Chapter 9
Ionic and Covalent Bonding
8–1
John A. Schreifels
Chemistry 211
Chapter 9-1
Overview
• Ionic Bonds
– Describing Ionic Bonds
– Electron Configuration of Ions
– Ionic Radii
• Covalent Bonds
–
–
–
–
–
Describing Covalent Bonds
Polar Covalent Bonds: Electronegativity
Writing Lewis Electron-Dot Formulae
Bond Length and Order
Bond Energy
John A. Schreifels
Chemistry 211
8–2
Chapter 9-2
IONIC BONDING
• Ionic bonds are electrically neutral groups
that are held together by the attraction arising
from the opposite charges of a cation and an
anion.
• Substances that have ionic bonds in a solid
form a salt having high melting point and high
crystallinity.
• Bonding thought of as the result of the
combination of neutral atoms with transfer of
one or more electrons from one atom to the
other.
John A. Schreifels
Chemistry 211
Chapter 9-3
8–3
LEWIS SYMBOLS AND THE OCTET
RULE
•
•
•
It was observed that the electron configuration of Na
many substances after ion formation was that of Mg
an inert gas  octet rule.
O
Octet rule: Main-group elements gain, lose, or
share in chemical bonding so that they attain a
valence octet (eight electrons in an atoms valence
shell).
E.g. The electron configuration of each reactant in
the formation of KCl gives:
Na+
Mg2+
O
2-
– K+ is that of [Ar]
– Cl is also that of [Ar].
• The other electrons in the atom are not as
important in determining the reactivity of that
substance.
• The octet rule is particularly important in
compounds involving nonmetals.
John A. Schreifels
Chemistry 211
8–4
Chapter 9-4
Energy in Ionic Bonding
• When potassium and chlorine atoms approach each
other we have:
K(g) K+(g) + e
Ei = +418 kJ
Cl(g)+ e Cl(g)
Eea = 349 kJ
K(g)+Cl(g) K+(g) + Cl(g)
E = + 69 kJ
• Positive E = energy absorbed  energetically not
allowed.
• Driving force must be the formation of the crystalline
solid.
K+(g) + Cl(g)  KCl(s)
8–5
John A. Schreifels
Chemistry 211
Chapter 9-5
Formation of Crystalline Lattice
• Energy of crystallization estimated from Coulomb’s
Law by
zz
E k 1 2
d
– assuming ions are spheres.
– Use ionic radii to determine charge separation. Coulomb’s Law
rK+ = 133x1012 m; rCl = 181x1012 m
d = 133x1012 m + 181x1012 m = 314x1012 m
z1 = z2 = 1.602x1012 C(oulombs); actually one is the negative
of the other.
k = 8.99x109 Jm/C2
E

