9. Chemical Bonding I: Basic Concepts
bond formation: valence electrons
transfer of electrons from one atom to another: cations and anions = ionic bond
strength of ionic bond
Coulomb potential energy
E∝
Q cat · Q an
r
r = distance between cation and anion, Q = charge
E is negative: formation of ionic bond is exothermic
the more negative E , the stronger the bond
the larger the charge, the more negative E
the smaller the ion, i.e., the smaller r , the more negative E
lattice energy = quantitative measure of ionic bond strength
lattice energy = energy required to completely separate one
mole of a solid ionic compound into gaseous ions
GChem I
9.1
the more positive the lattice energy, the stronger the bond
lattice energy is determined indirectly via the Born-Haber cycle,
which is based on Hess’s law and relates the lattice energy of an
ionic compound to ionization energies, electron affinities, and
other atomic and molecular properties
example: lattice energy of lithium fluoride
standard enthalpy of formation of lithium fluoride:
1
Li(s) + F2 (g) −→ LiF(s)
2
∆H f◦ = −594.1 kJ/mol
The formation reaction can be carried out in five steps instead:
(1) Convert solid lithium to lithium vapor (sublimation):
∆H1◦ = 155.2 kJ/mol
Li(s) −→ Li(g)
(2) Dissociate
1
2
1
F2 (g) −→ F(g)
2
mole of F2 gas into gaseous F atoms:
∆H2◦ = 75.3 kJ/mol
(3) Ionize one mole of gaseous Li atoms:
Li(g) −→ Li+ (g) + e−
GChem I
∆H3◦ = 520 kJ/mol
9.2
(4) Add one mole of electrons to one mole of gaseous F atoms:
F(g) + e− −→ F− (g)
∆H4◦ = −328 kJ/mol
+
(5) Combine one mole of gaseous Li and one mole of gaseous
F− to form one mole of solid LiF:
Li+ (g) + F− (g) −→ LiF(s)
∆H5◦ =? kJ/mol
◦
Note: lattice energy = −∆H5
Li(s) −→ Li(g)
1
F2 (g) −→ F(g)
2
Li(g) −→ Li+ (g) + e−
∆H1◦ = 155.2 kJ/mol
∆H2◦ = 75.3 kJ/mol
∆H3◦ = 520 kJ/mol
F(g) + e− −→ F− (g)
∆H4◦ = −328 kJ/mol
1
Li(s) + F2 (g) −→ LiF(s)
2
∆H f◦ = −594.1 kJ/mol
Li+ (g) + F− (g) −→ LiF(s)
∆H5◦ =? kJ/mol
Hess’s law:
∆H f◦ = ∆H1◦ + ∆H2◦ + ∆H3◦ + ∆H4◦ + ∆H5◦
GChem I
9.3
−594.1 kJ/mol = 155.2 kJ/mol + 75.3 kJ/mol + 520 kJ/mol
+ (−328 kJ/mol) + ∆H5◦
=⇒
∆H5◦ = −1017 kJ/mol
=⇒ lattice energy of LiF = 1017 kJ/mol
Compound
LiF
LiCl
LiBr
LiI
NaCl
NaBr
NaI
KCl
KBr
KI
MgCl2
Na2 O
MgO
GChem I
◦
Lattice energy ( kJ/mol)
Melting Point ( C)
1017
828
787
732
788
736
686
699
689
632
2527
2570
3890
845
610
550
450
801
750
662
772
735
680
714
Sub (1275)
2800
9.4
sharing of one or more pair(s) of electrons between atoms =
O
covalent bond
H
..
...
...
...
...
H
...
....
.....
.....
....
.
.
.
.
O
..
...
...
...
...
.....
.....
.....
.....
.....
.
..
...
...
...
...
H
transfer (ionic bond) or sharing (covalent bond) −→ each atom
achieves an especially stable electron configuration, often a no2
6
ble
..........................
.......................... ns np........:.......octet
...........
.........gas
................. configuration
H
C
O
C
water
...
...
...
...
...
...
H
Lewis symbol: chemical symbol (=nucleus + core electrons) +
·· ·
·
dots
electrons
· Si · for valence· Cl
·
·
example:
··
H
·
· Si ·
·
Si [Ne]3s 2 3p 2
acetic acid
Lewis symbols mostly for main-group elements
Lewis structure: combination of Lewis symbols to show cova··
lent bonds
H
Cl ··
..........................
··
··
· H + · Cl ·· −→ H·· Cl ··
··
··
↑
H
...
...
...
...
..
O
H
C
...
...
...
...
...
...
...
...
...
..........................
H
..........................
hydrogen molecule H
H
. .
... ... = covalent bond
shared pair
... ...
bonding pair :
..........................
··
...
...
...
lone pair:
C
··
Cl ·
..........................
·· ·
