Ionic bonding

BY
Dr. SURYAKANT B. BORUL
M. Sc., M.Phil., Ph. D.
Head & Assistant Professor,
Department of Chemistry,
Late Ku. Durga K. Banmeru Science
College, Lonar. Dist. Buldana. 443302.
LATE KU. DURGA K. BANMERU SCIENCE
COLLEGE, LONAR, DIST. BULDANA. 443302.
(Affiliated to Sant Gadage Baba Amravati University Amravati; 2 (f) & 12 B; NAAC Accredited with ‘C’ grade)

Ionic Bonding-
Defn- “The chemical bond which is formed by transfer of one or more
electrons from the valence shell of one atom to the valence shell of another
atom is called as ionic bond.”
Or “The chemical bond which is formed due to electrostatics force of
attraction in between opposite charge ions is called as ionic bond.”
For example – Bond in between Na and Cl
Na + Cl [Na+] [Cl-] or Na+Cl-
(2,8,1) (2,8,7) (2,8) (2,8,8)

Here, electrostatics force of attraction in between opposite charge
ions Na+ and Cl- in NaCl molecule.

The formation of an ionic bond is favoured when-
1) Metals has low ionization energy
2) Other elements has high electron affinity and
3) The resulting compound has lattice energy.

Types of Cations
The formation of an ionic compound is due to atom attaining electronic
configuration similar to that of inert gas element by an anion while the cation may
achieve any one of the following configurations.
1. No valence electron (Ex.- H+)
2. Ions with inert gas configuration–These ions have inert gas e. c. (ns2np6) in
their outermost shell. (Except foe n=1).
Cations Inert Gas
Li+ He (1s2)
Na +, Mg 2+, Al 3+ Ne (2s2 2p6)
K+, Ca+, Sc3+ , Ti4+ Ar (3s2 3p6)
Rb+ , Sr2+ , Y3+, Zr4+, Kr (4s2 4p6)
Cs+ , Ba2+ , La3+ , Ce4+ Xe (5s2 5p6)

3. Ions with pseudo inert gas configuration (ns2np6nd10)– These ions
have inert gas e.c. (ns2 np6 nd10) in their outermost shell.
Cations Electronic Configuration
Ag+ , Cd 2+, In3+, Sn4+ 4s2 4p6 4d10
Au+ , Hg 2+,Ti3+, Pb4+ 5s2 5p6 5d10
4. The inert s2 pair configuration ((n-1)s2 p6 d10 ns2)–
When the elements having valence shell configuration ns2 npx
(x=1,2,3) lose their p electrons only, cations with ns2 configuration are
formed.

Cations Electronic Configuration
Ga+ , Ge 2+ and As3+ 4s2
In+ , Sn 2+ and Sb3+ 5s2
Ti+ , Pb 2+ and Bi3+ 6s2
This is possible only when the energies of the ns & np electrons
differ sufficiently so as to result in the stepwise ionization during the
chemical bond formation. Therefore, only the post transition elements of
group IIA, IV A and VA give such ions.

5. The d and f ions:- The transition metal ions formed by the loss of the outer
valence shell electrons without the ionization of the d electrons have the
configuration of the outer shell as- ns2 np6 ndx (x= 1 to 9) and are classified
as the d ions.
Ex: Ti2+ , V2+ , Cr2+ , Co2+ etc.
The ions are derived from the inner transition elements by the loss of
the outer s and d electrons and have configuration. (n-1) s2 ,(n-1)p6, (n-
2)d10, (n-2)f1-13 ex.- Lanthanide and actinide ions.
6. Ions with Irregular configurations:- These are certain ions that cannot be
classified into any particular class
Ex: Ga4+

Energetics of Ionic Bond Formation
Their are three types of energies are involved in the ionic bond
formation. These are as follows
A) Ionization energy
B) Electron affinity or energy
C) Lattice energy
A) Ionization energy-
“The amount of energy required to remove the outer most electron
from an isolated gaseous atom of an element in its ground state to form
cation is called as ionization energy.”

M(g) M(g) + e- ; H = +I
 The energy required for this change is denoted by I.
 The energy is to be supplied in the process it is given a positive sign.
 The energy is measured in electron volts (eV) or Kcal / mole.
 The magnitude of ionization energy is a direct measure of ease of cation
formation.
 If its value is low. Cation is readily formed. Alkali and alkaline earth
metals have low values of ionization energy.

B) Electron affinity or energy-
“The amount of energy released when an electron is added to an
isolated neutral gaseous atom in its ground state to produce an anion is
called as electron affinity or energy.”
X(g) + e- X-
(g); H = -E
 It is denoted by E.
 It is the energy released, it is given a negative sign.
 The energy is measured in electron volts(eV) or Kcal/ mole.
 Anion formation will be favoured if more energy is released in above
process i.e. if electron affinity is high.