9
2

 8.99x10 J  m / C  1.602x10
314x1012 m
19
C
  6.02x1023mol1
2
 442 .3x103J / mol  442 .3kJ / mol
8–6
• This is related to the negative of the lattice energy, as discussed
later.
John A. Schreifels
Chemistry 211
Chapter 9-6
BORN-HABER CYCLE AND LATTICE
ENERGIES
•
Overall energetics for the formation of crystalline solids can be determined
from a Born-Haber cycle that accounts for all of the steps towards the
formation of solid salts from the elements. For the formation of KCl from
its elements we have:
K (s)  K(g)
1/ 2 Cl2(g)  Cl(g)
1. Sub of t he metal
2. Dissociation of ch lorine
K(g)  K   e
3. Ionizati on of pota ssium
4. Formatio n of chlor ide ions ( Eea)
5. Formatio n of solid KCl
Sum react ions and e nergies
•
•
•
Cl  e  Cl
K (g)  Cl(g)  KCl(s)
K(s)  1/ 2Cl2(g)  KCl(s)
 89.2 kJ
 122 kJ
 418 kJ
 348.6 kJ
 715 kJ
 434.4 kJ
Net energy change of 434 kJ/mol indicates energetically favored.
Energy for the fifth step is the negative of the
lattice energy: energy required to break ionic bonds and sublime (always
positive).
E.g. Determine the lattice energy of BaCl2 if the heat of sublimation of Ba
is 150.9 kJ/mol and the 1st and 2nd ionization energies are 502 and 966
8–7
kJ/mol, respectively. The heat for the synthesis of BaCl2(s) from its
elements is 806.06 kJ/mol.
John A. Schreifels
Chemistry 211
Chapter 9-7
Energy Level Diagram of Born Haber
Cycle
8–8
John A. Schreifels
Chemistry 211
Chapter 9-8
LATTICE ENERGIES AND PERIODICITY
zz
E k 1 2
d
• Lattice energy can also be
determined from Coulomb’s law:
Coulomb’s Law
–
Directly proportional to charge on
each ion.
– Inversely proportional to size of
compound (sum of ionic radii).
• Table (right) presents the lattice
energies for alkali and alkaline
earth ionic compounds. The lattice
energies
– decrease for compounds of a
particular cation with atomic
number of the anion.
– decrease for compounds of a
particular anion with atomic number
of the cation.
John A. Schreifels
Chemistry 211
Lat. E,
kJ/mol
Lat. E,
kJ/mol
LiF
1030
MgCl2
2326
LiCl
834
SrCl2
2127
LiI
730
MgO
3795
NaF
910
CaO
3414
NaCl
788
SrO
3217
NaBr
732
ScN
7547
NaI
682
KF
808
KCl
701
KBr
671
CsCl
657
CsI
600
8–9
Chapter 9-9
Ionic Radii
• Ionic radius – a measure of the size of
a spherical region around the nucleus
of an atom where electrons are most
likely to reside.
– Cation loses electrons from its valence
shell. Electrons from other valence
shells are closer to the nucleus.
– Cation also has more protons than
electrons which adds to the pull on the
remaining electrons and decreases the
radius.
– Anion has more electrons than protons; the pull of the nucleus is
less per valence electron. Also, the electron – electron repulsion is
greater. These lead to larger radius for an anion.
8–10
John A. Schreifels
Chemistry 211
Chapter 9-10
Ionic Radii - Trends
– Ionic radii increase down any column because of the addition
of electron shells.
– In general, across any period the cations decrease in radius.
When you reach the anions, there is an abrupt increase in
8–11
radius, and then the radius again decreases.
John A. Schreifels
Chemistry 211
Chapter 9-11
Ionic Radii - Isoelectronic Ions
• Isoelectronic substances have the same number of
electrons and electron configuration.
2

