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Notes
Part II
Fall 2010
Dr. Charles Berks
Periodic Properties
Covalent (or bonding) Atomic Radius: The covalent atomic radius is defined differently
for nonmetals and metals.
Nonmetals: One-half the distance between two atoms bonded together.
Metals: One half the distance between 2 atoms of the metal next to each other in a
crystal of the metal.
The covalent atomic radius is often expressed in picometers (pm).
Atomic Radius: An average value for the covalent atomic radius of an element. It is also
often expressed in pm. In this course we usually consider the atomic radius rather than
the covalent atomic radius.
Trends in Atomic Radius:
Atomic radii tend to decrease across a row in a period periodic table.
Period 2 Element
Li
Be
B
C
Atomic Radius (pm)
152
112
85
77
29
Across a period in the s and p blocks electrons are added to the same shell while
nuclear charge is increasing as protons are added to the nucleus. Electrons in the same
shell are not very effective at shielding each other from the nuclear charge. Thus the
effective nuclear charge, Zeff, is increasing. As a result the atomic radius tends to
decrease.
Compare the atomic structure diagrams of lithium and beryllium.
The effective nuclear charge of for the valence electron of lithium can be estimated.
Zeff of a specific electron = Z – S where Z is the charge of the nucleus and S is the
shielding effect of other electons.
Zeff must be estimated because while the charge of the nucleus is known the shielding
effect of other electrons, S, is in doubt. S depends on whether the other electrons are in
the same or different shell, subshell and orbital as the electron being considered and also
on the penetration ability of the electron being considered.
Penetration:
Penetration is the process by which an outer electron moves through the region
occupied by inner electrons.
In general for electrons of the same shell; the ability of an electron to penetrate
depends inversely on its quantum number l. Therefore s electrons penetrate more
than p electrons and p electrons more than d electrons.
For the outer electron of lithium we can approximate Zeff = Z – S = 3 – 2 = 1
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For the beryllium the atomic structure diagram would be
Thus for beryllium Zeff = Z – S = 4 – 2 = 2. Again this is an estimate of the effective
nuclear charge.
Therefore it is not surprising that the atomic radius of lithium is greater than that of
beryllium since the outer-most electrons of beryllium experience the greater effective
nuclear charge.
Atomic radius tends to increase down a column in the periodic table. This is because the
size of the orbitals increase as one moves down a column. The number of electron shells
has also increased and this results in greater repulsions between electrons.
Group 1A Element
Li
Na
K
Rb
Atomic Radius (pm)
152
186
227
248
Group 7A Element
F
Cl
Br
I
In general,
Atomic Radius (pm)
72
99
114
133
Atomic Radius ∞ Period Number
Zeff
That is; as the number of shells of electrons increases the radius of the atom increases.
Whereas as the effective pull (the effective nuclear charge) of the nucleus increases the
radius of the atom decreases. As one crosses a row in the periodic table the number of
shells of electrons remains constant so it is the effective pull of the nucleus that is the
major factor. As one proceeds down a column in the periodic table it is the number of
shells of electrons that is the major influence on the radius or size of the atom.
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Ionic Radius:
The ionic radius is the radius of a cation or anion. In order to compare the radius of
various ions it is important to be able to predict the electron configuration of the ion.
Consider the atoms sodium and fluorine and the monatomic ions they form.
Na ([Ne]3s1)  Na+ ([Ne]) + e- and F ([He]2s22p5) + e-  F- ([Ne])
The sodium cation, Na+, and the fluoride anion are isoelectronic with neon. That is, both
ions and the neon atom have the same electron configurations. Each species differs in
nuclear charge. In fact, most group A elements produce ions that are isoelectronic with a
noble gas. These ions are said to have s2p6 electron configurations.
Trends in Ionic Radii:
In general the factors affecting the radius of an ion are same factors determining the
radius of an atom. Thus ionic radius ∞ n of valence shell of the ion
Zeff
1. Consider a series of isoelectronic ions. For example consider the series from O2until Mg2+. Each of these ions is isoelectronic with neon. Thus in the general
relationship the value of n for the valence shell is constant at n = 2. However the
charge of the nucleus is increasing across the series and therefore Zeff is increasing
across the group. As the effective charge of the nucleus increases the radius of the
ion must decrease.
