In the previous section, we completed a discussion on dipole moments. In this section, we will see the 'partial covalent character' of ionic bonds
• In the same way, the ionic bonds will have some 'partial covalent character'
• Let us see the factors on which this ‘partial covalent character’ in ionic bonds, depend:
1. Consider an ionic bond between a cation and an anion
♦ The cation is the one which has lost electrons. It is the +ve ion
♦ The anion is the one which has gained electrons. It is the -ve ion
2. The anion has gained some ‘extra electrons’
• If the cation pulls those ‘extra electrons’, we can say these:
♦ Both the anion and the cation has now access to the 'extra electrons'
♦ The ‘extra electrons’ are being ‘shared’ by the two ions
• Because of this ‘sharing’, we can say that, some covalent character has been induced in that ionic bond
3. So what are the conditions which favors the ‘pulling of electrons by the cation’?
• There are four conditions, written as A, B, C and D below:
(A) Size of the cation
(i) Consider a cation. It has lost electrons
• With the loss of electrons, the nucleus is able to pull the remaining electrons towards the center with a greater force
• So the size of the cation becomes smaller than the parent atom
(Remember that, the positive charge of the nucleus remains the same)
(ii) Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
(iii) This indicates that, a smaller cation has greater ‘pulling capacity’
■ So we can write:
If the cation in the ‘ionic bond under consideration’ is smaller, it is a favourable condition for inducing ‘partial covalent character’
(B) Size of anion
(i) Consider an anion. It has gained electrons
• With the gain of electrons, the nucleus is not able to pull all the electrons towards the center with the original force
(Remember that, the positive charge of the nucleus remains the same)
• So the size of the anion is larger than the parent atom
(ii) Also, with the gaining of electrons, the electron-electron repulsion increases
• This also causes an increase in the size of the anion when compared to the parent atom
(iii) So in general, the size of the anion becomes larger
• This indicates that, a larger anion has lesser ‘pulling capacity’
• This is same as:
A larger anion has a greater tendency to donate electrons
■ So we can write:
If the anion in the ‘ionic bond under consideration’ is larger, it is a favourable condition for inducing ‘partial covalent character’
(C) Charge on the cation
(i) The charge of the cation may be +1, +2 or +3
(ii) The increase in charge indicates that, the remaining electrons are held more tightly than in the parent atom
(iii) Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
■ So we can write:
If the cation in the ‘ionic bond under consideration’ has a greater charge, it is a favourable condition for inducing ‘partial covalent character’
(D) Type of the cation
(i) We know that:
• Metallic elements lose electrons and become cations
• Non-metallic elements gain electrons and become anions
(ii) Here we consider the formation of cations
• We know that cations are formed from metallic elements
♦ The various metallic elements are:
✰ Alkali metals and Alkaline earth metals (s-block)
✰ Transition metals (d-block)
(iii) The s-block elements have the general electronic configuration: [core]ns1-2
• After losing electrons, the general electronic configuration becomes: [core]ns0
• That means, the general electronic configuration of 'cations formed from s-block elements' is: [core]ns0
■ But [core]ns0 is the general configuration of the nearest noble gases
• That means, 'cations formed from s-block elements' has the configuration of the nearest noble gas
• An example:
♦ The electronic configuration of Ca is [Ar]4s2
♦ The electronic configuration of Ca+2 is [Ar]
(iv) The transition elements have the general electronic configuration: [core](n-1)d1-10ns1-2
• After losing electrons, the general electronic configuration becomes: [core](n-1)d1-10ns0
• That means, the general electronic configuration of 'cations formed from s-block elements' is [core](n-1)d1-10ns0
(v) So we have two types of cations:
• Cations with general configuration: [core]ns0
♦ These are formed from s-block elements
• Cations with general configuration: [core](n-1)d1-10ns0
♦ These are formed from transition elements
(vi) We are given two cations:
• Suppose that:
♦ We are given two cations, one from each of the above two types
• Also suppose that:
♦ They have the same size
♦ They have the same positive charges
(vii) We want to compare the two cations and find:
Which cation can better 'pull the electrons of the anion'?
