L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p1
COVALENT BONDING
I.
Formation of Covalent Bond
Review
(1)
A covalent bond exists between two atoms when they share electrons because the two
nuclei attract the electrons. This electron-nucleus attraction outweighs the repulsion
between the nuclei and between the electrons.
(2)
Covalent bond is made by sharing of electron pairs, or overlapping of atomic orbitals
between 2 atoms.
(A) Covalent Bond in Terms of Electronic Sharing
(1) Atoms may attain the noble gas electronic structure by sharing electrons to give covalent
bonds.
Example: 2 fluorine atoms share electrons to achieve the electronic structure of neon.
In fluorine molecules, each fluorine atom has
six electrons (i.e. 3 pairs of electrons ) to itself in the outermost shell, and
-
a share of two electrons in the bond between them.
(2) In hydrogen chloride, a hydrogen atom and a chlorine atom share a pair of electrons so
that they attain the electronic structures of helium and argon respectively.
Electronic structure:
Some atoms achieve the noble gas electronic structure by the formation of double bonds
or triple bonds.
e.g.
Double bond:
( two pairs of electrons are shared)
Triple bond:
(three pairs of electrons are shared)
In covalent compound,
Bond pair refers to a pair of electrons shared between the bonded atoms. Lone pair
refers to a pair of electrons not used in bonding.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p2
(3) The atoms in the molecule are held together by the attractive force of the two nuclei for
the negative charge cloud (i.e. the bonding electrons) between them.
(B) Covalent Bond in terms of Overlapping of Orbitals ( Valence Bond Theory )
In the valence bond theory,
-
Covalent bond is formed by the overlap of 2 atomic orbitals each containing an
unpaired electron.
In other words, each atom contribute 1 electron in the bond, i.e. each atom has a
half-filled valence orbital.
-
The theory can be illustrated by using hydrogen molecule as an example:
(1) 2 isolated hydrogen atoms (A and B ) each has one electron in a 1s orbital, are very far apart.
(2) When the atoms are brought together, there will come a time when the region of space
covered by one orbital will merge with that covered by the other, i.e. they will overlap.
(3) When overlapping occurs, the electron belonging to A can move around the nucleus of B, and
vice versa. A’s electron and B’s electron can move into each other’s orbitals.
(4) The electrons are shared between the orbitals. The perfect sharing of electrons makes a
covalent bond. In the diagram shown below, the dot and cross are drawn at the points where
the circles drawn. This shows that the two electrons are shared between the two atoms.
The consequences of orbital overlapping are as follows:
-
The electronic configuration of each atom is similar to a noble gas electronic
configuration after orbital overlapping.
The region of overlapping corresponding to an increase in electron density
between the two nuclei. The electrostatic attraction between the electron density
and the bonded nuclei maintains the covalent bond.
(5) The larger the overlapping of the atomic orbitals, the stronger the covalent bond is.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p3
(C) ‘Dot and Cross’ Diagrams for Simple Molecules
In ‘dot and cross’ diagrams for simple molecules,
(1) only outermost shell electrons are shown as ‘dots’ and ‘crosses’.
(2) All electrons are the same.
(3) The bond pair of electrons may be represented by a ‘bond line’.
The ‘dot and cross’ diagram for simple molecules are drawn as follows:
Methane (CH4)
Phosphorus trichloride (PCl3)
(D) Octect Rule and its Limitation
Octect Rule:
Atoms form bond in an attempt to achieve the stable noble gas electronic structure, with 8
electrons in the valence shell.
The compounds encountered previously follow the octect rule in their formation. However,
octect rule has its limitations PCl5 and BF3 do not follow the octect rule.
PCl5
In PCl5, 5 chlorine atoms bond to the central phosphorus atom ( a Group V element ) as shown:
The outershell of the phosphorus atom has 10 electrons. The octect has expended to accommodate
10 electrons in the valence shell.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p4
BF3
In Boron trifluoride (BF3) , three fluorine atoms bond to the central boron atom ( a Group III
element ) as shown:
The boron atom only has 6 valence electrons. It need two more valence electrons to achieve the
octect structure.
