We are discussing the basics of VSEPR theory related to molecules with lone pairs. In the previous section 4.17, we completed the discussion on Case I (a) and (b). In this section, we will see Case II
Case II (a): AB3E
We will write it in steps:
1. In AB3E, there are 3 terminal atoms and 1 electron pair
• So there is a total of 4 items
• These 4 items are distributed around A
2. Suppose that, all the 4 items are 'sticks'
• Then there will be 4 sticks around A
• We have already seen such a case in the previous sections
• It is the molecule AB4. It has 4 sticks around A
3. When there are 4 sticks around A, the shape is tetrahedral
• This is shown in the fig.4.105(a) below:
4. But actually in our present case, we do not have 4 sticks
• We have only 3 sticks and 1 lone pair
• So from fig.4.105(a), we remove 1 stick
• In the place of that 'removed stick', we put a lone pair
• This is shown in fig.4.105(b)
♦ The stick corresponding to Bi 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
♦ Bi is written inside dashed circle because, that 'B atom' is not actually present in fig.b
5. Now there is a problem
• In the tetrahedral structure that we saw in AB4, we know that, the angle between any two bonds will be 109.5o
♦ That is., the angle between any two sticks will be 109.5o
• In our present case:
♦ We can write the '109.5o' between the two sticks: BiiA and BiiiA
✰ Because, they are two well defined sticks
♦ We can write the '109.5o' between the two sticks: BiiA and BivA
✰ Because, they are two well defined sticks
♦ We can write the '109.5o' between the two sticks: BiiiA and BivA
✰ Because, they are two well defined sticks
♦ But we cannot write the '109.5o' between the dashed blue line and any stick
■ The reason can be written in (iv) steps:
(i) The 'bonding electrons' are confined inside the stick
• So they occupy a definite space
(ii) But 'lone electrons' are not confined. They freely occupy a region. The region is an electron cloud
• So the dashed blue line indicate a cloud
(iii) We can easily measure an angle from a stick
• Because, a stick is well defined
(iv) But we cannot measure the angle from a cloud
• Because, the electron cloud does not have a well defined boundary
6. So there is no need to specify the 'angles in which one side is a blue dashed line'
■ We need to write only these: The angles between any two sticks
7. There is yet another problem:
• We had earlier obtained the 109.5o as follows:
The 4 sticks try to push each other as far away as possible, and they settle down with angles of '109.5'o between them
• But here there are no '4 sticks'. There are only '3 sticks' and 1 lone pair
♦ However, since they are all electrons, they will all be pushing each other
♦ But the pushing (repulsion) is different
■ We can write:
The following two repulsion are different:
(i) Repulsion between the 2 sticks
(ii) Repulsion between a stick and a blue dashed line
So how do we finalize the angle?
8. The VSEPR theory helps us in this situation
• We have seen the details in the previous section
• We saw that: (lp-bp repulsion) > (bp-bp repulsion)
9. Based on this, we can think about the angle
It can be written in (v) steps;
(i) The three sticks in fig.4.105(b), tries to maintain an angle of 109.5o between them
(ii) They try to maintain this angle, by using the bp-bp repulsion mentioned in (8) above
♦ This repulsion is indicated by the red double headed arrows in fig.4.106(a) below:
♦ There are three 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
(iii) Now, the sticks are acted upon by other repulsion forces also. They are also shown in fig.4.106(a)
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are three yellow arrows
✰ Readers are advised to recognize each of the yellow arrows individually
✰ Some of them are partially obstructed from view
✰ Even then, they must be fully recognized
(iv) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
♦ Each yellow arrow is pushing down on a stick
✰ So the sticks tend to bend downwards
♦ This downward bending is resisted by the red arrows
✰ But according to the VSEPR theory, yellow is stronger than red
♦ So the angle 109.5 will decrease
♦ This is shown in fig.4.106(b)