The following steps (viii), (ix) and (x) give the answer:
(viii) Consider the two configurations:
♦ [core]ns0
♦ [core](n-1)d1-10ns0
• We see that, in the 'cations from transition metals', there is an extra term: (n-1)d1-10
• That means there are electrons in the d orbitals
(ix) These 'electrons in the d orbitals' have lesser shielding effect
• So the outer electrons are attracted more strongly towards the nucleus
• Thus we can write:
In the 'cations from transition metals', the outer electrons are attracted more strongly towards the nucleus
• Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
(x) So we can write the conclusion in steps:
• We want to compare two cations
• They both have the same size and same charge
• In such a situation, the 'cations from transition metals' have greater ability for inducing ‘partial covalent character’
(A) If the cation in the ‘ionic bond under consideration’ is smaller, it is a favourable condition for inducing ‘partial covalent character’
(B) If the anion in the ‘ionic bond under consideration’ is larger, it is a favourable condition for inducing ‘partial covalent character’
(C) If the cation in the ‘ionic bond under consideration’ has a greater charge, it is a favourable condition for inducing ‘partial covalent character’
Partial Covalent character of ionic bonds
• We have seen that, the covalent bonds have some 'partial ionic character'• In the same way, the ionic bonds will have some 'partial covalent character'
• Let us see the factors on which this ‘partial covalent character’ in ionic bonds, depend:
1. Consider an ionic bond between a cation and an anion
♦ The cation is the one which has lost electrons. It is the +ve ion
♦ The anion is the one which has gained electrons. It is the -ve ion
2. The anion has gained some ‘extra electrons’
• If the cation pulls those ‘extra electrons’, we can say these:
♦ Both the anion and the cation has now access to the 'extra electrons'
♦ The ‘extra electrons’ are being ‘shared’ by the two ions
• Because of this ‘sharing’, we can say that, some covalent character has been induced in that ionic bond
3. So what are the conditions which favors the ‘pulling of electrons by the cation’?
• There are four conditions, written as A, B, C and D below:
(A) Size of the cation
(i) Consider a cation. It has lost electrons
• With the loss of electrons, the nucleus is able to pull the remaining electrons towards the center with a greater force
• So the size of the cation becomes smaller than the parent atom
(Remember that, the positive charge of the nucleus remains the same)
(ii) Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
(iii) This indicates that, a smaller cation has greater ‘pulling capacity’
■ So we can write:
If the cation in the ‘ionic bond under consideration’ is smaller, it is a favourable condition for inducing ‘partial covalent character’
(B) Size of anion
(i) Consider an anion. It has gained electrons
• With the gain of electrons, the nucleus is not able to pull all the electrons towards the center with the original force
(Remember that, the positive charge of the nucleus remains the same)
• So the size of the anion is larger than the parent atom
(ii) Also, with the gaining of electrons, the electron-electron repulsion increases
• This also causes an increase in the size of the anion when compared to the parent atom
(iii) So in general, the size of the anion becomes larger
• This indicates that, a larger anion has lesser ‘pulling capacity’
• This is same as:
A larger anion has a greater tendency to donate electrons
■ So we can write:
If the anion in the ‘ionic bond under consideration’ is larger, it is a favourable condition for inducing ‘partial covalent character’
(C) Charge on the cation
(i) The charge of the cation may be +1, +2 or +3
(ii) The increase in charge indicates that, the remaining electrons are held more tightly than in the parent atom
(iii) Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
■ So we can write:
If the cation in the ‘ionic bond under consideration’ has a greater charge, it is a favourable condition for inducing ‘partial covalent character’
(D) Type of the cation
(i) We know that:
• Metallic elements lose electrons and become cations
• Non-metallic elements gain electrons and become anions
(ii) Here we consider the formation of cations
• We know that cations are formed from metallic elements
♦ The various metallic elements are:
✰ Alkali metals and Alkaline earth metals (s-block)
✰ Transition metals (d-block)
(iii) The s-block elements have the general electronic configuration: [core]ns1-2
• After losing electrons, the general electronic configuration becomes: [core]ns0
• That means, the general electronic configuration of 'cations formed from s-block elements' is: [core]ns0
■ But [core]ns0 is the general configuration of the nearest noble gases
• That means, 'cations formed from s-block elements' has the configuration of the nearest noble gas
• An example:
♦ The electronic configuration of Ca is [Ar]4s2
♦ The electronic configuration of Ca+2 is [Ar]
(iv) The transition elements have the general electronic configuration: [core](n-1)d1-10ns1-2
• After losing electrons, the general electronic configuration becomes: [core](n-1)d1-10ns0
• That means, the general electronic configuration of 'cations formed from s-block elements' is [core](n-1)d1-10ns0
(v) So we have two types of cations:
• Cations with general configuration: [core]ns0
♦ These are formed from s-block elements
• Cations with general configuration: [core](n-1)d1-10ns0
♦ These are formed from transition elements
(vi) We are given two cations:
• Suppose that:
♦ We are given two cations, one from each of the above two types
• Also suppose that:
♦ They have the same size
♦ They have the same positive charges
(vii) We want to compare the two cations and find:
Which cation can better 'pull the electrons of the anion'?
