Section
Section 4B
4b
Covalent
bonding
Page
1
Covalent Bonding
A covalent bond is formed by sharing a pair of electrons between two atoms (overlapping of
atomic orbitals). The force holding two atoms that linked together by covalent bonding is
basically electrostatic attraction between the nuclei and the shared electrons. And
by sharing of electrons, each atom could gain the noble gas electronic structure.
Bond length and Covalent radii
The bond length of a covalent bond is the distance between the nuclei of the two
bonded atoms. It can be measured by X-ray analysis of crystals or diffraction of X-ray by
gases or vapours.
The results of measurements on a wide variety of compounds shows that the inter-nuclear
distance between two atoms A and B linked by a covalent bond is very nearly constant, and is
independent of the varied nature of the molecules in which the bond length is measured.
It is also found that the inter-nuclear distance A  B is equal to the arithmetic mean of the A 
A and B  B distances,
i.e.
A–B=
A-A + B-B
2
For example:
Hydrogen molecule H 2
Chlori ne molecule Cl2
Covelent radi us =
0.037 nm
Van der Waals radi us
= 0.120 nm
Covelent radi us
= 0.104 nm
Van der Waals radi us
= 0.180 nm
Hydrogen chloride molecule HCl
Bond distance = 0.141 nm
and because of this simple relationship, it is possible to allot what are called covalent radii to
the elements:
For any covalently bonded diatomic molecules consisting of identical atoms, their internuclear
distance divided by two is called the covalent radius.
This distance can be found by using electron diffraction technique.
As is to be expected, the heavier elements in any one group of the periodic table have the
larger radii since they contain more electrons.
For elements in the same period, those with higher atomic numbers have the smaller radii;
these elements have their outer electrons in the same orbits, but the electrons are more
strongly attracted by increasing positive charges on the nuclei.
The covalent radii of some common elements are given below:
C
0.077
N
0.070
O
0.066
F
0.064
Si
0.117
P
0.110
S
0.104
Cl
0.099
Section 4b
Covalent bonding
Page 2
For the same element, the covalent radius, and hence the internuclear distance (bond length) in
different covalent compounds depend on the number of covalent bonds linking the two
atoms together, i.e. depends upon the number of pairs of electrons shared.
C=
C

0.067
N=
0.060
N
0.060

O=
0.055
0.055
Since covalent bonds are electrostatic attraction between nuclei and shared electrons,
therefore an increase in number of sharing electrons will increase the electron density in
between the bonding nuclei. This in turn increase the electron-nucleus attraction and
hence causing an increase in bond strength. The increase in bond strength also tends to
pull the nuclei closer together with a consequent reduction in the internuclear distance, i.e.
decrease in bond length.
Again, the additivity of covalent radii to give the bond length will also break down in certain
cases.
For example:
Benzene ,
C6H6
Structure proposed by Kekule:
Actual
structure :
covalent radius of C
=
expected C  C bond length =
0.077 nm
________ nm
experimental determined bond length
=
0.139 nm
covalent radius of C =
=
expected C = C bond length =
0.067 nm
________ nm
_________
C=C
<
0.139 nm
C … C
<
_________
CC
Dative Covalent Bonds
It is a special example of covalent bond and this is formed by the sharing of a pair of electron
both of which came from one atom. It is formed from the overlapping of an empty orbital with
an orbital occupied by a lone pair of electrons. The resulting bonding is indistinguishable from
other covalent bonding.
For example: NH4+
NH3 and BF3
Compounds containing unshared pairs of electrons (lone pairs) form dative covalent bond easily.
The atom providing the two electrons to make up the dative bond is known as donor (donor
must contain lone pair of electrons). The atom sharing the pair of electrons from the donor is
called acceptor.
.
1
(a)
Use your knowledge of electron orbitals as probability pattern to explain why it is
impossible to give a value for the radius of an isolated atom.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
(b)
Why is the van der Waals radius always greater than the covalent radius for a
given element?
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Section
4b
Covalent
bonding
Page
3
Relation of shapes of simple molecules to electron distribution
Ionic bonds are the electrostatic attractions that exist between oppositely charged ions.
ions radiate a spherically symmetrical positive or negative field, ionic bonds are
directional.
