Chapter 4.23 - Orbital Overlap Concept

In the previous section 4.22, we saw the formation of the bond between two H atoms. Based on that, we will continue our discussion on VB theory

Orbital overlap concept

1. We have seen that, the H₂ molecule is formed when two H atoms come close to each other such that, the distance between their nuclei is 74 pm
2. We know that in hydrogen, the nucleus is situated at the center of the 1s orbital
• That is.,
    ♦ The single electron of H creates a cloud
    ♦ This cloud is spherical in shape
    ♦ We call this sphere as the ‘1s orbital’ of H
    ♦ The nucleus of H is situated at the center of the 1s sphere
3. The electron inside the 1s orbital can be represented using an arrow
• This is shown in fig.4.129(a) below
    ♦ The 1s orbital is shown in orange color
          ✰ It is given a bit of transparency so that, the nucleus is also visible
          ✰ The nucleus is shown as a small red sphere

4. When the distance is 74 pm, the two 1s orbitals penetrates into each other
• This is shown in fig.4.129(b) above 
• This inter penetration is called overlapping of orbitals
• When the overlapping occurs, the density of the cloud will be greater in the region between the two nuclei
• Both the electrons will be present in that overlapping region
5. Thus both the electrons will be attracted to both the nuclei
• This results in the bonding between the two H atoms

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• In this way, different orbitals can overlap to form bonds

• Let us see how CH₄ (methane) is formed. It can be written in steps:

1. We know the Lewis dot structure of CH₄. It is shown in the fig.4.130(a) below:
• The central atom is C. It forms single covalent bonds with four H atoms

Fig.4.130

2. The electronic configuration of C is 1s²2s²2p²
• We want the electrons in the valence shell. They are the electrons which take part in the chemical reactions. So we want 2s² and 2p²
• They are shown in fig.4.130(b)
3. The fig.b shows the energy levels also
• We see that:
    ♦ The two electrons in the 2p orbitals have a greater energy
    ♦ The two electrons in the 2s orbitals have a lesser energy
4. We have four H atoms ready to combine with the C atom
• We know that, an H atom has only one orbital: the 1s orbital  
• Let us connect the various orbitals of C with the 1s orbital of H atoms:
    ♦ The 1s orbital of the first H atom overlap with the p_x orbital of C
    ♦ The 1s orbital of the second H atom overlap with the p_y orbital of C
5. Now we have a problem. It can be written in 3 steps:
(i) The 2s orbital is completely filled. Because, there are two electrons in it
(Recall Pauli's exclusion principle)
(ii) So the 1s of the third H atom cannot overlap with the 2s of C
(iii) Also we see no place for the fourth H atom at all
6. This problem can be solved in the following way. It involves 2 steps:
(i) Give enough energy so that one electron in the 2s orbital of C gets excited and jumps to the 2p_z orbital
(ii) Now there are four half filled orbitals. This is shown in fig.c
    ♦ The 1s orbital of the first H atom overlap with the p_x orbital of C
    ♦ The 1s orbital of the second H atom overlap with the p_y orbital of C
    ♦ The 1s orbital of the third H atom overlap with the p_z orbital of C
    ♦ The 1s orbital of the fourth H atom overlap with the 2s orbital of C
7. Based on the information in (6) above, let us try to make a model. It can be written in 3 steps:
(i) Fig.4.131(a) below shows the three 2p orbitals of C
• Each of them contains one electron

Fig.4.131

(ii) The point of intersection of the three axes is the position of the 'nucleus of the C atom'
• We know that the center of the 2s sphere will be at the center of the atom
• So the 2s and 2p orbitals can be shown together as in fig.b
• The 2s also contains one electron
(iii) Now we attach the 1s orbital of a H atom to each of the half filled orbitals of C
• This is shown in fig.c
• The four H atoms are numbered as H_i, H_ii, H_iii, H_iv
• Note that, 'electron pairs' are shown at the overlapping regions of the various orbitals

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• The fig.4.131(c) appears to be a satisfactory model of CH₄. But there are four drawbacks

• They are discussed below:
Drawback 1 
• This can be written in 3 steps:
1. From the fig.4.131(c), we directly get the following  bond angles:
∠H_iCH_ii = 90^o, ∠H_iCH_iii = 90^o, ∠H_iiCH_iii = 90^o
2. These angles are readily available because, the p orbitals lie along the three coordinate axes
3. But the above 90^o values are not acceptable because, we know that in CH₄, all the bond angles are equal to 109.5^o

Drawback 2
• This can be written in 3 steps:

1. The 90^o angles obtained above, involves H_i, H_ii and H_iii only
• We want the angles related to H_iv also
2. The angles related to H_iv can be calculated using 3 steps:
(i) Imagine a line joining the centers of the 'H_iv orange' and 'C cyan' spheres
(ii) Write the angles which this line makes with CH_i, CH_ii and CH_iii
(iii) Thus we will get ∠H_iCH_iv, ∠H_iiCH_iv and ∠H_iiiCH_iv
3. But there is a problem. It can be written in steps:
(i) The 'H_iv orange' and 'C cyan' are both spheres
    ♦ They can overlap from any direction
    ♦ There are infinite number of possible directions
    ♦ Three of them are shown in fig.4.132 below:

Fig.4.132

(ii) So there are infinite number of possible values for the angles related to H_iv
(iii) This is not acceptable because, we know that, all bond angles in CH₄ are equal to 109.5

Drawback 3
• This can be written in 3 steps:
1. Let us consider the bond lengths
• We can easily see that, lengths of the following three bond are equal:
    ♦ Bond between H_i and C
    ♦ Bond between H_ii and C
    ♦ Bond between H_iii and C
• But the bond length between H_iv and C will be different
2. That means:
    ♦ Three out of the four bonds in CH₄ are of the same length
    ♦ The fourth bond is of a different length
3. This is not acceptable because experiments show that, all bond lengths in CH₄ are equal

Drawback 4
• This can be written in 5 steps:
1. Let us consider the bond energies
• We have:
Energy of the electron in 2p_x
= Energy of the electron in 2p_y
= Energy of the electron in 2p_z
≠ Energy of the electron in 2s
2. So we can write:
    ♦ The electrons of H_i, H_ii and H_iii will be combining with electrons of the same energy
    ♦ The electrons of H_iv will be combining with an electron of a different energy
3. That means:
• Energies of the following three bond are equal:
    ♦ Bond between H_i and C
    ♦ Bond between H_ii and C
    ♦ Bond between H_iii and C
• But the energy of the bond between H_iv and C will be different
4. That means:
    ♦ Three out of the four bonds in CH₄ have the same energy
    ♦ The fourth bond has a different energy
5. This is not acceptable because experiments show that, all bond lengths in CH₄ have the same energy

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■ Because of the above four drawbacks, the model shown in fig.4.131(c) is not acceptable
• In the next section, we will see how scientists solved this problem

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Chapter 4.21 - Postulates in the VSEPR Theory

We are discussing the basics of VSEPR theory related to molecules with lone pairs. In the previous section 4.20, we completed the discussion on the last Case IV. In this section, we will see the official definition and the various features of the theory

• The VSEPR theory was first put forward by Sidwick and Powell in 1940
• The theory helps us to predict the shapes of covalent molecules
• The procedure is very simple. It is based on the interactions between electron pairs in the valence shell of atoms
• Let us first explain what this 'interactions between electron pairs in the valence shell of atoms' is. It can be written in 7 steps:
1. There are so many electrons in an atom. We need not consider all of them
• We consider only those ‘electrons in the valence shells of the atoms’
• We know that, ‘electrons in the valence shell of an atom’ are called valence electrons of that atom
• We already know the method to find the 'number of valence electrons' of any atom  
2. Consider an atom A in a molecule
• Since A is part of a molecule, it would be bonded with other atom/atoms
• Consider the valence electrons of A
• Those valence electrons can be classified into two categories:
(i) Bonded electrons
(ii) Non-bonded electrons
Let us see some examples:
Example 1:
• Fig.4.122(a) below, shows the Lewis dot structure of BH₃
• We see 3 red dots around B
• They are the valence electrons of Boron (Recall that, we show only the valence electrons in Lewis dot structures)
• All those 3 electrons have entered into bonding
■ Electrons which have 'entered into bonding' are called bonded electrons