..........................
··
O
··
..........................
H
H
– Typeset by FoilTEX –
GChem I
acetic acid with lone pairs
9.5
··
··
chlorine molecule ·· Cl
nitrogen molecule ·· N
triple covalent bond
··
··
oxygen molecule O
..........................
..........................
..........................
..........................
··
Cl ··
··
N ··
octet rule
..........................
..........................
··
O
··
Lewis structure for O2 is problematic: experiments show that O2
is paramagnetic =⇒ O2 has unpaired electrons
·· N
bond order: single
..........................
..........................
..........................
N ··
; double
; triple
bond length: distance between the centers of two atoms joined
by a covalent bond
l >l >l
– Typeset by FoilTEX –
polar covalent bond: electron pair is not shared equally between
both atoms
the electrons are displaced towards the more nonmetallic element
– Typeset by FoilT X –
E
δ+
δ−
H Cl
GChem I
9.6
– Typeset by FoilTEX –
electronegativity E N : ability of an atom to attract electrons in
a compound; Pauling scale: most electronegative atom, F, has
EN = 4
electronegativity is related to electron affinity and ionization energy
¯
¯
if ∆E N = ¯E N1 − E N2 ¯ ≈ 0, the bond is covalent
if ∆E N ≈ 0.4–2.0, the bond is polar covalent
if ∆E N ' 2.0, the bond is ionic
Guidelines for Drawing Lewis Structures
Rules
1. Determine the skeleton structure.
(a) H is always a terminal atom and has only one bond.
(b) In simple compounds, O and X (halogens) are terminal
atoms. If X is bonded to O, O is terminal
(c) Oxygen atoms do not bond to each other except in (1) O2
and O3 molecules; (2) the peroxides, which contain the
O2 2− group; and the (3) the rare superoxides, which con−
tain the O2 group.
(d) In ternary acids (oxoacids), hydrogen usually bonds to an
O atom, not to the central atom. (Exceptions are phosphorous acid, H3 PO3 , where one H is bonded to P, and
GChem I
9.7
hypophosphorous acid, H3 PO2 , where two H are bonded
to P.)
(e) The least electronegative atom is usually the central
atom.
(f) For ions or molecules that have more than one central
atom, the most symmetrical skeletons possible are used.
2. All the valence electrons must be accounted for.
3. Usually, each atom in a Lewis structure acquires an electron
configuration with an outer-shell octet. (The outer shell of H
has only 2 electrons.) Nonmetals of the third period and beyond may be surrounded by more than eight electrons (expanded valence). In covalent compounds containing Be and
B, the number of electrons surrounding the central atom in a
stable molecule may be less than 8.
4. Usually, all the electrons in a Lewis structure are paired.
5. Often, both atoms in a bond pair contribute equal numbers of
electrons to the covalent bond, but sometimes both electrons
in a bond pair are derived from a single atom (coordinate
covalent bond).
6. Sometimes, it is necessary to represent double or triple covalent bonds in a Lewis structure (C, N, O, P, S).
7. Sometimes, it is impossible to draw a single Lewis structure
that is consistent with all the available data. In these instances the true structure can only be represented as a composite or hybrid of two or more plausible structures. This
situation is called resonance.
GChem I
9.8
Steps
1. Determine the number of available valence electrons, A . Add
up the total number of valence electrons for all the atoms in
the structure. For anions, add the charge of the ion. For
cations, subtract the charge of the ion. This is the number of
electrons that must appear in the Lewis structure.