The following steps (viii), (ix) and (x) give the answer:
(viii) Consider the two configurations:
♦ [core]ns0
♦ [core](n-1)d1-10ns0
• We see that, in the 'cations from transition metals', there is an extra term: (n-1)d1-10
• That means there are electrons in the d orbitals
(ix) These 'electrons in the d orbitals' have lesser shielding effect
• So the outer electrons are attracted more strongly towards the nucleus
• Thus we can write:
In the 'cations from transition metals', the outer electrons are attracted more strongly towards the nucleus
• Naturally, other electrons in the vicinity will also be attracted towards the nucleus of the cation
• That means, the 'extra electrons' of the anion will be 'pulled'
(x) So we can write the conclusion in steps:
• We want to compare two cations
• They both have the same size and same charge
• In such a situation, the 'cations from transition metals' have greater ability for inducing ‘partial covalent character’
Let us write a summary of the above four points A, B, C and D:
(B) If the anion in the ‘ionic bond under consideration’ is larger, it is a favourable condition for inducing ‘partial covalent character’
(C) If the cation in the ‘ionic bond under consideration’ has a greater charge, it is a favourable condition for inducing ‘partial covalent character’
(D) When comparing two cations with the same size and charge, the 'cations from transition metals' have greater ability for inducing ‘partial covalent character’
■ The above four rules were formulated by the Polish scientist Kazimierz Fajans
• They are known as Fajans' rules
Percent covalent character
• In some ionic bonds, the 'partial covalent character' will be small
♦ In such ionic bonds, we say:
✰ The percent covalent character is small
• In some ionic bonds, the 'partial covalent character' will be large
♦ In such ionic bonds, we say:
✰ The percent covalent character is large
■ Based on the discussions so far in this section, we can write:
• The percent covalent character in an ionic compound depends on 3 factors:
1. The 'polarising capacity' of the cation
♦ 'Polarising capacity' is the ability of the cation to pull the extra electrons of the anion
✰ Some cations have large polarising capacity
✰ Some cations have small polarising capacity
2. The 'polarisability' of the anion
♦ 'Polarisability' is the capacity of the anion to keep the extra electrons
✰ Some anions will allow the cation to pull the electrons by a large distance
✰ Some anions will allow the cation to pull the electrons only by a small distance
3. The 'extent of distortion' suffered by the anion
• 'Extent of distortion' is the distance by which the extra electrons of the anion are pulled
♦ Some anions suffer only a small extent of distortion
♦ Some anions suffer a large extent of distortion
• This can be explained using fig.4.79 below. The explanation can be written in 3 steps:
(i) Cations and anions:
♦ A+ and B+ are two cations
♦ C- is an anion
(ii) Formation of molecules:
♦ The first molecule is formed by A+ and C-
✰ This is shown in fig.4.79(a)
♦ The second molecule is formed by B+ and C-
✰ This is shown in fig.4.79(b)
(iii) Extent of distortion:
♦ In both molecules, the anion is the same C-
✰ Under the influence of A+, this anion suffers only a small distortion
✰ Under the influence of B+, this anion suffers a large distortion
■ We must remember that, the 'distortion' is actually, a shift in the 'electron cloud'. We cannot say that, discrete electron particles are pulled by the cation
♦ In such ionic bonds, we say:
✰ The percent covalent character is small
• In some ionic bonds, the 'partial covalent character' will be large
♦ In such ionic bonds, we say:
✰ The percent covalent character is large
■ Based on the discussions so far in this section, we can write:
• The percent covalent character in an ionic compound depends on 3 factors:
1. The 'polarising capacity' of the cation
♦ 'Polarising capacity' is the ability of the cation to pull the extra electrons of the anion
✰ Some cations have large polarising capacity
✰ Some cations have small polarising capacity
2. The 'polarisability' of the anion
♦ 'Polarisability' is the capacity of the anion to keep the extra electrons
✰ Some anions will allow the cation to pull the electrons by a large distance
✰ Some anions will allow the cation to pull the electrons only by a small distance
3. The 'extent of distortion' suffered by the anion
• 'Extent of distortion' is the distance by which the extra electrons of the anion are pulled
♦ Some anions suffer only a small extent of distortion
♦ Some anions suffer a large extent of distortion