Since
non-
Covalent bonds are formed by overlapping of atomic orbitals of the atoms involved. The
bonding electrons must have opposite spin, in accordance with the Pauli Exclusion Principle.
The more the atomic orbitals overlap, the more stable the covalent bond formed. The strongest
bonds will be formed if the atoms approach in such a way that there is maximum overlap
between the atomic orbitals. It follows that a covalent bond will have a preferred direction.
.
2.
Why is ionic bonding referred to as 'non-directional' ?
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The VSEPR (Valence Shell Electron Pair Repulsion) Theory
A covalent molecule will have a shape which is determined by the angles between the bond
joining the atoms together. The VSEPR Theory bases its result on minimizing the electrostatic
repulsions between electron pairs in the valence shell (outermost shell). Electron pairs can be
bond pair electrons or lone pair electrons. The arrangement of electron pairs (bonds) are in
such a way so as to minimize the electron-electron repulsion between them.
The magnitude of these repulsion depends on whether the electrons are lone pairs or bond
pairs. The lone pair electrons are closer to the nucleus than the bonded electrons (why?)
and hence the lone pair may be visualized as being short and fat, while the bond pair orbitals
are long and thin. Hence the lone pair exert a greater repelling effect than a bond pairs and
the repulsion between the electron pairs decrease in the following order:
lone pair - lone pair
> lone pair - bond pair
>
bond pair - bond pair
In order to predict the shape of the molecule, we should first find out the central atom in the
molecule and the total number of electron pairs present in the outermost shell.
Example
1
Methane
In the methane molecule, carbon is the central atom and four electron pairs surround the
central carbon atom.
H
H
C
C
H
H
H
H
H
H
In order to attain minimum repulsion, the four electron pairs are directed towards the corners
of a regular tetrahedron. The bond angle is 109.5.
Example
2
Ammonia
The central atom in ammonia is nitrogen and four electron pairs surround the central nitrogen
atom, three bond pairs and one lone pair.
107°
H
N
H
H
N
H
H
H
Since the repulsion between lone pair and bond pair is greater than that between bond pairs,
therefore the bond angle is compressed to 107 and the shape of ammonia is pyramidal.
Section
Example
3
4b
Covalent
bonding
Page
4
Water
The central atom is oxygen and it also contains four electron pairs in the valence shell, two
bond pairs and two lone pairs.
O
H
104.5°
O
H
H
H
Repulsion exerted by the lone pairs are greater than that between lone pair and bond pair and
in turn greater than that between bond pairs. Therefore the bond angle is further compressed
to 104.5 . The water molecule is V-shaped (or bent) .
Example
4
Boron trifluoride
The central atom is boron and it contains only three electron pairs. In order to attain
minimum repulsion, the electron pairs are directed towards the corners of a triangle.
F
B
F
120o
F
Therefore the shape of boron trifluoride is planar trigonal and the bond angle is 120.
.
3.
4.
5.
What shape would you expect for the following
answer and state the expected bond angle.
Explain
your
(a)
NH2__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
(b)
NH4+
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
(a)
How many pairs of bonding electrons are there in a molecule of beryllium
hydride, BeH2 ?
(b)
If the electron pairs repel each other so that they occupy regions of space as far
apart as possible, what shape must the molecules have?
State whether you expect the following molecules and ions to have an identical shape
or a very similar shape to the CH4 molecule. Give reasons.
(i) SiCl4
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__________________________________________________________________
(ii)
6.
molecules ?
POCl3
__________________________________________________________________
__________________________________________________________________
Explain why the molecule of xenon tetrafluoride, XeF4, is square planar in shape.