2. Determine the number of valence electrons needed for
octets, N . This number N equals the number of non-H atoms
times eight plus the number of H-atoms times two. (Note the
exceptions for Be and B.)
3. Determine the number of shared electrons, S .
S=N−A
The number of bond pairs is S/2.
4. Start with a plausible skeleton structure. This is a representation of the order in which atoms are bonded together. The
skeleton structure consists of one or more central atoms with
the other terminal atoms bonded to the central atom(s).
5. Place the bond pairs in the skeleton structure, starting with
one bond pair between each pair of atoms. If bond pairs remain, form double or triple bonds.
If the number of bond pairs is less than the number needed
to bond all atoms, then S is increased to the number of elecGChem I
9.9
trons needed. This generally indicates that the central atom
displays expanded valence.
6. Determine the number of remaining valence electrons, A −S .
Place these electrons as lone pairs, starting with the terminal atoms to complete the octets (exceptions!). If electrons
are left over, place them on the central atom (expanded valence).
7. Use the concept of formal charge to assess the plausibility of
the Lewis structure.
Formal Charge
Formal Charge: An atom in a molecule owns its lone pairs completely, but has an equal share in bonding electron pairs.
If this results in an atom having more electrons in the molecule
than as a free, neutral atom, we say that the atom has a negative formal charge in the Lewis structure. If it has less electrons
in the molecule, then it has a positive formal charge.
FC = number of valence electrons of free atom − number of lonepair electrons − 12 number of shared electrons
The sum of formal charges must equal zero for a neutral molecule and must equal the charge for ions.
GChem I
9.10
Lewis structures with the lowest formal charges are most plausible.
Exercise: Write the Lewis structure for CF2 Cl2 (Freon-12)
A = number of available valence electrons
A = 4 + 2 × 7 + 2 × 7 = 32
N = number valence electrons needed for octets
N = 5 × 8 = 40
S = N − A = number of shared electrons
S = N − A = 40 − 32 = 8
=⇒ 4 bond pairs
skeleton structure: halogens are terminal; C is the central atom
Cl
F
C
F
Cl
Freon-12 skeleton
place the 4 bond pairs
GChem I
9.11
Cl
...
...
..
...
....
.
F
..........................
C
..........................
F
Cl
Freon-12 skeleton
Cl
..
...
..
...
...
..
F
..........................
C
..........................
F
...
...
...
...
...
...
Cl
8Freon-12
electronsbonds
placed, leaves 32 − 8 = 24 valence electrons
C has octet; place 24 electrons as lone pairs on the halogens:
·· ·
·· Cl
·
·· ··
F
··
...
...
..
...
...
.
..........................
C
...
...
...
...
...
...
..........................
·· ·
F
·· ·
·· Cl ··
··
Lewis
–Freon-12
Typeset by Foil
TEX –
1
calculate formal charges:
2
F : 7−6− = 0
2
2
Cl : 7 − 6 − = 0
2
8
C : 4−0− = 0
2
Exercise: Write the Lewis structure for methanol CH3 OH
GChem I
9.12
Exercise: Write the Lewis structure for carbon disulfide CS2
Exercise: Write the Lewis structure for the carbonate ion CO3
2−
A = 1 × 4 + 3 × 6 + 2 = 24
N = 4 × 8 = 32
O
S = N − A = 32 − 24 = 8
=⇒
4 bonds
skeleton
O
Cstructure:
O
O
O
C
·O·
CO2 skeleton
· ·· ·
..
...
..
...
...
..
O
place the four bond pairs:
··
·· O
··
..........................
C
O
··
O
··
..........................
..........................
CO2 skeleton
....
...
..
...
...
..
O
..........................
C
..........................
..........................
O
O
CO224
Lewis
leaves
− 8 = 16 valence electrons:
O