• This can be explained using fig.4.79 below. The explanation can be written in 3 steps:
 |
| Fig.4.79 |
♦ A+ and B+ are two cations
♦ C- is an anion
(ii) Formation of molecules:
♦ The first molecule is formed by A+ and C-
✰ This is shown in fig.4.79(a)
♦ The second molecule is formed by B+ and C-
✰ This is shown in fig.4.79(b)
(iii) Extent of distortion:
♦ In both molecules, the anion is the same C-
✰ Under the influence of A+, this anion suffers only a small distortion
✰ Under the influence of B+, this anion suffers a large distortion
■ We must remember that, the 'distortion' is actually, a shift in the 'electron cloud'. We cannot say that, discrete electron particles are pulled by the cation
• Let us see some practical applications of Fajans' rules:
Example 1:
1. Consider the following two molecules:
♦ LiBr (Lithium bromide)
♦ KBr (Potassium bromide)
• We want to compare the 'percent covalent character' of the two molecules
2. Cations and Anions in each molecule:
♦ LiBr is formed from the cation Li+ and anion Br-
♦ KBr is formed from the cation K+ and anion Br-
• In both the molecules, the anion is the same Br-
• The comparison is to made between the cations
3. So we apply Fajans' first rule:
(i) The size of the given cations can be written in increasing order as:
Li+ < K+
(Recall the 'periodic trend in ionic size along a group')
(ii) According to Fajans' first rule, when the size of the cation decreases, the condition becomes more favourable for covalent character
• So we get:
♦ Li+ will give the most covalent character
♦ K+ will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
KBr < LiBr
• That is., among the given two molecules,
♦ LiBr will have the highest 'percent covalent character'
Example 2:
1. Consider the following molecules:
♦ LiI (Lithium iodide)
♦ LiBr (Lithium bromide)
♦ LiCl (Lithium chloride)
♦ LiF (Lithium fluoride)
• We want to compare the 'percent covalent character' of the above molecules
2. Cations and Anions in each molecule:
♦ LiI is formed from the cation Li+ and anion I-
♦ LiBr is formed from the cation Li+ and anion Br-
♦ LiCl is formed from the cation Li+ and anion Cl-
♦ LiF is formed from the cation Li+ and anion F-
• In all the molecules, the cation is the same Li+
• The comparison is to made between the anions
3. So we apply Fajans' second rule:
(i) The size of the given anions can be written in increasing order as:
F- < Cl- < Br-< I-
(Recall the 'periodic trend in ionic size along a group')
(ii) According to Fajans' second rule, when the size of the anion increases, the condition becomes more favourable for covalent character
• So we get:
♦ I will give the most covalent character
♦ F will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
LiF < LiCl < LiBr < LiI
• That is., among the given molecules,
♦ LiF will have the lowest 'percent covalent character'
♦ LiI will have the highest 'percent covalent character'
Example 3:
1. Consider the following molecules:
♦ NaCl (Sodium chloride)
♦ MgCl2 (Magnesium chloride)
♦ AlCl3 (Aluminium chloride)
• We want to compare the 'percent covalent character' of the above molecules
2. Cations and Anions in each molecule:
♦ NaCl is formed from the cation Na+ and anion Cl-
♦ MgCl2 is formed from the cation Mg2+ and two numbers of anion Cl-
♦ AlCl3 is formed from the cation Al3+ and three numbers of anion Cl-
• In all the molecules, the anion is the same Cl-
• The comparison is to made between the cations
■ Four important points to remember:
(i) Number of bonds in each molecule:
• In NaCl, one Na+ ion enters into an 'ionic bond' with one Cl- ion
♦ So we need to consider one bond only
• In MgCl2, one Mg2+ ion enters into an 'ionic bond' with each of the two Cl- ions
♦ So we need to consider two bonds
• In AlCl3, one Al3+ ion enters into an 'ionic bond' with each of the three Cl- ions
♦ So we need to consider three bonds
(ii) Analyzing each bond in the molecules:
• In NaCl, we need to consider one bond only. It is the bond between Na+ and Cl-
♦ So we need to analyse the interaction between one Na+ and one Cl-
• In MgCl2, we need to consider two bonds. But both are the same: the bond between Mg2+ and Cl-
♦ So we need to analyse the interaction between one Mg2+ and one Cl-
• In AlCl3, we need to consider three bonds. But all three are the same: the bond between Al3+ and Cl-
♦ So we need to analyse the interaction between one Al3+ and one Cl-
(iii) So in short, all we need to do is this:
♦ Analyse the ionic bond between one Na+ and one Cl-
♦ Analyse the ionic bond between one Mg2+ and one Cl-
♦ Analyse the ionic bond between one Al3+ and one Cl-
(iv) Comparing anions or cations?
