 Mutual
electrical attraction between the
nuclei and valence electrons of different
atoms that binds the atoms together
 Atoms are usually less stable by themselves
than combined with other atoms
 Ionic

Bonding
Chemical bonding resulting from the
attractions of cations and anions
 In pure ionic bonding, atoms give up
electrons completely to other atoms, which
accept them completely to form ions which
attract one another
 Covalent



Bonding
Chemical bond that shares the electrons
between two atoms
A pure covalent bond results in the electrons
being “owned” equally by two atoms
 Both nuclei have these electrons around them
an equal amount of time
Example of purely covalent bond : F2
 Both atoms attract each others’ electrons
equally (both have an electronegativity of 4.0,
so the pull on the electrons is exactly the same
from both atoms)
 Most
compounds do not have purely covalent
or ionic bonds

The bonds fall somewhere on a continuum
 Look
at the bond’s percentage of ionic
character
Bonds with 0 to 50% ionic
character are classified as
covalent
 Bonds with > 50% ionic
character are classified
as ionic

 Ionic

Electrons get transferred between atoms forming
ions that attract one another
 Polar

Bonds
Covalent Bond
Electrons shared between atoms unequally
 Nonpolar
Covalent Bond
Electrons shared between
atoms equally

 We
can predict the bond type between two
atoms by calculating the absolute value of
the electronegativity difference between
them.
Electronegativity
Difference
Bond Type
0- 0.3
Nonpolar Covalent
>0.3 – 1.7
Polar Covalent
>1.7
Ionic
 If
the absolute value of the electronegativity
difference between two atoms is > 1.7, then
the bond they form is probably ionic
 Example #1: NaCl



Na – electronegativity of 0.9
Cl – electronegativity of 3.0
Electronegativity difference = | 0.9 – 3.0 | = 2.1
 An electron moves from the Na atom to the Cl atom
causing Na to have a +1 charge and Cl to have a -1
charge
 The Na + ion is then attracted to the Cl- ion forming a
very strong bond

If the absolute value of the electronegativity
difference between two atoms is >0.3 - 1.7, then
the bond they form is probably polar covalent



Covalent because the electrons are shared between the
atoms
Polar because the electrons are shared unequally between
the atoms (spend more time around one nucleus)
Example: BF3



Boron – electronegativity of 2.5
Fluorine – electronegativity of 4.0
Electronegativity difference of | 4.0-2.5|= 1.5
 Polar covalent - electrons spend more time around the F nucleus
due to greater electronegativity
 Electron cloud is distorted towards the Fluorine atoms (more
electron density) resulting in a partial negative charge on the
fluorine atoms and a partial positive charge on the boron atom.
 If
the absolute value of the
electronegativity difference between
two atoms is 0.0- 0.3, then the bond
they form is probably nonpolar covalent
 Electrons are shared equally between
the nuclei of the bonded atoms,
resulting in a balanced distribution of
electrical charge
 Ionic character of 0 to 5% ionic
character

Example: F2
 Electronegativity difference of |4.0 – 4.0| = 0.0
 Molecules


Neutral group of atoms held together by covalent
bonds
Can exist on its own as an individual unit

Molecular formula
 Shows the relative numbers of atoms of each kind in
a chemical compound by using element symbols and
subscripts to show the number of atoms

Examples:
 H2O - water
 O2
- oxygen
 C12H22O11 - sucrose
 Energy
required to break a chemical bond
and form neutral isolated atoms


Units kj/mole
Energy required to break one mole of bonds
 Chemical
compounds tend to form so that
each atom by gaining, losing or sharing
electrons has an octet of electrons in its
highest occupied energy level
 Hydrogen
– needs total of 2 electrons
 Beryllium – needs total of 4 electrons
 Boron – needs 6 electrons
 Aluminum – needs 6 electrons
 Atoms of elements in groups 15, 16, 17, and
18 can sometimes form expanded octets


Can have more than 8 electrons surrounding the
nucleus
Usually not N, O, or F
 Symbols
represent the nucleus and inner
shell electrons
 Dot pairs between the symbol represent pairs
of shared electrons
 Dots adjacent to symbol represent unshared
pairs of electrons
Count up the number of valence electrons.
1.