Ion
O2FNa+
Mg2+
Radius (pm)
140
136
95
65
2. Cation formation from an atom results in a decrease in radius. When a group A
element forms a cation the cation is normally a s2p6 ion. That is, it has an electron
configuration that is isoelectronic with a noble gas. The ionization process results in
the valence shell electrons in the atom being removed. Thus the cation has a lesser
number of shells of electrons than the original atom. This results in a decrease in
electron repulsions. At the same time the nuclear charge is constant but the number
of shells of screening electrons has been reduced. Thus the effective nuclear charge
experienced by the outer electrons of the cation must be less than those experienced
by the outer electrons of the original atom. Therefore the radius of the cation is
smaller than that of the atom.
3. Anion formation results in an increase in radius. The addition of electrons results
in an increase in electron repulsions while the nuclear charge remains constant.
Thus the radius must increase.
32
4. As you go down a group ionic radius increases. The increasing number of shells
of electrons results in greater repulsions between electrons. Additionally the size of
the orbitals is also increasing as the value of n associated with the orbitals increases.
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Ionization Energy
First Ionization Energy (IE1): The first ionization energy is the energy required to remove
the most loosely held electron from an isolated gaseous atom in its ground state. The
process is endothermic and therefore is associated with a positive enthalpy change, H.
The chemical equation related to IE1 can be written as
A(g)  I+(g) + eIE1 can be understood in terms of the modified Bohr Theory. IE1 is the energy change
associated with the electron jump from the ground state to the outermost orbit. Therefore
IE1 = E∞ - Eground = Zeff2B
n2ground
(Since the energy of the n = ∞ orbit is zero.)
In the case of the sodium atom the value for IE1 = 496 kJ/mol and the chemical equation
is
Na(g)  Na+(g) + e-.
The value of nground for the sodium atom is 3. Thus we can rearrange the above equation
to calculate the value of Zeff. Doing so, gives a calculated value of 1.84 for the effective
nuclear charge.
Trends in First Ionization Energies
1. Across a period IE1’s tend to increase. That is, IE1 ∞ Zeff
Consider the period 2 elements.
Atom
IE1
(MJ/mol)
Zeff (calc)
Li
0.52
Be
0.90
1.3
1.7
B
0.80a
a
C
1.09
N
1.40
1.6
1.8
2.1
Exceptions to the general trend
O
1.31a
F
1.68
Ne
2.08
2.0
2.3
2.5
Exceptions to the general trend can be rationalized. The IE1 for boron is less than the
corresponding value for Be. This is because that in the case of B the ionized electron is
being removed from a 2p orbital whereas the ionizing electron for beryllium is being
removed from a 2s orbital. For boron the 2p electron being ionized is being more
efficiently screened from the nucleus by the electrons in the completed 2s subshell and
therefore is easier to remove than the 2s electron being ionized in boron. A similar
exception is observed in the values for the IE1’s of Al and Ga in periods 3and 4.
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Another exception occurs between N and O. N has a ½-filled outer 2p subshell from
which the ionized electron must be removed whereas for O ionization creates a ½-filled
subshell. For O electron-electron repulsion between electrons in the same subshell is
being reduced thus ionization of oxygen proceeds with a lower IE1 than the general trend
would predict. Similar exceptions are observed in the IE1’s of S and Se in periods 3 and
4.
2. Down a group it is the variation in the value of nground that makes a larger contribution
to the value of the IE1. Thus general trend follows the relation IE1 ∞ 1/nground.
Successive Ionization Energies
Successive ionization can occur for multi-electron systems. The general equation for a
second ionization can be represented as the following.
I+(g)  I2+(g) + eThus the second ionization energy can be defined as the energy required to remove the
most loosely held electron from a monatomic cation of 1+ charge. For any one particular
system the successive values for the ionization energies become more and more
endothermic as the electrons are being removed. Consider the following table illustrating
the successive ionization energies for sodium, magnesium and aluminum.
Element
Na
Mg
Al
Successive Ionization Energies (kJ/mol) for Na, Mg and Al
IE1
IE2
IE3
496
4560
6940
738
1450
7730
578
1820
2750
IE4
9540
10500
11600
Note that for each of the three elements the ionization energies increase markedly after
the removal of the valence electrons. Thus comparison of successive ionization energies
for an element can be used to predict the number of valence electrons an atom of an
element possesses.
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Electron Affinity
First Electron Affinity (EA1)
The first electron affinity is defined as the energy change associated with the addition of
an electron to an isolated gaseous atom.
A(g) + e-  I-(g)
If the anion formed by the reaction is more stable than the original atom then the process
is exothermic and the EA1 is a negative quantity. If the atom is more stable than the anion
then the process is endothermic and the EA1 is a positive quantity.
