Chapter 9
Chemical bonds- sharing or transfer of e- to
attain stable e- config (noble gas e- config with
8 valence e-)
Types of Chemical Bonding
1) Ionic- metal and non-metal, metal transfers
e- to non-metal
2) Covalent- non-metal and non-metal, share e3) Metallic- between two metals, lose e- to sea
of valence e- which holds metals together,
cations are attracted to sea of valence e-
-Valence e- are the ones involved in bonding
How many valence e- in the following?
1) I
2) P
3) Na
4) C
5) Aℓ
6) Sr
7) S
8) Ar
1) 7
2) 5
3) 1
4) 4
5) 3
6) 2
7) 6
8) 8
Lewis Dot Structures
-illustrates valence e- around an element
-Examples
All elements want to achieve the octet rule when
forming bonds
octet rule- all elements want to achieve the
stable e- config of a noble gas (8 valence e-)
Ionic Compounds
-most are crystalline solids at room temp
-arranged in repeating 3-D patterns
-have very high melting points
ex- NaCℓ melts at 801°C (1474°F)
-conduct electric current when dissolved in water
-are neutral in charge- one metal, one non-metal
-ions in ionic compounds are arranged in a
pattern called a crystal lattice
-every cation surrounded by anions; and
every anion surrounded by cations
-maximizes attractions between + and –
ions
lattice energy- the energy released when the
solid crystal forms from separate ions in the
gas state
-always exothermic
-hard to measure directly, but can be calculated
from knowledge of other processes
Born-Haber Cycle
-method for determining the lattice energy of an
ionic substance by using other reactions
-uses Hess’s Law to add up heats of other
processes
-Page 369-371
Trends in Lattice Energy
1) Ion Size
-larger ions mean the center of positive charge
(nucleus of the cation) is farther away from
negative charge (electrons of the anion)
-larger ion = weaker attraction = smaller lattice
energy
-as ion size decreases, lattice energy increases
-as ion size increases, lattice energy decreases
-indirectly proportional
Example:
1) Arrange these ions in order of increasing
lattice energy?
KCℓ, CsCℓ, LiCℓ, NaCℓ
CsCℓ, KCℓ, NaCℓ, LiCℓ
2) Ionic Charge
-larger charge means the ions are more strongly
attracted
-if same magnitude of charge go to periodic table
and look at placement
-same trend as ionic size (indirectly
proportional)
-larger charge = stronger attraction = larger
lattice energy
Examples:
1) Which compound has higher lattice energy?
NaCℓ
or
MgCℓ2
MgCℓ2
Na1+Cℓ1Mg2+Cℓ12) Arrange in order of increasing lattice energy.
CaO, KBr, KCℓ, SrO
KBr
KCℓ
SrO
CaO
K1+Br1K1+Cℓ1Sr2+O2Ca2+O2-
Molecular Compounds
-generally have low melting and boiling points
-most are gases or liquids at room temp
-do not conduct an electric current when
dissolved in water
-made up or two or more non-metals
Covalent Bonding
-electrons are shared
-each element still is trying to achieve stable econfig of noble gas
bonding pairs/shared pairs- pair of e- shared
between two atoms
lone pairs/unshared pairs- electrons not involved
in bonding
single covalent bond- two atoms share a pair of
electrons
-bonds are represented as dashes (─)
double covalent bond- bond that involves two
shared pairs of e- (═)
triple covalent bond- bond that involves three
shared pairs of e- (≡)
simple formula- shows # and type of atoms
ex- H2O
structural formula- shows arrangement of atoms
in a molecule
ex-
-Lewis structures make it seem as though e- are
shared equally between the elements of
covalent bonds
-this is not the case
Ex- HF
-H has a slight positive charge and F has a slight
negative charge
Does this make the bond ionic?
-No, these bonds are said to be polar- having a +
and – pole
polar covalent bond- intermediate in nature
between a pure covalent and an ionic bond
Electronegativity
-ability of an atom to attract e- to itself in a
chemical bond
-results in polar bonds
-increases as you move across a period
-decreases as you move down a group
-page 378 Figure 9.10
Ex- Put in order of increasing electronegativity
Li, N, F, Be, C
Li, Be, C, N, F
-the degree of polarity in bonds depends on
electronegativity difference (∆EN) between the
elements
-if identical in electronegativiy, the e- are shared
equally and is nonpolar or purely covalent
-large ∆EN means ionic bond
-intermediate ∆EN means polar covalent
Page 378
Effect of ∆EN on Bond Type:
∆EN
Bond Type
Small (0-0.4)
Covalent
Example
Cℓ2
Intermediate
(0.4-2.0)
Polar
Covalent
HCℓ
Large (2.0+)
Ionic
NaCℓ
Examples:
Determine whether the bonds formed are
covalent, polar covalent, or ionic.
1) Sr and F
2) N and Cℓ
3) N and O
4) I and I
5) Cs and Br
6) P and O
1) ∆EN= 3.0 = ionic
2) ∆EN= 0.0 = covalent
3) ∆EN= 0.5 = polar covalent
4) ∆EN= 0.0 = covalent 5) ∆EN= 2.1 = ionic
6) ∆EN= 1.4 = polar covalent
-the polarity of a bond can be quantified by the
size of its dipole moment (µ)
-a dipole moment occurs anytime there is a
separation of + and – charges
resonance structures- two or more valid Lewis
structures
example:
**Write resonance structures for nitrate and
nitrite ions
*One resonance structure may be better than
another because of formal charge
formal charge- charge an atom would have if all
bonding e- were shared equally between the
bonded atoms
-ignores electronegativity
formal charge = # of valence electrons - (# of
lone electrons + ½ (# of shared e-))
Rules for best resonance structures:
1) the sum of all formal charges in a neutral
molecule must be zero
2) the sum of all formal charges in an ion must
equal the charge of the ion
3) small or zero formal charges on individual
atoms are better than large ones
4) when formal charge cannot be avoided,
negative formal charge should reside on the
most electronegative atom
Ex- HF
H= 1-(0 + ½(2)) = 1-1 = 0
F= 7-(6 + ½(2)) = 7-7 = 0
*Obeys the first rule
*Draw the two skeletal structures for HCN
* Find formal charges to see which one is better
Read Section 9.9 on pages 387-391 on oddelectron species, incomplete octets, and
expanded octets
bond energy- the energy required to break one
of the bonds in the gas phase
*always positive because it takes energy to break
a bond
page 392 Table 9.3- Average Bond Energies
-average bond energies can be used to estimate
enthalpy change in a reaction
-bond breaking is endothermic:
∆H(breaking) = +
-bond making is exothermic:
∆ H(making) = −
∆Hrxn = ∑ (∆H(bonds broken)) + ∑ (∆H(bonds
formed))
Ex:
H3C─H(g) + Cℓ─Cℓ(g)  H3C─Cℓ(g) + H─Cℓ(g)
-use numbers from the chart
-remember reactants are + and products are –
-only find ∆H for the bonds shown
(414 + 243) + (-339 + -431) =
657kJ + -770kJ =
-113.00kJ
-Calculate heat of reaction for all of the bonds in
the following:
CH4(g) + 2H2O(g)  4H2(g) + CO2(g)
-must draw Lewis structures for each one
-find for each of the bonds- use coefficients and
number of each bond
Try For More Practice 9.10 on page 395