Bonding
Year 11 DP Chemistry
What is a bond?
A chemical bond is a force that holds atoms
together making a new substance


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Ionic Bonds result from electrostatic
attraction between oppositely
charged ions
Covalent Bonds result from
electrostatic attractions involving
electrons and positively charged
nuclei
Metallic Bonds result from
electrostatic attractions between
delocalised electrons and a lattice
of positively charged nuclei
What type of bond?
Ionic Bond: As a rule of thumb, we say that the difference
between the electronegativity values needs to be high (i.e.
greater than 1.7) to be ionic. They form between cations on
the left and anions on the right of the Periodic Table.
Covalent Bond: If the difference between the
electronegativity values of two highly electronegative atoms
is low, a covalent bond is formed. They tend to form between
non-metals, but sometimes metals are involved (eg Al2Cl6)
Metallic Bond: If the difference between the electronegativity
values of two highly electropositive atoms is low, a metallic
bond is formed. These form between metals of the same or
different type of atom
Electronegativity
The relative tendency of an atom to attract
bonding electrons to itself on the Pauling Scale
Formation of ions
Groups 1,2,3
Lose electrons
 Group 1 – loses one
e- to gain a full
valence shell
 Group 2 – loses 2 eto gain a full
valence shell
 Group 3 – loses ??
Groups 5,6,7
Gain electrons
 Group 5 – gains 3
e- to gain a full
valence shell
 Group 6 – gains 2
e- to gain a full
outer shell
 Group 7 - ??
Transition ions


Ions of the transition elements can form more
than one ion
This is due to s and d orbitals having similar
energy levels
For example – Fe forms two ions
[Ar]4s23d6
Can you deduce which two ions and why?
Fe(II) – losing two 4s electrons
F(III) – losing two 4s e- and one 3d e- to give a
half-filled d
Exercises
 Using
a Periodic Table, determine which
ions are formed from the following
elements? Na, Al, O, Ca, F, N
 Describe the size of each of the above
ions relative to their original atoms.
Ionic Bonding
Oppositely charged ions are formed by electron
transfer due to a large electronegativity
difference(> 1.7 difference)
Na has a low electronegativity relative to Cl, so ions are formed
by a transfer of an electron to achieve a full valence shell for
both atoms. These oppositely charged ions then form a bond.
Sodium Chloride Lattice
 Ionic
compounds form
a repeating crystal
lattice of positive and
negative ions. The
compound is neutral
 The electrostatic
attraction is very
strong so ionic
compounds form
solids at room
temperature
This shows a model of a NaCl
lattice with alternating positive
and negative ions
Ionic formulas
 The
formulas for ionic compounds are
found by balancing the overall charges
to zero (neutral)



First determine the charge on each ion
Second determine the number of each ion
needed to make the compound neutral
Write the positive ion first indicating the
number of each using subscripted numbers
Examples:
Li+ F-  LiF
Mg2+ Cl-  MgCl2
Polyatomic ions
Some ions contain more than one element and
the charge on the ion is spread (delocalised) over
the entire ion. They have specific names and act
as a single unit.
Important ones to know:
NH4+ (ammonium)
NO3- (nitrate)
OH- (hydroxide)
Ionic compounds form in the
same way with polyatomic ions.
2SO4 (sulfate)
Here we see Na SO . Notice the
sulfate did not change formula.
CO32- (carbonate)
PO43- (phosphate)
HCO3- (bicarbonate or hydrogen carbonate)
2
4
Exercises
 Write
the formulas formed by the following
pairs of elements: (Ca, S); (K, N); (Ca, P);
(Fe(II), O); (Fe(III), O)
 Write the formulas for the following
compounds: potassium hydroxide,
magnesium nitrate, sodium hydrogen
sulfate
 Use electronegativity values to determine
the type of bonding between Al and O;
Al and Br.
Covalent bonding
Electrons are shared
between two atoms.
These atoms are most
commonly non-metals.
The shared electrons are held in place
by the positively charged nuclei that are
sharing them.
Single bond = 1 shared pair
Double bond = 2 shared pairs
Triple bond = 3 shared pairs
How many bonds?
Single bonding:
Multiple bonding:
Multiple bonding
The more shared pairs the stronger the bond and
the shorter the bond. Note the trend in C-C bonds.
Note: bond energy is the amount of
energy required to break a bond
Coordinate (dative) bonds
In this type of covalent bond, the difference
is that one of the atoms in the pair donates
both of the electrons in the bond.
Examples: CO, NH4+, H3O+
Electron Dot
Diagrams
(Lewis)
The dots represent the
valence electrons for
each element
The Octet Rule
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.
Fluorine molecule
Hydrogen Chloride
Here, the octet rule is satisfied for Cl, but is irrelevant for H
which can only hold 2 e-.
Double Bonds - ethene
Two pairs of shared electrons
Draw the Lewis structure for Acetylene (ethyne) – C2H2
Formation of water
Notice the double bond in O2. No other configuration will
satisfy the octet rule. Why is H4O not formed?
Resonance


