewis structure for the carbonate ion is,
formal charge on an atom in a Lewis structure = 1 2
total number of bonding electrons ( )
total number of valence electrons in the free atom -
total number of nonbonding electrons

48. Lewis structure and formal charge of carbonate ion
The Lewis structure for the carbonate ion is,
C: =0
O(a): =0
O (b):6-6-1 = -1
O ©: = -1, Total of -2 for carbonate ion.
formal charge on an atom in a Lewis structure = 1 2
total number of bonding electrons ( )
total number of valence electrons in the free atom -
total number of nonbonding electrons

49. Resonanace
A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.
(+ and – are the formal charges)
O-O bond in O3 should be longer than the double bond, but both oxygen bonds are equal in length because of resonance as shown. O + - O + -

50. Delocalized Bonding: Resonance
According to theory, one pair of bonding electrons is spread over the region of all three atoms. O
This is called delocalized bonding, in which a bonding pair of electrons is spread over a number of atoms. 2

51. . What are the resonance structures of the
                                            .
What are the resonance structures of the
Benzene? carbonate (CO32-) ion?
benzene
Carbonate ion O C - O C - O C -

52. Lewis Structures and Exceptions to the Octet Rule
1. Incomplete Octet - less then eight electrons around an atom other than H.
Let’s look at BF3
2. Odd Electron - if there is an odd number of valence electrons it isn’t possible to give every atom eight electrons.
Let’s look at NO
3. Expanded Octet - elements in 3rd period and beyond may have 10 and 12 electrons around it.

53. Exceptions to the Octet Rule
The Incomplete Octet
BeH2 H Be F B
BF3

54. Exceptions to the Octet Rule
Odd-Electron Molecules
N – 5e-
O – 6e-
11e- NO N O
The Expanded Octet (central atom with principal quantum number n > 2)
Electrons can be accomodated in the d orbitals S F
S – 6e-
6F – 42e-
48e-
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
SF6

55. Exceptions to the Octet Rule
For example, the bonding in phosphorus pentafluoride, PF5, shows ten electrons surrounding the phosphorus.
: F : :
F : :
: F : P
F : : :
: F : 2

56. Exceptions to the Octet Rule
In xenon tetrafluoride, XeF4, the xenon atom must accommodate two extra lone pairs.
: F :
F : : : Xe
: F :
F : : : 2

57. Single bond < Double bond < Triple bond
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.
Bond Energy
H2 (g)
H (g) +
DH0 = kJ
Cl2 (g)
Cl (g) +
DH0 = kJ
HCl (g)
H (g) +
Cl (g)
DH0 = kJ
O2 (g)
O (g) +
DH0 = kJ O
N2 (g)
N (g) +
DH0 = kJ N
Bond Energies
Single bond < Double bond < Triple bond

59. Bond Energies (BE) and Enthalpy changes in reactions
Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. (Examples 9.13,9.14)
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)

60. H2 (g) + Cl2 (g) HCl (g)
2H2 (g) + O2 (g) H2O (g)

61. Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g) HF (g)
DH0 = SBE(reactants) – SBE(products)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ) H 1
436.4 F 1
156.9
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ) H F 2
568.2
1136.4
DH0 = – 2 x = kJ

62. Bond Energy
To illustrate, let’s estimate the DH for the following reaction.
In this reaction, one C-H bond and one Cl-Cl bond must be broken.
In turn, one C-Cl bond and one H-Cl bond are formed. 2

63. Bond Energy simple arithmetic yields DH.
DH0 = [BE (C-H) + BE (Cl-Cl)] –
[BE (C-Cl) + BE (H-Cl)]
=[( )] – [( )] kJ = -104 kJ 2