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Section
4b
Covalent
bonding
Page
5
A summary on the shape of simple molecules
Formula
AX2
AX3
AX4
AX5
AX6
Number of lone pairs
No lone pair
e.g. CO2 , N2O , BeCl2 , NO2+
One lone pair
e.g. SO2 , PbCl2
Two lone pairs
e.g. NO2 , NH2- , Cl2O , H2O
No lone pair
e.g. BF3 ,
CO32-,
NO3- ,
SO3
One lone pair
e.g. NH3 ,
PCl3
No lone pair e.g. NH4+ , CCl4 , SO42Two lone pairs e.g. XeF4
No lone pair
e.g. PCl5
No lone pair
e.g. SF6
Shape
linear
angular
bent / V-shaped
planar trigonal
(trigonal planar)
pyramidal
(trigonal pyramidal)
tetrahedral
square planar
trigonal bipyramidal
octahedral
Octet rule and its limitation
The requirement of eight valence shell electrons for atoms (except hydrogen and helium) in
forming chemical bonds is called the octet rule. Although the octet rule can apply to most
elements in the first two periods, there are always some exceptions to this rule.
Example 1
Boron trifluoride
In this compound, boron forms only three covalent
bonds with three fluorine atoms and hence only 6
electrons around the boron atom.
Boron trifluoride reacts violently with molecules such as
water or ammonia because the boron atom is electron
deficient.
Example
2
Phosphorus
It forms two chlorides with chlorine, i.e. phosphorus trichloride PCl 3 and phosphorus
pentachloride PCl5.
Both of these chlorides are covalent chlorides. (N.B. In solid PCl 5, it
consists of PCl6 and PCl4+ species.)
In case of trichloride PCl3, phosphorus just use the
three unpaired electrons in the 3p orbitals for bonding.
The molecule will be pyramidal in shape.
In case of pentachloride PCl5, phosphorus uses all its
five electrons in principal shell 3 to form covalent bonds
with five chlorine atoms. In order to attain minimum
repulsion, these bonds are trigonal bipyramidal in
shape.
F
Cl
Cl
P
Cl
F
Cl
F
S
F
F
Cl
PCl5
F
SF6
Section
Example
3
4b
Covalent
bonding
Page
6
Sulphur
Sulphur also has octet expansion, e.g. SF6. Sulphur
uses all its 6 valence electrons to form 6 covalent bonds
with 6 fluorine atoms. The molecule is trigonal
bipyramidal in shape.
No octet expansion in case of nitrogen or oxygen , although they belongs to the same group
as phosphorus and sulphur. The main reason for that is because the outermost shell of
nitrogen and oxygen can only hold 8 electrons, i.e. 2s and 2p orbitals.
Molecules containing multiple bonds
The VSEPR Theory can also be applied in determining the shapes of molecules containing
multiple bonds. In predicting the shape of molecules, electrons involved in a multiple bond is
treated as a single electron pair.
Example 1
Sulphur dioxide
Dot-and-cross diagram for sulphur dioxide :
Total number of electron pairs surround the central sulphur atom = __________
Shape of sulphur dioxide molecule :
_________________________________
Example 2
Ethene
Dot-and-cross diagram for ethene :
Total number of electron pairs surround the central carbon atom = __________
Shape of ethene molecule :
_________________________________
.
7.
Both molecules of ethene (C2H4)
planar , but with different HCH
H
H
117o
C=C
126o
H
H
and methanal (CH2O) are
bond angles. Explain why.
H
C=O
H
trigonal
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Section
4b
Covalent
bonding
Page
7
Giant Covalent Crystals
1.
Diamond
The carbon atoms in diamond are joined to four other carbon atoms by strong, rigid,
directional covalent bonds and result in the formation of a three-dimensional network .
Each carbon is bonded tetrahedrally to its 4 nearest neighbors by strong
and therefore the coordination number of carbon in diamond is 4.
2.
covalent bonds
Graphite
The structure of graphite contains of layers of carbon atoms. Each carbon atoms is bonded
planar trigonally to three other carbon atoms by single, rigid covalent bonds. Therefore
the coordination of carbon in graphite is 3.
The remaining electron is delocalized above and below the layer lattice, hence the C-C bond
is stronger in graphite than in diamond.
Individual layers are held together by van der waal's forces. This account for the slippery
nature and electrical conductivity (in the plane of layers) of graphite.
The bond length in graphite is 0.142 nm (c.f. diamond, 0.154nm) suggesting some multiple
bond character.
.
8.
(a)
Why is diamond so much harder than graphite ?