C) Lattice energy- It is related to the formation of an ionic solid from its
ions. Lattice energy of an ionic crystal M+ X- is defined in the following
two ways-
1. The energy released when exact number of gaseous cations M+
(g) and gaseous
anions X-
(g) come close together from infinity to from one mole of solid ionic
crystal, M+ X-
(S) is called as lattice energy.
M+
(g) + X-
(g) M+ X-
(S) + Energy released
2. The energy required for removing ions of one mole of solid ionic crystal from
their equilibrium positions in crystal to affinity is called as lattice energy.
M+ X-
(S) + Energy supplied M+
(g) + X-
(g)
 It is denoted as U

Factors favouring the Formation of Ionic
Bond
The formation of an ionic compound MX will be favoured if
i) The Ionization energy of element is low
ii) Electron affinity or energy of X is high
iii) Lattice energy of compound MX is high
 Calculation of lattice energy- Lattice energy may be calculated
theoretically using Madelung constant or it may be determined
experimentally using Born Haber cycle. Both the methods are
discussed as-

Theoretical Calculation of Lattice energy using Madelung
constant
In 1918 Max Born and Alfred Landé proposed that the lattice energy
could be derived from the electrostatic potential of the ionic lattice and a
repulsive potential energy term. Lattice energy can be theoretically calculated
using the Born-Lande equation.
Where e = Charge on electron (1.6022X 10-19C)
Z+ and Z- = Charge on cation and anion respectively
NA= Avogadro number (6.023 x 1023); n = Born Exponent
r = Distance between nuclei of cation and anion in cm.
U= Lattice energy of the ionic compound
M= Madelung (From name of Erwin Madelung, a German physicist)

Madelung constant-
It is a correction factor which takes into account the electrostatic forces
exerted by neighboring ion pair. It entirely depends upon the arrangement of
positive and negative ions in crystal, i.e. upon the geometry of the ionic crystal.
It does not depends upon the nature of ions present in the crystal.
It can be calculated by summing the mutual potential energies of all the
ions a lattice. Values of Modelung constants for some common crystal are as-
Crystal type Modelung constant
NaCl 1.747558
CaCl 1.762670
CaF2 5.03878
TiO2 4.816

Born exponent-
It is a repulsion exponent which allows for repulsive forced between the
electron clouds of oppositely charged ions. It can be evaluated from the results
of experimental measurements of the compressibility of the crystal.
It is found that for all the crystals n lies in the neighbouring of 9. It
depends upon configuration of ion.
For the different configuration the values are as –
He= 5, Ne=7, Ar=9, Kr=10, Xe=12
For a crystal having two ions of different electronic configuration average of the
values given above is used. For ex.-in case of NaCl, Na+ ion has configuration of
neon (n=7) and Cl - has configuration of argon (n=9). Thus evaluation of lattice
energy of NaCl the value of n to be used 8.

Experimental Determination of Lattice Energy using Born
Haber cycle
The lattice energy of an ionic solid like NaCl may be determined by
using Born-Haber Cycle. It is a thermo-chemical cycle and was devised by
Born and Haber in 1919.
The cycle first relates the lattice energy of crystalline solid
(unknown quantity) to other known thermo-chemical quantities. Then the
use of Hess’s law to evaluate the unknown quantity.

Lattice energy of sodium chloride may be determined by
using Born-Haber cycle as follows
Sodium chloride may be considered to be formed solid sodium metal
and gaseous chlorine by two different methods described below:
Method 1: It is the direct combination of solid sodium and gaseous chlorine
to give solid sodium chloride. The process may be represented by following
equation.
Na(s) + 1/2 Cl2(g)→ Na Cl (s); H, = -414.2 kJ/mol

This equation tells us that when one mole of solid sodium combines
with half mole of gaseous chlorine molecules, one mole of crystalline sodium
chloride is formed. During this process 414.2 kJ mol of energy is also
evolved. This energy is called heat of formation of sodium chloride and is
represented by the symbol H.
Method 2 : It involves five different steps described below
Step 1 : Sublimation of Sodium: In this process 1 mole of solid sodium
Na(s) changes to gaseous sodium Na(s). The energy required for this process
is SNa (Heat of sublimation of sodium) Its value is experimentally found out
to be 108.7 kJ/mol.
Na(s) → Na(g) ; SNa = 108.7 kJ/mol

