We are discussing the basics of VSEPR theory. In the previous section, we saw three examples
♦ Example 1 was related to the type AB
♦ Example 2 was related to the type AB2
♦ Example 3 was related to the type AB3
• In this section, we will see the type AB4
Example 4:
1. Consider the general form of molecule: AB4
• There is one atom of element A and four atoms of element B
2. We want to know the shape of the molecule AB4
• Let us assume that, the bonds are ‘covalent’
• A ball-and-stick model of AB4 is shown in fig.4.85(a) below
♦ A is the central atom
♦ The four B atoms are distributed around A:
3. So we have 5 atoms
• Let us assume that, all the five atoms lie on the same plane
• So whatever be the positions of the atoms in AB4, the ‘molecule AB4 as a whole’ will be planar
• It will be a 2D structure
4. The VSEPR theory tells us that:
Each stick will be trying to push the other sticks as far away as possible
• In our present case, there are four sticks
♦ All the four sticks push each other with equal forces
♦ When this happens, will the sticks settle down with an angle of 90o between them?
♦ The 90o is shown in fig.4.85(b) above
5. Let us try the calculations that we did in the previous cases
For our present case, the calculation will be as follows:
(i) In AB4, there are four sticks
(ii) The central angle of a circle is 360o as shown in fig.4.85(c) above
(iii) We divide this 360o by the 'number of sticks'
♦ We get: 360⁄4 = 90
(iv) This 90o is the angle between the bonds in fig.(b)
6. All sticks appear to be happy. But this is not the case. The sticks are not at all happy
■ They are not happy because, an 'alternate arrangement' is possible
• In that 'alternate arrangement', the angles between the sticks will be greater than 90o
• Naturally, the sticks will prefer that 'alternate arrangement'
• Let us see the details of that 'alternate arrangement'. It can be written in (viii) steps:
(i) We have five atoms in AB4
♦ A is the central atom. The four B atoms are the terminal atoms
(ii) In fig.4.85(b), all the five atoms lie on the same plane
• From that arrangement in fig.4.85(b), two terminal atoms will be kicked out of the plane
[Remember that, a plane can be made to pass through any three given points (Details here)]
(iii) So the molecule is no longer planar
• The molecule no longer has a 2D structure
• When 2 atoms are 'out of plane', the molecule will have a 3D structure
• It is shown in fig.4.86(a) below:
(iv) The four terminal atoms are named as Bi, Bii, Biii and Biv
• We see that, the central atom A and two terminal atoms Bi and Bii lie on a plane
• Biii and Biv are kicked out
♦ Biii is now at the front of the plane
♦ Biv is now at the rear of the plane
• While drawing in 2D, the plane in fig.4.86(a), represents the 'plane of paper'
♦ In the fig., it is given a bit of transparency
♦ In this way, the atom Biv at the rear will also become visible
(v) The whole model is placed on a wooden platform
• We see that, the bottom three terminal atoms form a perfect base
(vi) Let us join the four terminal atoms with 'imaginary lines'
• Since they are imaginary lines, we will use dashed lines
♦ They are shown in magenta color in fig.4.86(b) above
• We can draw six magenta dashed lines:
♦ Dashed line joining Bi and Bii
♦ Dashed line joining Bi and Biii
♦ " " " Bi and Biv
♦ " " " Bii and Biii
♦ " " " Bii and Biv
♦ " " " Biii and Biv
(vii) When the six dashed lines are drawn, the shape becomes clear. It is a triangular pyramid
• It has a base and three lateral faces
♦ The base is an equilateral triangle
♦ The lateral faces are also equilateral triangles
• All the four equilateral triangles are exactly identical (congruent)
• Bi is situated at the apex of the pyramid
• Bii, Biii and Biv are situated at the corners of the base
• The central atom A is situated at the 'centroid of the pyramid'
(We have learnt about 'centroid of a triangle' in our math classes. But here, it is 'centroid of the pyramid')
(viii) In the fig.4.86(b), the pyramid is placed on a platform
• We can roll it over in any direction we like
♦ It will always land on a triangular base
✰ Since all the 'B atoms' are identical, the new orientation will be exactly same as the previous
• That means, the structure in fig.4.86(b) is symmetric in all directions
• But there are two major differences:
(i) NH3 molecule is of the type AB3
♦ But here we are discussing the type: AB4
(ii) In NH3, the central atom N is situated at the apex of the pyramid
♦ But in AB4, the central atom A is situated at the 'centroid of the pyramid'
7. Now we know about the 'alternate arrangement'
• The sticks are happy because, the angles have become greater than 90o
■ So what are the angles now?
