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:
4. But actually in our present case, we do not have 5 sticks
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 |
• 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 Biv 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
♦ Biv 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ Two of them is shown in orange color in fig.4.111(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BiiiABv = 120o
✰ 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 Biv?
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:
♦ ∠BiABiii =∠BiiABiii = ∠BiABv = ∠BiiABv = 90o
✰ These are the angles between axial bonds and equatorial bonds
♦ ∠BiiiABv = 120o (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 90o and 120o as follows:
The 5 sticks try to push each other as far away as possible, and they settle down with angles of '120o' and '90o' 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 Biv?
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 120o 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:
(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 120o
(v) So we can write:
The ∠BiiiABv must be 120o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 120o
♦ 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 90o
♦ 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 90o
(vii) So we can write:
The ∠BiABii must be (90+90) = 180o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 180o
♦ 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, Biii-A-Bv forms the pivot:
10. Note that, a 'seesaw shape' is a 3D structure. it can be explained based on fig.4.114(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, Bii and A
(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.114(a), the plane is given a bit of transparency so that, Bv 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 180o
♦ The equatorial angle will be less than 120o
♦ 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 173o and 102o
• This is shown in fig.4.114(c) above
• 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 Biv 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
♦ Biv 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ Two of them is shown in orange color in fig.4.111(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BiiiABv = 120o
✰ 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 Biv?
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:
♦ ∠BiABiii =∠BiiABiii = ∠BiABv = ∠BiiABv = 90o
✰ These are the angles between axial bonds and equatorial bonds
♦ ∠BiiiABv = 120o (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 90o and 120o as follows:
The 5 sticks try to push each other as far away as possible, and they settle down with angles of '120o' and '90o' 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 Biv?
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 120o 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 |
♦ 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 120o
(v) So we can write:
The ∠BiiiABv must be 120o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 120o
♦ 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 90o
♦ 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 90o
(vii) So we can write:
The ∠BiABii must be (90+90) = 180o in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 180o
♦ 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, Biii-A-Bv forms the pivot:
 |
| Fig.4.113 |
 |
| Fig.4.114 |
♦ In our present case, the three points are: Bi, Bii and A
(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.114(a), the plane is given a bit of transparency so that, Bv 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 180o
♦ The equatorial angle will be less than 120o
♦ 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 173o and 102o
• 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:
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 Bi and Bii 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
♦ Bi and Bii 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv =∠BiABvi = ∠BiiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.115(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 Bi and 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 4 angles only:
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ These are the angles between equatorial bonds
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 '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 Bi and 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 (v) steps:
(i) The 4 equatorial sticks in fig.4.115(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.4.116(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 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 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.4.116(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.4.116(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 90o will not change
• Also, since the yellow arrows in fig.4.116(b) are vertical, they have no effect on the violet 90o
• 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 90o
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 90o in all the molecules of the type AB4E2
• An example is shown in fig.4.117 below:
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 |
• 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 Bi and Bii 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
♦ Bi and Bii 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:
♦ ∠BiABiii = ∠BiiABiii = ∠BiABiv = ∠BiiABiv = ∠BiABv = ∠BiiABv =∠BiABvi = ∠BiiABvi = 90o
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.115(b)
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ 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 Bi and 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 4 angles only:
♦ ∠BiiiABiv = ∠BivABv = ∠BvABvi = ∠BviABiii = 90o
✰ These are the angles between equatorial bonds
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 '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 Bi and 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 (v) steps:
(i) The 4 equatorial sticks in fig.4.115(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.4.116(a) below:
 |
| Fig.4.116 |
♦ 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 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.4.116(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.4.116(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 90o will not change
• Also, since the yellow arrows in fig.4.116(b) are vertical, they have no effect on the violet 90o
• 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 90o
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 90o in all the molecules of the type AB4E2
• An example is shown in fig.4.117 below:
 |
| Fig.4.117 |
• In the next section, we will see case IV
No comments:
Post a Comment