Fig.4.122

Example 2:
• Fig.4.122(b) shows the Lewis dot structure of NH₃
• We see 5 red dots around N
• They are the valence electrons of Nitrogen
• Out of the five valence electrons, three have entered into bonding
    ♦ Two electrons have not entered into any bonding
■ Electrons which have 'not entered into bonding' are called non-bonded electrons
3. Now we know 'bonded electrons' and 'non-bonded electrons'. The next two items that we have to see are:
    ♦ Bonded electron pair
    ♦ Non-bonded electron pair
They can be easily explained:
(i) Bonded electron pair:
• Consider any 'bonded electron'. It will be always present at one end of a '—'
(Recall that '—' indicates a bond)
• In a '—', there will be two electrons, one at each end
    ♦ Both those electrons are 'bonded electrons'
• Since there are two electrons in a bond, we can call it 'a pair of electrons'
    ♦ To be precise: 'a pair of bonded electrons'
■ So we can write:
    ♦ Whenever we see a '—', we are looking at a Bonded electron pair
    ♦ In 'ball and stick models', we represent the '—' using sticks
          ✰ So a 'stick' represent a bonded electron pair
(ii) Non-bonded electron pair:
• We have seen what 'non-bonded electrons' are
    ♦ Such electrons are always seen in pairs
■ So we can write:
    ♦ Non-bonded electrons are always seen in 'groups of two'
    ♦ Such a group is called a: Non-bonded electron pair
          ✰ A non-bonded electron pair is also called a: lone pair
4. Next we want to see what is meant by 'interactions'
We can easily guess. There will be 4 types of interactions:
(i) A 'bonded electron pair' will interact with any other 'bonded electron pair' which is close by
(ii) A 'bonded electron pair' will interact with any other 'non-bonded electron pair' which is close by
(iii) A 'non-bonded electron pair' will interact with any other 'bonded electron pair' which is close by
(iv) A 'non-bonded electron pair' will interact with any other 'non-bonded electron pair' which is close by
But (iii) is same as (ii). So we can write:
■ There will be 3 types of interactions:
(i) A 'bonded electron pair' will interact with any other 'bonded electron pair' which is close by
(ii) A 'bonded electron pair' will interact with any other 'non-bonded electron pair' which is close by
(iii) A 'non-bonded electron pair' will interact with any other 'non-bonded electron pair' which is close by
5. But why do they interact?
• The answer is simple: They interact because they are all negatively charged
• We know that, like charges repel each other
(We can expect a condition of 'no interaction', only if the particles are 'charge less'. If two charged particles are brought close to each other, interaction will definitely occur)   
6. So we can write:
The interactions are all repulsive
7. Note that, the interactions are between 'pairs'
We do not consider the interactions between individual electrons

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• Now we have a basic idea about what the 'interactions between electron pairs in the valence shell of atoms' is

• The VSEPR theory is based on these interactions
• Now we will see the VIII postulates of the theory and their explanations
I. The shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom
Explanation of this postulate can be written in 2 steps:
(i) First we have to count the number of electron pairs (in the valence shell of the central atom)
    ♦ We have to count all the bonded pairs
    ♦ We have to count all the lone pairs 
    ♦ Then we take the sum
(ii) The shape of the molecule will depend on this sum
Application of this postulate can be written in 2 steps:
(i) We have already seen the application. Let us see two examples: (a) and (b)
(a) AB₄
    ♦ Number of bonded pairs = 4
    ♦ Number of lone pairs = 0
    ♦ Sum = (4+0) = 4
• So there are 4 items around A. The basic shape will be tetrahedral
(b) AB₃E
    ♦ Number of bonded pairs = 3
    ♦ Number of lone pairs = 1
    ♦ Sum = (3+1) = 3
• So there are 4 items around A. The basic shape will be tetrahedral
(ii) The 'sum' which is the 'total number of items around A' helps us to decide the basic shape

II Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
Explanation of this postulate:
This postulate do not need much explanation. We obviously know that, there will be repulsion between like charges. So the pairs repel one another
Application of this postulate:
This postulate has tremendous application. The various shapes attained by the various molecules is a consequence of these repulsions

III These pairs of electrons tend to occupy such positions in space that minimize repulsion and thus maximize distance between them
Explanation of this postulate can be written in 3 steps:
(i) Consider some particles which experience repulsion between each other
    ♦ Naturally, each particle will try to push the other particles ‘as far away as possible’
(ii) ‘As far away as possible’ can be achieved only by ‘increased distances’ between the particles
    ♦ When the distance increases, the particles begin to experience lesser repulsion from each other
(iii) So the particles will try to achieve ‘maximum possible distances’ from each other
Application of this postulate can be written in 2 steps:
(i) Tendency to achieve ‘maximum possible distances’ will obviously influence the final positions of the atoms
(ii) Final positions of the atoms will give the shape of the molecules
(iii) In the final positions, the atoms will be 'as far way from each other as possible' 

