Chemical Bonding
Chapter 8 Sections 1-8
• A chemical bond is: a strong
electrostatic force of attraction
between atoms in a molecule or
compound.
• Bonding between atoms occurs
because it creates a more stable
arrangement for the atoms.
Lewis Symbols – Dot Diagrams
• Convenient way to show the valence electrons
Three types of bonding
• Metallic bonding – results from the attraction
between metal atoms and the surrounding
sea of electrons
• Ionic bonding – results from the electrical
attraction between positive and negative ions.
• Covalent bonding – results from the sharing
of electron pairs between two atoms
Ionic Bonding
• Many atoms transfer electrons and
other atoms accept electrons,
creating cations (positive metal ions)
and anions (negative nonmetal ions).
• The resulting ions are attracted to
each other by electrostatic force.
Ionic Bonding
• The ions closely pack together in a crystal
lattice.
• This arrangement maximizes the attractive
forces among cations and anions while
minimizing repulsive forces.
• Because force is proportional to the charge on
each ion, larger charges lead to stronger
interactions.
• Because force is inversely proportional to the
square of the distance between the centers of
the ions, smaller ions lead to stronger
interactions.
Ionic bonding between Na and Cl
Sodium atom
1s2 2s2 2p6 3s1
Chlorine atom
1s2 2s2 2p6 3s2 3p5
1s2 2s2 2p6 3s1
1s2 2s2 2p6 3s2 3p5
Sodium ion Na1+
1s2 2s2 2p6
Chlorine ion Cl11s2 2s2 2p6 3s2 3p6
Covalent bonding
• In many cases electrons do not
completely transfer from one
atom to another.
• The electrons between atoms are
shared.
Covalent bonding between H2
• Hydrogen’s electron configuration is 1s1
• Because both H atoms need 1 more
electron to become isoelectronic with
He, it is unlikely that either will give up
an electron.
Covalent bonding between H2
↑
1s
↓
1s
They share the two electrons.
H· + H ·  H : H
Types of Covalent Bonds
• When electrons are shared equally the
bond is called a NONPOLAR covalent
bond. (i.e. H2)
• Sometimes the electrons between two
atoms are NOT shared equally. The bond
created is called a POLAR covalent bond.
..
..
H· + ·Cl:
. .  H:Cl:
..
Polar Covalent Bonding
• An example of this would be HCl.
HCl molecule
Hydrogen atom
↑
1s
Chlorine atom
[Ne] ↑↓
3s
↑↓ ↑↓ ↓
3p
How to classify bond types
• Electronegativity – ability of an atom in a
molecule to attract shared electrons to it
• Each element on the periodic table is assigned
an electronegativity value (see page 353) that
ranges from 0.7 to 4.0.
• The difference in the electronegativity
determines the bonding type (ionic, polar
covalent, or nonpolar covalent).
Electronegativity Values
If the electronegativity difference is:
1.7 and higher
ionic
0.3 to 1.7
polar
covalent
0.0 to 0.3
nonpolar
covalent
What if I get an electronegativity
difference that is 0.3 or 1.7?
• These cut-off numbers are guidelines.
• It is a gradual change not stair-step.
Ionic Character
• As the electronegativity difference increases,
the ionic character increases as well!
Practice Problems
What type of bond will occur between
iodine and the following elements:
cesium, iron, and sulfur?
Bonding
between I and:
Cesium
Iron
Sulfur
Electronegativity
difference
Bond Type
Determine the type of bond
between the following pairs.
Bonding
between
Li & Cl
S & O
Ca & Br
P & H
Si & Cl
S & Br
Electronegativity
difference
Bond type
Other ways to determine bonding
• Electronegativity is not the only factor in
determining bonding.
• Generally, bonds between a metal and
nonmetal are ionic, and between two
nonmetals the bonds are covalent.
• Examination of the properties of a compound
is the best way to determine the type of
bonding.
Ionic Bonding
• Ionic compounds are formed to maximize
stability.
• Nonmetal - will gain electrons to become
isoelectronic with nearest noble gas; called an
anion
• Metal – will lose electrons to become
isoelectronic with noble gas; called a cation
Transition Metals
Zinc
Electron configuration is 1s22s22p63s23p64s23d10
When it forms the +2 ion, it loses the 2 valence
electrons in the 4s sublevel.
Zn2+ configuration is 1s22s22p63s23p63d10
Practice Problems
• Write the electron configurations for the
following ions.
