In the previous section, we saw the limitations of octet rule. In this section, we will see ionic bonds and lattice enthalpy
• We know that, an ionic bond is formed between a positive ion (cation) and a negative ion (anion)
1. The formation of a cation can be represented as:
$\mathbf\small{\rm{M(g)\longrightarrow M^+(g)\,+e^-}}$
• It indicates that, one electron is removed from an atom of element 'M'
• We have seen this process in the previous chapter (Details here)
♦ We have seen these following points also:
✰ Energy is involved in the 'removal of an electron' in this way
✰ This energy is always endothermic
✰ This energy is called ionization enthalpy (ΔH)
2. The formation of an anion can be represented as:
$\mathbf\small{\rm{X(g)+e^-\longrightarrow X^-(g)}}$
• It indicates that, one electron is added to an atom of element 'X'
• We have seen this process in the previous chapter (Details here)
♦ We have seen these following points also:
✰ Energy is involved in the 'removal of an electron' in this way
✰ This energy may be endothermic or exothermic
✰ This energy is called electron gain enthalpy (ΔegH)
3. Once a cation and an anion are formed, we get the ionic compound
• It can be represented as:
$\mathbf\small{\rm{M^+(g)+M^+(g)\longrightarrow MX(s)}}$
4. So it is obvious that, for the formation of the ionic compound,
♦ We must be able to get the cation easily
♦ We must be able to get the anion also easily
5. 'Getting the cation easily' means that, we need to supply only a small amount of energy to obtain the cation
■ In other words, to get the cation easily, the ionization enthalpy must be low
6. 'Getting the anion easily' means that, we get a large amount of energy when the anion is formed
■ In other words, to get the anion easily, the electron gain enthalpy must be highly negative
7. The conditions mentioned in (5) and (6) enable us to make two predictions:
(i) The cations will be formed from metals
♦ In other words, metallic ions will be cations
♦ This is because, metals have low ionization enthalpies
(ii)The anions will be formed from non-metals
♦ In other words, non-metallic ions will be anions
♦ This is because, non-metals have high negative electron gain enthalpies
The cation NH4+ is formed from two non-metals N and H
• Some basic details about this ion can be written in 4 steps:
1. NH4+ acts as a single unit
2. This unit acts as the cation in many ionic compounds
3. An example is NH4Cl (ammonium chloride)
♦ The cation is NH4+
♦ The anion is Cl-
♦ The two ions are held together by electrostatic force of attraction
♦ Thus NH4Cl is formed
4. This is just like the formation of NaCl from Na+ and Cl- ions
Solved example 4.2
Write Lewis dot symbols for the following atoms and ions:
S and S2- ; Al and Al3+ ; H and H-
Solution:
The required Lewis dot symbols are shown in fig.4.32 below:
Part (a):
S and S2- :
(i) S has 6 valence electrons
(ii) It needs 2 more electrons to attain octet
(iii) When those two electrons are added, the S atom gains a charge of -2 and becomes S2- ion
(iv) So the 'S with 8 valence electrons' is written inside square brackets and -2 is written at top right
Part (b):
Al and Al3+ :
(i) Al has 3 valence electrons
(ii) It needs 5 more electrons to attain octet. But it is easier to lose the 3 electrons
(iii) When those three electrons are lost, the Al atom gains a charge of +3 and becomes Al3+ ion
(iv) So the 'Al with zero valence electrons' is written inside square brackets and +3 is written at top right
Part (c):
H and H- :
(i) H has 1 valence electron
(ii) It can either lose this electron or gain an extra electron to attain duplet. In our present case, it gains one electron
(iii) When that electron is added, the H atom gains a charge of -1 and becomes H- ion
(iv) So the 'H with 2 valence electrons' is written inside square brackets and -1 is written at top right
Solved example 4.3
Use Lewis dot symbols to show electron transfer between the following atoms to form cations and anions:
(a) K and S (b) Ca and O (c) Al and N
Solution:
Part (a): K and S
(i) One K atom loses it's valence electron to become K+ ion
• This is shown in fig.4.33(a) below:
(ii) So from two K atoms, we get two K+ ions and two electrons
(iii) The S atom is in need of two electrons. It accepts the two electrons lost by K atoms
• The S atom thus becomes S2- ion
• This is shown in fig.4.33(b)
