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 4.20 - Square Pyramidal Shape

We are discussing the basics of VSEPR theory related to molecules with lone pairs. In the previous section 4.19, we completed the discussion on Case III (a) and (b). In this section, we will see Case IV

Case IV: AB₅E
We will write it in steps:
1. In AB₅E, there are 5 terminal atoms and 1 electron pair
• So there is a total of 6 items
• These 6 items are distributed around A
2. Suppose that, all the 6 items are 'sticks'
• Then there will be 6 sticks around A
• We have already seen such a case in the previous sections
• It is the molecule AB₆. It has 6 sticks around A
3. When there are 6 sticks around A, the shape is octahedral
• This is shown in the fig.4.118(a) below:

Fig.4.118

4. But actually in our present case, we do not have 6 sticks

• We have only 5 sticks and 1 lone pair
• So from fig.4.118(a), we remove 1 stick
• In the place of that 'removed stick', we put a lone pair
• This is shown in fig.4.118(b)
    ♦ The stick corresponding to B_ii is removed
    ♦ That 'removed stick' is indicated by a dashed blue line
          ✰ The blue dashed line help us to remember the position of the lone pair
    ♦ B_ii is written inside dashed circle because, that 'B atom' is not actually present in fig.b
5. Now there is a problem
• In the octahedral structure that we saw in AB₆, we know the values of various angles:
    ♦ ∠B_iAB_iii  = ∠B_iiAB_iii = ∠B_iAB_iv = ∠B_iiAB_iv = ∠B_iAB_v = ∠B_iiAB_v =∠B_iAB_vi = ∠B_iiAB_vi = 90^o
          ✰ These are the angles between axial bonds and equatorial bonds
    ♦ ∠B_iiiAB_iv  = ∠B_ivAB_v = ∠B_vAB_vi = ∠B_viAB_iii = 90^o
          ✰ These are the angles between equatorial bonds
■ There are a total of 12 angles. Do we need to mention all these angles even after removing B_ii?
6. In the previous cases, we saw that:
We need not write any angles in which, one side is a blue dashed line
• So in our present case, we need to mention the following 8 angles only:
    ♦ ∠B_iAB_iii  = ∠B_iAB_iv = ∠B_iAB_v =∠B_iAB_vi = 90^o
          ✰ These are the angles between axial bonds and equatorial bonds
          ✰ One of them is shown in orange color in fig.4.118(b)
    ♦ ∠B_iiiAB_iv  = ∠B_ivAB_v = ∠B_vAB_vi = ∠B_viAB_iii = 90^o
          ✰ These are the angles between equatorial bonds
          ✰ One of them is shown in violet color in fig.4.118(b)
7. There is yet another problem:
• We had earlier obtained the 90^o as follows:
The 6 sticks try to push each other as far away as possible, and they settle down with angles of '90^o' between them
• But here there are no '6 sticks'. There are only '5 sticks' and 1 lone pair
    ♦ However, since they are all electrons, they will all be pushing each other
    ♦ But the pushing (repulsion) are different
■ Would the 'angles mentioned in (6)' change, due to the removal of B_ii?
8. We know that:
(lp-lp repulsion) > (lp-bp repulsion) > (bp-bp repulsion)
• Based on this, we can think about the angles. It can be written in (vi) steps:
(i) The 4 equatorial sticks in fig.4.118(b), tries to maintain an angle of 90^o between them
(ii) They try to maintain this angle, by using the bp-bp repulsion
    ♦ This repulsion is indicated by the red double headed arrow in fig.4.119(a) below:

