Chapter Nine
Chemical Bonding I
1
The Ionic Bond
and
Lattice Energies
2
Lewis Dot Symbols
Consists of atomic symbol surrounded by 1 dot for each
valence electron in the atom
Only used for main group elements
# valence electrons = group number
3
Ionic Bonding
Electrons are transferred from 1 atom to another
Metal atoms:
Lose electrons to form cations
Nonmetal atoms: Gain electrons to form anions
Electrostatic force bonds ions into an ionic compound
Form an ionic salt with repeating structure: NaCl, LiF
Ionic Bonds follow the octet rule
Atoms lose or gain valence e- to make an octet (8e-)
8 valence e- = Noble gas configuration
Li
Li+ + e-
e- +
F
F -
Li+ +
F -
Li+ F -
1s22s1→ 1s2
1s22s22p5 →1s22s22p6
4
The Born Haber Cycle
Formation of a salt can be tracked by individual steps
Enthalpies associated with each step are additive
Lattice energies are exothermic: salt is more stable than ions
(∆H) associated with gaseous atoms, so convert to gas
Ionization energies: Electron loss
Electron affinity: Electron gain
Lattice energy: Energy released when forming a crystal
Formation energy (∆Hf) not based on gaseous atoms
Energy associated with forming 1 mole of a compound from its
elements in their normal states
Sum of energy transfers called “Born-Haber cycle”
Can use the cycle to the enthalpy of any step in cycle
5
Energy Changes in the Born-Haber Cycle
Li (s)  Li(g)
∆Hsubl = +155 kJ
1/2 F2(g)  F(g)
∆Hdiss = +75 kJ
Li (g)  Li+(g)+ e∆H IE = + 520kJ
F(g) + e-  F-(g)
∆HEA = - 328 kJ
Li+ (g) + F-(g)  LiF (s) ∆Hlattice E = ? kJ
Li(s) + 1/2F2(g)  LiF(s) ∆H °f = - 594 kJ
Hess’s Law:
∆H °f =∆Hsubl+ ∆Hdiss + ∆H IE + ∆HEA + ∆Hlattice E
-594kJ =155kJ + 75kJ + 520kJ - 328kJ + ∆Hlattice E
∆Hlattice E = -1016kJ
6
Lattice Energies of Common Salts
7
The Covalent Bond and
Electronegativity
8
Lewis Structures
Lewis structures represent covalent bond formation
Shared pairs of electrons give both atoms an octet
F
7e-
+
1s22s22p5 →1s22s22p6
F
F F
7e-
8e- 8e-
Bonding Pairs: Shared electrons count for both atoms
Represented by a dash (-) between bonding atoms
Lone Pairs: Non-shared electrons count for 1 atom
Represented by a pair of dots (••) around atom
lone pairs
F
F
single covalent bond
lone pairs
9
10
Multiple Bonds
More than one pair of electrons is shared between atoms
so each atom can form an octet.
Single Bond:
Double bonds:
Triple bonds:
1 shared pair:
2 shared pairs:
3 shared pairs:
1 dash (-)
2 dashes (=)
3 dashes (≡)
Allows atoms in the Lewis structure to share extra
electrons if there are not enough for the central atom
O
C
O
double bonds
N
N
triple bond
Electronegativity and Polar Covalent Bonds
Electronegativity
The ability of an atom to attract electrons
F is the most electronegative atom
Nonmetals high electronegativies
H
Polar Covalent Bonds
F
Differences in electronegativity result in unequal sharing
of electrons between atoms
More electronegative atom has a partial negative charge
Percent Ionic Character
Measure of polarity of bond
100% ionic is full transfer of electron, no sharing
100% covalent is equal sharing, H2, Cl2, etc.