We can write the answer in (iii) steps:
(i) In fig.4.85(c), ∠BiABiv is marked. It is an angle at A. It is equal to 109.5o
(ii) In this way, we can mark five more angles at A. They are:
∠BiABii, ∠BiABiii, ∠BiiABiii, ∠BiiABiv, and ∠BiiiABiv
• All of them are equal to 109.5o
(iii) This central angle of 109.5o is a property of a tetrahedron. It is called:
Angle between lines joining tetrahedron centroid to vertices
8. Thus we see that, the planar 2D structure in the previous fig.4.85(b) expanded into a 3D structure in fig.4.86 (c)
• In this way, the sticks achieved greater separation between themselves
■ This is called tetrahedral structure
9. Now, there is a problem with 3D structures. We cannot draw them easily in our notebooks and records. So we adopt a different 'drawing method'. We saw it in the case of dipole moment in NH3 molecule (see fig.4.73 of section 4.11)
• Based on that method, our present AB4 molecule can be represented as shown in fig.4.87(a) below:
10. Let us see some real life examples:
• CH4 (Methane) and NH4+ (Ammonium ion) have tetrahedral structures
• They are shown in the fig.4.87 (b) and (c) above
11. We will now write a basic concept that we must all keep in our minds. It can be written in 3 steps:
(i) We want a structure which satisfies the following three conditions:
♦ There are four sticks
♦ All of them start from a common point
♦ Each stick pushes the other three as far away as possible
(ii) There is only one structure in the whole universe that can satisfy all the above three conditions:
♦ It is the tetrahedral structure
(iii) So naturally, the molecules of the form AB4 will prefer that structure
All molecules of the form AB4 will have a tetrahedral structure
• But this is not true
♦ Some AB4 molecules, which have 'lone pairs', are not exactly tetrahedral
♦ We will see them in later sections
♦ Example 1 was related to the type AB
♦ Example 2 was related to the type AB2
♦ Example 3 was related to the type AB3
• In this section, we will see the type AB4
Example 4:
1. Consider the general form of molecule: AB4
• There is one atom of element A and four atoms of element B
2. We want to know the shape of the molecule AB4
• Let us assume that, the bonds are ‘covalent’
• A ball-and-stick model of AB4 is shown in fig.4.85(a) below
♦ A is the central atom
♦ The four B atoms are distributed around A:
 |
| Fig.4.85 |
• Let us assume that, all the five atoms lie on the same plane
• So whatever be the positions of the atoms in AB4, the ‘molecule AB4 as a whole’ will be planar
• It will be a 2D structure
4. The VSEPR theory tells us that:
Each stick will be trying to push the other sticks as far away as possible
• In our present case, there are four sticks
♦ All the four sticks push each other with equal forces
♦ When this happens, will the sticks settle down with an angle of 90o between them?
♦ The 90o is shown in fig.4.85(b) above
5. Let us try the calculations that we did in the previous cases
For our present case, the calculation will be as follows:
(i) In AB4, there are four sticks
(ii) The central angle of a circle is 360o as shown in fig.4.85(c) above
(iii) We divide this 360o by the 'number of sticks'
♦ We get: 360⁄4 = 90
(iv) This 90o is the angle between the bonds in fig.(b)
6. All sticks appear to be happy. But this is not the case. The sticks are not at all happy
■ They are not happy because, an 'alternate arrangement' is possible
• In that 'alternate arrangement', the angles between the sticks will be greater than 90o
• Naturally, the sticks will prefer that 'alternate arrangement'
• Let us see the details of that 'alternate arrangement'. It can be written in (viii) steps:
(i) We have five atoms in AB4
♦ A is the central atom. The four B atoms are the terminal atoms
(ii) In fig.4.85(b), all the five atoms lie on the same plane
• From that arrangement in fig.4.85(b), two terminal atoms will be kicked out of the plane
[Remember that, a plane can be made to pass through any three given points (Details here)]
(iii) So the molecule is no longer planar
• The molecule no longer has a 2D structure
• When 2 atoms are 'out of plane', the molecule will have a 3D structure
• It is shown in fig.4.86(a) below:
 |
| Fig.4.86 |
• We see that, the central atom A and two terminal atoms Bi and Bii lie on a plane
• Biii and Biv are kicked out
♦ Biii is now at the front of the plane
♦ Biv is now at the rear of the plane
• While drawing in 2D, the plane in fig.4.86(a), represents the 'plane of paper'
♦ In the fig., it is given a bit of transparency
♦ In this way, the atom Biv at the rear will also become visible
(v) The whole model is placed on a wooden platform
• We see that, the bottom three terminal atoms form a perfect base
(vi) Let us join the four terminal atoms with 'imaginary lines'
• Since they are imaginary lines, we will use dashed lines
♦ They are shown in magenta color in fig.4.86(b) above
• We can draw six magenta dashed lines:
♦ Dashed line joining Bi and Bii
♦ Dashed line joining Bi and Biii
♦ " " " Bi and Biv
♦ " " " Bii and Biii
♦ " " " Bii and Biv
♦ " " " Biii and Biv
(vii) When the six dashed lines are drawn, the shape becomes clear. It is a triangular pyramid
• It has a base and three lateral faces
♦ The base is an equilateral triangle
♦ The lateral faces are also equilateral triangles
• All the four equilateral triangles are exactly identical (congruent)
• Bi is situated at the apex of the pyramid
• Bii, Biii and Biv are situated at the corners of the base
• The central atom A is situated at the 'centroid of the pyramid'
(We have learnt about 'centroid of a triangle' in our math classes. But here, it is 'centroid of the pyramid')
(viii) In the fig.4.86(b), the pyramid is placed on a platform
• We can roll it over in any direction we like
♦ It will always land on a triangular base
✰ Since all the 'B atoms' are identical, the new orientation will be exactly same as the previous
• That means, the structure in fig.4.86(b) is symmetric in all directions
Note: We have seen triangular prism in an earlier section where we discussed the dipole moment in NH3 molecule (Details here)
(i) NH3 molecule is of the type AB3
♦ But here we are discussing the type: AB4
(ii) In NH3, the central atom N is situated at the apex of the pyramid
♦ But in AB4, the central atom A is situated at the 'centroid of the pyramid'
• The sticks are happy because, the angles have become greater than 90o
■ So what are the angles now?