....
...
..
...
...
··O·
·· O
·
CO2
bonds



 ··
 ·· O
C
··
..........................
C
...
...
..
...
...
..
..........................
CO2 bonds
CO2 Lewis
GChem I
2−
..........................
..........................
..........................
..........................



·· 
O
··
9.13
however: the double bond could also be placed between the
central C atom any of the other two O atoms:




 ··
 ·· O
··
2−
·· ·
·· O
·
...
...
..
...
...
.
..........................
C
..........................
..........................




·· 
O
··



↔
 ··
 O
··


↔
 ··
 ·· O
··
...
...
..
...
....
.
C
·· O ··
...
...
..
...
...
.
..........................
...
...
..
...
...
.
C
..........................
··
O ··
··





↔
2−
·· ·
·· O
·
..........................
..........................
2−
..........................



·· · 
O· 
··
resonance
formal charges:
8
C : 4−0− = 0
2
4
O : 6−4− = 0
2
2
O : 6 − 6 − = −1
2
Exercise: Write the Lewis structure for phosphorus pentafluoride
GChem I
– Typeset by FoilTEX –
9.14
3
PF5
A = 1 × 5 + 5 × 7 = 40
N = 6 × 8 = 48
S = N − A = 48 − 40 = 8
=⇒
4 bonds
skeleton structure:
F
F
F
F
P
F
P
F
FF
F
F
?only four bonds? need at least five
increase S to 10; leaves 30 electrons for lone pairs
F
F
F
.....
.....
.....
.....
.....
.
.....
.....
.....
.....
.....
.
...
.....
.....
.....
.
.
.
....
F
.
.....
.....
.....
.....
.....
.....
.
.....
.....
...
.....
.....
.....
.....
....
.....
.....
.
.....
....
.
.
.
.
.
.
.
.
.
..... ...
...
.
.
.
.
.
.
.
.
.
......
...
...
.
.
.
.
.
.
.....
....
....
...
....
...
...
...
F
P
F
P
FF
F
F
place the remaining 30 electrons as lone pairs on the F atoms:
GChem I
9.15
·· ··
F
··
·· ··
F
··
.....
.....
.....
.....
.....
.
.
.....
.....
.....
.....
.....
P
...
...
...
...
...
...
..
.....
.....
.....
....
.....
.....
.....
.....
.....
.....
.
·· F ··
··
·· ·
F
·· ·
·· ·
F
·· ·
P has an expanded octet (extended valence)
origin: d -orbitals are involved
2
6
octet rule: ns + np −→ 8e
−
Exercise: Write the Lewis structure for the triiodide ion I3
−
A = 3 × 7 + 1 = 22
N = 3 × 8 = 24
S = N − A = 24 − 22 = 2
=⇒
1 bonds
one bond pair? need two bond pairs: increase S to 4
skeleton structure:
!!
II
II
II
""−−
−
−
I
skeleton
I
33 skeleton
place
the two bond pairs:
!!
""−−
II– TypesetIIby FoilT XII–
....................................................
....................................................
−
bonds
II−
33 bonds
GChem I
!! ····
···· II
····
−
−
2
E
....................................................
9.16
····
II
····
....................................................
···· ·· ""−−
II ·
···· ·
I−
3 skeleton
!
"−
I
I "−
! I
I
I
I
I−
3− bonds
place
lone pairs to fulfill octet rule:
I3 skeleton
! ··
··
··" · "−
! ·· I
I
I−
I ··
I ··
I ·· ·
..........................
..........................
..........................
..........................
..........................
..........................
−
I−
bonds
check
if allLewis
electrons are accounted for:
33 almost
!
·· · "−−
6 +··2··I+ 4 + 2 ··
+
6
=
20
I
···I···
··
·· ·
..........................
..........................
..........................
..........................
two
− electrons left over; place the two remaining electrons as a
I−
almost Lewis
33 Lewis
lone pair on the central atom:
!
·· ··I
··
..........................
··
·· I ··
..........................
·· · "−
I
·· ·
I−
3 Lewis octet (extended valence)
expanded
Exercise: Write the Lewis structure for sulfuric acid H2 SO4
Exceptions to the Octet Rule