(iv) The two K+ ions get attached to the S2- ion because of the electrostatic force of attraction
• Thus one molecule of K2S (potassium sulfide) is formed
• This is shown in fig.4.34 below:
Part (b): Ca and O
(i) One Ca atom loses it's two valence electrons to become Ca2+ ion
• This is shown in fig.4.35(a) below:
(ii) The O atom is in need of two electrons. It accepts the two electrons lost by Ca atom
• The O atom thus becomes O2- ion
• This is shown in fig.4.35(b)
(iv) The Ca2+ ion get attached to the O2- ion because of the electrostatic force of attraction
• Thus one molecule of CaO (Calcium oxide) is formed
• This is shown in fig.4.36 below:
Part (c): Al and N
(i) One Al atom loses it's three valence electrons to become Al3+ ion
• This is shown in fig.4.37(a) below:
(ii) The N atom is in need of three electrons. It accepts the three electrons lost by Al atom
• The N atom thus becomes N3- ion
• This is shown in fig.4.37(b)
(iv) The Al3+ ion get attached to the N3- ion because of the electrostatic force of attraction
• Thus one molecule of AlN (Aluminium nitride) is formed
• This is shown in fig.4.38 below:
• We know that, an ionic bond is formed between a positive ion (cation) and a negative ion (anion)
1. The formation of a cation can be represented as:
$\mathbf\small{\rm{M(g)\longrightarrow M^+(g)\,+e^-}}$
• It indicates that, one electron is removed from an atom of element 'M'
• We have seen this process in the previous chapter (Details here)
♦ We have seen these following points also:
✰ Energy is involved in the 'removal of an electron' in this way
✰ This energy is always endothermic
✰ This energy is called ionization enthalpy (ΔH)
2. The formation of an anion can be represented as:
$\mathbf\small{\rm{X(g)+e^-\longrightarrow X^-(g)}}$
• It indicates that, one electron is added to an atom of element 'X'
• We have seen this process in the previous chapter (Details here)
♦ We have seen these following points also:
✰ Energy is involved in the 'removal of an electron' in this way
✰ This energy may be endothermic or exothermic
✰ This energy is called electron gain enthalpy (ΔegH)
3. Once a cation and an anion are formed, we get the ionic compound
• It can be represented as:
$\mathbf\small{\rm{M^+(g)+M^+(g)\longrightarrow MX(s)}}$
4. So it is obvious that, for the formation of the ionic compound,
♦ We must be able to get the cation easily
♦ We must be able to get the anion also easily
5. 'Getting the cation easily' means that, we need to supply only a small amount of energy to obtain the cation
■ In other words, to get the cation easily, the ionization enthalpy must be low
6. 'Getting the anion easily' means that, we get a large amount of energy when the anion is formed
■ In other words, to get the anion easily, the electron gain enthalpy must be highly negative
7. The conditions mentioned in (5) and (6) enable us to make two predictions:
(i) The cations will be formed from metals
♦ In other words, metallic ions will be cations
♦ This is because, metals have low ionization enthalpies
(ii)The anions will be formed from non-metals
♦ In other words, non-metallic ions will be anions
♦ This is because, non-metals have high negative electron gain enthalpies
■ However, there is one exception:
• Some basic details about this ion can be written in 4 steps:
1. NH4+ acts as a single unit
2. This unit acts as the cation in many ionic compounds
3. An example is NH4Cl (ammonium chloride)
♦ The cation is NH4+
♦ The anion is Cl-
♦ The two ions are held together by electrostatic force of attraction
♦ Thus NH4Cl is formed
4. This is just like the formation of NaCl from Na+ and Cl- ions
• Now we know how the cations and anions are formed
• Next we have to learn some basics about crystal and lattice. It can be written in 5 steps:
1. Take a small quantity of any substance. We want to know whether it is a crystal
2. To call that substance a crystal, the following two conditions must be satisfied:
(i) That substance must be a solid
(ii) The particles (atoms, molecules or ions) that make up that solid must be arranged in a regular pattern
3. Since there is a regular pattern, we will be able to 'identify the smallest unit' in the crystal
• Continuous repetition of that 'smallest unit' will give rise to the crystal
4. The 'continuous repetition' takes place in all directions. So it is a 3D structure
5. When we look at a crystal, we will see a regular arrangement of particles (atoms, molecules or ions)