E.g. in CCl4 there is 1 C which has 4 valence
electrons and 4 Cl atoms which each have 7 valence
electrons


Therefore for C 1 x 4 = 4 and for Cl 4 x 7 = 28
Adding 4 and 28 together gives you 32 electrons
Identify the central atom and put the other
atoms around the central atom.
2.




If C is present, put C in the center.
Otherwise, the least electronegative atom goes in
the center.
H never goes in the center.
Halogens don’t usually go in the center unless they
are bonded to other halogens
Put a pair of electrons (dots) between the
central atom and the outside atoms.
Start distributing the remaining electrons
3.
4.
Subtract those electrons placed in step 3 from the
total in step 1 to find electrons remaining
 Put the remaining electrons around the outside
atoms first until all atoms either have an octet or
are satisfied (Exceptions to the Octet Rule for the
exceptions).

5.
If there are electrons left over after
satisfying the octet rule, put them as
unshared pairs on the central atom.
If there are too few electrons for all
atoms (including the central atom) to
have an octet (or be satisfied if it is an
exception), then double and or triple
bonds should be considered.
6.

C, N, and O are most atoms likely to form
double or triple bonds.
 To
draw the Lewis dot structure for a
polyatomic ion (a covalently bonded
molecule that has a charge), you follow
the same steps except the following:


When counting up the valence electrons in step
1, add the absolute value of the charge if the
charge is negative to the total of the valence
electrons. If the charge is positive, subtract
the absolute value of the charge from the
total valence electrons.
After completing the Lewis dot structure, put
the entire structure in brackets with the
charge on the right upper corner on the
outside of the right bracket.
 Some
elements, especially C, N, and O can
share more than one electron pair.


Double bond – 2 pairs of electrons are shared
Triple bond – 3 pairs of electrons are shared
 Double
bonds have a greater bond energy
and shorter bond lengths than single bonds
 Triple bonds have a greater bond energy and
shorter bond lengths than single bonds
 Triple bonds stronger than double
 Double bonds stronger than single
Bond
Avg. Bond
Length (pm)
Avg bond energy
(kj/mole)
C-C
154
346
C=C
134
612
C= C
120
835
C-N
147
305
C=N
132
615
C=N
116
887
 Bonding
in molecules or ions that cannot be
correctly represented by a single Lewis
structure

Draw possible Lewis structures with doubleheaded arrow in between
Bond between oxygens is a hybrid of a single and
double bond
 Chemical
formula indicates the relative number
of atoms of each kind in a chemical compound.

Molecular formula indicates the number of atoms of
each element contained in a single molecule


E.g. C8H18
Ionic compounds have formula units, not molecules
(Remember, these ions are located in a lattice of
positively and negatively charged ions – are not
found as individual units)
 One formula unit is used to designate the
simplest ratio of the cations to anions

E.g. in a crystal of NaCl, the formula NaCl is
used to represent one unit
 Prefix
system of naming – must learn prefixes
Number
Prefix
1
Mono-
2
Di-
3
Tri-
4
Tetra-
5
Penta-
6
Hexa-
7
Hepta-
8
Octa-
9
Nona-
10
Deca-
Element with the smaller group number is given
first in the name
 If both elements are in same group, element whose
period number is greater is given first.
 If first element subscript is one, no prefix is given.
 If subscript is greater than 1, then the prefix
corresponding to the subscript is added to the
beginning of the element name.
 The second element is always given a prefix
indicating the number of atoms and the element
root + ide is used.
 The –o or –a at the end of a prefix is usually
dropped when the word following the prefix begins
with another vowel i.e. monoxide instead of
monooxide or pentoxide instead of pentaoxide

Formula
Prefix-system name
N2O
Dinitrogen monoxide
NO
Nitrogen monoxide
NO2
Nitrogen dioxide
N2O3
Dinitrogen trioxide
N2O4
Dinitrogen tetroxide
N2O5
Dinitrogen pentoxide
 Theory
used to determine the shape of
molecules