– Fe2+
– S2– Mg2+
• Use electron configurations to explain why the
most probable charge on the strontium ion is
+2.
Size of Ions
Li,152 pm
3e and 3p
Does+ the size go
up+ or down
Li , 60 pm
when
an
2e and 3losing
p
electron to form
a cation?
Size of Ions
+
Li,152 pm
3e and 3p
Li + , 78 pm
2e and 3 p
Forming
a cation.
• CATIONS are SMALLER than the atoms
from which they come.
• The electron/proton attraction has
gone UP and so size DECREASES.
Size of Ions
Does the size go up or down
when gaining an electron
to form an anion?
Size of Ions
F, 71 pm
9e and 9p
F- , 133 pm
10 e and 9 p
Forming
an anion.
• ANIONS are LARGER than the atoms from
which they come.
• The electron/proton attraction has gone
DOWN and so size INCREASES.
• Trends in ion sizes are the same as atom
sizes.
Trends in Ion Sizes
Figure 8.13
Which is Bigger?
•
•
•
•
Cl or Cl- ?
K+ or K ?
Ca or Ca+2 ?
I- or Br- ?
Which is Bigger?
•
•
•
•
Cl or Cl- ? ClK+ or K ?
K
Ca or Ca+2 ? Ca
I- or Br- ? I-
Lattice Energy Effects
• The change in energy when
separated gaseous ions are packed
together to form an ionic solid.
M+(g) + X-(g)  MX(s)
• Lattice energy is negative
(exothermic) from the point of view
of the system.
Lattice Energy
• To determine which compound will have the highest
lattice energy, take into consideration the following:
– The size of the ions in the compound
• The smaller the size, the greater the
lattice energy
– The charge of the ions in the compound
• The greater the charge, the greater the
lattice energy
Calculating ∆Hf
• We can take advantage of the fact the
energy is a state function and break the
reaction into steps, the sum of which is
the overall reaction.
• Let’s do #41 Na(s) + ½ Cl2 (g)  NaCl(s)
Given the following:
Lattice energy
-786 kJ/mol
Ionization energy for Na
495 kJ/mol
Electron affinity for Cl
-349 kJ/mol
Bond energy of Cl2
239 kJ/mol
Enthalpy sublimation for Na
109 kJ/mol
Process
Step 1: Sublimation of Na
Na(s)  Na(g)
Step 2: Ionization of Na
Na (g)  Na+ (g) + eStep 3: Dissociation of Cl2
½ Cl2 (g)  Cl(g)
Step 4: Formation of Cl- (Electron Affinity)
Cl (g) + e-  Cl-(g)
Step 5: Formation of NaCl
Na+(g) + Cl-(g)  NaCl(s)
Na(s) + ½ Cl2 (g)  NaCl(s)
109 kJ/mol
495 kJ/mol
119.5 kJ/mol
-349 kJ/mol
-786 kJ/mol
-411.5 kJ/mol
Drawing Lewis Structures
Valence Electron Review
• Valence electrons are in outermost level
• You can use periodic table or electron
configuration to determine valence electrons
Example: Phosphorus
–Located in Group 15 or 5A
–Electron configuration is 1s22s22p63s23p3
–Contains 5 valence electrons
Complete Exercise 1 on worksheet
Drawing Lewis Structures
Octet Rule
•
•
•
•
Most useful rule for creating Lewis structures
Every atom usually has 8 valence electrons
Exception: hydrogen is good with 2 (like He)
Lines are used to link atoms together (same as
using 2 dots)
Same as
Steps to Drawing Lewis Structures
1. Count valence electrons.
2. Connect atoms together with bonds. In
molecules with a single atom of one element
and several atoms of another element, the
single atom is generally in the center with the
other atoms attached to it.
3. Add electrons around outside of atoms to
give each atom 8 electrons (or 2 in the case
of hydrogen).
4. Count electrons used. This number must be
the same as valence electrons.
5+(3*7)=26 e-
PCl3
Complete
Exercise 2.
Bonding Pairs
Lone Pairs
(a.k.a. nonbonding electrons)
Helpful Hints
•
•
•
•
•
•
Carbon atoms form 4 bonds.
Nitrogen atoms form 3 bonds.
Oxygen atoms form 2 bonds.
Hydrogen atoms form 1 bond.
Fluorine atoms form 1 bond.