• This regular arrangement is called the 'lattice of that crystal'
• Each crystal will have it's own lattice
♦ For example, the lattice of rock salt (common salt) will consist of 'repeating cubes'
• When we see a 'substance which is a crystal', we say this:
♦ 'That substance is a crystal'
♦ OR
♦ 'That substance is a crystalline substance'
• Both statements are correct
• Next we have to learn some basics about crystal and lattice. It can be written in 5 steps:
1. Take a small quantity of any substance. We want to know whether it is a crystal
2. To call that substance a crystal, the following two conditions must be satisfied:
(i) That substance must be a solid
(ii) The particles (atoms, molecules or ions) that make up that solid must be arranged in a regular pattern
3. Since there is a regular pattern, we will be able to 'identify the smallest unit' in the crystal
• Continuous repetition of that 'smallest unit' will give rise to the crystal
4. The 'continuous repetition' takes place in all directions. So it is a 3D structure
5. When we look at a crystal, we will see a regular arrangement of particles (atoms, molecules or ions)
• This regular arrangement is called the 'lattice of that crystal'
• Each crystal will have it's own lattice
♦ For example, the lattice of rock salt (common salt) will consist of 'repeating cubes'
• Now we know what a crystal is
♦ 'That substance is a crystal'
♦ OR
♦ 'That substance is a crystalline substance'
• Both statements are correct
• Now, rock salt (NaCl) is a crystalline substance
• Let us analyse it's structure. It can be written in steps:
1. Rock salt is a crystalline substance. It is made up of Na+ and Cl- ions
2. The smallest unit of the 'lattice of NaCl' is a cube
• The Na+ and Cl- ions occupy the corners of that cube
• This is shown in fig.4.31 below:
3. At first glance, the structure will appear to be a mixture of cyan and red balls
• But by taking a closer look, we will be able to detect a pattern
• The coordinate axes x, y and z, will help us to detect the pattern easily
• The pattern can be written in 3 steps:
(i) Put your finger tip on any cyan ball
• Move the finger tip in the x direction (forward or backward)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
• Move the finger tip in the y direction (towards left or right)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
• Move the finger tip in the z direction (upwards or downwards)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
(ii) Put your finger tip on any red ball
• Move the finger tip in the x direction (forward or backward)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
• Move the finger tip in the y direction (towards left or right)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
• Move the finger tip in the z direction (upwards or downwards)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
(iii) That is., when moving in the x, y or z directions:
You will meet cyan and red balls alternately and that too, at regular intervals
(iv) Another interesting point:
• Put your finger tip on any cyan ball
♦ Consider the x-direction
✰ You will see two red balls. One at the front and the other at the back
♦ Consider the y-direction
✰ You will see two red balls. One at the left and the other at the right
♦ Consider the z-direction
✰ You will see two red balls. One at the top and the other at the bottom
■ In short, any cyan ball will be surrounded by six red balls
■ In the same way, we can prove that, any red ball will be surrounded by six cyan balls
4. Let us analyse the 'energies absorbed and released' during the formation of NaCl
• The analysis can be written in steps:
(i) We want a Na+ ion. That is., we want the following reaction to take place:
$\mathbf\small{\rm{Na(g)\longrightarrow Na^+(g)\,+e^-}}$
• For this reaction to take place, we have to supply energy (called ionization enthalpy)
• The ionization enthalpy in this case is 495.8 kJ/mol
• We have to supply this energy. So it is positive energy
(ii) We want a Cl- ion. That is., we want the following reaction to take place:
$\mathbf\small{\rm{Cl(g)+e^-\longrightarrow Cl^-(g)}}$
• When this reaction takes place, we receive energy (called electron gain enthalpy)
• The electron gain enthalpy in this case is -348.7 kJ/mol
• We receive this energy. That is the reason for the -ve sign