General Trends in First Electron Affinities
1. Across a period first electron affinities become more negative in value. This is
understandable since effective nuclear charge for the outer-most electrons tends to
increase as one crosses a period. However exceptions to the general trend can be
observed. Consider the following table for the period 2 elements.
Element
EA1
Li
-60
First Electron Affinities (kJ/mol) for Period 2 Elements
Be
B
C
N
O
F
>0
-27
-122
>0
-141
-328
Ne
>0
Lithium as a typical alkali (group IA) element has a rather exothermic EA1. This is
because the added electron completes the s subshell of the valence shell. Elements such
as Be, N and Ne which already have rather stable electron configuration form anions of
lesser stability and therefore have endothermic EA1’s.
2. First electron affinities tend to become less exothermic down a group. Down a group
the added electron is being added to shells which are further and further away from the
positively charged nucleus. Period 2 elements often have less exothermic first electron
affinities than the general trend would predict. This is because the period 2 elements are
of relatively small radius and electron-electron repulsions tend to resist the addition of an
extra electron.
Li
-60
Na
-53
K
-48
Rb
-47
First Electron Affinites (kJ/mol)
O
-141
S
-200
Se
-195
Te
-190
F
-328
Cl
-349
Br
-325
I
-295
Similar to ionization energies successive electron affinities are possible. Second and
higher electron affinities are always endothermic processes. For example, the second
electron affinity of oxygen is 844 kJ/mol.
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Ionic Bonding and the Born-Haber Cycle
Ionic Bonding
Binary ionic bonding forms between a metal and a nonmetal whose electronegativities
widely differ. This type of bonding produces clusters or aggregates of ions that are more
stable than the isolated atoms from which they are formed. This stability is due to the
electrostatic attractions that exist between the cations and the anions of opposite charge.
Ionic bonding is associated with crystalline solids of high melting and boiling points.
When dissolved in a solvent the ions can conduct electricity.
Crystal Lattice and Lattice Energy.
Sodium metal reacts with chlorine gas to form the most famous example of an ionic
compound, sodium chloride. Whereas metallic sodium and chlorine gas are very reactive
chemicals sodium chloride is very stable. Let us see if we can discover the source of this
stability of sodium chloride.
Consider the reaction between an unstable gaseous sodium atom and an unstable chlorine
atom to produce an ion-pair of gaseous sodium chloride. Is this an exothermic process?
The overall reaction is
Na(g) + Cl(g)  Na+(g) + Cl-(g)
Can we predict the enthalpy change for this reaction? Yes we can if we use the Law of
Hess and realize that the above process can be thought of happening in two stages. What
we are doing is creating a thermochemical cycle leading to the formation of a sodium
chloride ion pair.
Stage 1: IE1 of sodium
Na(g) +  Na+(g) + e-; ΔH = IE1 = 496 kJ
Stage 2: EA1 of chlorine
Cl(g) + e-  Cl-(g); ΔH = EA1 = -349 kJ
Combining Stage 1 with Stage 2
Na(g) + Cl(g)  Na+(g) + Cl-(g); ΔH = IE1 + EA1 = 496 kJ + (-349) kJ = 147 kJ
Thus we can see that the process of forming an ion-pair from the gaseous atoms is
endothermic! Where then does the great stability of the ionic compound come from?
To truly appreciate the stability of sodium chloride we must construct a realistic cycle of
its formation from sodium and chlorine.
37
The true thermochemical cycle we require must feature 3 stages each containing 1 or
more parts.
Stage 1: Preparation of the gaseous metal cation: Metallic sodium is a metallic solid! The
solid must first be sublimed to form the sodium atoms in the gaseous state. These gaseous
atoms must then be ionized to form sodium cations.
Stage 2: Preparation of the gaseous nonmetal anion: Chlorine as an element is of course
bimolecular chlorine gas. The chlorine molecule must be first converted into individual
gaseous chlorine atoms. Then the chlorine atoms can be converted to gaseous chloride
ions.
Stage 3: Formation of the Crystal Lattice: Lastly the sodium cations and the chlorine
anions combine together not to form gaseous ion-pairs of sodium chloride but combine to
form the crystalline solid sodium chloride. It is the crystalline sodium chloride that is the
chemical we call sodium chloride.
The preparation of any binary ionic compound can be evaluated in terms of the three
above stages. The stages when combined result in a thermochemical cycle leading from
the individual elements in their standard states to the binary ionic compound in its
standard state, the crystal lattice. This thermochemical chemical cycle is often called a
Born-Haber cycle.