Other halogens (Cl, Br, and I) frequently form 1
bond (but not always).
Determining the Central Atom
• In a molecule, the atom that typically forms
the greatest number of bonds is in the center,
with other atoms attached to it.
• Example: CH3Cl
– Carbon forms 4 bonds
– Hydrogen forms 1 bond
– Chlorine forms 1 bond
SO CARBON IS IN THE MIDDLE WITH HYDROGEN
AND CHLORINE AROUND IT! Don’t forget electrons
on chlorine to make 8!
• Complete exercise 3.
Covalent Bonding
Multiple Bonds
• It is possible for more than one pair of electrons to
be shared between two atoms (multiple bonds):
• One shared pair of electrons = single bond (e.g. H2);
• Two shared pairs of electrons = double bond (e.g. O2);
• Three shared pairs of electrons = triple bond (e.g. N2).
H H
O O
N N
Octet in each case
• Generally, bond distances shorten with multiple
bonding.
Resonance
• Occurs when more than one valid Lewis
structure can be written for a particular
molecule.
Odd Number of Electrons…
NO
Number of valence electrons = 11
N O
N O
Resonance occurs when more than one valid Lewis structure can be
written for a particular molecule (i.e. rearrange electrons)
NO2
Number of valence electrons = 17
O N O
O N O
O N O
Molecules and atoms which are neutral (contain no formal charge) and with an
unpaired electron are called Radicals
Beyond the Octet
• Elements in the 3rd period or higher can have
more than an octet if needed.
• Atoms of these elements have valence d
orbitals, which allow them to accommodate
more than eight electrons.
More than an Octet…
Elements from the 3rd period and beyond, have ns, np and
unfilled nd orbitals which can be used in bonding
Cl
PCl5
P : (Ne) 3s2 3p3 3d0
Number of valence electrons = 5 + (5 x 7) = 40
Cl
P
Cl
SF4
Cl
Cl
S : (Ne) 3s2 3p4 3d0
Number of valence electrons = 6 + (4 x 7) = 34
F
The Larger the central atom, the more atoms
you can bond to it – usually small atoms such
as F, Cl and O allow central atoms such as P
and S to expand their valency.
F
S
F
F
Formal Charge
Difference between the # of valence electrons in the free atom and the # of
electrons assigned to that atom in the Lewis structure.
1

FC  G.N. -  # BE  # LPE 
2

FC = formal charge; G.N. = Group Number
#BE = bonding electrons; #LPE = lone pair electrons
If Step 4 leads to a positive formal charge on an inner atom beyond the second row,
shift electrons to make double or triple bonds to minimize formal charge, even if
this gives an inner atom with more than an octet of electrons.
Formal Charge
O C O
(-1)
(0)
(+1)
Not as good
O C O
(0)
(0)
Better
(0)
Molecular Shapes
• Lewis structures give atomic connectivity:
they tell us which atoms are physically
connected together. They do not tell us the
shape.
• The shape of a molecule is determined by its
bond angles.
• Consider CCl4: experimentally we find all ClC-Cl bond angles are 109.5.
Therefore, the molecule cannot be planar.
All Cl atoms are located at the vertices of a
tetrahedron with the C at its center.
Molecular Shape of CCl4
VSEPR Theory
 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.
Why is VSEPR Theory Important?
• Gives a specific shape due to the number
of bonded and non-bonded electron pairs
in a molecule
• Tells us the actual 3-D structure of a
molecule
• In bonding, electron pairs want to be as
far away from each other as possible.
VSEPR and Resulting Geometries
How does VSEPR THEORY work?
We can use VSEPR theory using 4 steps.
1. Draw the Lewis Structure for the
molecule.
Example: SiF4
F
F-Si-F
F
How does VSEPR THEORY work?
We can use VSEPR theory using 4 steps
1. Draw the Lewis Structure for the
molecule.
2. Tally the number of bonding pairs and
lone (non-bonding) pairs on the center
atom.
F
F-Si-F
F
Bonding pairs: 4
Lone pairs on central atom: 0
How does VSEPR THEORY work?
We can use VSEPR theory using 4
steps
1. Draw the Lewis Structure for the
molecule
2. Tally the number of bonding pairs
and lone pairs on the center atom.
3. Arrange the rest of the atoms so
that they are as far away from
each other as possible.
F
F
F
Si
F
How does VSEPR THEORY work?