(iii) So the energy transactions are as follows:
• We supply 495.8 kJ/mol
• We receive 348.7 kJ/mol
• The net effect appears to be a 'supply of (495.8-348.7) = 147.1 kJ/mol'
(iv) But in reality, the net effect is that, we 'receive energy'. Let us see the reason:
• The Na+ that we obtain, is in the gaseous state
• The Cl- that we obtain, is also in the gaseous state
• But the resulting NaCl is in the solid state
• The solid NaCl is a crystal having a definite lattice structure that we saw in fig.4.31 above
• A lattice is a very stable structure. So, when a lattice is formed, a lot of energy is released
• When one mole of NaCl is formed, we receive 788 kJ of energy
• In other words, the 'enthalpy of lattice formation' of NaCl is -788 kJ/mol
• So the net energy transaction is: (495.8-348.7-788) = -640.9 kJ/mol
• That means, we receive 640.9 kJ/mol of energy during the formation of NaCl
1. Consider the formation of an ionic solid
2. Energy is supplied to obtain the cation
3. Energy is received when the anion is obtained
4. The 'net transaction' based on (2) and (3) may be +ve
+ve energy indicates that energy is to be supplied
5. But (2) and (3) are not the only transactions
• There is one more item:
♦ The energy released during the formation of the lattice
♦ This energy is called 'enthalpy of lattice formation'
6. So we have to consider the net of (2), (3) and (5)
• This net will be -ve
• -ve energy indicates that, energy is released
7. So, during the formation of an ionic solid, we will receive energy
The definition can be written in 7 steps:
1. Consider the +ve and -ve ions (in gaseous form) which are initially at infinite distances apart
2. Bring them close to each other so that electrostatic attractions come into effect between the oppositely charged ions
• While bringing the ions close together:
♦ The oppositely charged ions will naturally come close to each other
✰ So energy will be released
♦ The similarly charged ions tend to repel each other
✰ So we will need to supply energy
3. The ions will 'settle down into a lattice form' to give a solid ionic compound
4. In this process, we will get a 'net energy release'
5.Calculate the 'energy released per mole of the resulting compound'
6. This energy is called lattice enthalpy
7. Based on the above steps, we are receiving energy
♦ So by this definition, the lattice enthalpy is -ve
We can write the definition in a 'reverse' manner also. This can be done in 6 steps:
1. Take one mole of a solid ionic compound
2. Supply enough energy so that, the +ve and -ve ions break away from the lattice
• While doing this:
♦ The similarly charged ions will naturally repel away from each other
✰ So energy will be released
♦ The oppositely charged ions will tend to stick together
✰ So we will need to supply energy
3. The energy must be just sufficient to separate the ions into infinite distances apart so that, there will not be any attraction or repulsion between them
4. In this process, we will need to provide a 'net energy supply'
5. The energy required for this process is called lattice enthalpy
6. Based on the above steps, we are supplying energy
♦ So by this definition, the lattice enthalpy is +ve
■ We can write any one of the above definitions. The numeric value of the energy will be the same in both definitions. But the signs will be opposite
• But in reality, there are many more factors like bond length, bond angle, bond enthalpy etc.,
We will see those factors in the next section
• Let us analyse it's structure. It can be written in steps:
1. Rock salt is a crystalline substance. It is made up of Na+ and Cl- ions
2. The smallest unit of the 'lattice of NaCl' is a cube
• The Na+ and Cl- ions occupy the corners of that cube
• This is shown in fig.4.31 below:
 |
| Fig.4.31 |
• But by taking a closer look, we will be able to detect a pattern
• The coordinate axes x, y and z, will help us to detect the pattern easily
• The pattern can be written in 3 steps:
(i) Put your finger tip on any cyan ball
• Move the finger tip in the x direction (forward or backward)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
• Move the finger tip in the y direction (towards left or right)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
• Move the finger tip in the z direction (upwards or downwards)
♦ The next ball you meet will be red
♦ The ball after that will be cyan. so on . . .