Fig.4.119

(iii) But the sticks are acted upon by other repulsive forces also
    ♦ There are a total of 12 repulsive forces. So there will be 12 double headed arrows
          ✰ 4 of them lie on a horizontal plane (the equatorial plane of the bipyramid)
          ✰ The remaining 8 lie on various vertical planes
(iv) First we will see the 4 double headed arrows in the horizontal plane
They are shown in fig.4.119(a)
• The bp-bp repulsion is indicated by a red double headed arrow
    ♦ There are 4 such arrows
• We see an interesting situation here:
    ♦ All arrows are red. There are no yellow or cyan arrows
          ✰ So magnitudes of the repulsion are the same
    ♦ All the reds are acting symmetrically on each other
          ✰ So they will cancel each other
    ♦ Thus, the violet 90^o will not change
• Also, since the red arrows in fig.4.116(a) are horizontal, they have no effect on the orange 90^o
• So the forces in fig.4.119(a) have no effect on either orange or violet 90^o angles
(v) Next we will see the 8 double headed arrows in the vertical planes. They are shown in fig.4.119(b)
• We see that:
    ♦ All the arrows above the equator are red
          ✰ They are acting symmetrically
    ♦ All the arrows below the equator are yellow
          ✰ They are acting symmetrically
• We know that yellow is stronger than red
    ♦ So the reds will be compressed
    ♦ So the orange 90^o will decrease
(vi) So we can write:
The orange angles must be 90^o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 90^o
    ♦ This is shown in fig.4.119(c)
    ♦ This fig.4.119(c) shows the final shape
9. Now we can write about the final shape
(i) The lone pairs influence the shape of the atoms but they are invisible
(ii) So the final shape is determined by the positions of the atoms
(iii) In the fig.4.119(c), we have four 'B atoms' and one 'A atom'
• Together, they resemble the 'square pyramid'
    ♦ B_iii, B_iv, B_v and B_vi are at the four corners of the square base
    ♦ The four magenta dashed lines help us to visualize the square base
    ♦ Bi is at the apex
■ So we call it: Square pyramidal shape
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.119(c)
10. Some facts about the angle:
• In a perfect square pyramid, the orange angle will be exactly 90^o
• In fig.119(c), the orange angle is only 'slightly less than 90^o'
    ♦ So it almost looks like a square pyramid
• In real life AB₅E molecules, the orange angle is indeed 'slightly less than 90^o'
• If the angle is far less than 90^o, the shape will be very different from a square pyramid
• Fig.4.120 below, shows a comparison:

Fig.4.120

• In both figs. (a) and (b), B_iii, B_iv, B_v and B_vi lies on the corners of a square. But:
    ♦ In fig.(a), the central atom A lies in the same plane of the square
    ♦ In fig.(b), the central atom A lies below the plane of the square
• It is interesting to note that, in fig.b, the violet angle is still exactly 90^o
11. Note that, a 'square pyramid' is a 3D structure. It can be explained based on fig.4.121(a) below:

Fig.4.121

(i) We know that, a plane can be made to pass through any three given points in space (Details here)
    ♦ In our present case, the three points are: B_i, B_iv and A
    ♦ Bvi also lies in that same plane
(ii) So B_iii and B_v will be 'out of plane'
    ♦ B_iii will be infront of the plane
    ♦ B_v will be behind the plane
(iii) In the fig.121(a), the plane is given a bit of transparency so that, B_v also becomes visible
■ So the 2D representation of the square pyramidal shape will be as shown in fig.4.121(b) above 
12. The actual value of the angle:
• In the general case, we write that:
    ♦ The axial angle will be less than 90^o
    ♦ The equatorial angle will be 90^o
    ♦ We do not write the exact value of the axial angle
• This is because, in real life situations, the angle varies from molecule to molecule
    ♦ For example:
          ✰ In BrF₅, the axial angle is 84.8^o
• This is shown in fig.4.121(c) above

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We have completed the discussion on all the four cases. Now we can try to answer the two questions that we saw at the beginning of the discussion on 'molecules with long pairs'. We saw them in section 4.17. We will write them again:
(i) The smallest cases mentioned above (case I with two B atoms) are AB₂E and AB₂E₂
Why is it that, there are no smaller cases like ABE and ABE₂?
(ii) The largest case mentioned above (case IV with five B atoms) is AB₅E
Why is it that, there is no larger cases like AB₅E₂?
• We will now see the answers:
ABE:
1. There are 2 items around the central atom A
• So we consider AB₂
• Then the basic shape will be 'linear'
2. From that linear shape, we remove 1 stick and put a lone pair in it's place
3. Now, there is one B atom and one lone pair
    ♦ But the lone pairs are invisible
    ♦ So we consider only one A and one B
4. Obviously, the shape will be linear
    ♦ Because, we can draw only a straight line between two points
5. So there is not much to discuss about ABE

ABE₂:
1. There are 3 items around the central atom A
• So we consider AB₃
• Then the basic shape will be 'triangular planar'. We saw them in section 4.17
2. From that shape, we remove 2 sticks and put lone pairs in their place
3. Now, there is one B atom and two lone pairs
    ♦ But the lone pairs are invisible
    ♦ So we consider only one A and one B
4. Obviously, the shape will be linear
    ♦ Because, we can draw only a straight line between two points
5. So there is not much to discuss about ABE₂