♦ This fig.4.106(b) shows the final shape
10. Now we can write about the final shape. It can be written in 3 steps
(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) We have three 'B atoms' and one 'A atom'
♦ A atom is at the apex
♦ The three 'B atoms' are at the corners of the base of a triangular pyramid
♦ So we call it: Trigonal pyramidal structure
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.106(b)
11. We note an interesting point. It can be written in steps:
(i) Suppose that, in fig.106(a), there is no lone pair
(ii) Then, there will be no yellow arrows. There will only be three red arrows
(iii) If there are no yellow arrows, the sticks will not be bent downwards
♦ They will remain in the same plane as the atom A
♦ That means, the three 'B atoms' will be in the same plane as A
(iv) The result will be a planar structure
♦ A will be at the centroid of an equilateral triangle
♦ Each of the 'B atoms' will be at the corners of that triangle
♦ In effect, it will be a trigonal planar structure that we saw in the earlier section 4.13
(v) So we can write:
♦ The lone pair pushes the three sticks downwards
♦ The the trigonal planar is converted into trigonal pyramidal due to this pushing
12. The actual value of the angle:
• In the general case, we write that: The angle will be less than 109.5o
♦ We do not write the exact value
• This is because, in real life situations, the angle varies from molecule to molecule
♦ For example:
✰ In NH3, the angle is 107.8o
• This is shown in fig.4.107 below:
Case II (a): AB3E
We will write it in steps:
1. In AB3E, there are 3 terminal atoms and 1 electron pair
• So there is a total of 4 items
• These 4 items are distributed around A
2. Suppose that, all the 4 items are 'sticks'
• Then there will be 4 sticks around A
• We have already seen such a case in the previous sections
• It is the molecule AB4. It has 4 sticks around A
3. When there are 4 sticks around A, the shape is tetrahedral
• This is shown in the fig.4.105(a) below:
 |
| Fig.4.105 |
• We have only 3 sticks and 1 lone pair
• So from fig.4.105(a), we remove 1 stick
• In the place of that 'removed stick', we put a lone pair
• This is shown in fig.4.105(b)
♦ The stick corresponding to Bi 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
♦ Bi is written inside dashed circle because, that 'B atom' is not actually present in fig.b
5. Now there is a problem
• In the tetrahedral structure that we saw in AB4, we know that, the angle between any two bonds will be 109.5o
♦ That is., the angle between any two sticks will be 109.5o
• In our present case:
♦ We can write the '109.5o' between the two sticks: BiiA and BiiiA
✰ Because, they are two well defined sticks
♦ We can write the '109.5o' between the two sticks: BiiA and BivA
✰ Because, they are two well defined sticks
♦ We can write the '109.5o' between the two sticks: BiiiA and BivA
✰ Because, they are two well defined sticks
♦ But we cannot write the '109.5o' between the dashed blue line and any stick
■ The reason can be written in (iv) steps:
(i) The 'bonding electrons' are confined inside the stick
• So they occupy a definite space
(ii) But 'lone electrons' are not confined. They freely occupy a region. The region is an electron cloud
• So the dashed blue line indicate a cloud
(iii) We can easily measure an angle from a stick
• Because, a stick is well defined
(iv) But we cannot measure the angle from a cloud
• Because, the electron cloud does not have a well defined boundary
6. So there is no need to specify the 'angles in which one side is a blue dashed line'
■ We need to write only these: The angles between any two sticks
7. There is yet another problem:
• We had earlier obtained the 109.5o as follows:
The 4 sticks try to push each other as far away as possible, and they settle down with angles of '109.5'o between them
• But here there are no '4 sticks'. There are only '3 sticks' and 1 lone pair
♦ However, since they are all electrons, they will all be pushing each other
♦ But the pushing (repulsion) is different
■ We can write:
The following two repulsion are different:
(i) Repulsion between the 2 sticks
(ii) Repulsion between a stick and a blue dashed line
So how do we finalize the angle?