2Ca

K

Ar

Cl

S
20
19
18
17
16
All have 18 electrons
• Largest radius = ion with smallest number of protons.
• Smallest radius = ion (atom) with largest number of
protons.
8–12
John A. Schreifels
Chemistry 211
Chapter 9-12
The Covalent Bond
• Repulsive forces of the electrons offset by the attractive forces
between the electrons and the two nuclei.
• Most stable bond energy and bond distance characterizes
bonds between two atoms.
• Strengths of Covalent Bonds:
• Bonds form because their formation produces lower energy
state than when atoms are separated.
• Breaking bonds increases the overall energy of the system.
Energy for breaking bonds has a positive sign (negative means
that energy is given off).
H - H (g)  2H(g) H = 436 kJ.
• Ionic vs. Covalent Bonds
– Ionic compounds have high melting and boiling points and tend to
be crystalline;
– Covalently bound compounds tend to have lower melting points
since the attractive forces between the molecules are relatively
weak.
John A. Schreifels
Chemistry 211
8–13
Chapter 9-13
Lewis Structures
• Lewis structure valence electrons represented by
dots and are placed where they would be in any
bonding that might exist.
• Lewis structures of second row elements:
– H2
– CH4
– H 2O
BH3
NH3
HF
• Each has 8 electrons around the central atom; thus
we can predict the number of bonds that will form
from the position in the periodic table.
.. ..
E.g. The structure of chlorine is :Cl
.. :Cl
.. :
– Bonding electrons = shared electrons.
– Non-bonding or lone pair = unshared electrons
John A. Schreifels
Chemistry 211
8–14
Chapter 9-14
Lewis Structures(cont’d)
•
•
•
Octet can be filled by donation of electrons from each atom or
one atom can supply both electrons.
E.g. H+ + NH3  NH
. 4
"co-ordinate covalent bond".
E.g.2 BF3  F   BF4
Multiple bonds may form as a result when the two atoms
forming the bond do not have enough electrons.
– O=O
– NN
Multiple bonds are shorter and stronger than single bonds
because of the extra electrons holding the two atoms together.
8–15
John A. Schreifels
Chemistry 211
Chapter 9-15
Polar Bonds: Electronegativity
•
Electronegativity is a measure of the atom’s ability to gain or lose
electrons. It is directly related to its ionization tendency and its ability to
form the inert gas configuration. Obtained by:   Ei  Eea
2
where Ei = ionization energy and Eea = the electron affiinity.
E.g. Li has a very low ionization energy and electron affinity, while Cl
has a both a high ionization energy and high electron affinity.
Electronegativity will be high for Cl and low for Li.
– Fluorine has the highest electronegativity of 4.0.
•
Electronegativities (see Fig. 9.15)
– increase from bottom to top of periodic table and
– increase to a maximum towards the top right.
•
•
•
•
•
Combination of elements with intermediate electronegativities forms
bonds that are intermediate between covalent and ionic.
can provide an insight as to the type of bond that would be expected.
Ionic bonds formed when   2
covalent bonds forms when   1.
Polar covalent forms when 1    2, the bonding is "intermediate"
between the two.
John A. Schreifels
Chemistry 211
8–16
Chapter 9-16
Polar Bonds: Electronegativity2
E.g.1 Determine the polarity of the N – H in NH3.
E.g. 2 Predict the type of bond formed in CCl4.
• The magnitude of  indicates if electrons are polarized around
one element in preference to the other.
• Polar bond polar. With intermediate , a small charge on the
atom due to that bond develops. + and  designates which is
the positive and negative side respectively.
E.g.3 Determine the relative polarities of HF, HCl, HBr and HI.
8–17
John A. Schreifels
Chemistry 211
Chapter 9-17
Lewis Structures of Polyatomic
Molecules
• Procedure for more complicated molecules:
– Determine the total number of valence electrons from each
atom.
– Distributed atoms around the central atom (least
electronegative. Hydrogen atoms are usually attached to
any oxygen.
– Satisfy the octet of the atoms bonded to the central atom.
– Satisfy the octet of the central atom by distributing the
remaining electrons as electron pairs around it. (multiple
bonds may be necessary)
E.g. Determine the Lewis structure of H2SO4.
E.g. Draw the Lewis dot structures of NCl3, CSe2, and CO.
John A. Schreifels