IV The valence shell is taken as a sphere with the electron pairs localizing on the spherical surface at maximum distance from one another
Explanation of this postulate can be written in 4 steps:
(i) We have seen that, in the final positions, the terminal atoms will be 'as far way from each other as possible'
(ii) All terminal atoms will be equidistant from the central atom A
(iii) This is similar to the definition of a sphere:
    ♦ The sphere will have a center point
    ♦ All points on the surface of the sphere will be equidistant from the center
(iv) In the case of a molecule:
    ♦ The central atom is the center of the sphere
    ♦ All terminal atoms will lie on the surface of a sphere
    ♦ This is because, the terminal atoms are equidistant from the central atom
    ♦ The surface of the sphere can be considered as the valence shell of the central atom
Application of this postulate:
• The application can be explained with the help of an example. It can be written in 3 steps:
(i) Consider a molecule with the tetrahedral shape
(ii) The center of the tetrahedron will be the center of a sphere
(iv) The four terminal atoms will lie exactly on the surface of that sphere

V A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair
Explanation of this postulate can be written in 4 steps:
(i) The central atom is bonded to a number of terminal atoms
• The bonds may be single, double or triple
(ii) In the VSEPR theory:
    ♦ A double bond is considered as a single bond
    ♦ A triple bond is also considered as a single bond
(iii) So we can write:
• In the VSEPR theory,
    ♦ A '=' is considered as a '—'
    ♦ A '≡' is also considered as a '—'
(iv) But what about the pairs?
A '=' will contain two pairs. How can we consider them as '—' ?
A '≡' will contain three pairs. How can we consider them as '—' ?
■ The solution is that:
• The ‘two pairs’ in a '=' is considered as a single pair
    ♦ Not just an ‘ordinary single pair’
    ♦ But a ‘single super pair’
• The ‘three pairs’ in a '≡' is also considered as a single pair
    ♦ Not just an ‘ordinary single pair’
    ♦ But a ‘single super pair’
Application of this postulate:
• The application can be explained with the help of an example. It can be written in 2 steps:
(i) In CO₂, there are two double bonds 
(ii) But while applying the VSEPR theory, we considered them as single bonds
(See fig.4.82 of section 4.13)

VI Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure
Explanation of this postulate can be written in 3 steps:
(i) We know that, some molecules can be represented by two or more resonance structures (Details here)
(ii) The VSEPR theory is applicable to any of those structures
(iii) We will get the same result because, VSEPR theory treats double and triple bonds as single bonds
Application of this postulate:
• The application can be explained with the help of an example. It can be written in 3 steps:
(i) The resonance structures of CO₂ can be seen in fig.4.51 in section 4.9
(ii) In all the structures, C is the central atom and the two O are the terminal atoms
(iii) We can apply the theory to any one of those structures. The result will be a linear shape
(See fig.4.82 of section 4.13)

VII The repulsive interactions between electron pairs decrease in the order:
(lp-lp) > (lp-bp) > (bp-bp)
• That means:
    ♦ The force of repulsion between two lone pairs will be the greatest
    ♦ The force of repulsion between two bond pairs will be the least
    ♦ The force of repulsion between a lone pair and a bond pair will have an intermediate value
The explanation for this postulate was given by Nyholm and Gillespie in 1957. It can be written in 3 steps:
(i) The bond pairs have a definite space within the molecule
    ♦ This is because, a bond pair will be belonging to two atoms
    ♦ So that bond pair will lie on the line between the two owner atoms
(ii) But a lone pair belongs to only one atom (the central atom)
    ♦ So they are spread out into a greater space than bond pairs
    ♦ In other words, the lone pairs occupy a greater space than bond pairs
(iii) So the lone pairs are able to apply a greater ‘push’ on others
    ♦ That is the reason why we get: (lp-lp) > (lp-bp) > (bp-bp)
Application of this postulate:
• The application can be explained with the help of an example. It can be written in 3 steps:
(i) We used the following double headed arrows:
    ♦ Cyan double headed arrow to indicate lp-lp repulsion
    ♦ Yellow double headed arrow to indicate lp-bp repulsion
    ♦ Red double headed arrow to indicate bp-bp repulsion
(ii) We know that:
    ♦ Cyan is stronger than yellow
    ♦ Yellow is stronger than red
(iii) Now consider the shape of H₂O molecule
(See fig.4.103 in section 4.17)
• If all the arrows in the fig.4.103(a) are of the same color, there will be a perfect symmetry
    ♦ The angle will not become lesser than 109.5
• That means, using the same colored arrows, we will not be able to explain the lesser angle in a water molecule
• That is., if all repulsions are considered to be of the same magnitude, we will not be able to explain the lesser angle in a water molecule
(iv) In this way, this postulate has tremendous applications in explaining the shape of various molecules
• It explains why the actual shape deviates from the expected shape
    ♦ For example, in the case of water:
          ✰ It explains why the actual angle is less than the expected value of 109.5^o
VIII For predicting the shape of a molecule using VSEPR theory, it is convenient to divide the molecules into two categories:
(i) Molecules in which central atom has no lone pair
(ii) Molecules in which central atom has one or more lone pairs
Explanation of this postulate:
• Whenever we apply the VSEPR theory, we have to first look whether the central atom has lone pairs or not
• Method of application will be different for the two categories:
    ♦ Molecules in which central atom has no lone pair
    ♦ Molecules in which central atom has one or more lone pairs
Application of this postulate:
• We have already seen the application of this postulate
• We have applied the theory separately for the two categories
(i) Molecules in which central atom has no lone pair
    ♦ were discussed in sections 4.13, 4.14, 4.15 and 4.16
(ii) Molecules in which central atom has one or more lone pairs
    ♦ were discussed in sections 4.17, 4.18, 4.19 and 4.20