(1) Incomplete Octet
electron deficient compounds: in covalent compounds containing Be and Group IIIA elements, esp. B, the number of electrons
– Typeset
by FoilTEX – the central atom in a stable molecule may be 1less
surrounding
than 8
examples:
BeClby2 FoilTEX –
– Typeset
GChem I
1
9.17
··
·· Cl
··
..........................
Be
Be··: 4e−
·· Cl
··
Be
·· Cl
B
··
··
BCl3
·· Cl ··
··
··
·· Cl
B
··
..........................
..........................
..........................
..........................
..........................
...
...
...
...
...
...
..........................
...
...
...
...
...
...
..........................
··
Cl ··
··
·· ·
··
Cl
Cl
· · ···
··
··
Cl ··
··
·· Cl ··
··
B:
6e−
Be and B fulfill the octet rule in other species, e.g, BeF4 2− and
BF4 −
(2) Expanded Octet (Extended Valence)
3rd or higher period elements can be surrounded by more than
4 valence pairs in some compounds: d -subshell
examples of compounds with five valence pairs on the central
−
atom: PF5 , I3
examples of compounds with six valence pairs on the central
atom:
GChem I
– Typeset by FoilTEX –
9.18
1
·· ··
F
··
SF6
·· ··
F
··
·· ··
F
··
·· ··
F ··
.....
.....
.....
.....
.....
.
.
.....
.....
.....
.....
.....
....
...
..
....
...
S
...
...
...
...
...
...
...
.....
.....
.....
.
.
.
....
.....
.....
.....
.....
.....
.
·· F ··
··
·· ·
F
·· ·
·· ·
F
·· ·
XeF4
·· ··
F
··
·· ··
F
··
.
.....
.....
.....
.....
.....
S
...
...
...
...
...
...
.....
.....
.....
.....
.....
.
·· F ··
··
.....
.....
.....
.....
.....
.
.
.....
.....
.....
.....
.....
··
Xe
··
...
.....
.....
.....
.
.
.
.
....
.....
.....
.....
.....
.....
.
·· ·
F
·· ·
·· ·
F
·· ·
·· ·
F
·· ·
(3) Odd-Electron Compounds
NO
·· ··
F
··
.....
.....
.....
.....
.....
.
....
.....
.....
.....
.
.
.
....
.
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.
·· ·
F
·· ·
··
Xe
A = 1 × 5 + 1· ·× 6 = 11
·· ·
·· ··
F
F
· = 16 · · ·
N = 2 ×·8
S = N − A = 16 − 11 = 5 ?!
·
N
··
..........................
..........................
··
O
··
formal charges:
4
N : 5−3− = 0
2
4
O : 6 −H4 − = 0
2
– Typeset by FoilTEX –
..
...
..
...
...
..
H
C
H
stable odd-electron
molecules like NO are rare
·
..........................
..........................
GChem I
9.19
– Typeset by FoilTEX –
1
·
N
··
··
O
··
..........................
..........................
odd-electron species are called radicals and are normally very
H
reactive
....
...
..
...
...
examples:
H
..........................
C
·
..........................
H
H
.....
...
..
...
...
CH3 (methyl radical)
H
OH (hydroxyl radical)
··
·O
··
..........................
··
··
..........................
C
·
..........................
H
H
·O
H most symmetrical skeleton
If more than one central atom,
use
..........................
example 1:
C2 H4 ethene
A = 2 × 4 + 4 × 1 = 12
N = 2 × 8 + 4 × 2 = 24
S = N − A = 24 − 12 = 12
skeleton structure:
GChem I
– Typeset by FoilTEX –
=⇒
6 bonds
– Typeset by FoilTEX –
9.20
H
H
H
C
H
C
C
H
C
H
H
H
place the six bond pairs:
H
H
..
...
..
...
...
..
..
...
..
...
...
..
H
C
...
...
..
...
......
...
...
...
...
...
C
H
..........................
..........................
..........................
..........................
...
...
...
...
...
...
H
H
C
...
...
..
...
......
...
...
...
...
...
C
H
...
...
...
...
...
...
H
example 2:
P2 O7 4−
A = 2 × 5 + 7 × 6 + 4 = 56
N = 2 × 8 + 7 × 8 = 72
S = N − A = 72 − 56 = 16
=⇒
8 bonds
skeleton structure:
– Typeset by FoilTEX –
1
GChem I
– Typeset by FoilTEX –
9.21
1