(ii) Put your finger tip on any red ball
• Move the finger tip in the x direction (forward or backward)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
• Move the finger tip in the y direction (towards left or right)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
• Move the finger tip in the z direction (upwards or downwards)
♦ The next ball you meet will be cyan
♦ The ball after that will be red. so on . . .
(iii) That is., when moving in the x, y or z directions:
You will meet cyan and red balls alternately and that too, at regular intervals
(iv) Another interesting point:
• Put your finger tip on any cyan ball
♦ Consider the x-direction
✰ You will see two red balls. One at the front and the other at the back
♦ Consider the y-direction
✰ You will see two red balls. One at the left and the other at the right
♦ Consider the z-direction
✰ You will see two red balls. One at the top and the other at the bottom
■ In short, any cyan ball will be surrounded by six red balls
■ In the same way, we can prove that, any red ball will be surrounded by six cyan balls
4. Let us analyse the 'energies absorbed and released' during the formation of NaCl
• The analysis can be written in steps:
(i) We want a Na+ ion. That is., we want the following reaction to take place:
$\mathbf\small{\rm{Na(g)\longrightarrow Na^+(g)\,+e^-}}$
• For this reaction to take place, we have to supply energy (called ionization enthalpy)
• The ionization enthalpy in this case is 495.8 kJ/mol
• We have to supply this energy. So it is positive energy
(ii) We want a Cl- ion. That is., we want the following reaction to take place:
$\mathbf\small{\rm{Cl(g)+e^-\longrightarrow Cl^-(g)}}$
• When this reaction takes place, we receive energy (called electron gain enthalpy)
• The electron gain enthalpy in this case is -348.7 kJ/mol
• We receive this energy. That is the reason for the -ve sign
(iii) So the energy transactions are as follows:
• We supply 495.8 kJ/mol
• We receive 348.7 kJ/mol
• The net effect appears to be a 'supply of (495.8-348.7) = 147.1 kJ/mol'
(iv) But in reality, the net effect is that, we 'receive energy'. Let us see the reason:
• The Na+ that we obtain, is in the gaseous state
• The Cl- that we obtain, is also in the gaseous state
• But the resulting NaCl is in the solid state
• The solid NaCl is a crystal having a definite lattice structure that we saw in fig.4.31 above
• A lattice is a very stable structure. So, when a lattice is formed, a lot of energy is released
• When one mole of NaCl is formed, we receive 788 kJ of energy
• In other words, the 'enthalpy of lattice formation' of NaCl is -788 kJ/mol
• So the net energy transaction is: (495.8-348.7-788) = -640.9 kJ/mol
• That means, we receive 640.9 kJ/mol of energy during the formation of NaCl
Based on the above discussion, we can arrive at an important conclusion. It can be written in 7 steps:
2. Energy is supplied to obtain the cation
3. Energy is received when the anion is obtained
4. The 'net transaction' based on (2) and (3) may be +ve
+ve energy indicates that energy is to be supplied
5. But (2) and (3) are not the only transactions
• There is one more item:
♦ The energy released during the formation of the lattice
♦ This energy is called 'enthalpy of lattice formation'
6. So we have to consider the net of (2), (3) and (5)
• This net will be -ve
• -ve energy indicates that, energy is released
7. So, during the formation of an ionic solid, we will receive energy
Now we are in a position to define Lattice enthalpy
1. Consider the +ve and -ve ions (in gaseous form) which are initially at infinite distances apart
2. Bring them close to each other so that electrostatic attractions come into effect between the oppositely charged ions
• While bringing the ions close together:
♦ The oppositely charged ions will naturally come close to each other
✰ So energy will be released
♦ The similarly charged ions tend to repel each other
✰ So we will need to supply energy
3. The ions will 'settle down into a lattice form' to give a solid ionic compound
4. In this process, we will get a 'net energy release'
5.Calculate the 'energy released per mole of the resulting compound'
