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:
4. But actually in our present case, we do not have 6 sticks
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 |
• 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 Bii 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
♦ Bii 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv =∠BiABvi = ∠BiiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 Bii?
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:
♦ ∠BiABiii = ∠BiABiv = ∠BiABv =∠BiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.118(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 90o as follows:
The 6 sticks try to push each other as far away as possible, and they settle down with angles of '90o' 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 Bii?
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 90o 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:
(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 90o will not change
• Also, since the red arrows in fig.4.116(a) are horizontal, they have no effect on the orange 90o
• So the forces in fig.4.119(a) have no effect on either orange or violet 90o 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 90o will decrease
(vi) So we can write:
The orange angles must be 90o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 90o
♦ 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'
♦ Biii, Biv, Bv and Bvi 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 90o
• In fig.119(c), the orange angle is only 'slightly less than 90o'
♦ So it almost looks like a square pyramid
• In real life AB5E molecules, the orange angle is indeed 'slightly less than 90o'
• If the angle is far less than 90o, the shape will be very different from a square pyramid
• Fig.4.120 below, shows a comparison:
• In both figs. (a) and (b), Biii, Biv, Bv and Bvi 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 90o
11. Note that, a 'square pyramid' is a 3D structure. It can be explained based on fig.4.121(a) below:
(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: Bi, Biv and A
♦ Bvi also lies in that same plane
(ii) So Biii and Bv will be 'out of plane'
♦ Biii will be infront of the plane
♦ Bv will be behind the plane
(iii) In the fig.121(a), the plane is given a bit of transparency so that, Bv 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 90o
♦ The equatorial angle will be 90o
♦ 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 84.8o
• This is shown in fig.4.121(c) above
• 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 Bii 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
♦ Bii 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv =∠BiABvi = ∠BiiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 Bii?
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:
♦ ∠BiABiii = ∠BiABiv = ∠BiABv =∠BiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.118(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 90o as follows:
The 6 sticks try to push each other as far away as possible, and they settle down with angles of '90o' 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 Bii?
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 90o 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 |
♦ 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 90o will not change
• Also, since the red arrows in fig.4.116(a) are horizontal, they have no effect on the orange 90o
• So the forces in fig.4.119(a) have no effect on either orange or violet 90o 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 90o will decrease
(vi) So we can write:
The orange angles must be 90o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 90o
♦ 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'
♦ Biii, Biv, Bv and Bvi 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 90o
• In fig.119(c), the orange angle is only 'slightly less than 90o'
♦ So it almost looks like a square pyramid
• In real life AB5E molecules, the orange angle is indeed 'slightly less than 90o'
• If the angle is far less than 90o, the shape will be very different from a square pyramid
• Fig.4.120 below, shows a comparison:
 |
| Fig.4.120 |
♦ 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 90o
11. Note that, a 'square pyramid' is a 3D structure. It can be explained based on fig.4.121(a) below:
 |
| Fig.4.121 |
♦ In our present case, the three points are: Bi, Biv and A
♦ Bvi also lies in that same plane
(ii) So Biii and Bv will be 'out of plane'
♦ Biii will be infront of the plane
♦ Bv will be behind the plane
(iii) In the fig.121(a), the plane is given a bit of transparency so that, Bv 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 90o
♦ The equatorial angle will be 90o
♦ 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 84.8o
• This is shown in fig.4.121(c) above
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
(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
• We have seen how the VSEPR theory is applied in the various situations
• In the next section, we will see the official definition
• In the next section, we will see the official definition
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