Using the above information we can now calculate the enthalpy of formation of solid
sodium chloride to be -411 kJ per mol of sodium chloride formed.
38
In order to appreciate the Born-Haber cycle we must review the concept of enthalpy of
formation. By definition, the enthalpy of formation is the enthalpy change associated with
the formation of 1 mole of a chemical in a particular state from its elements in their
standard states. The enthalpy of formation thus is a comparison of the relative stability of
the desired chemical in a given state to its elements in their standard states. As we can see
by the large negative value for the enthalpy of formation of solid sodium chloride it is
much more stable than its elements. The cycle allows us to see where this stability
actually comes from. Most steps in the cycle are actually endothermic! Only 2 steps are
in fact exothermic; the first electron affinity of chlorine and the final step leading to the
formation of the crystal lattice. It is this final step that is by far the more exothermic step
and is the real source of the stability of the solid sodium chloride!
Lattice Energy (U)
Energy released when isolated gaseous ions combines to form an ionic crystal lattice.
M+(g) + X-(g)  MX(s)
The lattice energy is by far the most exothermic stage of the Born-Haber cycle. The
stability of ionic compounds is directly attributable to the highly exothermic nature of the
lattice energy.
There exists some confusion concerning the term lattice energy. Some chemists define
the lattice energy as the amount of energy required to separate the ions in one mole of
crystal from their positions in the lattice to an infinite separation in the gaseous state.
That is
MX(s)  M+(g) + X-(g).
Both definitions result in a lattice energy of the same magnitude but of opposite sign. The
definition used by our textbook is that given above and always results in a negative value
for the lattice energy.
The value for the lattice energy is hard to measure experimentally. Very often the value is
obtained with the aid of the Born-Haber cycle. However, in order to do so the other
required thermochemical values such as electron affinities must be available.
In 1956 Kapustinskii published an equation which allows for the calculation of the lattice
energy for certain ionic compounds. The Kapustinskii equation has the form
where K = 1.2025x10-4 J m mol-1; d = 3.45x10-11 m; ν is the # of ions in the empirical
formula; z+ is the charge of the cation; z- is the charge of the anion; r+ is the radius of the
cation in m; r- is the radius of the anion.
Applying the Kapustinskii equation to sodium chloride gives a calculated value of -762
kJ/mol for the lattice energy. The Born-Haber cycle value is sometimes reported as -771
kJ/mol.
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Standard States of the Halogens
Many important ionic compounds are halides. Therefore knowledge of the different
standard states for fluorine, chlorine, bromine and iodine is crucial in order to correctly
prepare a Born-Haber cycle for these compounds. Both fluorine and chlorine and gases
and therefore stage 1 for sodium fluoride would be analogous to stage 1 in the cycle of
sodium chloride provided above.
Bromine is a liquid in its standard state. Therefore the first stage must include a
thermochemical step for the vapourization of the liquid bromine. This enthalpy change
can be referred to as the enthalpy of vapourization of liquid bromine or as the enthalpy of
formation of gaseous bromine.
Br2(l)  Br2(g); ΔH = ΔHvap
Iodine is solid. The Born0-Haber cycle of an iodide must therefore feature a step where
the solid is sublimed. This thermochemical step can also be thought of as the enthalpy of
formation of gaseous iodine.
I2(s)  I2(g); ΔH = ΔHsub
Properties Associated with Lattice Energy
As the magnitude of the lattice energy increases the following properties become more
and more apparent in ionic compounds.
1. Melting points become higher and higher
2. Solubility in water often decreases ( Review solubility rules)
The relative magnitude of the lattice energy, [U], for two ionic compounds can be
predicted using the following relationship. This equation is based on the Kapustinskii
equation.
[U] ∞ z+z-/r
where z+ and z- represent the charge magnitudes of the ions and r is the sum of the ionic
radii.
That is r = r+ + rLet’s compare lithium chloride (LiCl) and calcium oxide (CaO).
Chemical
LiCl
CaO
z+
1
2
z1
2
r+ (pm)
60
99
r- (pm)
181
140
r (pm)
241
239
For both chemicals the value of r is roughly the same. CaO however can a much larger
charge factor. Thus we can predict that calcium oxide has the lattice energy of greater
magnitude. The following properties have been reported for these two chemicals.
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Chemical
LiCl
CaO
U (kJ/mol)
Melting Point (oC)
Solubility
-845
614
vs
-3460
2850
ss*
*Slightly soluble with the formation of calcium hydroxide
Drawing Lewis Structures
Example 1: Carbon Dioxide, CO2
1.
