Basic concepts of chemical bonding
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Chapter 8
• Chemical bond: attractive force holding two or more
atoms together.
• Covalent bond - sharing e- between the atoms, e.g.
nonmetals.
• Ionic bond - transfer of e- from a metal to a nonmetal.
• Metallic bond: attractive force holding pure metals
together.
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Chapter 8
Lewis Symbols
• Represent the e- as dots around the symbol for the
element.
• No. of e- available for bonding are indicated by unpaired
dots.
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Chapter 8
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Chapter 8
The Octet Rule
• All noble gases except He has an s2p6 configuration.
• Octet rule: atoms tend to gain, lose, or share electrons
until they are surrounded by 8 valence electrons (4
electron pairs).
• Exceptions: molecules with an odd no. of e-, e.g. NO
and central atom with >8 e-, e.g. PF5
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Chapter 8
Ionic Bonding
Consider the reaction between sodium and chlorine:
Na(s) + ½Cl2(g) NaCl(s)
Hºf = -410.9 kJ
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Chapter 8
• The reaction is violently exothermic.
• NaCl is more stable than its constituent elements. Why?
• Both Na+ and Cl have an octet of e-
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Chapter 8
• NaCl forms a very regular structure in which each Na+
ion is surrounded by 6 Cl ions & each Cl ion is by six
Na+ ions.
• There is a regular arrangement of Na+ and Cl in 3D.
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Chapter 8
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Chapter 8
Energetics of Ionic Bond Formation
• The reaction NaCl(s) Na+(g) + Cl (g) is endothermic
( H = +788 kJ/mol).
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Chapter 8
The formation of a crystal lattice from the ions in the gas
phase is exothermic:
Na+(g) + Cl-(g) NaCl(s) H = -788 kJ/mol
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Chapter 8
• Lattice energy: the energy required to completely
separate an ionic solid into its gaseous ions.
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Chapter 8
Lattice energy depends on the charges on the ions and the
sizes of the ions:
El
Q1Q2
d
k is a constant (8.99 x 10 9 J·m/C2), Q1 and Q2 are
the charges on the ions, and d is the distance between
ions.
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Chapter 8
Lattice energy increases as
• The charges on the ions increase
• The distance between the ions decreases.
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Chapter 8
Electron Configurations of Ions
• Electron configurations can predict stable ion formation:
•
•
•
•
•
Mg: [Ne]3s2
Mg+: [Ne]3s1
not stable
Mg2+: [Ne]
stable
Cl: [Ne]3s23p5
Cl : [Ne]3s23p6 = [Ar] stable
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Chapter 8
Transition-metal ions
• Lattice energies are generally large enough to
compensate for the loss of up to only 3 e- from atoms
• Most transition metal have >3 e- beyond a noble-gas
core…meaning the octet rule is limited
• E.g. silver has a [Kr]4d105s1 electron configuration.
In forming Ag+, the 5s electron is lost, leaving a
completely filled 4d subshell.
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Chapter 8
Fe has an electron configuration [Ar]3d64s2. In forming Fe2+,
the two electrons are lost, leading to an [Ar]3d6 configuration.
Removal of an additional electron gives the Fe3+ ion with a
configuration [Ar]3d5.
Thus, transition metals generally do not form ions with
noble-gas configurations…the octet rule is limited in that
scope.
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Chapter 8
Polyatomic Ions
• Formed when there is an overall charge on a compound
containing covalent bonds. E.g. SO42 , NO3
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Chapter 8
Covalent Bonding
• When similar atoms bond, they share pairs of e- to each
obtain an octet.
• Each pair of shared electrons constitutes one chemical
bond.
• E.g.: H + H H2
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Chapter 8
Covalent Bonding
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Electron density concentrates between nuclei,
thus overall electrostatic attractions are
attractive Chapter 8
Lewis Structures
• Covalent bonds can be represented by the Lewis symbols
of the elements:
Cl + Cl
Cl Cl
• Each pair of electrons in a bond is represented by a single
line:
H
H O
H N H
Cl Cl
H F
H C H
H
H
H
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Chapter 8
Multiple Bonds
• More than one pair of e- can be shared between two
atoms (multiple bonds):
H H
O O
N N
• Generally, bond distances decrease as we move from
single through double to triple bonds.