II.
Dative Covalent Bond ( Coordinate Covalent Bond )
-
Dative covalent bond is formed by the overlapping of a empty orbital with an
orbital occupied by a lone pair of electrons.
In other words, a dative coordinate bond is a covalent bond in which one of the
atoms supplies both electrons being shared.
Example:
For the H3NBF3 molecule, a dative covalent bond is formed between N and B at which N
donates both electrons for sharing.
Other Examples:
Al2Cl6 : Two of the aluminium-chlorine bonds are dative bonds, which attract two AlCl3 together.
The dative covalent bond is made from chlorine atoms (donor of a pair of electrons) to
aluminium atoms (an acceptor of a pair of electrons).
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
BeCl2 :
Chpt7:p5
In solid state, beryllium chloride consists of chain of atoms held by dative covalent bond
made from chlorine atoms to beryllium atoms.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
III.
Chpt7:p6
Bond Enthalpies
Review
(1)
(2)
(3)
Most chemical reactions occur with bond breaking and bond formation. Bond breaking is
endothermic ( energy absorbing ) and bond formation is exothermic ( energy releasing ).
The overall enthalpy change of a reaction is the sum of the enthalpy changes during the
processes of bond breaking and bond formation.
There are two types of ‘bond energy’ : bond dissociation enthalpy and bond enthalpy .
Bond enthalpies can be regarded as a comparison of the strength of covalent bonds
( forces holding atoms together in a covalently bonded molecule).
The greater the bond strength, the higher the bond enthalpy, and vice versa.
The smaller the bond strength, the lower the bond enthalpy, and vice versa.
The energy released in forming a bond is numerically equal to the bond dissociation enthalpy
but the sign is reversed.
(A) Bond Dissociation Enthalpy
Bond dissociation enthalpy is the enthalpy change when one mole of a particular
bond in a specific compound is broken.
It is not an average value.
Example,
H2 (g) 
2H(g)
H0 = +436 kJ mol-1
The bond dissociation enthalpy of H2(g) is the enthalpy change when 1 mole of H2(g) is dissociated.
Note:
(1) Bond dissociation enthalpy is twice the value of enthalpy of atomization of hydrogen, which
is defined as the enthalpy change when 1 mole of gaseous hydrogen atom is formed from
hydrogen in its stable physical state under standard conditions.
1/2 H2(g)  H(g) H0 = +218 kJ mol-1
(2) Bond dissociation is always on endothermic process because energy is needed to overcome
the attraction between the bonded nuclei.
(3) For a given diatomic molecule (e.g. hydrogen, chlorine), the bond dissociation enthalpy is
always constant.
(4) For a polyatomic molecule (e.g. ammonia, methane), the bond dissociation enthalpy of any
particular bond also depends on specific bond environment (e.g. the presence of other
bonding atoms and their nature).
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p7
Process
CH4(g)  CH3(g) + H(g)
Enthalpy change kJ mol-1
435
CH3(g)  CH2(g) + H(g)
CH2(g)  CH(g) + H(g)
CH(g)  C(g) + H(g)
365
523
339
CH4(g)  C(g) + 4H(g)
1662 (total)
The for C—H bond have different bond dissociation enthalpies. To generate the four values, bond
enthalpy is introduced.
(B) Average Bond Enthalpy
Bond enthalpy is the average of the bond dissociation enthalpies required to break
a particular type of bond in generally most compounds containing the bonding
atoms.
Alternatively, this is the average standard enthalpy change for the breaking of a mole of a
particular type of bond in a gaseous molecule to form gaseous atoms.
Example:
average bond enthalpy (C—H) = 1662 / 4 = 415.5 kJ mol-1
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p8
Name:____________________ Class No.:____ Date:_______________ Marks:______________
1.
Determine the bond enthalpy (C—H) through a Born-Haber cycle.