AB₅E₂:
• There are 7 items around the central atom A
• So we consider AB₇
• Then we need a 'basic shape' for 7 sticks around the central atom A
• But such a basic shape is not available
• So we need not discuss AB₅E₂ at present

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• We have seen how the VSEPR theory is applied in the various situations
• In the next section, we will see the official definition

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Chapter 4.19 - See Saw and Square Planar shapes

We are discussing the basics of VSEPR theory related to molecules with lone pairs. In the previous section 4.18, we completed the discussion on Case II (a) and (b). In this section, we will see Case III (a) and (b)

Case III (a): AB₄E
We will write it in steps:
1. In AB₄E, there are 4 terminal atoms and 1 electron pair
• So there is a total of 5 items
• These 5 items are distributed around A
2. Suppose that, all the 5 items are 'sticks'
• Then there will be 5 sticks around A
• We have already seen such a case in the previous sections
• It is the molecule AB₅. It has 5 sticks around A
3. When there are 5 sticks around A, the shape is trigonal bipyramidal
• This is shown in the fig.4.111(a) below:

Fig.4.111

4. But actually in our present case, we do not have 5 sticks

• We have only 3 sticks and 1 lone pair
• So from fig.4.111(a), we remove 1 stick
• In the place of that 'removed stick', we put a lone pair
• This is shown in fig.4.111(b)
    ♦ The stick corresponding to B_iv is removed
    ♦ That 'removed stick' is indicated by a dashed blue line
          ✰ The blue dashed line help us to remember the position of the lone pair
    ♦ B_iv is written inside dashed circle because, that 'B atom' is not actually present in fig.b
5. Now there is a problem
• In the trigonal bipyramidal structure that we saw in AB₅, we know the values of various angles:
    ♦ ∠B_iAB_iii  = ∠B_iiAB_iii = ∠B_iAB_iv = ∠B_iiAB_iv = ∠B_iAB_v = ∠B_iiAB_v = 90^o
          ✰ These are the angles between axial bonds and equatorial bonds
          ✰ Two of them is shown in orange color in fig.4.111(b)
    ♦ ∠B_iiiAB_iv  = ∠B_ivAB_v = ∠B_iiiAB_v = 120^o
          ✰ These are the angles between equatorial bonds
          ✰ One of them is shown in violet color in fig.4.111(b)
■ There are a total of nine angles. Do we need to mention all these angles even after removing B_iv?
6. In the previous three sections, we saw that:
We need not write any angles in which, one side is a blue dashed line
• So in our present case, we need to mention the following five angles only:
    ♦ ∠B_iAB_iii  =∠B_iiAB_iii = ∠B_iAB_v = ∠B_iiAB_v = 90^o
          ✰ These are the angles between axial bonds and equatorial bonds
    ♦ ∠B_iiiAB_v = 120^o (This angle lies in the equatorial plane)
          ✰ This is an angle between two equatorial bonds
7. There is yet another problem:
• We had earlier obtained the 90^o and 120^o as follows:
The 5 sticks try to push each other as far away as possible, and they settle down with angles of '120^o' and '90^o' between them
• But here, there are no '5 sticks'. There are only '4 sticks' and 1 lone pair
    ♦ However, since they are all electrons, they will all be pushing each other
    ♦ But the pushing (repulsion) are different
■ Would the 'angles mentioned in (6)', change due to the removal of B_iv?
8. We know that:
(lp-lp repulsion) > (lp-bp repulsion) > (bp-bp repulsion)
• Based on this, we can think about the angles. It can be written in (v) steps:
(i) The two equatorial sticks in fig.4.111(b), tries to maintain an angle of 120^o between them
(ii) They try to maintain this angle, by using the bp-bp repulsion
    ♦ This repulsion is indicated by the red double headed arrow in fig.4.112(a) below:

Fig.4.112

(iii) But the sticks are acted upon by other repulsive forces also
    ♦ There are a total of 9 repulsive forces. So there will be 9 double headed arrows
          ✰ 3 of them lie on a horizontal plane (the equatorial plane of the bipyramid)
          ✰ The remaining 6 lie on various vertical planes
(iv) First we will see the 3 double headed arrows in the horizontal plane
They are shown in fig.4.112(a)
• The lp-bp repulsion is indicated by yellow double headed arrows
    ♦ There are two yellow arrows
• The bp-bp repulsion is indicated by a red double headed arrow
    ♦ There is only one such arrow
• We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
    ♦ The yellow arrows are exerting a push on red arrow
    ♦ We know that yellow is stronger than red
          ✰ So the red will get compressed
          ✰ As a result, the angle will become less than 120^o
(v) So we can write:
The ∠B_iiiAB_v must be 120^o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 120^o
    ♦ This is shown in fig.4.112(c)
    ♦ This fig.4.112(c) shows the final shape
(vi) Next we will see the 6 double headed arrows in the vertical planes
They are shown in fig.4.112(b)
• The bp-bp repulsion is indicated by red double headed arrows
    ♦ There are four red arrows
          ✰ Readers are advised to recognize each of the red arrows individually
          ✰ Some of them are partially obstructed from view
          ✰ Even then, they must be fully recognized 
• The lp-bp repulsion is indicated by yellow double headed arrows
    ♦ There are two such arrows
(vii) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
    ♦ The top ends of the 'two red arrows above the equator' are pushed by the yellow arrow above equator
          ✰ We know that, red is weaker than yellow
          ✰ So the two red arrows at top will be compressed
          ✰ As a result, the angle (shown in orange color) will become less than 90^o
    ♦ The bottom ends of the 'two red arrows below the equator' are pushed by the yellow arrow below equator
          ✰ We know that, red is weaker than yellow
          ✰ So the two red arrows at bottom will be compressed
          ✰ As a result, the angle (shown in orange color) will become less than 90^o
(vii) So we can write:
The ∠B_iAB_ii must be (90+90) = 180^o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 180^o
    ♦ This is shown in orange color in fig.4.112(c)
    ♦ This fig.4.112(c) shows the final shape
9. Now we can write about the final shape
(i) The lone pairs influence the shape of the atoms but they are invisible
(ii) So the final shape is determined by the positions of the atoms
(iii) In the fig.4.112(c), we have four 'B atoms' and one 'A atom'
• Together, they resemble the 'seesaw' at children's parks and playgrounds. So we call it: Seesaw-shape
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.112(c)
• The 'seesaw motion' of the molecule can be seen in the animation in fig.4.113 below. Note that, we have to place the molecule on the table in such a way that, B_iii-A-B_v forms the pivot:

Fig.4.113

10. Note that, a 'seesaw shape' is a 3D structure. it can be explained based on fig.4.114(a) below:

Fig.4.114

(i) We know that, a plane can be made to pass through any three given points in space (Details here)
    ♦ In our present case, the three points are: B_i, B_ii and A
(ii) So B_iii and B_v will be 'out of plane'
    ♦ B_iii will be infront of the plane
    ♦ B_v will be behind the plane
(iii) In the fig.114(a), the plane is given a bit of transparency so that, B_v also becomes visible
■ So the 2D representation of the seesaw shape will be as shown in fig.4.114(b) above 
11. The actual value of the angle:
• In the general case, we write that:
    ♦ The axial angle will be less than 180^o
    ♦ The equatorial angle will be less than 120^o
    ♦ We do not write the exact value
• This is because, in real life situations, the angle varies from molecule to molecule
    ♦ For example:
          ✰ In SF₄, the angles are 173^o and 102^o
• This is shown in fig.4.114(c) above

---

Next we consider Case III (b): AB₄E₂
We will write it in steps:
1. In AB₄E₂, there are 4 terminal atoms and 2 electron pairs
• So there is a total of 6 items
• These 6 items are distributed around A
2. Suppose that, all the 6 items are 'sticks'
• Then there will be 6 sticks around A
• We have already seen such a case in the previous sections
• It is the molecule AB₆. It has 6 sticks around A
3. When there are 6 sticks around A, the shape is octahedral
• This is shown in the fig.4.115(a) below:

Fig.4.115

4. But actually in our present case, we do not have 6 sticks
• We have only 4 sticks and 2 lone pairs
• So from fig.4.115(a), we remove 2 sticks
• In the place of those 'removed sticks', we put lone pairs
• This is shown in fig.4.115(b)
    ♦ The sticks corresponding to B_i and B_ii are removed
    ♦ Those 'removed sticks' are indicated by dashed blue lines
          ✰ The blue dashed lines help us to remember the positions of the lone pairs
    ♦ B_i and B_ii are written inside dashed circles because, those 'B atoms' are not actually present in fig.b
5. Now there is a problem
• In the octahedral structure that we saw in AB₆, we know the values of various angles:
    ♦ ∠B_iAB_iii  = ∠B_iiAB_iii = ∠B_iAB_iv = ∠B_iiAB_iv = ∠B_iAB_v = ∠B_iiAB_v =∠B_iAB_vi = ∠B_iiAB_vi = 90^o
          ✰ These are the angles between axial bonds and equatorial bonds
          ✰ One of them is shown in orange color in fig.4.115(b)
    ♦ ∠B_iiiAB_iv  = ∠B_ivAB_v = ∠B_vAB_vi = ∠B_viAB_iii = 90^o
          ✰ These are the angles between equatorial bonds
          ✰ One of them is shown in violet color in fig.4.115(b)
■ There are a total of 12 angles. Do we need to mention all these angles even after removing B_i and B_ii?
6. In the previous cases, we saw that:
We need not write any angles in which, one side is a blue dashed line
• So in our present case, we need to mention the following 4 angles only:
    ♦ ∠B_iiiAB_iv  = ∠B_ivAB_v = ∠B_vAB_vi = ∠B_viAB_iii = 90^o
          ✰ These are the angles between equatorial bonds
7. There is yet another problem:
• We had earlier obtained the 90^o as follows:
The 6 sticks try to push each other as far away as possible, and they settle down with angles of '90^o' between them
• But here there are no '6 sticks'. There are only '4 sticks' and 2 lone pairs
    ♦ However, since they are all electrons, they will all be pushing each other
    ♦ But the pushing (repulsion) are different
■ Would the 'angles mentioned in (6)' change, due to the removal of B_i and B_ii?
8. We know that:
(lp-lp repulsion) > (lp-bp repulsion) > (bp-bp repulsion)
• Based on this, we can think about the angles. It can be written in (v) steps:
(i) The 4 equatorial sticks in fig.4.115(b), tries to maintain an angle of 90^o between them
(ii) They try to maintain this angle, by using the bp-bp repulsion
    ♦ This repulsion is indicated by the red double headed arrow in fig.4.116(a) below:

Fig.4.116

(iii) But the sticks are acted upon by other repulsive forces also
    ♦ There are a total of 12 repulsive forces. So there will be 12 double headed arrows
          ✰ 4 of them lie on a horizontal plane (the equatorial plane of the octahedron)
          ✰ The remaining 8 lie on various vertical planes
(iv) First we will see the 4 double headed arrows in the horizontal plane
They are shown in fig.4.116(a)
• The bp-bp repulsion is indicated by a red double headed arrow
    ♦ There are 4 such arrows
• We see an interesting situation here:
    ♦ All arrows are red. There are no yellow or cyan arrows
          ✰ So magnitudes of the repulsion are the same
    ♦ All the reds are acting symmetrically on each other
          ✰ So they will cancel each other
    ♦ Thus, the violet 90^o will not change
• Also, since the red arrows in fig.4.116(a) are horizontal, they have no effect on the orange 90^o
• So the forces in fig.4.116(a) have no effect on either orange or violet 90^o angles
(v) Next we will see the 8 double headed arrows in the vertical planes
They are shown in fig.4.116(b)
• The lp-bp repulsion is indicated by yellow double headed arrows
    ♦ There are 8 such arrows
(vii) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
    ♦ All arrows are yellow. There are no red or cyan arrows
          ✰ So magnitudes of the repulsion are the same
    ♦ All the yellows are acting symmetrically on each other
          ✰ So they will cancel each other
    ♦ Thus, the orange 90^o will not change
• Also, since the yellow arrows in fig.4.116(b) are vertical, they have no effect on the violet 90^o
• So the forces in fig.4.116(b) have no effect on either orange or violet 90 angles
(vii) So we can write:
In AB₄E₂, the basic angles of the octahedron are preserved. All the bond angles are 90^o
9. Now we can write about the final shape
(i) The lone pairs influence the shape of the atoms but they are invisible
(ii) So the final shape is determined by the positions of the atoms
(iii) In the fig.4.116(c), we have four 'B atoms' and one 'A atom'
• The four 'B atoms' are the four corners of a square
    ♦ This is clear from the magenta dashed lines
• So the shape is: Square planar
    ♦ The equatorial bonds form the diagonals of the square
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.116(c)
10. In this case we need not discuss much about whether the shape is 2D or 3D
• We know that, a 'square' is a planar shape. It is 2D
11. The actual value of the angle:
• Since it is a perfect square, the angles will always be 90^o in all the molecules of the type AB₄E₂
• An example is shown in fig.4.117 below:

Fig.4.117

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• In the next section, we will see case IV

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