Chemistry 211
8–18
Chapter 9-18
FORMAL CHARGES
• Formal Charge (of an atom in a Lewis formula) the hypothetical
charge obtained by assuming that bonding electrons are equally
shared between the two atoms involved in the bond. Lone pair
electrons belong only to the atom to which they are bound.
E.g. determine the formal charge on all elements: PCl3, PCl5,
and HNO3.
• formal charge (FC) allows the prediction of the more likely
resonance structure.
• To determine the more likely resonance structure:
– FC should be as close to zero as possible.
– Negative charge should reside on the most electronegative and
positive charge on the least electronegative element.
E.g. draw the resonance structures of H2SO4; determine the
formal charge on each element and decide which is the most
likely structure.
John A. Schreifels
Chemistry 211
8–19
Chapter 9-19
Lewis Structures and Resonance
• Quantum theory indicates that any position possible
for an electron.
• Equivalent electron positions often possible:
• E.g. SO2: O=S-O and :O-S=O.
– Each structure equally likely.
– the true form of the molecule is a hybrid of these and is
called resonance and the hybrid form is called a resonance
hybrid.
8–20
John A. Schreifels
Chemistry 211
Chapter 9-20
Exceptions to the Octet Rule
• Although many molecules obey the octet rule,
there are exceptions where the central atom
has more than eight electrons.
– Generally, if a nonmetal is in the third period or
greater it can accommodate as many as twelve
electrons, if it is the central atom.
– These elements have unfilled “d” subshells that
can be used for bonding.
E.g determine the Lewis dot structure of XeF4,
ICl3, and SF4
8–21
John A. Schreifels
Chemistry 211
Chapter 9-21
Bond Dissociation Enthalpies
• Bond dissociation energy, D – the energy required
to break one mole of a type of bond in an isolated
molecule in the gas phase.
• Useful for estimation of heat of unknown reactions.
• Average bond energies listed in tables (e.g. C – H
bond); rest pf structure not very important
• HO-H bond in H2O and CH3O-H bond are 492 and
435 kJ/mol.
• Hess’s law can be used with bond dissociation
energies to estimate the enthalpy change of a
reaction. The breaking in a C – H bond would be C –
H(g)  C(g) + H(g)
H = D = 410 kJ.
– Sign always positive since energy must be supplied to break
8–22
bond.
John A. Schreifels
Chemistry 211
Chapter 9-22
Using Bond Dissociation Enthalpies
• E.g. Estimate the heat of formation of H2O(g) from bond
dissociation energies. Thus determine:
• H2(g) + ½ O2(g)  H2O(g)
Hof = ?
• From the book (Table 9.5):
H – H (g)  2H(g)
½ O=O  O(g)
2H(g) + O(g)  H – O – H (g)
H2(g) + ½ O2(g)  H2O(g)
H = D1 = 436 kJ
H = D2 = 494/2 = 247 kJ
H = 2D3 = 2*459 kJ
Hof = 235 kJ
Actual = 241.8 kJ
• Can be determined by suming all the energies for the bonds
broken and subtract from if the sum of the energies for the bonds
formed.
E.g. 2 Estimate the energy change for the chlorination of ethylene:
– CH2=CH2(g) + Cl2(g) CH2ClCH2Cl
8–23
John A. Schreifels
Chemistry 211
Chapter 9-23
Using Bond Dissociation Enthalpies
• It may be necessary to include a phase change since many
reactions or reactants are not in the gas phase.
E.g. Determine the heat of formation of CCl4(l).
• Solution: The reaction is:
Hof = ?
C(gr) + 2Cl2(g)  CCl4(l)
• Write the reactions and sum energies:
C(gr)  C(g)
H1 = 715 kJ
2Cl – Cl(g)  4Cl(g)
C(g) + 4Cl(g)  CCl4(g)
CCl4(g)  CCl4(l)
C(gr) + 2Cl2(g)  CCl4(l)
H2 = 486
H3 = 1320
H4 = 43
H = 162 kJ
Actual is  139 kJ.
John A. Schreifels
Chemistry 211
8–24
Chapter 9-24
Electronegativities
9_12
IA
IIA
Li
1.0
Be
1.5
Na
0.9
Mg
1.2
K
0.8
H
2.1
VIIIB
IIIB
IVB
VB
VIB
VIIB
Ca
1.0
Sc
1.3
Ti
1.5
V
1.6
Cr
1.6
Mn
1.5
Fe
1.8
Co
1.8
Rb
0.8
Sr
1.0
Y
1.2
Zr
1.4
Nb
1.6
Mo
1.8
Tc
1.9
Ru
2.2
Cs
0.7
Ba
0.9
La–Lu
1.1–1.2
Hf
1.3
Ta
1.5
W
1.7
Re
1.9
Os
2.2
Fr
0.7
Ra
0.9
Ac–No
1.1–1.7
IIIA
IVA
VA
VIA
VIIA
B
2.0
C
2.5
N
3.0
O
3.5
F
4.0
Al
1.5
Si
1.8
P
2.1
S
2.5
Cl
3.0
IB
IIB
Ni
1.8
Cu
1.9
Zn
1.6
Ga
1.6
Ge
1.8
As
2.0
Se
2.4
Br
2.8
Rh
2.2
Pd
2.2
Ag
1.9
Cd
1.7
In
1.7
Sn
1.8
Sb
1.9
Te
2.1
I
2.5
Ir
2.2
Pt
2.2
Au
2.4
Hg
1.9
Tl
1.8
Pb
1.8
Bi
1.9
Po
2.0
At
2.2
Return to slide 16
8–25
John A. Schreifels
Chemistry 211
Chapter 9-25
8–26
Return to Slide 23
John A. Schreifels
Chemistry 211
Chapter 9-26