Calculate the number of valence electrons.
Carbon = 4
2 Oxygen = 2x6=12
Total = 16
Note:
If the example is an anion then increase the number by the charge.
If the example is a cation then decrease the number by the charge.
2.
Octet Rule: The number of valence electrons needed for octets =
2(# H atoms in the formula) + 8(# of other atoms in the formula)
8(3) = 24
3.
Number of valence electrons shared = (result of calculation 2) – (result of
calculation 1)
= 24 – 16 = 8
4.
Number of Covalent Bonds = (Result of calculation 3)/2
= 8/2 = 4
5.
Construct a Skeleton Structure
a) Central atom tendency increases towards the center and towards the bottom of
the periodic table.
b) Oxyacids such as nitric acid or sulfuric acid have their acidic protons bonded to
oxygen.
c) Avoid O-O single bonds since peroxides tend to be unstable.
O C O
6.
Complete the octets about each atom by adding non-bonding electrons (lone pairs).
O C O
7.
Calculate the formal charge for each atom. If the formal charge for an atom is not
zero then the value (including the sign) must be added to the atom in the Lewis
structure.
41
Formal Charge
Formal charge is used as a guide in drawing Lewis structures.
Let x = # of valence electrons associated with the isolated atom of an element
And let y = # of valence electrons associated with an atom of the element in a
bonding situation
Then the formal charge of the atom in the bond = x – y
When the formal charge of an atom is not zero than the magnitude and the sign of
the formal charge are added to the atom in the Lewis structure.
Consider the ammonium cation.
H
H N H
H
The formal charge of the N = 5 – 4 = 1+. This is because each N atom has 5 valence
electrons and the N in the cation has 4 valence electrons associated with it.
The formal charge of each H = 1 – 1= 0. This because a hydrogen atom has 1
valence electron and each H is associated with only 1 valence electron.
Thus the Lewis structure of the ammonium cation is
H
H N H
H
Formal charge calculation assumes that bonding electrons are equally shared.
For a neutral molecule the sum of all the formal charges = 0.
For an ion the sum of all the formal charges = the charge of the ion.
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Principle of Electroneutrality
Formal charges ought to be avoided, if possible, especially those of magnitude greater
than one.
For example, hydrogen cyanide, could have two possible skeleton structures but only one
is correct.
H C N
H N C
more likely
Resonance
Lewis structures were developed following very simplistic rules. It is a wonder that such
simple rules allow for the writing of structures that could give us any kind of valuable
insight into the structure of real molecules or polyatomic ions. Yet Lewis structures do
have real value!
It must not be surprising that there are molecules and polyatomic ions for which Lewis
structures do not give a very complete understanding of the chemistry of the chemical. A
Lewis structure is like a photograph. For a very simple molecule a single photograph
might be sufficient to give us some idea into its character. For many molecules, much
like humans, a single photo is not adequate. Such a molecule or polyatomic ion is called a
resonance hybrid and the resonance hybrid must be represented by several photos to get a
better understanding of its nature just as an individual may place several photos into a
photo album to fully represent the person’s interests and character.
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Failure of the Octet Rule
- 3 cases
1.
Species having incomplete octets.
Central atom has less than 4 valence electrons, such as, Be, B or Al.
Consider boron trifluoride. Boron coming from Group III A has 3 valence electrons.
First try to arrive at a Lewis structure following the rules assuming the octet rule is
followed.
1. 24 valence electrons; 2. 4(8) = 32 valence electrons; 3. 32-24 = 8 shared
4. 4 bonds
However, 4 bonds are not reasonable since fluorine normally only forms 1 bond.
Thus the boron must have an incomplete octet!
In the Lewis structure it is important to remember that non-central atoms must
follow the octet rule. Thus a reasonable structure would be
F
F
B
F
Note that the correct structure still contains 24 valence electrons!
44
2
Octet Expansion
Only may occur for central atoms having atomic number equal to or greater than
14 (Si). That is, the central atom must have a d subshell in its valence shell.
The number of valence electrons in an expanded octet is usually limited to a
maximum of 12 (sometimes 14). Thus the maximum number of covalent bonds is 6
(or 7).
Consider each of the following examples.
a) Bromine trifluoride 1. 28
2. 32 3. 4
4. 2 bonds
But at least 3 bonds are required! Therefore expand the octet of the central atom.