Step 2: Dissociation of chlorine: In this process half mole of chlorine is
dissociated into 1mole of chlorine atoms. The energy required for this process
is 1/2 DCl2 (where DCl2, is the heat of dissociation of one mole of chlorine).
Experimental value of 1/2 DCl2 is 112.95 kJ/mol
Cl2(g) Cl(g) ; 1/2 DCl2 = 112.95 kJ/mol
Step 3: Formation of sodium ions: 1 mole of gaseous sodium atoms are
converted to sodium ions by removal of an electron from each of them.
Energy required for this process is INa .(Ionization energy of sodium) Its
experimental value is 489.5 kJ/mol.
Na(g) Na+
(g) + e- INa = 489.5 kJ/ mol

Step 4: Formation of chloride ions: One mole of chlorine atoms (formed in
step 2) take up electrons given by sodium and are converted to negatively
charged chloride ions. The process is accompanied by release of energy. By
definition the energy released in this process is electron affinity of chlorine
(ECl). Its experimental value is -351.4 kJ/mol
Cl(g) + e-→ Cl-
(g) ; ECl2= -351 kJ/ mol
Step 5: Formation of ionic crystal Na+Cl-
(s) : Gaseous sodium and chloride
ions formed instep (3) and (4) above combine to give solid sodium chloride
crystal Na+Cl-
(s). Energy is related in this process also and by definition of
lattice energy of NaCl.

It is represented by UNaCl. Its value is to be determine fro other
values.
Na+
(g) + Cl-
(g) → Na+Cl- ; UNaCl = ?
According to Hess’s law the energy change in method (1) must be
equal to total of energy changes of all steps in method (2) i.e.
Hf = SNa + ½ DCl2 + INa + ECl2 + UNaCl
Putting the actual values we gets.
-414.2 = +108.7 +1/2*225.9 + 489.5 - 351.4 + UNaCl
Therefore
UNaCl = -414.2 -108.7 – 112.95- 489.5 + 351.4
UNaCl = -773.95KJ/Mol

Representation of Born –Haber Cycle for the formation of NaCl ionic
crystal

Problem based on Lattice Energy
1) To calculate the lattice energy of NaCl crystal the data is-
Sublimation energy of Na (SNa)=108.710 Kj/ Mol
Dissociation energy for Cl2(DCl2)=225.9 Kj/ Mol
Ionization energy for Na(g) (ECl2)=489.5 Kj/ Mol
Electron affinity for Cl(g) (INa)=-351.4 Kj/ Mol
Heat of formation of NaCl (Hf ) = -414.2 Kj/ Mol
Ans-
We know that,
Hf = SNa + ½DCl2 + INa + ECl2 + UNaCl
Na+ + Cl- → Na+Cl- ; UNaCl = ?
UNaCl = -414.2 -108.7 – 112.95- 489.5 + 351.4
UNaCl = -773.95KJ/Mol

2) To calculate the heat of reaction of KF from its elements from the following
data by use of Born Haber cycle. Sublimation energy of K (SK)=87.8 Kj/
Mol
Dissociation energy for F2(DF2)=158.9 Kj/ Mol
Ionization energy for K(g) (IK)= 414.2 Kj/ Mol
Electron affinity for F(g) (EF2)= -334.7 Kj/ Mol
Lattice energy for KF(UKF)= -807.5 Kj/ Mol
Heat of formation of KF (Hf ) = ?
Ans- We know that,
Hf = SK + ½DF2 + IK + EF2 + UKF
K(S) + ½ F2(g) → KF (S) ; Hf =?
Hf = 87.8 + ½(158.9) + 414.2 + (-334.7) + (-807.5)
Hf = 87.8 + 79.45 + 414.2 + (-1142.2)= 581.45-1142.2=-560.75
Heat of formation of KF (Hf ) = -560.75KJ/Mol

3) To calculate the heat of reaction of MgF2 from its elements by using of Born
Haber cycle. Thermochemical data as-
Sublimation energy of Mg (SMg)=146.4 Kj/ Mol
Dissociation energy for F2(DF2)=158.9 Kj/ Mol
Ionization energy for Mg(g) (IMg)= 2184.0 Kj/ Mol
Electron affinity for F(g) (EF2)= -334.7 Kj/ Mol
Lattice energy for MgF2(U)= -2922.5 Kj/ Mol
Heat of formation of MgF2 (Hf ) = ?
Ans- We know that,
Hf = SMg + DF2 + IMg + EF2 + UMgF2
Mg(S) + F2(g) → MgF2(S) ; Hf =?
Hf = 146.4 + 158.9 + 2184 + 2(-334.7) +(-2922.5)
Hf = 146.4 + 158.9 + 2184 -669.4 -2922.5=2489.3-3591.9
Heat of formation of MgF2 (Hf ) = -1102.6KJ/Mol