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We have seen the VIII postulates. Now we will see some solved examples

Solved example 4.7  
Discuss the shape of the following molecules using the VSEPR model:
BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃
Solution:
(i) The Lewis dot structure of BeCl₂ is shown in fig.4.123(a) below:

Fig.4.123

• The central atom Be has no lone pairs. So it is of the type AB₂
• Molecules of the type AB₂ have a linear structure (See fig.4.82 in section 4.13)
• So BeCl₂ is linear
(ii) The Lewis dot structure of BCl₃ is shown in fig.4.123(b) above
• The central atom B has no lone pairs. So it is of the type AB₃
• Molecules of the type AB₃ have a trigonal planar structure (See fig.4.84 in section 4.13)
• So BCl₃ is trigonal planar
(iii) The Lewis dot structure of SiCl₄ is shown in fig.4.124(a) below:

Fig.4.124

• The central atom Si has no lone pairs. So it is of the type AB₄
• Molecules of the type AB₄ have a tetrahedral structure (See fig.4.82 in section 4.14)
• So SiCl₄ is tetrahedral
(iv) The Lewis dot structure of AsF₅ is shown in fig.4.124(b) above
• The central atom As has no lone pairs. So it is of the type AB₅
• Molecules of the type AB₅ have a trigonal bipyramidal structure
• So AsF₅ is trigonal bipyramidal
(v) The Lewis dot structure of H₂S is shown in fig.4.125(a) below:

Fig.4.125

• The central atom S has two lone pairs. So it is of the type AB₂E₂
• Molecules of the type AB₂E₂ have a bent shape
• So H₂S is of bent shape
(vi) The Lewis dot structure of PH₃ is shown in fig.4.125(b) above
• The central atom P has one lone pair. So it is of the type AB₃E
• Molecules of the type AB₃E have a bent shape
• So PH₃ is of trigonal pyramidal shape

Solved example 4.8
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss
Solution:
The solution can be written in 5 steps:
1. Shapes of both H₂O and NH₃ are obtained from the basic shape: Tetrahedron
2. In the tetrahedron, the angle is 109.5^o. But:
    ♦ In NH₃, the angle is 107^o
    ♦ In H₂O, the angle is 104.5^o
• We want to know why the angles are different from 109.5^o
• We also want to know why the angle in H₂O is lesser than that in NH₃
3. Diagrams that we used in our discussions:
    ♦ We obtained the structure of H₂O using the fig.4.103 in section 4.17
    ♦ We obtained the structure of NH₃ using the fig.4.106 in section 4.18
4. In fig.106, we see that, there is one lone pair in N atom
    ♦ This gives rise to lp-bp repulsions
• The lp-bp repulsion (yellow) is stronger than bp-bp repulsion (red)
    ♦ So the red arrows will get compressed
■ Thus the basic angle of 109.5^o will decrease to 107^o in NH₃ 
5. In fig.103, we see that, there are two lone pairs in H atom
• This gives rise to lp-lp repulsions (in addition to lp-bp repulsions)
• The lp-lp repulsion (cyan) is stronger than lp-bp repulsion (yellow)
• The lp-bp repulsion (yellow) is stronger than bp-bp repulsion (red)
    ♦ So the cyan will compress the yellows
    ♦ The yellows will further compress the red
          ✰ As a result, the red is compressed more
■ So the angle reduces to 104.5^o in H₂O

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• In the next section, we will see valence bond theory

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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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