11
The Electronegativities of Common Elements
12
9.5
Writing Lewis Structures
13
Writing Lewis Structures: General Information
Electronegativity
Central atom usually has the lowest electronegativity
(atom lower or to the left in periodic table)
Terminal atoms (except H) have higher electronegativities
Terminal Atoms
Bonded to only one other atom
Hydrogen atoms are terminal atoms
Bonding
Hydrogen atoms are bonded to oxygen atoms in oxoacids
Make the molecule as symmetrical as possible
14
Write the Lewis Structure of HNO
1. Add up the valence electrons in the structure
1 (H) + 5 (N) + 6 (O) = 12 valence electrons
2. Arrange the atoms & place bonding electrons
H - N – O nitrogen less electronegative, put in the center
3. Place e- pairs around terminal atoms
••
H − N − O:
••
4. Place remaining electron pairs on central atom
••
••
H− N−O:
5. Add double bond to finish nitrogen octet
••
••
••
H−N =O
••
15
Formal Charge
and
Resonance Theory
16
Formal Charge
17
Difference between the # of valence electrons in a free atom &
the # of electrons assigned to that atom in a Lewis structure.
F.C. = Group # - (# of lone e- + # bonds)
Molecule is most stable if formal charge is 0 for each atom.
••
••
••
••
••
H−N =O
:O − S = O
H: = 1-(0 +1)= 0
N: = 5-(2+3)= 0
O: = 6-(4+2)= 0
O: = 6-(6 +1)= -1
S: = 6-(2 +3)= +1
O: = 6-(4 +2)= 0
••
••
••
The most likely Lewis structure has the lowest formal charges
Negative formal charge must be on more electronegative atom
Sum of formal charges must =0 for molecules or = ionic charge
Resonance Theory
If a molecule or ion can be represented by 2 or more Lewis
structures that differ only in electron location, the true
structure is a composite of them.
••
••
••
••
••
••
0
0
••
••
:O − S = O ↔ O = S − O:
••
-1
+1
••
+1
-1
Resonance Structures
Equivalent Lewis structures that can be drawn for a molecule
Formal charges will usually be present
Delocalization
Electrons are spread out over several atoms
Stabilizes molecule
18
Exceptions
to the Octet Rule
19
20
Incomplete Octet
Not enough electrons to cover central atom
BeH2
Be – 2e2H – 2x1e4e-
H
Be
H
Terminal atoms unwilling to donate electrons
Destabilizes terminal atoms: creates formal charge
BF3
B – 3e3F – 3x7e24e-
F
B
F
F
Coordinate Covalent Bonds
21
1 atom provides both electrons
Electrons are then shared between 2 atoms
Ex: BF3 and NH3
B needs 2 electrons to fill octet
N has a lone pair to share
F
F
B
F
H
N
H
H
Free Radicals and Expanded Octets
Free Radicals:
Molecules with an odd number of valence electrons (N)
Extremely reactive, odd electron wants to be part of a pair
••
•
••
:O − N = O
••
••
Compounds with expanded valence shells
Central atom has more than eight electrons
May have lone pair electrons as well as bonding pairs
22
Bond Enthalpy
23
Bond Enthalpy
Bond Enthalpy
Energy required to break a particular bond
in a molecule in the gas phase.
Enthalpy change for the Reaction (∆H)
∆H = ∆Hbond breaking + ∆H bond formation
Enthalpy change: 2H2(g) + O2(g) → 2 H2O(g)
∆Hbond breaking = 2∆H H-H + ∆H O=O
= 2(436kJ) + 499kJ
= +1371kJ (endothermic)
= 4∆H H-O =4(460kJ)
∆Hformation
= -1840kJ (exothermic)
∆Hreaction
=+1371kJ-1840kJ = -469kJ
24
Estimate the enthalpy change for the
combustion of 1 mole of methane
25
1. Write the reaction: CH4(g)+ 2 O2(g) → CO2(g)+ 2 H2O(g)
2. Calculate energy needed to break the bonds in reactants and
energy produced when the bonds of products form.
Bonds broken
∆H
Bonds formed
∆H
4C-H: 4 x 414 kJ =
20=0: 2 x 498 kJ=
Edissociation=
+1656kJ
+ 996kJ
+2652 kJ
Bond-breaking:endothermic (+)
2C=O: 2 x 736 kJ =
4H-O: 4 x 464 kJ =
Eformation =
-1472kJ
-1856kJ
-3328kJ
Bond-forming:exothermic (-)
3. Calculate ∆H for the reaction
∆H approx = +2652kJ - 3328kJ = -676kJ
Representative Bond Enthalpies
26