Based on the premise that repulsion between the
sets of valence electrons surrounding an atom
causes the electron pairs to be oriented as far
apart as possible.
Looks at the bonded pairs and lone pairs on the
central atom only

Bonded pairs are set as far apart from one another as
possible
 Some
representative
molecular
geometries – need
to learn molecular
geometries
located on VSEPR
Summary sheet in
Chapter 6 folder

Must determine molecule type from Lewis Dot
structure – must draw this first




Let A = central atom (in example below, N)
Let B = outside atoms regardless of whether they are the
same element or not (3 outside atoms )
Let E = represent the # of lone pairs on the central atom
(only) (one lone pair on N)
Let subscripts for B and E represent the number of
outside atoms and number of lone pairs on the central
atom




Example: NF3 is AB3E
N is central atom or A,
the 3 Fs are represented as B3
and there is one lone pair on the
central atom or E
 Molecular geometry - from VSEPR
Summary sheet is trigonal pyramidal
 Angles
between the bonded pairs of atoms
around the central atom
 Bond angles between bonded pairs tend to
be larger for those atoms without lone pairs
on the central atom

Lone pairs cause increased repulsion and force
bonded pairs closer together
 Electrons
are unevenly distributed
throughout the molecule


Because of this there are areas of partial positive
charge and areas of partial negative charge
Forms dipoles
Equal but opposite
charges separated
by short distances

More positive
Partial Partial +
more negative
 Polar
Bonds present in molecule
 Symmetry of molecule

Do dipoles when taken together, cancel out or
form a dipole in one direction?
If cancel out - then nonpolar
 If overall dipole is
present – polar

 If
molecule type is AB2,AB3,AB4, AB5, AB6,
AB2E3, or AB4E2, and the outside atoms are
the same element, then the molecule is
nonpolar
 If molecule type is AB and A and B are the
same element, then the molecule is nonpolar
 All others are polar

E.g. CH4 is AB4 all of the B’s are the same (H),
therefore the molecule is nonpolar

CHCl3 is AB4 but not all of the B’s are H, (1 H and 3
Cls), therefore the molecule is polar
 When
bonding occurs, orbitals from the
bonding atoms become mixed and form new
orbitals that are equivalent energies (i.e.
hybrid orbitals)

Hybrid orbitals

Orbitals of equal energy produced by the combination
of two or more orbitals on the same atom
 # of orbitals produced = # of orbitals that have
combined
 All

single bonds are sigma bonds – σ
Atomic orbitals overlap and form hybrid orbitals
 Double

In addition, double bonds also have one π bond


and Triple bonds have one sigma bond
Pi bonds (π) are overlapping atomic p orbitals on
adjacent atoms
Triple bonds also have two π bonds
# of
Hybrid
bonding
Orbital
pairs and Formed
lone pairs
around
central
atom
# of hybrid Geometry of
orbitals
electron Pairs
(not geometry 
of atoms)
2
sp
2
Linear
3
sp2
3
Trigonal planar
4
sp3
4
Tetrahedral
5
dsp3
sp3d
or 5
Trigonal
bipyramidal
6
d2sp3 or 6
sp3d2
octahedral
Count
double and
triple bonds
as one
bonding pair
since there
is only one
sigma bond

Intermolecular forces


Between molecules
3 types



Dipole-dipole interaction
Hydrogen Bonding
London Dispersion
Forces
Weaker

Intramolecular Forces


Within molecules
True bonding

Covalent bonding
Stronger
Dipole-dipole interactions
1.


Attraction between molecules that occurs
between oppositely charged ends of partially
charged molecules
Occurs between polar molecules
Hydrogen bonding – special case of dipoledipole interaction
2.
NOT Bonding!!!!! – intermolecular force
 Occurs when highly polarized H is attracted to
the lone pair of a highly electronegative N, O,
and F
 Occurs with molecules having N, O, or F directly
bonded to a H
 Strongest intermolecular
force

London Dispersion Forces
3.