We can use VSEPR theory using 4 steps
1. Draw the Lewis Structure for the molecule
2. Tally the number of bonding pairs and
lone pairs on the center atom.
3. Arrange the rest of the atoms so that they
are as far away from each other as
possible
4. Give the type of geometry the molecule
has:
Tetrahedral
Another Example:
To determine the electron pair geometry:
1) draw the Lewis structure;
2) count the total number of electron pairs around the
central atom.
3) arrange the electron pairs in one of the geometries
to minimize e--e- repulsion.
4) multiple bonds count as one bonding pair for VSEPR
The VSEPR Model
Predicting Molecular Geometries
The VSEPR Model
Predicting Molecular Geometries
The VSEPR Model
Difference between geometry and shape
Geometry:
We determine the geometry only looking at electrons.
All the atoms that obey the octet rule have the same
tetrahedral-like geometry.
Shape:
We name the shape by the positions of atoms.
We ignore lone pairs in the shape.
The VSEPR Model
Predicting Shape
Shape
The VSEPR Model
Predicting Shape
Shape
The 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.
The 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
C O
124.3o
The VSEPR Model
Molecules with Expanded Valence Shells
Atoms that have expanded octets have AB5 (trigonal
bipyramidal) or AB6 (octahedral) electron pair
geometries.
Examples:
PF5 trigonal bipyramidal
SCl6 octahedral
The VSEPR Model
Molecules with Expanded Valence Shells
The VSEPR Model
Molecules with Expanded Valence Shells
The VSEPR Model
Molecules with More than One Central Atom
In acetic acid, CH3COOH, there are three central
atoms.
We assign the geometry about each central atom
separately.
Hybrid Orbitals
• In bonding, s and p orbitals are used in
bonding. It is easy to tell which ones are
used by looking at our molecule.
• For example, CH4. Looking again at the
Lewis structure, we see that there are 4
bonds. We call this sp3 hybridized.
Hybrid Orbitals
• Regions of electron density-EACH BOND AND
LONE PAIR OF ELECTRONS ON THE CENTRAL
ATOM IS KNOWN AS A REGION OF ELECTRON
DENSITY.
• 2 regions of electron density-sp hybridized
• 3 regions of electron density-sp2 hybridized
• 4 regions of electron density-sp3 hybridized
Hybridization
sp Hybrid Orbitals
The two lobes of an sp hybrid orbital are 180 apart.
Hybrid Orbitals
sp2 Hybrid Orbitals
Important: when we mix n atomic orbitals we must get
n hybrid orbitals.
sp2 hybrid orbitals are formed with one s and two p
orbitals. (Therefore, there is one unhybridized p orbital
remaining.)
The large lobes of sp2 hybrids lie in a trigonal plane.
All molecules with trigonal planar electron pair
geometries have sp2 orbitals on the central atom.
Hybridization
Hybridization
sp3 Hybrid Orbitals
sp3 Hybrid orbitals are formed from one s and three p
orbitals. Therefore, there are four large lobes.
Each lobe points towards the vertex of a tetrahedron.
The angle between the large lobes is 109.5
All molecules with tetrahedral electron pair geometries
are sp3 hybridized.
Hybridization
Hybrid Orbitals
Hybrid Orbitals
Summary
To assign hybridization:
1.
2.
3.
4.
Draw a Lewis structure.
Assign the geometry using VSEPR theory.
Use the geometry to determine the hybridization.
Name the shape by the positions of the atoms.
Hybridization and Multiple Bonds
•Multiple bonds overlap differently and are
called s-bonds and p-bonds
•All single bonds are s
•Double bonds contain
1 s and 1 p bond
•Triple bonds contain
1 s and 2 p bonds
Bond Energy
Covalent Bonding & Orbital
Overlap
•As two nuclei approach each other their
atomic orbitals overlap.
•As the amount of overlap increases, the
energy of the interaction decreases.
•At some distance the minimum energy is
reached.
•The minimum energy corresponds to the
bonding distance (or bond length).
Covalent Bonding & Orbital
Overlap
•As the two atoms get closer, their nuclei
begin to repel and the energy increases.
•At the bonding distance, the attractive
forces between nuclei and electrons just
balance the repulsive forces (nucleus-nucleus,
electron-electron).
Bond Energies
• Bond breaking requires energy (endothermic).
• Bond formation releases energy (exothermic).
H = D(bonds broken) - D(bonds formed)
energy required
energy released
Bond Energies