F
Br F
F
Note that the accepted Lewis structure still contains 28 valence electrons.
b) Sulfate anion 1. 32 2. 40 3. 8
4. 4 bonds
Based on the octet rule the Lewis structure would be the following.
O
2
O S O
O
The Principle of Electroneutrality must now be applied. S has an atomic number greater
than 14 and therefore the octet of the sulfur can be expanded. Expansion of the
sulfur can take place by converting S-O single bonds into double bonds. This is
accomplished by converting a lone pair on one of the oxygen into the electrons of a
second bond between that oxygen and the sulfur. Note this lowers the formal charge
of the sulfur by one unit while increasing the formal charge of the oxygen by one
unit.
O
2
O S O
O
O
S
O
O
45
O
O
O
S
O
O
3.
Odd-Electron Species (Free Radicals)
Free radicals are normally unstable but a few examples may have a limited lifetime
in nature. Common free radicals include nitrogen monoxide and nitrogen dioxide.
The Lewis structure for nitrogen monoxide is the following.
N O
Electronegativity
Electronegativity is the ability of a bonded atom to attract the electrons of the bond. Thus
electronegativity is different from the electron affinity of an atom, which is a
characteristic of an isolated gaseous atom.
Since electronegativity is a property of a bonded atom its value depends on the identity of
the atom and the identity of the atom to which the bond is formed.
Linus Pauling:
Established the Pauling Scale of electronegativity values.
Arbitrarily set the electronegativity value of fluorine at 4.0.
Periodic trends in electronegativity values:
Generally increase across a period.
Tend to decrease down a column.
Fluorine is the most electronegative element and cesium (or francium) is the least.
Most metals have electronegativity values less than 2.
Most nonmetals have electronegativity values greater than 2.
Metalloids tend to have electronegativity values between 1.8 and 2.
Electronegativity Differences and Bond Type.
a) Covalent bond. Electronegativity difference less than 0.5
Cl2
EN(Cl) = 3.0. Obviously the electronegativity difference between
any two identical atoms must be zero. Thus for any homonuclear
diatomic molecule the bond type must be covalent.
b) Polar Covalent Bond. Electronengativity difference between 0.4 and 2.0
HCl EN(Cl) = 3.0; EN(H) = 2.1. Thus ΔEN = 0.9
c) Ionic Bond. Electronegativity difference is greater than 2.0.
NaCl; EN(Cl) = 3.0; EN(Na) = 0.9. Thus ΔEN = 2.1
46
Percent Ionic Character
A pure covalent bond only exists between two identical atoms such as in a homonuclear
diatomic molecule. Most covalent bonds are at least somewhat polar covalent. Thus most
covalent bonds have some degree of ionic character. The extent of the ionic character of a
bond can be expressed in terms of the bond’s percent ionic character.
Consider the H-Cl bond in hydrogen chloride. As shown on the previous page this bond
can be classified as a polar covalent bond based on the electronegativity difference
between the two atoms.
It is possible to measure the Dipole Moment for the H-Cl bond in hydrogen chloride.
The experimentally determined value for the dipole moment is 1.03 D. The debye unit, D,
is 3.34x10-34 C.m.
If hydrogen chloride were ionic it would contain the H+ cation and the Cl- anion. The
dipole moment of this imaginary ionic compound can be calculated since the bond length
is known to be 1.27x10-10 m.
Thus the dipole moment of imaginary ionic HCl = (charge) (distance)
= (1.60x10-19 C) (1.27x10-10 m)
= (2.03x10-29 C.m) (1 D/3.34x10-34 C.m)
= 6.08 D
The ionic character of the bond = (1.03 D / 6.08 D) = 0.169 and
The percent ionic character is 16.9%.
In general the percent ionic character is the ratio of the measure dipole moment of a bond
to the dipole moment of the imaginary ionic bond expressed as a percentage.
Percent ionic character of a bond increases as the dipole moment difference between the
bonded atoms increases. No bond is completely ionic. Sodium chloride, for example, has
a percent ionic character of approximately 75%. In general, bonds with greater than 50%
ionic character are classified as ionic.
It is conceivable that bond type as predicted by electronegativity difference be in
disagreement with the type of bond predicted by percent ionic character. For example the
bond in LiI is classified as polar covalent based upon the electronegativity difference
between Li and I ΔEN = 1.5) whereas the percent ionic character is above 50%. In such
cases more credence is usually given the percent ionic character since it is calculated
using data specific to the particular bond.