O




 O





4−
O
P
O




O





P
O
O
place the eight bond pairs, then fulfill octet rule, then check if all
electrons are accounted for:




 ··
 ·O
 · ··




·· ·
·· O
·
...
...
..
...
...
.
..........................
P
...
...
...
...
...
...
·· O ··
··
4−
·· ·
·· O
·
..........................
··
O
··
...
...
..
...
...
.
..........................
P
...
...
...
...
...
...
·· O ··
··
..........................
··
O ··
··










8 bond pairs + 20 lone pairs = 56 electrons
the bonds between the P’s and the terminal O’s are coordinate
covalent bonds
GChem I
– Typeset by FoilTEX –
9.22
2
formal charges:
8
P : 5−0− = 1
2
2
terminal O : 6 − 6 − = −1
2
formal charges sum to 4−
consider alternative structure with expanded valence on the P’s
by moving a lone pair, from say each of the top O’s, into a bond
pair
2
6
6
6
6 ··
6 ·O
6 · ··
6
6
6
4
·· O ··
·· O ··
..
...
..
...
...
..
..
...
..
...
...
..
..........................
P
34°
..........................
...
...
...
...
...
...
··
O
··
..
...
..
...
...
..
..
...
..
...
...
..
P
..........................
..........................
...
...
...
...
...
...
·· O ··
··
·· O ··
··
··
O ··
··
7
7
7
7
7
7
7
7
7
5
(resonance)
formal charges:
10
=0
2
4
top O : 6 − 4 − = 0
2
P : 5−0−
GChem I
9.23
other terminal O :
2
6 − 6 − = −1
2
formal charges sum to 4−
Bond Energies
bond dissociation energy D = enthalpy change for breaking one
mole of bonds in molecules
H3 C CH3 (g) −→ H3 C(g) + CH3 (g) ∆H = D = +347 kJ/mol
H2 C CH2 (g) −→ H2 C(g) + CH2 (g) ∆H = D = +611 kJ/mol
HC CH(g) −→ HC(g) + CH(g) ∆H = D = +837 kJ/mol
!: D > 0, bond breaking = endothermic process, requires energy
=⇒ bond formation = exothermic
H3 C(g) + CH3 (g) −→ H3 C CH3 (g) ∆H = −D = −347 kJ/mol
bond energy D : (i) tabulated values are averages (over different
molecules); (ii) defined for the gas phase
D
>D
>D
bond energies −−→ estimate for enthalpy of reaction
∆Hrxn =
GChem I
X
D(bonds broken) −
X
D(bonds formed)
9.24
example 1:
CH4 (g) + Cl2 (g) −→ CH3 Cl(g) + HCl(g) ∆H =?
(1) bonds broken
C H:
D = +414 kJ/mol
Cl Cl :
D = +243 kJ/mol
(2) bonds formed
C Cl :
D = +330 kJ/mol
H Cl :
D = +431 kJ/mol
∆H = (414 kJ/mol + 243 kJ/mol) − (330 kJ/mol + 431 kJ/mol)
∆H = −104 kJ/mol
example 2:
CH4 (g) + 2 O2 (g) −→ CO2 (g) + 2 H2 O(g) ∆H =?
(1) bonds broken
4×C H :
4 × 414 kJ/mol
2×O O :
2 × 498 kJ/mol
GChem I
9.25
(2) bonds formed
2×C O :
2 × 803 kJ/mol
4×O H :
4 × 464 kJ/mol
∆H = (4 × 414 kJ/mol + 2 × 498 kJ/mol) − (2 × 803 kJ/mol + 4 × 464 kJ/mol)
∆H = −810 kJ/mol
estimate!
check:
o n
o
◦
◦
∆H =
− ∆H f [CH4 (g)] + 2∆H f [O2 (g)]
©
ª
∆H ◦ = 2 × (−242 kJ/mol) + (−394 kJ/mol)
©
ª
− (−74.8 kJ/mol) + 2 × 0 kJ/mol
◦
n
2∆H f◦ [H2 O(g)] + ∆H f◦ [CO2 (g)]
∆H ◦ = −803 kJ/mol
GChem I
9.26