When more than one possible Lewis structure can
be drawn, it indicates resonance
The actual structure is an average of the
resonance structures
Notice that the bond lengths
would be different in benzene
unless there is a resonance
structure like the one
represented above
The Octet Rule
2nd row elements C, N, O, F observe the
octet rule.
2nd row elements B and Be often have fewer
than 8 electrons around themselves - they are
very reactive.
3rd row and heavier elements CAN exceed
the octet rule using empty valence d orbitals.
When writing Lewis structures, satisfy octets
first, then place electrons around elements
having available d orbitals.
Lewis Diagrams
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
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Determine the number of valence e- for each
element
Determine the number of bonds for each element
(usually the number of e- needed to fill the
valence shell – exceptions AHL)
Central atom is the one with the most bonds
Join the atoms so that all have the correct number
of bonds
Each bond has one e- from each atom (unless
dative bond), so a pair makes a bond
Once all bonds are made, place the remainder of
e- around the appropriate atoms as non-bonding
pairs
Check that the octet rule has been satisfied for
each element
Completing a Lewis Structure -CH3Cl


All octets (ignoring H)
have been satisfied.
H
..
H
C
..
H
..
Cl
..
..

Valence e- (C = 4, H = (3)(1), Cl = 7 Total = 14 )
Number of bonds (C = 4, H = 1, Cl = 1)
Carbon is the central atom (most bonds)
Bonds are arranged and extra e- are added around Cl
as non-bonding pairs.
..