8. The VSEPR theory helps us in this situation
• We have seen the details in the previous section
• We saw that: (lp-bp repulsion) > (bp-bp repulsion)
9. Based on this, we can think about the angle
It can be written in (v) steps;
(i) The three sticks in fig.4.105(b), tries to maintain an angle of 109.5o between them
(ii) They try to maintain this angle, by using the bp-bp repulsion mentioned in (8) above
♦ This repulsion is indicated by the red double headed arrows in fig.4.106(a) below:
♦ There are three 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
 |
| Fig.4.106 |
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are three yellow arrows
✰ Readers are advised to recognize each of the yellow arrows individually
✰ Some of them are partially obstructed from view
✰ Even then, they must be fully recognized
(iv) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
♦ Each yellow arrow is pushing down on a stick
✰ So the sticks tend to bend downwards
♦ This downward bending is resisted by the red arrows
✰ But according to the VSEPR theory, yellow is stronger than red
♦ So the angle 109.5 will decrease
♦ This is shown in fig.4.106(b)
♦ This fig.4.106(b) shows the final shape
10. Now we can write about the final shape. It can be written in 3 steps
(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) We have three 'B atoms' and one 'A atom'
♦ A atom is at the apex
♦ The three 'B atoms' are at the corners of the base of a triangular pyramid
♦ So we call it: Trigonal pyramidal structure
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.106(b)
11. We note an interesting point. It can be written in steps:
(i) Suppose that, in fig.106(a), there is no lone pair
(ii) Then, there will be no yellow arrows. There will only be three red arrows
(iii) If there are no yellow arrows, the sticks will not be bent downwards
♦ They will remain in the same plane as the atom A
♦ That means, the three 'B atoms' will be in the same plane as A
(iv) The result will be a planar structure
♦ A will be at the centroid of an equilateral triangle
♦ Each of the 'B atoms' will be at the corners of that triangle
♦ In effect, it will be a trigonal planar structure that we saw in the earlier section 4.13
(v) So we can write:
♦ The lone pair pushes the three sticks downwards
♦ The the trigonal planar is converted into trigonal pyramidal due to this pushing
12. The actual value of the angle:
• In the general case, we write that: The angle will be less than 109.5o
♦ We do not write the exact value
• This is because, in real life situations, the angle varies from molecule to molecule
♦ For example:
✰ In NH3, the angle is 107.8o
• This is shown in fig.4.107 below:
 |
| Fig.4.107 |
Next we consider Case II (b): AB3E2
We will write it in steps:
1. In AB3E2, there are 3 terminal atoms and 2 electron pairs
• 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.4.108(a) below:
4. But actually in our present case, we do not have 5 sticks
• We have only 3 sticks and 2 lone pairs
• So from fig.4.108(a), we remove 2 sticks
• In the place of those 'removed sticks', we put lone pairs
• This is shown in fig.4.108(b)
♦ The sticks corresponding to Biii and Biv 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
♦ Biii and Biv are written inside dashed circles because, those 'B atoms' are 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
♦ ∠BiiiABiv = ∠BivABv = ∠BiiiABv = 120o
✰ These are the angles between equatorial bonds
■ There are a total of 9 angles. Do we need to mention all these angles even after removing Biii and Biv?
♦ So in our present case, we need to mention the following two angles only:
♦ ∠BiABv = ∠BiiABv = 90o (These are the angles related to axial bonds)
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.108(b)
7. There is yet another problem:
• We had earlier obtained the 90o 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 '3 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 Biii and Biv?
8. We saw that:
(lp-lp repulsion) > (lp-bp repulsion) > (bp-bp repulsion)
• Based on this, we can think about the angle. It can be written in (v) steps:
(i) The three sticks in fig.4.108(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.4.109(b) 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.4.109(a)
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are two yellow arrows
• The lp-lp repulsion is indicated by a cyan 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 cyan arrow is exerting a push on two yellow arrows
♦ We know that cyan is stronger than yellow
✰ So the two yellows will get compressed
♦ But the compression of the yellow arrows have no effect on the 90o angle
■ So we can write:
The three double headed arrows in the horizontal plane have no effect on the 90o angle. We can totally ignore the three repulsive forces in fig.4.109(a)
(v) Next we will see the 6 double headed arrows in the vertical planes
They are shown in fig.4.109(b)
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are four yellow arrows
✰ Readers are advised to recognize each of the yellow arrows individually
✰ Some of them are partially obstructed from view
✰ Even then, they must be fully recognized
• The bp-bp repulsion is indicated by red double headed arrows
♦ There are two such arrows
(vi) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