Standard enthalpy change of formation of methane = -75 kJ mol-1
Standard enthalpy change of atomization of hydrogen = +218 kJ mol-1
Standard enthalpy change of atomization of carbon (graphite ) = + 715 kJ mol-1
Determine the bond enthalpy (C—H ) through a Born-Haber cycle.
(2) Given the following thermochemical data at 298 K:
standard enthalpy change of formation of ethane = -86 kJ mol-1
standard enthalpy of atomization of hydrogen = + 218 kJ mol-1
standard enthaipy of atomization of carbon(graphite) =+715 kJ mol-1
bond enthalpy (C—H) = +415 kJ mol-1
Determine the bond enthalpy (C—C)
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
The bond enthalpies of some bonds are listed:
Bond
Bond Enthalpy / kJ mol-1
H—H
436
H—Cl
431
C—C
348
C—H
413
N—H
388
O—H
463
Using bond enthalpies, the enthalpy of atomization of a substance.
(3) Given the following bond enthalpy values:
C—C
= 348 kJ mol-1
C—H
= 412 kJ mol-1
Calculate the enthalpy of atomization of ethane, C2H6(g)
Chpt7:p9
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p10
3.
(C) Additive Rule of Bond Enthalpy
For many reactions. the change in bond enthalpies would be approximately equal to the enthalpy
change for the reaction.
Example: For the reaction X2(g) + Y2(g)  2XY(g)
2X(g)
Energy
+ 2Y(g)
Sum of bond energies of X—Y(g)
Sum of bond energies of
X—X and Y—Y
H
X—X (g) + Y—Y (g)
Note
<1> Bond breaking requires energy whereas bond forming releases energy.
<2> The enthalpy change is the sum of the energies required to break bonds in the
reactants plus the energies released from the formation of new bonds in products.
<3> bond enthalpies are average values only. Hence, the enthalpy of reaction is only an estimate.
Exercises
1, For the reaction given below
H2(g) + C2H4(g) —*C2H6(g)
(a)
calculate the enthalpy change of the above reaction by two different methods.
Method (1)
the enthalpy change of formation at 298K
Method (2)
H0f [C2H4(g)] = +52.3 kJ mol-1
H0f[C2H6(g)] = -84.6 kJ mol-1
the average bond energies at 298K
H—H
+436 kJ mol-1
C—H
C=C
+520 kJ mol-1
c-c
+414 kJ mol-1
+347 kJ mol-1
(b) account for the difference between the enthalpy of reaction calculated using the two
methods.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p11
Name: ____________________ Class No.:_____ Date: _______________ Marks :______________
2.
Given the following average bond enthalpies:
CC 813 kJ mol-1
C—C
346 kJ mol-1
C—H 413 kJ mol-1
H—H
436 kJ mol-1
Determine the standard enthalpy of hydrogenation of ethyne:
H—CC—H
3.
+ 2H2(g) 
CH3CH3
Catalytic reforming of hexane to cyclohexane and hydrogen is part of the process for the
industrial production of petrol:
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
IV.
Chpt7:p12
Relationship between Covalent Bond Enthalpies, Bond length and Covalent radii.
The strength of a covalent bond is measured by the bond enthalpy , the enthalpy change when a
bond is broken. The greater the bond enthalpy of a covalent, the stronger the covalent bond is.
Bond enthalpy
(1) decreases as the bond length increases.
(2) Increases with the difference in electronegativity or polarity of the bond.
(3) Increase as the number of electrons making up the bond increases (multiple bonds).
(A)
Effect of Bond length on Bond Enthalpies
(1) Bond length is the internuclear distance between 2 atoms making up a covalent bond.
(2) Consider the bond lengths and the standard bond enthalpies of hydrogen halides:
Hydrogen halides
Bond length / nm
Standard bond enthalpy / kJ mol-1
H—F
0.109
+565
H—Cl
0.135
+431
H—Br
0.151
+364
H—I
0.171
+297
It can be generalized that:
A covalent bond with long bond length tends to have small bond enthalpy.