Exercises
1. N2 has a triple bond. Using dot diagrams, show why a single
or double bond is incorrect.
2. Illustrate how CO and H3O+ contain dative bonds.
3. Draw a Lewis diagram for HCN
4. Draw the two resonance structure of ozone O3
5. Use a drawing to show how many resonance structures are
possible for the nitrate ion (NO3-)
6. Compare the bond lengths and strengths of the two C,O
bonds in the carboxyl group below.
Octet rule exceptions
There are 3 ways the octet rule breaks
down:
1.
2.
3.
Molecules with odd number of electrons
Molecules where an atom has < octet
Molecules where an atom has > octet
Odd number of electrons
NO (nitrous oxide)
 Total e- = 6+5=11
 Octet can be
achieved around
the O with a single
bond between
 Remaining earound the N
 What about a
double or triple
bond? Try it.
Less than an octet
This mostly occurs with H, B and Be
BF3 (boron trifluoride)
 Total e- = (3x7)+3 = 24
Octets to outer atoms
Extra e- (24-24=0) to B
Therefore no extra
electrons to add. B has
only 6 e- (< octet).
What about double
bonds??? see next…
Less than a octet (cont’d)
Add a double bond
to BF3 for a possible
octet…
This would give
3 resonance
structures. What
would they look
like?
The above structure would lead to a δ+ on F and a δ- on B.
NO
Is this likely considering the electronegativites?
Because B has only 6 valence
electrons, BF3 reacts strongly with
compounds that have unshared
pairs of electrons
Greater than an octet
This is the most likely exception to the octet rule.
PCl5
Starting in period
3, expanded
valence shells are
possible.
The octet rule is based upon valence orbitals containing an s and p orbital.
This gives 2 + 6 = 8 e- (an octet). In the third shell (n=3) d orbitals become
available. P is below. A 3s can be exited to the 3d, which allows for 5
valence shell bonding electrons.
Promote one e-
Greater than an octet (cont’d)
Sulfur (also in group III) can expand it’s octet to
have more than 8 electrons as well. Sulfur can
form SF2, SF4, SF6
Go to this website to see an
animation on expanded
octets in sulfur:
http://www.saskschools.ca/
curr_content/chem20/cov
molec/exceptns.html
Greater than an octet (cont’d)
Other notable expanded octets…
XeF4 (12 e-)
PF6- (12 e-)
Polarity
 Polar
covalent bonds
 Electrons
are unequally shared by atoms
 Electronegativity difference is 0.5>__<1.7
 Nonpolar
covalent bonds
 Electrons
are equally shared
 Electronegativity difference is <0.5
Polar covalent bonds
Some bonds are not purely ionic, but they still
have a significant difference in electronegativity
that leads to one atom pulling the electrons more
strongly than the other.
Polarity
This electronegativity
difference leads to a
partially positive end δ+ of
a molecule and a partially
negative end δ- (note: δ is
the Greek letter delta and
means partial)
F is more electronegative, so the
electrons spend more time
around the F nucleus
Non-polar covalent
Small or no difference in electronegativity
values leads to non-polar substances.
Cl is a highly electronegative
element, but there is no
difference when one Cl atom
bonds to another Cl atom
C and H have very little difference
in electronegativity, so methane is
non-polar
Non-polar compounds
Some compounds contain polar bonds, but
the polarity is cancelled out due to the
structure.
O is more electronegative than C meaning each bond is
polar towards the O atom, but due to its linear shape,
these polarities cancel each other resulting in a non-polar
molecule.
Exercise
1.
2.
3.
Show how S can form 3 compounds with F
Predict if these substances contain polar
bonds. H2, CCl4, H20
Use arrows to show the overall polarity of the
compounds below?
Molecular geometry VSEPR
To determine the shape of covalent
molecules, we use the Valence Shell
Electron Pair Repulsion Theory (VSEPR)
which states:
“The geometric arrangement of atoms around a central
atom is determined by the repulsion between electron pairs
in the valence shell of the central atom.”
Stay away!
I am
repulsed
by you
In other words…
Ditto!!
ee-
ee-
What shape?
So, VSEPR theory says that molecular
geometry is determined by the shape that
keeps e- pairs as far apart as possible
Consider CO2
We know C forms four bonds and O forms 2 bonds
The Lewis
structure
looks like this:
What arrangement will allow the
valence e- around the central atom
(C) to be as far apart as possible?
Linear
Linear (1800)
Linear molecules have two areas of
high electron density around the
central atom.
Other examples of linear molecules :
Ethene (C2H4)
Ethyne (acetylene) C2H2
Molecular chlorine (Cl2)
3 pairs around the central atom
3 e- pairs around the
central atom leads to a
trigonal planar shape as
in BF3
Trigonal planar - Angles are1200