47
Modern Theories of Covalent Bonding
1. Valence Shell Electron Pair Repulsion Theory (VSEPR)
2. Valence Bond Theory (Hybridization)
1. Valence Shell Electron Pair Repulsion Theory (VSEPR)
Basic assumption: Valence shell electron pairs are arranged about central atoms so that
repulsions between the electron pairs are minimized.
3 types of electron pair repulsions:
a) Lone pair-lone pair repulsions. The strongest type.
b) Lone pair-bonded pair repulsions.
c) Bonded pair-bonded pair repulsions. The weakest type.
Molecular geometry: The arrangement of atoms about a central atom. The molecular
geometry is also known as the shape.
Central atom: Any atom bonded to 2 or more other atoms.
Bond angle. Angle between 2 atoms bonded to the same central atom.
Procedure for applying VSEPR:
a) Draw or have access to a proper Lewis structure. For a resonance hybrid use an
important contributing structure.
b) Identify central atom(s).
c) About each central atom sum the number of bonds and the number of lone pairs.
Count multiple bonds as single bonds for this calculation. This sum of the
number of bonds and lone pairs is sometimes called the steric number.
d) Consult the following table to assign the orientation of electrons (orbital
orientation), molecular geometry and the bond angles about each central atom.
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Predicting Molecular Shape from VSEPR Theory
Steric Basic Geometry
1 lone pair
2 lone pairs
No.
0 lone pair
2
3 lone pairs
Linear
3
Trigonal planar
Bent
4
Bent
Tetrahedral
Trigonal pyramid
Trigonal bipyramid
Seesaw
T-shaped
Octahedral
Square pyramid
Square planar
5
6
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Linear
Applying VSEPR
Steric number
# of bonded
pairs
2
2
Orbital
Orientation
Ideal Bond Angle
Molecular
geometry
Example
180o
Linear
BeH2
Trigonal
Planar
BH3
SO2
Linear
o
120
3
3
Trigonal
Planar
3
2
Trigonal
Planar
Bent
3
1
Trigonal
Planar
Linear
4
4
Tetrahedral
4
3
4
4
109.5o
Tetrahedral
CH4
Tetrahedral
Trigonal
Pyramidal
NH3
2
Tetrahedral
Bent
H2O
1
Tetrahedral
Linear
5
5
Trigonal
Bipyramidal
5
*
4
5
120o
900
Trigonal
Bipyramidal
PCl5
Trigonal
Bipyramidal
See-Saw
SF4
3
Trigonal
Bipyramidal
T-shape
ClF3
5
2
Trigonal
Bipyramidal
Linear
I3 -
5
1
Trigonal
Bipyramidal
Linear
6
6
Octahedral
6
5
6
4
900
Octahedral
SF6
Octahedral
Square
Pyramidal
BrF5
Octahedral
Square Planar
XeF4
*When adding lone pairs to a trigonal bipyramidal orientation add lone pairs initially to
the equatorial positions in order to minimize 90o repulsions.
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Molecular Geometry and Dipole Moment
A dipole exists if there is a separation of charge. The magnitude of the dipole depends on
the magnitude of the charge being separated and the distance of the separation.
Consider the simple binary molecule, hydrogen chloride.
The electronegativity of chlorine (3.0) is greater enough than that of hydrogen (2.1) for
this bond to be classified as polar covalent (ΔEN = 0.9). The bond dipole moment may
now be calculated as
bond dipole moment = δ x (bond distance)
The molecular dipole moment is the vector sum of the bond dipole moment. For a
binary compound such as hydrogen chloride where there is only a single bond dipole
moment the molecular dipole moment is the same as the bond dipole moment.
For structures containing more than one bond the individual dipole moments must be
estimated and then the vector sum taken. This requires knowledge of molecular shape as
well as knowledge of electronegativity.
In our course we usually do not calculate the bond dipole moments. We simply use
electronegativity differences between bonded atoms to estimate whether covalent bonds
can be classified as polar or nonpolar. Obviously nonpolar bonds do not have a bond
dipole moment.
For example let us consider water. Water is known to be a polar molecule. This is
obvious if one considers that each bond dipole moment would add together to produce a
vector sum in the direction of the oxygen.
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Sometimes bond dipole moments oppose each other. When this occurs the effect of the
individual bond dipole moments cancel each other out. Thus carbon dioxide is a nonpolar
molecule even though each bond is polar covalent.
We all know the old adage like dissolves like is often to predict solubility of one chemical
in another. Thus it is not surprising to learn that the nonpolar gas carbon dioxide has very
limited so lubility in water.