(i.e. weak bond strength)
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p13
Exercise 1
Consider the data given below for the hydrogen halides:
Hydrogen halide Standard enthalpy of formation /kJ mol-1 Bond dissociation enthalpy / kJ mol-1
H—F
-269.4
+562
H—Cl
-92.8
+430
H—Br
-36.8
+367
H—I
+26.1
+298
(a) Explain briefly the trend in the bond dissociation energy of hydrogen halides.
(b) At temperatures above 400 K , hydrogen halide decomposes into constituent elements,
but hydrogen chloride and hydrogen bromide do not decompose.
Briefly account for this phenomenon.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p14
(B) Effect of Electronegativity Difference on Bond Enthalpies
(1) The greater the difference in electronegativity values between the bonding atoms, the greater
the bond enthalpy (i.e. high bond strength)
(2) Consider the bond strengths of hydrogen halides with the difference in electronegativity
values (in Pauling’s scale):
Hydrogen halides
Electronegativity difference
between atoms
Standard bond enthalpy / kJ mol-1
H-F
1.9
+565
H-Cl
0.9
+431
H-Br
0.7
+364
H-I
0.3
+298
It can be deduced that:
A great difference in the electronegativities of two atoms bonded together is associated with a
strong ionic character in the bond between them. This results in a higher bond enthalpy.
Note: Relation between Bond strength and Bond length:
As the bond length increases, the distance between the nuclear protons and the bond-paired
electrons increases. The attractive forces between the atoms is reduced, resulting in a
weaker
bond.
The above generalization is not always true. See the following example:
(a)
Bond
Bond length / nm
Standard bond enthalpy / kJ mol-1
F-F
0.142
+158
Cl-Cl
0.199
+242
Br-Br
0.288
+193
I-I
0.266
+151
Cl—Cl bond is stronger than the Br—Br bond because
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
(b)
Cl—Cl bond is stronger than the F—F bond even though Cl—Cl bond has a longer bond
length because
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p15
(C) Effect of Number of Electrons in a Bond on Bond Enthalpy
In general, bond enthalpies increase with the number of electrons forming a bond.
(D) Covalent Radii
Covalent radius :
It is one half the distance between two atoms of the same kind held together
by a single covalent bond.
H
0. 030
Li
0.123
Be
0.089
Na
0.157
Mg
0.136
B
0.080
C
0.077
N
0.070
0
0.066
Al
0.125
Si
0.117
P
0.110
S
0.104
F
0.064
Cl
0.099
Exercise
State and explain the trends in the covalent radius on going
(a) down any group, and
(b) across a short period of the Periodic Table.
(E) Additivity of Covalent—Radii
For a covalent bond A—B
Bond length = covalent radius of atom A + covalent radius of atom B
Example
Bond
C—C
C—O
C—Cl
H—Cl
C—H
Note:
Calculated bond length/nm
0.077 + 0.077=0.154
0.077 + 0.066=0.143
0.077 + 0.099=0.176
0.030 + 0.099=0.129
0.077 + 0.030=0.107
Experimental bond length/nm
0.154
0.143
0.177
0.127
0.109
The bond length calculated from the additivity of individual covalent radius is close to the
experimental bond length.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
V.
Chpt7:p16
The Shapes of Covalent Molecules and Polyatomic Ions
(A) Valence Shell E1ectron Pair Repulsion Theory
<1> Covalent bonding is directional and leads to molecules with specific shapes.
<2> The shapes of simple covalent molecules and polyatomic ions are determined by the
number of electron pairs in the outershell of the central atom.
Valence shell electron pair repulsion theory
(1) Electron pairs in the central atom tend to get as apart as possible to minimize the electrostatic
repulsions between electron pairs.in the valence shell
(2) The electron pairs may be bond pairs involving single bonds or mutiple bonds, or lone pairs not
invoved in bonding
Note:
When all the electrons pairs are bond pairs, and there is an identical atom attached to each
one, then the actual shape of the molecule is the same as the predicted one.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chpt7:p17
<4> Electron pairs in the central atom tend to get as far apart as possible.