6. This energy is called lattice enthalpy
7. Based on the above steps, we are receiving energy
♦ So by this definition, the lattice enthalpy is -ve
We can write the definition in a 'reverse' manner also. This can be done in 6 steps:
1. Take one mole of a solid ionic compound
2. Supply enough energy so that, the +ve and -ve ions break away from the lattice
• While doing this:
♦ The similarly charged ions will naturally repel away from each other
✰ So energy will be released
♦ The oppositely charged ions will tend to stick together
✰ So we will need to supply energy
3. The energy must be just sufficient to separate the ions into infinite distances apart so that, there will not be any attraction or repulsion between them
4. In this process, we will need to provide a 'net energy supply'
5. The energy required for this process is called lattice enthalpy
6. Based on the above steps, we are supplying energy
♦ So by this definition, the lattice enthalpy is +ve
■ We can write any one of the above definitions. The numeric value of the energy will be the same in both definitions. But the signs will be opposite
• From the above definitions, we get the impression that, the lattice enthalpy depends only on the attractive and repulsive forces between the ions
We will see those factors in the next section
Now we will see some solved examples
Solved example 4.2
Write Lewis dot symbols for the following atoms and ions:
S and S2- ; Al and Al3+ ; H and H-
Solution:
The required Lewis dot symbols are shown in fig.4.32 below:
 |
| Fig.4.32 |
S and S2- :
(i) S has 6 valence electrons
(ii) It needs 2 more electrons to attain octet
(iii) When those two electrons are added, the S atom gains a charge of -2 and becomes S2- ion
(iv) So the 'S with 8 valence electrons' is written inside square brackets and -2 is written at top right
Part (b):
Al and Al3+ :
(i) Al has 3 valence electrons
(ii) It needs 5 more electrons to attain octet. But it is easier to lose the 3 electrons
(iii) When those three electrons are lost, the Al atom gains a charge of +3 and becomes Al3+ ion
(iv) So the 'Al with zero valence electrons' is written inside square brackets and +3 is written at top right
Part (c):
H and H- :
(i) H has 1 valence electron
(ii) It can either lose this electron or gain an extra electron to attain duplet. In our present case, it gains one electron
(iii) When that electron is added, the H atom gains a charge of -1 and becomes H- ion
(iv) So the 'H with 2 valence electrons' is written inside square brackets and -1 is written at top right
Solved example 4.3
Use Lewis dot symbols to show electron transfer between the following atoms to form cations and anions:
(a) K and S (b) Ca and O (c) Al and N
Solution:
Part (a): K and S
(i) One K atom loses it's valence electron to become K+ ion
• This is shown in fig.4.33(a) below:
 |
| Fig.4.33 |
(iii) The S atom is in need of two electrons. It accepts the two electrons lost by K atoms
• The S atom thus becomes S2- ion
• This is shown in fig.4.33(b)
(iv) The two K+ ions get attached to the S2- ion because of the electrostatic force of attraction
• Thus one molecule of K2S (potassium sulfide) is formed
• This is shown in fig.4.34 below:
 |
| Fig.4.34 |
(i) One Ca atom loses it's two valence electrons to become Ca2+ ion
• This is shown in fig.4.35(a) below:
 |
| Fig.4.35 |
• The O atom thus becomes O2- ion
• This is shown in fig.4.35(b)
(iv) The Ca2+ ion get attached to the O2- ion because of the electrostatic force of attraction
• Thus one molecule of CaO (Calcium oxide) is formed
• This is shown in fig.4.36 below:
 |
| Fig.4.36 |
(i) One Al atom loses it's three valence electrons to become Al3+ ion
• This is shown in fig.4.37(a) below:
 |
| Fig.4.37 |
• The N atom thus becomes N3- ion
• This is shown in fig.4.37(b)
(iv) The Al3+ ion get attached to the N3- ion because of the electrostatic force of attraction
• Thus one molecule of AlN (Aluminium nitride) is formed
• This is shown in fig.4.38 below:
 |
| Fig.4.38 |
In the next section, we will see bond length, bond angle and bond enthalpy
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