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 4.6 - Lattice Enthalpy

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 (Δ_egH)
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

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■ However, there is one exception:

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

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• 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'

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• Now we know what a crystal is

• 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

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• 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:

Fig.4.31

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

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Based on the above discussion, we can arrive at an important conclusion. It can be written in 7 steps:

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

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Now we are in a position to define Lattice enthalpy

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

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• From the above definitions, we get the impression that, the lattice enthalpy depends only on the attractive and repulsive forces between the ions

• 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

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Now we will see some solved examples

Solved example 4.2
Write Lewis dot symbols for the following atoms and ions:
S and S^2- ; Al and Al³⁺ ; H and H^-
Solution:
The required Lewis dot symbols are shown in fig.4.32 below:

Fig.4.32

Part (a):
S and S^2- :
(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 S^2- 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

(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 S^2- ion
• This is shown in fig.4.33(b)
(iv) The two K⁺ ions get attached to the S^2- 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

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:

Fig.4.35

(ii) The O atom is in need of two electrons. It accepts the two electrons lost by Ca atom
• The O atom thus becomes O^2- ion
• This is shown in fig.4.35(b)
(iv) The Ca²⁺ ion get attached to the O^2- 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

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:

Fig.4.37

(ii) The N atom is in need of three electrons. It accepts the three electrons lost by Al atom
• The N atom thus becomes N^3- ion
• This is shown in fig.4.37(b)
(iv) The Al³⁺ ion get attached to the N^3- 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

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In the next section, we will see bond length, bond angle and bond enthalpy

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