Example
In methane. the 4 electron pairs surround the central carbon atom, tbe
electron pairs move to the corners of a single tetrahedron so that they are
furthest apart.
(B) Interactions between Lone pairs and Bond pairs Leading to Different Molecular Shapes
<1> Lone pair of electrons (around the central atom) is attracted by 1 nucleus only. There is a
greater electron density around the central atom. Lone pairs are more diffuse than bonding
pairs are, and hence repel other electron pairs from a bigger region of space.
Lone pair of electrons exert a stronger repulsive force than do bond pair of electron
<2> The directions of the electron clouds are governed by the repulsion between the electron
clouds. The order of repelling power is
As a result, bonding pairs tend to move as far apart as possible from lone pairs, though this
might result in their moving slightly closer to other bonding pairs.
<3> The number of lone pairs = (no.of e in valence shell — no. of e in bonding pairs)/2
(C) The shapes of NH3. H2O, NH4+, NH2NH3
Shape: Trigonal Pyramidal
electron pairs
bond pair
lone pair
:4
:3
:1
The lone pair is closer to the Nitrogen nucleus than the bond pairs
and so it repels them strongly. i.e. it pushes the N—H bonds closer
to each other. The N—H bond closer to each other. The H-N-H
angle is therefore 107.30 which is smaller than 109.50.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
NH4+
Chpt7:p18
shape : tetrahedral
electron pairs
bond pairs
lone pair
4
:4
:0
NH4+ and CH4 are isoelecronic so they have
the same shape. The H—N—H angle is 109.5°
H2O
shape : V—shaped (bent, angular)
electron pairs
:4
bond pairs
:2
lone pair
:2
The two bond pairs escape from the lone pair repulsion by moving together slightly. The
H—O—H angle is 104.5°
NH2-
shape : V—shaped (bent, angular)
electron pairs : 4
bond pairs
:2
lone pairs
:2
NH2- and H2O are isoelectronic. NH2 have the similar shape as H2O.
Exercise 1
For each of the following molecules, draw a three-dimensional structure and state the molecular
geometry.
(a) SF4
(b) PCl5
(c) OF2
SOLUTION
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
VI.
(A)
Chpt7:p19
Multiple Bonds
Strength of Multiple Bonds
In multiple bonds. there are more shared electrons (i.e. higher electron density) in between the
bonded atoms. There exists stronger electrostatic attraction between the nuclei and electrons.
Therefore,
Example
Bond
C-C
C=C
CC
NN
O=O
F—F
Bond enthalpy /kJmol-1
348
612
837
944
496
158
Bond length /nm
0.154
0.134
0.120
0.110
0.121
0.142
Exercise
Explain why a C=C bond is
(a) stronger than a C—C bond, but
(b) not twice as strong as a C—C bond
SOLUTION
(B)
The Effect of Multiple Bonding on Shape
A multiple bond is treated as though it. was a single electron pair.
Ethene C2H4
In an ethene molecule, each carbon atom is treated as having three pairs of electrons. Therefore,
the arrangement around each carbon atom is trigonal planar The bond angles (H—C—H) and (H—C—C) are both 120°. Note that the double bond is rigid
to rotation, i.e. cannot be twisted, so that ethene is a planar structure.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Ethyne (C2H2)
Chpt7:p20
In an ethyne molecule
The carbon atoms are held together by triple bonds. The triple bond is treated as an electron pair.
Each carbon atom is treated as having 2 electron pairs one shared with the other carbon atom
and another with a hydrogen atom. This gives a linear arrangement.
Carbon Dioxide
Electron pairs around each carbon atom : 2 (2 double bonds)
Shape : linear
(repulsion between the multiple bonds is minimized)
Sulphur Dioxide
Electron pairs around the sulphur atom : 2 bond pairs
1 lone pair
Shape : V—shaped
(S=O bond)