Chapter 9.3 - Chemical Properties of Water

In the previous section, we saw hydrides. In this section, we will see water.

Physical properties of water

This can be written in 5 steps:
1. Water is a colorless liquid. It is also tasteless.
2. We have seen that hydrogen bonding is present between water molecules. This hydrogen bonding is responsible for the unusual properties of water. Let us see some of those properties:
(i) High freezing point of water
• Suppose that, a sample of water and a sample of ethanol are placed side by side.
• If we decrease the surrounding temperature, water will be the first one to freeze.
    ♦ Water will freeze when the surrounding temperature becomes 0 ^oC.
    ♦ We will have to decrease the temperature to a very low value (-114.1 ^oC) to achieve freezing of ethanol.
• So compared to other common liquids, water has a high freezing point. This is because of the hydrogen bonding between water molecules.
• Due to this bonding, the molecules in the liquid water are already associated together. It is easier to freeze those molecules.
• In ethanol, there is no hydrogen bonding. So freezing point is very low.
(ii) High boiling point of water
• Due to the hydrogen bonding, the liquid water molecules are associated together.
• We will have to break all those bonds before we can boil the liquid water.
• For that, large amount of heat has to be supplied. So the boiling point is high.
(iii) High heat of vaporization
• We will need to supply a huge amount of heat in order to change liquid water to water vapor.
• This is due to the hydrogen bonding. The molecules in liquid water are associated together. The high energy is required to break the bonds. Liquid will change to vapour only if those bonds are broken.
• Life on Earth is possible mainly due to this high heat of vaporization of water. Rivers, lakes and oceans are able to retain water because water do not vaporize easily.
(iv) High heat of fusion
• The water in solid state (ice) will not melt easily because, large amount of heat is required for it's fusion. This is due to the hydrogen bonding between water molecules in the solid state.
• Huge amounts of energy is required to break those bonds. That is why we do not see rapid melting of ice caps of mountains even when summer sets in.
3. Some other useful properties of water:
    ♦ Water has high specific heat.
    ♦ It has high thermal conductivity.
    ♦ It has high surface tension.
    ♦ It has high dipole moment.
    ♦ It has high dielectric constant.
• These properties play a major role in sustaining life on Earth.
4. Water can dissolve the nutrients and minerals which are essential for plant and animal life. Once dissolved, they can be transported easily to various parts of the body.
5. Covalent compounds usually do not dissolve in water. But some important covalent compounds like alcohol and carbohydrates dissolve in water. This is because of the polar nature of water molecules. We will see more details about this solubility in later chapters. Solubility of these important covalent compounds plays a major role in easing many of our day to day activities.

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Structure of water

This can be explained in steps:
1. When water is in the gaseous form, there is not much hydrogen bonding between molecules.
• So when water is in gaseous form, we will be able to study individual molecules.
◼ Water molecule has a bent shape. We have already seen the reason for the bent. See fig.4.163 of section 4.28.
2. When water is in liquid state, the molecules are associated together due to hydrogen bonding.
• However, in this state, the molecules have some freedom of movement. The hydrogen bonds will be broken and new bonds will be formed continuously.
3. As the temperature becomes lower and lower, the kinetic energy of water molecules become lower and lower. They begin to move less.
• Now the hydrogen bonding become predominant. The molecules begin to arrange themselves in a hexagonal shape.
• When the water freezes, the molecules get fixed in the hexagonal crystal shape.
• There is lot of free space inside each hexagonal shape (remember that hexagon is a shape with six sides).
4. In the liquid state, there is not much free space between molecules because, there is no hexagonal arrangement.
◼ Since in solid state, there is much space between molecules, ice is lighter than liquid water.
• For more details about ice, see fig.11.5 of section 11.2 in physics lessons.

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Chemical properties of water

• Water enters into chemical reaction with a large number of substances. Let us see some important reactions:
A. Amphoteric nature
• Water is an amphoteric substance.
• That means, it can act as both acid and base.
• We have seen the details when we discussed Brönsted-Lowry Acids and Bases in section 7.12.

B. Redox reactions involving water
This can be written in steps:
1. Consider a metal which is above hydrogen in the reactivity series.
• If that metal is made to react with water, the hydrogen will be displaced from water.
2. The H atoms thus released will combine to form dihydrogen.
• So water is a good source of dihydrogen.
3. Let us see an example:
2H₂O (l) + 2Na (s) → 2NaOH (aq) + H₂ (g)
    ♦ In H₂O, the oxidation number of H is +1
    ♦ In H₂, the oxidation number of H is 0
4. So we can write:
Water can be reduced to dihydrogen by highly electropositive metals.
• But such reactions are explosive. They can be done only in labs with advanced safety equipment.
5. Consider a non-metal which is highly electronegative. For example, F (fluorine).
• If F is made to react with water, oxygen will be released.
2F₂ (g) + 2H₂O (l) → 4H⁺ (aq) + 4F^- (aq) + O₂ (g)
    ♦ In H₂O, the oxidation number of O is -2
    ♦ In O₂, the oxidation number of O is 0
6. So we can write:
Water can be oxidized to dioxygen by highly electronegative elements.
• But such reactions are explosive. They can be done only in labs with advanced safety equipment.

C. Hydrolysis reaction
• We have seen the mechanism of hydrolysis reactions in an earlier section 7.20.
• We will see more advanced details in later sections.