• In all the molecules, the anion is the same Li+
• The comparison is to made between the cations
♦ The 'cations to be compared' have different charges
3. So we apply Fajans' third rule:
(i) The charge of the given cations can be written in increasing order as:
Na+ < Mg2+ < Al3+
(ii) According to Fajans' third rule, when the charge of the cation increases, the condition becomes more favourable for covalent character
• So we get:
♦ Al3+ will give the most covalent character
♦ Na+ will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
NaCl < MgCl2 < AlCl3
• That is., Among the given molecules,
♦ NaCl will have the lowest 'percent covalent character'
♦ AlCl3 will have the highest 'percent covalent character'
Example 4:
1. Consider the following two molecules:
♦ KCl (Potassium chloride)
♦ AgCl (Silver chloride)
• We want to compare the 'percent covalent character' of the two molecules
2. Cations and Anions in each molecule:
♦ KCl is formed from the cation K+ and anion Cl-
♦ AgCl is formed from the cation Ag+ and anion Cl-
• In both the molecules, the anion is the same Cl-
• The comparison is to made between the cations
♦ The two cations have approximately the same size
♦ The two cations have the same charge
3. So we apply Fajans' fourth rule:
(i) Electronic configurations:
♦ The electronic configuration of K+ is [Ar]
♦ The electronic configuration of Ag+ is [Kr]4d10
(ii) Effect in transition metals:
• Ag is a transition metal
♦ The Ag+ have inner d-orbitals
♦ These d-orbitals will reduce the shielding effect
• So the outer electrons are attracted more strongly in Ag+ than in K+
• The 'extra electron in the anion' will also be attracted more strongly by Ag+ than K+
4. Fajans' fourth rule states that:
• When comparing two cations with the same size and charge, the 'cations from transition metals' have greater ability for inducing ‘partial covalent character’
• So we get:
♦ Ag+ will give the most covalent character
♦ K+ will give the least covalent character
• So we can write:
Arrangement of molecules in the increasing order of covalent character is:
KCl < AgCl
• That is., among the given two molecules,
♦ AgCl will have the highest 'percent covalent character'
Example 1:
1. Consider the following two molecules:
♦ LiBr (Lithium bromide)
♦ KBr (Potassium bromide)
• We want to compare the 'percent covalent character' of the two molecules
2. Cations and Anions in each molecule:
♦ LiBr is formed from the cation Li+ and anion Br-
♦ KBr is formed from the cation K+ and anion Br-
• In both the molecules, the anion is the same Br-
• The comparison is to made between the cations
3. So we apply Fajans' first rule:
(i) The size of the given cations can be written in increasing order as:
Li+ < K+
(Recall the 'periodic trend in ionic size along a group')
(ii) According to Fajans' first rule, when the size of the cation decreases, the condition becomes more favourable for covalent character
• So we get:
♦ Li+ will give the most covalent character
♦ K+ will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
KBr < LiBr
• That is., among the given two molecules,
♦ LiBr will have the highest 'percent covalent character'
Example 2:
1. Consider the following molecules:
♦ LiI (Lithium iodide)
♦ LiBr (Lithium bromide)
♦ LiCl (Lithium chloride)
♦ LiF (Lithium fluoride)
• We want to compare the 'percent covalent character' of the above molecules
2. Cations and Anions in each molecule:
♦ LiI is formed from the cation Li+ and anion I-
♦ LiBr is formed from the cation Li+ and anion Br-
♦ LiCl is formed from the cation Li+ and anion Cl-
♦ LiF is formed from the cation Li+ and anion F-
• In all the molecules, the cation is the same Li+
• The comparison is to made between the anions
3. So we apply Fajans' second rule:
(i) The size of the given anions can be written in increasing order as:
F- < Cl- < Br-< I-
(Recall the 'periodic trend in ionic size along a group')
(ii) According to Fajans' second rule, when the size of the anion increases, the condition becomes more favourable for covalent character
• So we get:
♦ I will give the most covalent character
♦ F will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
LiF < LiCl < LiBr < LiI
• That is., among the given molecules,