Weakest intermolecular force
Occurs due to random motion of electrons in
molecules forming temporary dipoles (areas of
partial charge) which attract each other
(oppositely charged ends attract)
Due to temporary nature of the dipoles,
attractions are fairly weak but when added
together can be significant
Occurs in all molecules (polar and nonpolar)
 Hydrogen
Bonding > Dipole-Dipole
Interactions> London Dispersion Forces
 Strength of intermolecular forces determines
whether molecules have



High surface tension
High boiling point
Capillary action
Is the molecule polar or nonpolar
If polar, is there
An N,O, or F directly
Bonded to an H
in the structure?
If yes, then the molecule has
hydrogen bonding and
London Dispersion
Forces
If nonpolar,
London Dispersion Forces only
If no, then the molecule has a
dipole-dipole interaction and
London Dispersion
Forces
 Surface


Tension
Force that pulls adjacent parts of a liquid’s
surface together, thereby decreasing surface
area to the smallest possible size
Molecules at the surface do not have other like
molecules on all sides of them and consequently
they cohere more strongly to those directly
associated with them on the surface forming a
surface "film” or “skin”.
The greater the strength of the intermolecular
forces between molecules in a substance the
greater the surface tension.
 Hexane
and
Water


Which will have the greatest surface tension?

Hexane is nonpolar and only has London dispersion
forces whereas water is very polar and has hydrogen
bonding as well as London Dispersion Forces

Water has higher surface tension
 The
greater the temperature of the
substance, the lower the surface tension.

The greater the temperature, the greater the
average kinetic energy of the molecules, the
further apart they are from one another and the
lower the force of attraction between them.
 Boiling

Change of a liquid to bubbles of vapor that
appear throughout the liquid (not just from
surface like evaporation)
 Boiling
points are determined by the strength
of the intermolecular forces between
molecules

The greater the strength of the intermolecular
forces between molecules, the higher the boiling
point.

In Groups 15, 16, and
17 there is a more
elevated boiling point
than expected for
compounds with N, O,
and F directly bonded
to a H due to the very
strong hydrogen
bonding that occurs.

Much larger energy than
normal is required to
overcome the attractive
force between
molecules to allow the
liquid to escape into the
gas phase.
Higher than expected B.P.
 Example

Hexane
 Which


1:
or
Water
substance has the higher Boiling Point
Hexane has only London Dispersion Forces
whereas water has hydrogen bonding and London
Dispersion forces
Due to strength of hydrogen bonding, water will
have higher boiling point
 Example


2: Isopropyl alcohol vs. water
Both polar with hydrogen bonding and London
dispersion forces
Which has the higher boiling point?



Water has stronger intermolecular forces due to
compact shape and very polar bonds
Isopropyl alcohol is still polar but due to carbons and
hydrogen areas of the molecule that are more
nonpolar in nature
Due to stronger intermolecular forces, water has the
higher boiling point
Highly polar molecule
Nonpolar areas
Polar
area
 Example

3: C8H18 (octane) vs. C4H10 (butane)
Which will have a higher boiling point?
 Both have only London Dispersion Forces but
octane has stronger intermolecular forces
 More randomly moving electrons that can
form temporary dipoles in more places in the
molecule
 Therefore octane has a higher boiling point
than butane
Octane
Butane
Attraction of the surface of a liquid to the
surface of a solid
 Responsible for meniscus seen with water in a
graduated cylinder

Water molecules are attracted to the side of the
graduated cylinder and these water molecules
attract more water molecules so that water
“travels” up the side
 If tube is small enough in diameter (capillary
tube), then water will rise against gravity due to
attractive forces of water molecules

 The
stronger the intermolecular forces the
more capillary action will be seen and the
larger the meniscus.


Hexane – nonpolar – only weak London Dispersion
Forces will not show a meniscus in a graduated
cylinder
Water - very polar – strong hydrogen bonds –
there will be a pronounced meniscus in a
graduated cylinder and will readily flow up a
capillary tube against gravity