If one of those pairs is a
non-bonding or lone pair
of electrons, the shape is
described as bent or vshaped as in SO2
Bent – angles are less than 1200
due to lone pairs taking up more
space than bonding pairs. These
angles are 1170
4 pairs
If we look at the Lewis
structure for CCl4, we might
assume a flat structure with
900 bond angles
However, in 3-D space,
it is possible to allow the
electrons to be further
apart using a
tetrahedral shape with
bond angles of 109.50
4 pairs (cont’d)
Tetrahedral:
No lone
pairs
109.50
Methane
Trigonal
Pyramidal:
One lone
pair
1070
Ammonia
Bent:
Two lone
pairs
1040
Water
5 pairs (AHL)
Trigonal
bipyramidal
(0 lone pairs)
See saw
(1 lone pair)
Angles = 900 & 1200
5 pairs (cont’d)
T-shaped
2 lone pairs
Linear
3 lone pairs
6 pairs (AHL)
Octahedral
(0 lone pairs)
Square pyramid
(1 lone pair)
Square planar
(2 lone pairs)
Justify the shape of XeF4. Why are the lone pairs at 1800?
What other two shapes are possible with 6 pairs?
Summary of geometries
Allotropes of Carbon
Each carbon atom in graphite is
bonded to 3 other carbon atoms
forming flat sheets of carbon rings.
These layers are loosely bonded to
each other making graphite soft
In diamond, each carbon is bonded
to 4 other carbons in a giant
repeating lattice. This lattice is nonpolar and very strong, making
diamond the hardest mineral on
Earth. It’s m.p. is over 35000C!
The fullerene contains 60 carbons
arranged like a soccer ball with
alternating 5 and 6-member rings.
Graphite
Carbon bonded to three
other carbon atoms leaves
one valence electron per
carbon atom. These electrons
are delocalised allowing
graphite to conduct electricity
The individual layers contain strong
covalent bonds, but are only loosely
bonded to other layers. This allows
them to easily slide over one another
making graphite useful in pencils and
as a solid lubricant
Fullerenes and nanotubes
The third allotrope of carbon was
discovered in 1985 and includes
some weird and wonderful shapes.
The first was known as a “Bucky Ball”
which has a structure like a soccer
ball and contains 60 carbons- C60
There are other fullerenes that contain more or
less than 60 carbons. These have been
detected in space leading scientists to suggest
that they could be the origin of life in the
universe
Nanotubes are another area of current research
by materials scientists. These tubes have high
tensile strength and are able to conduct
electricity. These have potential medical
applications by attaching to resistant bacteria or
cancer cells. They are also being researched for
the proposed “space elevator” to be used as
cables.
Bonding in Silicon
Silicon atoms contain 4 valence
electrons and form regular
repeating covalent bonds. It s
chemically similar to C. The
picture below show it’s crystalline
form
Silicon dioxide (SiO2), which is the chemical
component of sand. has a structure similar to
diamond with a repeating giant lattice of
tetrahedral shapes. This makes it very hard.
However, it has polar bonds allowing it to be
dissolved slowly by some alkaline solutions and it
also has a much lower m.p. than diamond (17700C)
This shows a
computer generated
model of the
complicated giant
lattice of silicon
dioxide.
Metallic bonding
Metallic bonding can be described
as
A repeating lattice of positive metal
ions in a sea of delocalised
electrons
Delocalised electrons are not
attached to any particular metallic
nucleus and are free to move
about the lattice.
Metallic Properties
Conductivity
This is due to the electrons
being able to move freely
through the lattice.
This means the electrons can
act as charge carriers for
conducting electricity and
energy carriers for conducting
heat
Alkali
Metals
Alkali Earth
metals
0.108
0.210
0.139
0.077
0.313
0.226
0.298
0.076
Cu
Fe
Electrical Conductivity of
Metals
Group
III
Transition Metals
0.017
0.016
0.023
0.023
0.048
0.069
0.077
0.187
0.006
0.067
0.093
0.137
0.172
0.211
0.143
0.095
0.596
0.630
0.166
0.138
0.377
0.067
0.116
Metallic Properties
Malleability and Ductility
The delocalised electrons in
the 'sea' of electrons in the
metallic bond, enable the
metal atoms to roll over each
other when a stress is applied.
References
Source of animations and diagrams:
http://intro.chem.okstate.edu/1314f00/lecture/
chapter10/vsepr.html
www.google.com (images search)
Sundin, C.; (n.d.); found at
http://www.uwplatt.edu/ ; accessed 04/2011
Dynamic Science; (n.d.); found at
http://www.dynamicscience.com.au/tester/s
olutions/chemistry/chemistry%20index.htm ;
accessed 04/2011
Courtland, R.; New Scientist (online);
28/10/2010; found at
http://www.newscientist.com/blogs/shortshar
pscience/2010/10/buckyballs-abound-inspace.html ; accessed 04/2011