Section
Section 4A
4a
Ionic bonding
Page
1
Ionic Bonding
Chemical Bonding
A chemical bond is a force which holds together two or more atoms, ions, molecules or any
combination of these.
The ability of an atom to form chemical bonds is called its valency. The electrons which
form chemical bonds are called valence electrons. These occur in the highest energy
orbitals of an atom. The outermost shell of an atom which contains these orbitals is called
the valence shell.
When atoms form bonds they try to achieve the most stable electronic configuration by
1.
gaining or losing electrons . If they again electrons they become anions. If they
lose electrons they become cations. This type of bond is called an ionic bond
(electrovalent bond).
2.
sharing electrons . This is called covalent
bonding .
Ionic Bonding
An ionic bond is resulted from the coulombic attraction between the oppositely
charged ions. The strength of an ionic bond is roughly proportional to the product of the
charges on the ions and inversely proportional to the square of the distance between the
centers of the ions.
F

Q1 Q 2
(r1  r2 ) 2
An ionic bond could be formed as a result of the transfer of one or more electrons from
one atom to another until the outer electron shell is identical with that of the nearest inert
gas.
.
1.
Why is ionic bonding referred to as ‘non - directional’ ?
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Criteria for ion formation
The ease of an atom to form ion is affected by the following factors.
1.
Electronic configuration
It would be easier to form ion if the resulting electronic structure of the ion is stable,
e.g. noble gas configuration.
2.
The charge of the ion
In general, ions of small charge is more stable than that of higher charge.
3.
For cations :
in order to be stable, the lattice energy must be large enough to
compensate for the energy involved in ionization and this is commonly
achieved for M + and M2+, but not for higher charged ions.
For anions :
formation of X - is exothermic, but the formation of X 2 - is usually
slightly exothermic ( or endothermic ) and the formation of X 3 - is
endothermic.
Size of the atom
For the formation of cations, those atoms with a larger size will be easier because the
ionization energies will be smaller. And for the formation of anions, atoms of smaller
sizes are more preferable because the electron affinity are greater.
Stoichiometry of ionic compounds
Section
4a
Ionic bonding
Page
2
So far we have predict the formula for an ionic compound by considering their electronic
configuration. But we can predict the formula of an ionic compounds from energy
consideration.
The method is that we could try to calculate the enthalpy of formation for an ionic
compound with different possible formulae. For example, when magnesium reacts with
chlorine, the product may have the following possible formulae:
MgCl
;
MgCl2
;
MgCl3
By drawing Born-Haber cycles we can calculate the enthalpy of formation for each possible
compound and we would expect that the compound which has the most negative
standard enthalpy of formation would be the most stable product, i.e. the most
probable product.
Given the following data, calculate the enthalpy of formation of the three possible
magnesium chloride.
1st I.E. of Mg
2nd I.E. of Mg
3rd I.E. of Mg
lattice energy of MgCl
lattice energy of MgCl2
lattice energy of MgCl3
=
=
=
=
=
736 kJmol-1
1448 kJmol-1
7740 kJmol-1
- 753 kJmol-1
- 2502 kJmol -1
=
sublimation energy of Mg
dissociation energy of Cl2
electron affinity of Cl
=
=
=
- 5440 kJmol -1
146 kJ mol-1
242 kJ mol-1
- 364 kJ mol-1
Enthalpy of Formation of
Enthalpy of Formation of
MgCl
MgCl2
=
=
Enthalpy of Formation of
MgCl3
=
__________ kJ mol -1
__________ kJ mol -1
__________ kJ mol -1
From the values of enthalpies of formation, MgCl3 is very unstable with respect to its
elements while MgCl is stable with respect to its elements but it is unstable with respect to
MgCl2.
Therefore, if MgCl is formed in the reaction, it would undergoes disproportionation to give
MgCl2 and Mg.
i.e. The product from the reaction between magnesium and chlorine has the formula of
MgCl2.
Ionic Radii
Study of ionic crystals by using x-ray diffraction or other methods, the equilibrium
distance between two nearest adjacent ions in a crystal, the internuclear distance, could be
determined.
The internuclear distance for the halides of sodium and potassium are given below in nm:
KF
KCl
KBr
KI
0.266
0.314
0.329
0.353
NaF
NaCl
NaBr
NaI
0.231
0.281
0.298
0.323
Difference
If the ions in a crystal are regarded as spheres, the internuclear distance between two ions
will be made up of the sum of the ionic radii of the two ions. Moreover, the constancy
of the difference between the internuclear distances of sodium and potassium halides, as
shown above, indicates that the ionic radii of anions and cations must be reasonably
constant in a series of such compounds.
But on x-ray analysis, or other methods, the internuclear distance does not give a value for