3. Valence Bond Theory (Hybridization)
VSEPR provides a simple and fast method for predicting molecular shape and bond
angles. However it does not explain anything concerning the nature of covalent bonding.
There is no consideration of the types of atomic orbitals that overlap to form covalent
bonds.
A more satisfactory theory is valence bond theory (VB theory). VB considers that a
covalent bond is formed by the overlap of atomic orbitals containing unpaired electrons.
The covalent bond is formed due to orbital overlap.
Consider the hydrogen molecule. VB Theory considers the covalent bond in hydrogen to
be formed by the orbital overlap of the two 1s orbitals of the hydrogen atoms. Remember
that the ground state electron configuration of the hydrogen atom is 1s1. Thus each atom
is donating one electron to the bond.
To picture how the bond would form let us start by considering the two atoms as being
independent of one another. In such a situation the two atoms would exhibit no orbital
overlap. As the atoms approach each other orbital overlap would commence. This results
in a net stabilization of the system as the positively charged nuclei become attracted to
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the negatively charged electron density between them. Note however, that stabilization
can only come about if the atomic orbitals overlap in such as way so that spins of the
electrons are spin-paired. That is, the paired electrons have opposite signs for ms.
As the degree of orbital overlap increases the net stabilization increases until the nuclei
begin to come too close to each other. Then net stabilization begins to decline. The above
graph shows that for hydrogen maximum stabilization occurs when the internuclear
distance is 74 pm. Thus the reported bond length for the H-H bond is 74 pm.
VB theory is simple to apply and easy to understand. Unfortunately the basic model
works with only a few examples. Consider the molecule BeF2. The central atom in this
molecule is beryllium and the ground state of beryllium contains no unpaired electrons (a
requirement for covalent bond formation according to VB).
Be: [He]2s2 – no unpaired electron
In terms of an orbital box diagram we can represent beryllium in its ground state as
Yet we know that the molecule BeF2 does exist and moreover it is found that both Be-F
bonds are identical (that is they have the same bond energy. For both of the bonds to be
identical implies that the bonds are formed by the orbital overlap of the same type of
atomic orbitals.
A solution to this dilemma comes from the notion of hybridization (an add-on to VB
theory). Hybridization is based on the idea that covalent bonds are sometimes formed by
the orbital overlap of non-ground state atomic orbitals. The excited state atomic orbitals
are formed on central atoms as they enter into bonding and are not observed in isolated
gaseous atoms. Hybridization involves the mixing of atomic orbitals of the central atom
to produce new hybrid atomic orbitals. These new orbitals are part of a single hybrid
atomic orbital subshell and are therefore degenerate.
# of hybrid atomic orbitals formed = # of atomic orbitals combined
= steric number (SN) of the central atom
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It is important to remember that hybridization is part of the VB theory and that VB theory
requires that every atom forming a bond with another atom must provide one electron.
Therefore we sometimes most promote one or more ground state electrons to ensure the
correct number of unpaired electrons. Remember that promotion takes place prior to
hybridization.
In the case of BeF2 promotion is required.
The beryllium must form 2 bonds yet the ground state of beryllium has no unpaired
valence electrons.
Thus we must promote one of the 2s electrons into one of the vacant 2p orbitals. This
would create an excited state beryllium atom having two unpaired valence electrons.
Now that we have the correct number of unpaired electrons you might think that this is
the excited state that the beryllium uses to form its bond. However, we know that the
beryllium must form two equivalent bonds resulting in a bond angle of 180 degrees
(as predicted by VSEPR). Thus hybridization is still necessary.
The steric number of the beryllium is 2. Hybridization requires us to combine 2 atomic
orbitals. We thus combine one 2s orbital and one 2p orbital to form a hybrid sp subshell
containing 2 hybrid atomic orbitals. We then use the electron that were in these orbitals in
the excited state (formed due to electron promotion) to populate the sp subshell. The
orbitals of the hybrid atomic orbital subshell are filled following Hund’s rule.
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The two sp hybrid orbitals centered about the central beryllium result in the desired
bond angle of 180 degrees.
Properties of Hybrid Atomic Orbitals
Hybrid orbitals are made up of 2 lobes of which one lobe is much larger than the other. It
is the larger lobe that is used to form a bond. (The smaller lobe is so small that it is often
ignored in the drawing of the orbital.) The hybrid orbital only overlaps well in one
direction and therefore is the ideal covalent bond forming orbital.
The two sp orbitals about the beryllium in BeF2 occupy a linear geometry about the
central. Each overlaps with a 2p orbital of fluorine to create a 180 degree bond angle.
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