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Chapter 8
Bond Polarity and
Electronegativity
• Sharing of electrons to form a covalent bond does not
imply equal sharing of those electrons.
• In some covalent bonds e- are located closer to one atom
than the other.
• Unequal sharing of electrons results in polar bonds.
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Chapter 8
Electronegativity
• Electronegativity: The ability of one atom in a molecule
to attract electrons to itself.
• Pauling set electronegativities on a scale from 0.7 (Cs) to
4.0 (F).
• Electronegativity increases
• across a period and
• up a group.
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Chapter 8
• Difference in electronegativity is a gauge of bond
polarity:
• EN difference around 0 result in equal or almost equal sharing
of electrons;
• EN difference around 2 result in polar covalent bonds (unequal
sharing of electrons);
• EN difference around 3 result in ionic bonds (transfer of
electrons).
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Chapter 8
• The positive end (or pole) in a polar bond is represented
+ and the negative pole -.
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Chapter 8
Dipole Moments
• Consider HF:
• Since there are two different “ends” of the molecule, we call HF
a dipole.
• Dipole moment, , is the magnitude of the dipole:
Qr
where Q is the magnitude of the charges; r is the distance
between charges.
• Dipole moments are measured in debyes, D.
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Chapter 8
Example: The bond length between H and Cl atoms in the HCl is 1.27
Å. Calculate the dipole moment, in D, that would result if the charges
on the H and Cl atoms were 1+ and 1-, respectively.
Note: Charge on each atom is the electronic charge, e: 1.60 x 10-19C; 1
Å = 10-10 m and 1 D = 3.34 x 10-30 C-m
Ans: 6.08 D
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Chapter 8
Exercise: The iodine monobromide molecule, IBr,
has a bond length of 2.49 Å and a dipole moment of
1.21 D. Calculate the effective charges on the I and
Br atoms, in units of electronic charge e
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Chapter 8
•
•
•
•
Bond Types and Nomenclature
In general, the least electronegative element is named
first.
The name of the more electronegative element ends in
–ide.
Ionic compounds are named according to their ions,
including the charge on the cation if it is variable.
Molecular compounds are named with prefixes.
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Chapter 8
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Chapter 8
Ionic
Molecular
MgH2
Magnesium hydride H2S
Hydrogen sulfide
FeF2
Iron(II) fluoride
Oxygen difluoride
Mn2O3 Manganese(III)
oxide
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OF2
Cl2O3 Dichlorine trioxide
Chapter 8
Drawing Lewis Structures
1. Add the valence electrons (groups for elements).
2. Write symbols for the atoms and show which atoms are
connected to which (connect with single bonds).
3. Complete the octet for the central atom the complete the
octets of the other atoms (H has only 2e-).
4. Place leftover electrons on the central atom.
5. If there are not enough electrons to give the central
atom an octet, try multiple bonds.
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Chapter 8
Example: Draw the Lewis structure for phosphorus
trichloride, PCl3
NB: There are 26 valence electrons and P is the central
atom
Exercise: Draw the Lewis structure for HCN and NO+,
NO3Prentice Hall © 2003
Chapter 8
Exercise: A liquid compound used in dry cleaning contains
14.5% C and 85.5% Cl by mass and has a MW of 166 amu.
Write the Lewis formula for the molecule
Exercise: An ionic compound has the following composition (by
mass): Ca, 30.3%; N, 21.2%; O, 48.5%. What is the formula and
the name of the compound? Write the Lewis structures for the
ions
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Chapter 8
Exceptions to the octet rule:
•
•
•
Molecules with an odd number of electrons e.g. NO;
Molecules in which one atom has less than an octet e.g. BF3;
Molecules in which one atom has more than an octet e.g. PCl5,
i.e. expanded valence shells. Note that elements from the third
period and beyond have ns, np and unfilled nd orbitals that can
be used for bonding
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Chapter 8
Formal Charge
• To determine which structure is most reasonable, we use
formal charge.
• Formal charge is the charge on an atom that it would
have if all the atoms had the same electronegativity.
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Chapter 8
Formal Charge
• Formal charge on a particular atom is:
valence electrons - number of bonds - lone pair electrons
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Chapter 8
• Consider:
C N
• For C:
•
•
•
There are 4 valence electrons (from periodic table).
In the Lewis structure there are 2 nonbonding electrons and 3
from the triple bond. There are 5 electrons from the Lewis
structure.