an ionic radius unless we know any one ionic radius by other method.
Section
4a
Ionic bonding
Page
3
Pauling, by considering isoelectronic compounds such as sodium fluoride or potassium
chloride, made a calculation and adopt the ionic radii of fluoride ion, F –, to be 0.136 nm. By
adopting this ionic radius as a value for F –, we can find out the values for other ions. (What
have you assumed in doing this?)
N.B.
The size of the ion of an element varies slightly, depending upon the
compound it is part of; ions are slightly soft, compressible, and deformable.
By adopting different standard values, different authorities may have different results on
the ionic radius.
For example, ionic radii (in nanometer) according to Pauling:
Li+
0.060
Be
2+
0.031
F
Na+
0.095
Mg 2+
0.065
Cl
–
0.181
K+
0.133
Ca
2+
0.099
Br
–
0.195
Rb+
0.148
Sr 2+
0.113
I
–
0.216
Cs+
0.169
Ba
0.135
O2-
0.146
2+
0.136
–
These values illustrate a number of conclusions :
1.
Ions in any one vertical group of the periodic table increase in size as the relative
atomic mass of the element increase.
2.
For a series of isoelectronic ions, for example, F –, Na+, Mg
decrease as atomic number increase. (Why?)
3.
Cations are smaller than the atom from which they are formed whilst anions
larger.
4.
When an element form more than one cations, the ion with the lower charge is larger
than the more highly charged ions. Lead, for example, Pb2+ of radius 0.121 nm is larger
than Pb4+, with radius of 0.084 nm. (Why?)
2+
and Al
3+
size of ion
are
Trends for ionic radii
1.
Down the group
The ionic radii increase down the group. When descending the group, both the number
of electrons and the number of protons increase, the increase in screening effect is
greater than that in nuclear charge. Therefore the effective nuclear charge decrease
as descending down a group and result in an increase in ionic radii.
2.
Across a period
For cations, as they are isoelectronic, they have the same number of electrons but
the number of protons increases when passing across the period, therefore the effective
nuclear charge increases as passing across the period and hence the ionic radii
decreases.
For anions, the anions formed are also isoelectronic. On passing across the period,
the number of electrons added to the anions decreases gradually (decreases from 3 to 1)
while the number of protons increases, hence the effective nuclear charge increase
across the period. Therefore the resulting effect is making the ionic radii decrease on
passing across the period.
.
2.
Do you expect the radius of (a) a sodium ion, Na+, and (b) a chloride ion,
Cl-, to be greater or smaller than the radius of the parent atom? Explain your
answer.
Section
4a
Ionic bonding
Page
4
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3.
(a)
X+
An atom X can form a cation
species has:
(i)
the smallest radius;
(ii)
(b)
and an anion
X - . Which of the three
the largest radius ?
In descending group I from lithium to caesium, how do ionic radii change
and how do you explain the trend ?
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Ionic Crystal Structures
1.
Sodium
Chloride
(rock-salt structure)
The structure can be considered as being made up of two interpenetrating facecentered cubic lattices, one made up by the chloride ions, i.e. the cube ABCDEFGH,
and the other made up by the sodium ion, i.e. the cube IJKLMNOP.
P
E
I
F
J
H
O
G
Chloride ion
A
D
L
C
K
B
N
Crystal
sodium ion
M
structure
of
NaCl
The coordination number is the number of nearest neighbours for a particular type
of atom or ion. Within the NaCl lattice, each Na+ is surrounded by six Cl- ions,
giving a coordination number of 6. Also, each Cl- is surrounded by six Na+ ions, and
have a coordination number of 6.
In the structure the coordination for sodium and chloride are both 6 and this is called
6 : 6 coordination.
Stoichiometry
of
Chloride ion:
sodium chloride
8 at corner :
6 on faces
Total number of chloride in 1 unit cell =
Sodium ion:

Na+
:
Cl-
=
4:4
=1
=3
12  1/4
11
=3
=1
4
12 on edges
1 at centre
Total number of sodium in 1 unit cell =





8  1/8
6  1/2
4
=
1:1
Stoichiometry of sodium chloride
=
NaCl
and the unit cell of sodium chloride contains 4 ion-pairs of sodium chloride
Section
4a
Ionic bonding
Page
5
(i.e. 4 sodium ions and 4 chloride ions).
A unit cell is the smallest basic portion of the crystal lattice that, when repeat at
various directions, can reproduce the entire crystal structure.
2.
Caesium Chloride (CsCl)
The structure can be considered as being made up of two interpenetrating simple cubic
lattices of caesium and chloride ions.
Caesium chloride has a coordination number of 8 for both caesium and chloride, i.e.
8:8 coordination. The unit cell of caesium chloride is similar to the body-centred cubic
structure of the alkali metals.
Caesium chloride adopt a structure different from that of sodium chloride because the
caesium ion is much bigger than the sodium ion, therefore caesium chloride has a
more open structure.
Stoichiometry
of
Caesium ion :
Chloride ion:

Cs+

caesium
chloride
8 at corner:
1 at centre
:
Cl-
8 x 1/8 = 1
1x1
= 1
= 1:1
Stoichiometry of caesium chloride
=
CsCl
and the unit cell of caesium chloride contains 1 ion-pair of sodium chloride
(i.e. 1 caesium ion and 1 chloride ion).
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.
4.
[a]
Sketch the unit cell of the caesium chloride structure, using marks of
different colour or shape to represent the centres of ions.
[b]
Sketch the sodium chloride structure in the same way.
[c]
Write down the co-ordination numbers for each ion in the two structures.
[d]
Explain why sodium chloride could not adopt the caesium chloride structure.
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[e]
Look up the relevant ionic radii and predict which of these two structures
(NaCl or CsCl] would be adopted by :
[i]
potassium fluoride
KF
[ii]
sodium bromide
NaBr
[iii] magnesium oxide
MgO
A value for Avogadro Constant can be calculated from the information related to the
crystal structure.
Section
4a
Ionic bonding
Page
6
Example: The density of sodium chloride is 2.165 g cm-3 and 1 mole of it weighs
58.448 g. The distance between centres of adjacent Na+ and Cl ions
is 2.819  10- 8 cm. Find a value for the Avogadro Constant.
END
-
OF
1.
SECTION EXERCISE
Solid sodium bromide has the same cubic structure as sodium chloride.
It has a
-3
density of 3.203 g cm . The distance between adjacent ions in the crystal
is 0.294 nm. Taking the formula weight of sodium bromide as 102.9, what
value do these data give for Avogadro Constant.
[6.321  1023]
-
2.
In what ways are a sodium ion and a neon atom alike? How do they differ?
-
3.
Hydrogen can form both an anion and a cation. Explain why this is so, and give
examples of compounds containing hydrogen anions and those containing
hydrogen cations.
-
4.
Suggest why magnesium oxide has a much higher melting point than sodium
fluoride.
-
5.
In a crystal of potassium chloride, the internuclear distance is 0.314 nm,
measured by x-ray analysis. It has the same structure as sodium chloride with a
density of 2004 kgm- 3.
(a)
What is the number of formula units in one of this unit cell?
(b)
Calculate the volume occupied by a formula unit in this unit cell.
(c)
From the density given, calculate the volume occupied by one mole of
formula units of potassium chloride.
(d)
Determine a value of Avogadro Constant from your results in (c and (b).
[ 4 ion-pairs ; 6.19 x 10- 29 m3 ; 3.72 x 10- 5 m3 ;
6.01 x 1023 ]
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Making Scheme
Section
1.
Density
= 3.203
Avogadro‘s
constant
Ionic bonding
Page
7
102.9
Molar volume of NaBr
=
Molar volume
=
Volume of 1 unit cell =
Molar volume of NaBr =
32.126
4a
32.126 cm3
(2  2.94  10-8)3 which contains 4 NaBr.
(Volume of 1 NaBr)  (Avogrdro‘s constant)
=
(2  2.94  10 -8 ) 3
4
=
6.32  1023

Avogadro‘s
constant
2.
They are isoelectronic, both contain 10 electrons (1s2 2s2 2p6). But sodium
ion has 11 protons, so it has greater effective nuclear charge and
smaller size in turn. By contrast, neon has 10 protons, which has
lower effective nuclear charge and hence larger size.
3.
Hydrogen atom gain 1 electron to form anion H- which has completely
filled 1s - orbital. Example : sodium hydride , NaH.
Hydrogen atom lose 1 electron to form cation H+ which has no
electron. Example : H+ in hydrochloric acid.
4.
Mg2+ and O2- are more highly charged than Na+ and F-.
Also, Mg2+ is
smaller than Na+. Hence, we expect the lattice energy of MgO is much
larger than that of NaF. MgO therefore has a higher melting point
than NaF.
5.
(a)
4
(b)
(2 x 0.31 x 10 9 ) 3
4
(c)
Molar
(d)
=
6.19  10-29 m3
2004
=
(39.1  35.5) / 1000
Molar volume
volume
=
3.72  10-5 m3
Avogadro‘s
constant =
3.72  10 5
6.19  10 29
=
6.01  1023