We can write the answer in (iii) steps:
(i) In fig.4.85(c), ∠BiABiv is marked. It is an angle at A. It is equal to 109.5o
(ii) In this way, we can mark five more angles at A. They are:
∠BiABii, ∠BiABiii, ∠BiiABiii, ∠BiiABiv, and ∠BiiiABiv
• All of them are equal to 109.5o
(iii) This central angle of 109.5o is a property of a tetrahedron. It is called:
Angle between lines joining tetrahedron centroid to vertices
8. Thus we see that, the planar 2D structure in the previous fig.4.85(b) expanded into a 3D structure in fig.4.86 (c)
• In this way, the sticks achieved greater separation between themselves
■ This is called tetrahedral structure
9. Now, there is a problem with 3D structures. We cannot draw them easily in our notebooks and records. So we adopt a different 'drawing method'. We saw it in the case of dipole moment in NH3 molecule (see fig.4.73 of section 4.11)
• Based on that method, our present AB4 molecule can be represented as shown in fig.4.87(a) below:
 |
| Fig.4.87 |
• CH4 (Methane) and NH4+ (Ammonium ion) have tetrahedral structures
• They are shown in the fig.4.87 (b) and (c) above
11. We will now write a basic concept that we must all keep in our minds. It can be written in 3 steps:
(i) We want a structure which satisfies the following three conditions:
♦ There are four sticks
♦ All of them start from a common point
♦ Each stick pushes the other three as far away as possible
(ii) There is only one structure in the whole universe that can satisfy all the above three conditions:
♦ It is the tetrahedral structure
(iii) So naturally, the molecules of the form AB4 will prefer that structure
• Based on the above discussion, we are inclined to think that:
• But this is not true
♦ Some AB4 molecules, which have 'lone pairs', are not exactly tetrahedral
♦ We will see them in later sections
■ At this stage, a doubt may arise in our minds. It can be stated in (ii) steps:
(i) In the case of the previous AB3, we saw that, the structure is trigonal planar
♦ The three sticks in AB3 are happy with the 120o angle
(ii) We know that, trigonal planar is a 2D structure
♦ Is there any 'possibility for increasing the 120o angle' by changing to a 3D structure?
The answer can be written in 5 steps:
1. Consider fig.4.88(a) below:
• It shows a molecule of the form AB3
• The structure is trigonal planar
• The molecule is lying flat on a platform
2. We know that, in fig.4.88(a), the angle between any two sticks is 120o
• But this 120o is obtained when we measure the angles parallel to the platform
3. If we measure the angles perpendicular to the platform, we get:
Angle between any two sticks is 180o
4. In fig.4.88(b), one of the 'B atoms' is trying to rise up
• It is trying to move 'out of plane'
• But when it is 'out of plane', the angle mentioned in (3) will become less than 180o
• According to the VSEPR theory, the sticks will not want such a 'reduction in angle'
5. So we can write:
AB3 will be always planar. It can never become a 3D structure
(i) In the case of the previous AB3, we saw that, the structure is trigonal planar
♦ The three sticks in AB3 are happy with the 120o angle
(ii) We know that, trigonal planar is a 2D structure
♦ Is there any 'possibility for increasing the 120o angle' by changing to a 3D structure?
The answer can be written in 5 steps:
1. Consider fig.4.88(a) below:
 |
| Fig.4.88 |
• The structure is trigonal planar
• The molecule is lying flat on a platform
2. We know that, in fig.4.88(a), the angle between any two sticks is 120o
• But this 120o is obtained when we measure the angles parallel to the platform
3. If we measure the angles perpendicular to the platform, we get:
Angle between any two sticks is 180o
4. In fig.4.88(b), one of the 'B atoms' is trying to rise up
• It is trying to move 'out of plane'
• But when it is 'out of plane', the angle mentioned in (3) will become less than 180o
• According to the VSEPR theory, the sticks will not want such a 'reduction in angle'
5. So we can write:
AB3 will be always planar. It can never become a 3D structure
• Now we continue the discussion on examples which do not have lone pairs
• The next example is of the type: AB5
• We will see it in the next section
• The next example is of the type: AB5
• We will see it in the next section
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