Formal charge: 4 - 5 = -1.
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Chapter 8
• Consider:
C N
• For N:
•
•
•
There are 5 valence electrons.
In the Lewis structure there are 2 nonbonding electrons and 3
from the triple bond. There are 5 electrons from the Lewis
structure.
Formal charge = 5 - 5 = 0.
• We write:
C N
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Chapter 8
Although the concept of formal charge helps us in
choosing between alternative Lewis structures,
formal charges do not represent real charges on
atoms…EN differences are essential in determining
the actual charges
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Chapter 8
Resonance Structures
Some molecules are not well described by Lewis Structures.
Typically, structures with multiple bonds can have similar
structures with the multiple bonds between different pairs of
atoms
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Chapter 8
• The most stable structure has:
•
•
the lowest formal charge on each atom,
the most negative formal charge on the most electronegative
atoms.
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Chapter 8
• Example: experimentally, ozone has two identical bonds
whereas the Lewis structure requires one single (longer)
and one double bond (shorter).
O
O
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Chapter 8
O
Resonance Structures
• Resonance structures are attempts to represent a real
structure that is a mix between several extreme
possibilities.
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Chapter 8
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Chapter 8
• Example: in ozone the extreme possibilities have one
double and one single bond. The resonance structure
has two identical bonds of intermediate character.
O
O
O
O
O
O
• Common examples: O3, CO32-, NO3-, SO42-, NO2, and
benzene.
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Chapter 8
High charges on S –
unstable!
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Hall © 2003
Low charge on S –
most stable!
Chapter 8
Resonance in Benzene
• Each pair of C atoms and the 6 additional electrons are
delocalized over the entire ring:
or
• Benzene is an aromatic compound
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Chapter 8
The resonance structures differ only in the
arrangement of the valence electrons in the
molecule. No atoms are moved.
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Chapter 8
Strengths of Covalent
Bonds
• The energy required to break a covalent bond is called
the bond dissociation enthalpy, D. That is, for the Cl2
molecule, D(Cl-Cl) is given by H for the reaction:
Cl2(g) 2Cl(g).
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Chapter 8
We define the A-B bond energy (D) as the average enthalpy
change for the breaking of an A-B bond in a molecule in the
gas phase
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Chapter 8
Example:
CH4(g) C(g) + 4H(g) H = 1660 kJ
the bond enthalpy is a fraction of
atomization reaction:
D(C-H) = ¼ H = ¼(1660 kJ) = 415 kJ
H for the
The bond enthalpy is positive as energy is required to
break bonds
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Chapter 8
Bond Enthalpies and the Enthalpies of
Reactions
• In any chemical rxn bonds are broken and new ones are
formed
• Enthalpy for the reaction = ∑(bond enthalpies for bonds
broken) - sum of ∑(bond enthalpies for bonds formed).
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Chapter 8
• Mathematically, if Hrxn is the enthalpy for a reaction,
then
H rxn
D bonds broken
D bonds formed
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Chapter 8
Consider the rxn:
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
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Chapter 8
Hrxn = ?
H rxn
D C-H
D Cl - Cl
D C - Cl
D H - Cl
104 kJ
From Table 8.4 above, the bond enthalpies (D) are:
C-H = 413 kJ; Cl-Cl = 242 kJ; H-Cl = 431 kJ and C-Cl = 328 kJ
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Chapter 8
• The overall reaction is exothermic which means than the
bonds formed are stronger than the bonds broken.
• The above result is consistent with Hess’ law.
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Chapter 8
Exercise: Calculate the enthalpies of reaction for:
1. N2H4(g) → N2(g) + 2H2(g)
2. HCN (g) → H(g) + C(g) + N(g)
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Chapter 8
EXPLOSIVES
Alfred Nobel – inventor of dynamite
Dynamite consists of nitroglycerin that is mixed with
diatomaceous earth or cellulose
Fossilised material containing silica
Another common explosive: trinitrotoluene (TNT)
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Chapter 8
Required properties of an explosive:
• must decompose very exothermically
• products must be gaseous and allow for pressure build-up
• rapid decomposition
• it must be stable so as to be detonated predictably
Should have weak chemical bonds and decompose into
molecules with very strong bonds e.g. N2, CO, CO2
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Chapter 8
Bond Enthalpy and Bond Length
• Multiple bonds are shorter than single bonds.