D. Hydrates formation
• We have learned about aqueous solutions. For example, when we dissolve NaCl in water, we get an aqueous solution of NaCl.
• Many salts occur in nature in such aqueous solutions. Such salts can be extracted into solid forms from those solutions.
• When such salts are extracted, they will contain some water molecules also.
• The water molecules can be present in three different forms:
(i) Coordinated water
(ii) Interstitial water
(iii) Hydrogen-bonded water
We will now see each one of them in detail
(i) Coordinated water
This can be explained using an example. It can be written in 5 steps:
1. Consider the ion: [Cu(H₂O)₆]²⁺
• There are six H₂O molecules around the Cu²⁺ ion.
2. Each of the six H₂O molecules is bonded to the central Cu²⁺ through a coordinate bond.
• Recall that in an ordinary covalent bond between two atoms, one electron belongs to one of the two atoms and the other electron belongs to the second atom.
• But in a coordinate bond, both the electrons in the bond belong to one of the two atoms.
3. In our present case, consider any one coordinate bond.
• Both the electrons in that bond will be supplied by the O atom of H₂O.
• The structure is shown in fig.9.3 below:

Fig.9.3

4. The structure shown in the above fig. is a positive ion with a charge of +2. It can combine with two Cl^- negative ions.
• Then the crystal lattice will be composed of [Cu(H₂O)₆]²⁺ ions and Cl^- ions.
    ♦ This is just like the crystal lattice of Na⁺Cl^-.
5. So we can write:
In the crystal lattice formed by [Cu(H₂O)₆]²⁺ ions and Cl^- ions, there are six coordinated water molecules.
(ii) Interstitial water.
This can be explained in 3 steps:
1. A crystal lattice is said to contain interstitial water if:
The H₂O molecules are present in the interstitial spaces of the crystal lattice.
2. One example is: MgSO₄.7H₂O
• The MgSO₄ crystal lattice is formed by Mg²⁺ ions and SO₄^2- ions.
    ♦ The H₂O molecules are present in the interstitial spaces of this crystal lattice.
• Seven H₂O molecules are available for every Mg²⁺ ion and every SO₄^2- ion.
• The H₂O molecules are independent units inside the crystal lattice. They are not chemically bonded to Mg, S or O atoms.   
3. Another example is (CdSO₄)₃.8H₂O
• The CdSO₄ crystal lattice is formed by Cd²⁺ ions and SO₄^2- ions.
    ♦ The H₂O molecules are present in the interstitial spaces of this crystal lattice.
• Eight H₂O molecules are available for every three Cd²⁺ ions and every three SO₄^2- ions.
• The H₂O molecules are independent units inside the crystal lattice. They are not chemically bonded to Cd, S or O atoms.
(iii) Hydrogen-bonded water
This can be explained using an example. It can be written in steps:
1. Consider CuSO₄.5H₂O
• This is obtained from the [Cu(H₂O)₆]²⁺ that we saw in case (i): coordinated water.
2. Out of the six H₂O molecules around the Cu²⁺ ion, two are removed. (one at the top and the other at the bottom)
• So [Cu(H₂O)₆]²⁺ becomes [Cu(H₂O)₄]²⁺
3. Two sulfate ions (SO₄^2-) take those positions previously occupied by H₂O molecules.
4. But the new SO₄^2- ions are not attached by coordinate bonding to the central Cu²⁺ ion.
• They are attached by electrostatic force between the two opposite charges of [Cu(H₂O)₄]²⁺ and SO₄^2-
• This is just like the attraction between Na⁺ and Cl^- in Na⁺Cl^-
• So our present salt can be written as: [Cu(H₂O)₄]²⁺SO₄^2-.
5. But there is one more H₂O molecule. This molecule is attached to the sulphate ion by hydrogen bonding.
• So the final structure will be as shown in fig.9.4 below:

Fig.9.4

6. Arrangement shown in the above fig. is one unit. There will an array of such units in the final crystal lattice.
• We can represent the crystal in two forms:
    ♦ CuSO₄.5H₂O
        ✰ This is just like writing the sodium chloride crystal as NaCl.
    ♦ [Cu(H₂O)₄]²⁺SO₄^2-.H₂O
        ✰ This is just like writing the sodium chloride crystal as Na⁺Cl^-.
• The second one is more accurate because, it gives us the following details:
    ♦ The +ve and -ve ions that make up the crystal lattice.
    ♦ Four H₂O molecules are attached by coordinate bonds.
    ♦ One H₂O molecule is attached by hydrogen bonding.

Solved example 9.2
How many hydrogen bonded water molecule(s) are associated in CuSO₄.5H₂O ?
Solution:
1. CuSO₄.5H₂O is better written as: [Cu(H₂O)₄]²⁺SO₄^2-.H₂O
2. We see that:
• Four water molecules are attached to Cu²⁺ by coordinate bonds.
• There is only one water molecule outside the coordination sphere. This water molecule is attached by hydrogen bond.

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In the next section, we will see hard and soft water.

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Chapter 4.12 - Partial Covalent Character in Ionic Bonds

In the previous section, we completed a discussion on dipole moments. In this section, we will see the 'partial covalent character' of ionic bonds

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]ns^1-2
• 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)d^1-10ns^1-2
• After losing electrons, the general electronic configuration becomes: [core](n-1)d^1-10ns⁰
• That means, the general electronic configuration of 'cations formed from s-block elements' is [core](n-1)d^1-10ns⁰
(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)d^1-10ns⁰
    ♦ 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)d^1-10ns⁰
• We see that, in the 'cations from transition metals', there is an extra term: (n-1)d^1-10
• 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’

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Let us write a summary of the above four points A, B, C and D:

(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’

(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:

Fig.4.79

(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

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• 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'

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In the next section, we will see VSEPR theory

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