♦ LiF will have the lowest 'percent covalent character'
♦ LiI will have the highest 'percent covalent character'
Example 3:
1. Consider the following molecules:
♦ NaCl (Sodium chloride)
♦ MgCl2 (Magnesium chloride)
♦ AlCl3 (Aluminium chloride)
• We want to compare the 'percent covalent character' of the above molecules
2. Cations and Anions in each molecule:
♦ NaCl is formed from the cation Na+ and anion Cl-
♦ MgCl2 is formed from the cation Mg2+ and two numbers of anion Cl-
♦ AlCl3 is formed from the cation Al3+ and three numbers of anion Cl-
• In all the molecules, the anion is the same Cl-
• The comparison is to made between the cations
■ Four important points to remember:
(i) Number of bonds in each molecule:
• In NaCl, one Na+ ion enters into an 'ionic bond' with one Cl- ion
♦ So we need to consider one bond only
• In MgCl2, one Mg2+ ion enters into an 'ionic bond' with each of the two Cl- ions
♦ So we need to consider two bonds
• In AlCl3, one Al3+ ion enters into an 'ionic bond' with each of the three Cl- ions
♦ So we need to consider three bonds
(ii) Analyzing each bond in the molecules:
• In NaCl, we need to consider one bond only. It is the bond between Na+ and Cl-
♦ So we need to analyse the interaction between one Na+ and one Cl-
• In MgCl2, we need to consider two bonds. But both are the same: the bond between Mg2+ and Cl-
♦ So we need to analyse the interaction between one Mg2+ and one Cl-
• In AlCl3, we need to consider three bonds. But all three are the same: the bond between Al3+ and Cl-
♦ So we need to analyse the interaction between one Al3+ and one Cl-
(iii) So in short, all we need to do is this:
♦ Analyse the ionic bond between one Na+ and one Cl-
♦ Analyse the ionic bond between one Mg2+ and one Cl-
♦ Analyse the ionic bond between one Al3+ and one Cl-
(iv) Comparing anions or cations?
• In all the molecules, the anion is the same Li+
• The comparison is to made between the cations
♦ The 'cations to be compared' have different charges
3. So we apply Fajans' third rule:
(i) The charge of the given cations can be written in increasing order as:
Na+ < Mg2+ < Al3+
(ii) According to Fajans' third rule, when the charge of the cation increases, the condition becomes more favourable for covalent character
• So we get:
♦ Al3+ will give the most covalent character
♦ Na+ will give the least covalent character
(iii) So we can write:
• Arrangement of molecules in the increasing order of covalent character is:
NaCl < MgCl2 < AlCl3
• That is., Among the given molecules,
♦ NaCl will have the lowest 'percent covalent character'
♦ AlCl3 will have the highest 'percent covalent character'
Example 4:
1. Consider the following two molecules:
♦ KCl (Potassium chloride)
♦ AgCl (Silver chloride)
• We want to compare the 'percent covalent character' of the two molecules
2. Cations and Anions in each molecule:
♦ KCl is formed from the cation K+ and anion Cl-
♦ AgCl is formed from the cation Ag+ and anion Cl-
• In both the molecules, the anion is the same Cl-
• The comparison is to made between the cations
♦ The two cations have approximately the same size
♦ The two cations have the same charge
3. So we apply Fajans' fourth rule:
(i) Electronic configurations:
♦ The electronic configuration of K+ is [Ar]
♦ The electronic configuration of Ag+ is [Kr]4d10
(ii) Effect in transition metals:
• Ag is a transition metal
♦ The Ag+ have inner d-orbitals
♦ These d-orbitals will reduce the shielding effect
• So the outer electrons are attracted more strongly in Ag+ than in K+
• The 'extra electron in the anion' will also be attracted more strongly by Ag+ than K+
4. Fajans' fourth rule states that:
• When comparing two cations with the same size and charge, the 'cations from transition metals' have greater ability for inducing ‘partial covalent character’
• So we get:
♦ Ag+ will give the most covalent character
♦ K+ will give the least covalent character
• So we can write:
Arrangement of molecules in the increasing order of covalent character is:
KCl < AgCl
• That is., among the given two molecules,
♦ AgCl will have the highest 'percent covalent character'
In the next section, we will see VSEPR theory
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