Solvation of Ions and Solvation Energy
Q.-Explain the term Solvation energy.
Ans- “The interaction that takes place when a substance is introduced in a
solvent is called as solvation and the energy associated with this is called as
solvation energy.”
Water is called a polar solvent because in its molecule the oxygen
atom is partly negatively charged and each hydrogen atom is partly
positively changed as -

When sodium chloride is introduced in such a solvent, the negative
end of water molecule attract the positive ions, and the positive end attract
the negative ions of the crystal.
These attraction forces exerted by the water molecules weaken the
attractions existing among the ions in the crystal.
Hence some of the ions in the crystal are pulled away from their positions in
crystal lattice as -

Once the Na+ and Cl- ions are broken away from the ionic lattice,
following two processes occur same time.
1. Each sodium ion is surrounded by a definite but unknown number of
water molecules, say ‘x’, with their negative ends (oxygen ends)
pointing towards it, as shown in figure.
Na+ + xH2O = [Na(H2O)x]+
This process is called solvation of sodium ion and the energy change
associated with it is called as solvation energy of sodium ion, (Hs)Na+.
The chemical species [Na(H2O)x]+ is called solvated or acquated sodium
ion and may also be represented as [Na(aq)]+ .

2) Each chloride ion is surrounded by definite but unknown number of water
molecules, say ‘y’ with their positive ends (hydrogen ends) pointing
towards it, as shown
Cl- + yH2O = [Cl(H2O)y]-
The process is called solvation of chloride ion and the energy change
associated with it is called solvation energy of chloride ion, (Hs)Cl-. The
chemical species [Cl(H2O)y]- is called solvated or aquated chloride ion and
may also be represented as [Cl(aq)]-

The total process may be written as:
NaCl(s) + (x+y)H2O [Na (H2O)x]+ + [Cl(H2O)y]-
or NaCl(s) + aq Na+
(aq) + Cl-
aq)
fig- Born-Haber cycle for determination of salvation energy.

Calculation of Solvation Energy
The energy changes during solvation of sodium and chloride ions may be
calculated using a Born-Haber type cycle as given in figure 1.6
Here L is the heat of solution of NaCl at infinite dilution (i.e. the
total amount of heat evolved or absorbed when one mole of sodium
chloride dissolved in such a large excess of water, that further addition of
water does not produce any heat change).
UNaCl is the lattice energy of NaCl.
(HS)Na+ and (Hs)Cl- are the solvation energies of sodium and
chloride ions.

Since heat of solution of NaCl at infinite dilution (L) and lattice
energy of NaCl (UNaCl) are experimentally known, the solvation energies
Na+ and Cl- ions can be calculated from following relation.
L = UNaCl + (HS)Na+ + (HS)cı-
It gives us the sum of solvation energies of sodium and chloride
ions. (There is no purely thermochemical way to separate this sum into
two parts corresponding to sodium and chloride ions.

Factors affecting solvation and solvation energy
i) Solvation energy and lattice energy: The dissolution of an ionic
compound in polar solvent is favoured if the attraction between solvent
molecules and ions, exceeds the attraction among the ions in a crystal
lattice or in other words if the energy of solvation of ions exceeds the
lattice energy of the crystal.
ii) Dielectric constant and solvation energy: For a given ion and the
solvent the dielectric constant and the solvent energy are related by
following equation, called Born equation.

Here,
H = Solvation energy of gaseous ion,
r = ionic radius and or
C = charge on the ion,
D = dielectric constant of the solvent.
From this equation it is evident that increase in the magnitude of
dielectric constant increases the solvation energy.
iii) Ionic size: Both solvation energy and lattice energy are increased by
decreases in cation and anion size. It is therefore difficult to relate solubility to
size of ion.

iv) However the two opposite charges are not of the same magnitude and in
general other factors being equal solubility increases with increase in
cation or anion size.
v) Ionic charge: With increasing cation or anion charge, the lattice energy
increases much more rapidly than the solvation energy. This results in
decrease of solubility.
vi) Electronic configuration of cations and their polarising effect:
a) If the anion is more readily polarized by the cation, than is the solvent,
the lattice energy will increase more than solvation energy and the
solubility will decrease.
b) If the solvent is more readily polarized by the cation, the solubility will
increase.

The ions having pseudo inert gas configuration Ag+, Pb++, Hg++,
etc. have high anion polarizing effect, hence their salts (AgCl, PbCl, HgCl)
have lower solubility in water, As compared to these, the alkaline earth
cations (Ca+, Ba++ etc.) having inert gas type configuration, have low anion
polarizing effect, hence their halides CaCl2, BaCl2, are readily soluble in
water.

Ionic bonding

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