♦ The top ends of the 'two yellow arrows at top' are pushing on the top red arrow
✰ We know that, red is weaker than yellow
✰ So the top red will be compressed
✰ As a result, the angle will become less than 90o
♦ The bottom ends of the 'two yellow arrows at bottom' are pushing on the bottom red arrow
✰ We know that, red is weaker than yellow
✰ So the bottom red will be compressed
✰ As a result, the angle will become less than 90o
(vii) So we can write:
The ∠BiABii must be (90+90) = 180 in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 180o
♦ This is shown in fig.4.109(c)
♦ This fig.4.109(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.4.109(c), we have three 'B atoms' and one 'A atom'
• Together, they resemble the letter 'T'. So we call it: T-shape
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.109(c)
10. Note that, a 'T-structure' is a 2D structure
All the four atoms lie in a plane
11. The actual value of the angle:
• In the general case, we write that: The angle will be less than 180o
♦ We do not write the exact value
• This is because, in real life situations, the angle varies from molecule to molecule
♦ For example:
✰ In ClF3, the angle is 175o
• This is shown in fig.4.110 below:
We will write it in steps:
1. In AB3E2, there are 3 terminal atoms and 2 electron pairs
• 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.4.108(a) below:
 |
| Fig.4.108 |
• We have only 3 sticks and 2 lone pairs
• So from fig.4.108(a), we remove 2 sticks
• In the place of those 'removed sticks', we put lone pairs
• This is shown in fig.4.108(b)
♦ The sticks corresponding to Biii and Biv 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
♦ Biii and Biv are written inside dashed circles because, those 'B atoms' are 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
♦ ∠BiiiABiv = ∠BivABv = ∠BiiiABv = 120o
✰ These are the angles between equatorial bonds
■ There are a total of 9 angles. Do we need to mention all these angles even after removing Biii and Biv?
6. In the previous two 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 two angles only:
♦ ∠BiABv = ∠BiiABv = 90o (These are the angles related to axial bonds)
✰ These are the angles between axial bonds and equatorial bonds
✰ One of them is shown in orange color in fig.4.108(b)
7. There is yet another problem:
• We had earlier obtained the 90o 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 '3 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 Biii and Biv?
8. We saw that:
(lp-lp repulsion) > (lp-bp repulsion) > (bp-bp repulsion)
• Based on this, we can think about the angle. It can be written in (v) steps:
(i) The three sticks in fig.4.108(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.4.109(b) below:
 |
| Fig.4.109 |
♦ 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.4.109(a)
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are two yellow arrows
• The lp-lp repulsion is indicated by a cyan 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 cyan arrow is exerting a push on two yellow arrows
♦ We know that cyan is stronger than yellow
✰ So the two yellows will get compressed
♦ But the compression of the yellow arrows have no effect on the 90o angle
■ So we can write:
The three double headed arrows in the horizontal plane have no effect on the 90o angle. We can totally ignore the three repulsive forces in fig.4.109(a)
(v) Next we will see the 6 double headed arrows in the vertical planes
They are shown in fig.4.109(b)
• The lp-bp repulsion is indicated by yellow double headed arrows
♦ There are four yellow arrows
✰ Readers are advised to recognize each of the yellow arrows individually
✰ Some of them are partially obstructed from view
✰ Even then, they must be fully recognized
• The bp-bp repulsion is indicated by red double headed arrows
♦ There are two such arrows
(vi) We see a clear and definite pattern here. The reader must be able to appreciate the pattern:
♦ The top ends of the 'two yellow arrows at top' are pushing on the top red arrow
✰ We know that, red is weaker than yellow
✰ So the top red will be compressed
✰ As a result, the angle will become less than 90o
♦ The bottom ends of the 'two yellow arrows at bottom' are pushing on the bottom red arrow
✰ We know that, red is weaker than yellow
✰ So the bottom red will be compressed
✰ As a result, the angle will become less than 90o
(vii) So we can write:
The ∠BiABii must be (90+90) = 180 in the normal case. But due to the compression by the yellow arrows, the angle becomes less than 180o
♦ This is shown in fig.4.109(c)
♦ This fig.4.109(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.4.109(c), we have three 'B atoms' and one 'A atom'
• Together, they resemble the letter 'T'. So we call it: T-shape
• Once we finalize the shape, we no longer need to show the blue dashed line. So it is not shown in fig.4.109(c)
10. Note that, a 'T-structure' is a 2D structure
All the four atoms lie in a plane
11. The actual value of the angle:
• In the general case, we write that: The angle will be less than 180o
♦ We do not write the exact value
• This is because, in real life situations, the angle varies from molecule to molecule
♦ For example:
✰ In ClF3, the angle is 175o
• This is shown in fig.4.110 below:
 |
| Fig.4.110 |
• In the next section, we will see case III
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