• Multiple bonds are stronger than single bonds.
• As the number of bonds between atoms increase, the
atoms are held closer and more tightly together.
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Chapter 8
Chapter 9
Molecular Geometry and
Bonding Theories
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Chapter 8
Molecular Shapes
• Lewis structures give atomic connectivity: they tell us
which atoms are physically connected to which.
• The shape of a molecule is determined by its bond angles.
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Chapter 8
Consider CCl4: experimentally we find all Cl-C-Cl
bond angles are 109.5 .
Therefore, the molecule cannot be planar.
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Chapter 8
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Chapter 8
• In order to predict molecular shape, we assume the
valence electrons repel each other. Therefore, the
molecule adopts whichever 3D geometry minimized this
repulsion.
• We call this process Valence Shell Electron Pair
Repulsion (VSEPR) theory.
• There are simple shapes for AB2 and AB3 molecules.
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Chapter 8
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Chapter 8
There are five fundamental geometries for molecular shapes
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Chapter 8
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Chapter 8
Molecular Shapes
• When considering the geometry about the central atom,
we consider all electrons (lone pairs and bonding pairs).
• When naming the molecular geometry, we focus only on
the positions of the atoms.
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Chapter 8
VSEPR Model
• To determine the shape of a molecule, we distinguish
between lone pairs (or non-bonding pairs, those not in a
bond) of electrons and bonding pairs (those found
between two atoms).
• We define the electron domain geometry by the positions
in 3D space of ALL electron pairs (bonding or nonbonding).
• The electrons adopt an arrangement in space to minimize
e -e repulsion.
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Chapter 8
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Chapter 8
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Chapter 8
VSEPR Model
• To determine the electron pair geometry:
• draw the Lewis structure,
• count the total number of electron pairs around the central atom,
• arrange the electron pairs in one of the above geometries to
minimize e -e repulsion, and count multiple bonds as one
bonding pair.
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Chapter 8
VSEPR
Model
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Chapter 8
VSEPR Model
•
•
•
•
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
We determine the electron pair geometry only looking at
electrons.
We name the molecular geometry by the positions of
atoms.
We ignore lone pairs in the molecular geometry.
All the atoms that obey the octet rule have tetrahedral
electron pair geometries.
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Chapter 8
VSEPR Model
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
• By experiment, the H-X-H bond angle decreases on
moving from C to N to O:
H
H C H
H
109.5O
H N H
H
107O
O
H
H
104.5O
• Since electrons in a bond are attracted by two nuclei, they do
not repel as much as lone pairs.
• Therefore, the bond angle decreases as the number of lone pairs
increase.
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Chapter 8
VSEPR Model
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
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Chapter 8
VSEPR Model
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
• Similarly, electrons in multiple bonds repel more than
electrons in single bonds.
Cl
111.4o
Cl
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C O
124.3o
Chapter 8
VSEPR Model
Molecules with Expanded Valence
Shells
• Atoms that have expanded octets have AB5 (trigonal
bipyramidal) or AB6 (octahedral) electron pair
geometries.
• For trigonal bipyramidal structures there is a plane
containing three electrons pairs. The fourth and fifth
electron pairs are located above and below this plane.
• For octahedral structures, there is a plane containing four
electron pairs. Similarly, the fifth and sixth electron pairs
are located above and below this plane.
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Chapter 8
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Chapter 8
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Chapter 8
VSEPR Model
Molecules with Expanded Valence
Shells
• To minimize e e repulsion, lone pairs are always
placed in equatorial positions.
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Chapter 8
VSEPR Model
Molecules with Expanded Valence
Shells
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Chapter 8
VSEPR Model
Shapes of Larger Molecules
• In acetic acid, CH3COOH, there are three central atoms.
• We assign the geometry about each central atom
separately.
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Chapter 8
• When there is a difference in electronegativity between
two atoms, then the bond between them is polar.
• It is possible for a molecule to contain polar bonds, but
not be polar.
• For example, the bond dipoles in CO2 cancel each other
because CO2 is linear.
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Chapter 8
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Chapter 8
• In water, the molecule is not linear and the bond dipoles
do not cancel each other.
• Therefore, water is a polar molecule.
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Chapter 8
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Chapter 8
The overall polarity of a molecule depends on its
molecular geometry
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Chapter 8