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_____________________________________________________________________
(b)
Why does graphite conduct electricity while diamond does not ?
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_____________________________________________________________________
(c)
Why does diamond have a high melting point (3550 oC) ?
_____________________________________________________________________
(d)
Graphite contains weak van der waals‘ forces , but nevertheless has
a very high melting point - even higher than diamond. Explain.
_____________________________________________________________________
_____________________________________________________________________
(e)
Why is graphite used in “lead” pencils , and also as a lubricant ?
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Section
3.
4b
Covalent
bonding
Page
8
Quartz
This is a crystalline form of solid silica (SiO 2). Each silicon atom is covalently bonded to
four oxygen atoms while each oxygen atom is shared by two silicon atoms. The structure is
similar to that of diamond.
silicon atom
oxygen atom
Molecular Crystals
Molecular crystals consist of molecules held in lattice sites by weak intermolecular forces such
as van der Waals’ forces or hydrogen bonds.
The following diagrams show the crystal structures of iodine and carbon dioxide.
iodine crystal
carbon dioxide crystal
Both solid iodine and carbon dioxide are crystallize in the face-centred cubic structures.
.
9.
Carbon dioxide is a gas at room temperature while silicon(IV) oxide is a solid which
melts at about 1700。. Explain this difference.
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Section
4b
Covalent
bonding
Page
9
Past Paper
-
1.
(i)
Define the covalent radius of an atom.
(ii)
State and explain the trends in the covalent radius on going down
and going across a short period of the periodic table.
any group
(iii) The covalent radius of carbon is
0.077 nm. The measured carbon- carbon
bond length in benzene is
0. 139 nm.
Estimate the carbon-carbon bond length in ethane. Explain any difference in
the carbon-carbon bond lengths in these two molecules.
(6 marks) (92 IIA 3(b))
2.
For each of the following molecules, draw a three-dimensional structure and state the
molecular geometry.
i)
SiF4
ii) OF2
(4 marks) (93 IA 1(b))
3.
i)
Sketch the shape you would expect NH4+ and NH3 molecules to have. Explain
the difference, if any, in the H–N–H bond angles in ammonium ion and ammonia.
ii)
Show by means of a sketch the shape you would expect the amide ion, NH 2- , to
have. Make an estimation of the likely values of the bond angles.
iii)
The ammonia molecule can form the positive ion, NH4+. Would you expect
methane, CH4, to form an ion CH5+? Give reason for your answer.
Making Scheme
92IIA3(b)
(i)
(ii)
(iii)
The covalent radius is defined as one-half the distance between
atoms of the same kind held together by covalent bond.
Going down
increases.
any
Going
any
group
of
the
periodic
table
the
covalent
two
1
radius
1
across
because
the
effective
nuclear
period
elements
charge
the
become
covalent
larger
increase
radius
down
across
the
decreases.
1
group
1
the
and
period
1
C  C bond length of CH3CH3 = 2 x 0.77 = 1.54 nm
C  C bond in ethane is a single bond which is longer than the
carbon-carbon
bond
in
benzene.
In
benzene,
it
is
because
of
delocalization electrons in the ring. The average bond order is larger
than 1. (That is , benzene has double bond character.)
93IA1(b)
(i)
SiF4
(ii)
2
2
2
2
1
2
OF2
F
O
Si
F
F
F
V - shaped /
F
bent
F
tetrahedral
3.
(i)
2
2
NH4+
tetrahedral
HNH bond angle = 109.5o
NH3
trigonal pyramidal
HNH bond angle = 107o
NH3 has smaller bond angle as NH3 has 1 lone pair ( & 3 bond
which has greater repulsive forces than 4 bond pairs in NH 4+.
(ii)
NH2- which has 2 lone pairs
HNH bond angle = 104.5o
&
2
bond
pairs
thus
it
(iii)
NH3 can form NH4+ ion as NH3 has unshared lone pair which can
dative covalent bond with H+ resulting NH4+ ion.
CH4 has no unshared lone pair thus it cannot form dative bond
H+. Therefore CH5+ cannot be formed.
pairs)
is v-shaped.
form
with