In the previous section 2.14, we completed a discussion on quantum numbers. In that discussion, we saw the shapes of various orbitals. In this section we will see how those shapes are derived
• We want to know how scientists confirmed this ‘spherical shape’
• The 11 steps below, will give the answer:
1. Consider the first main-shell (n=1)
• We know that, the one and only sub-shell in the (n=1) is the s sub-shell
• We know that, there will be only two electrons in any orbital
2. So where are those two electrons in that orbital situated ?
• We cannot specify the exact location of electrons
♦ We saw the reason when we discussed the 'uncertainty principle'
• We can only speak in terms of ‘probability’
■ We can say this:
There are certain regions where there is a greater probability of finding the electrons
3. Detailed calculations carried out by Schrodinger revealed the following three facts:
(i) The electrons in the orbital in s sub-shell, spend greater time near the nucleus
(ii) As the distance from the nucleus increases, the ‘probability of finding the electrons’ decreases
(iii) This ‘decrease in probability’ is same in ‘all directions from the nucleus’
♦ This 'same probability' is shown in fig.2.42(a) below:
(Right click and choose 'open in new tab' to get an enlarged view)
4. By analyzing the three facts, scientists arrived at the following conclusions:
(i) The two electrons are continuously moving around the nucleus
• So they will appear as a cloud
(ii) Since they spend more time at the nucleus, the cloud will appear to be denser near the nucleus
• Such a cloud is shown in fig.2.42(b) above
(iii) ‘Same probability in all directions from the nucleus’ indicates a spherical shape
• A 'spherical shape' whose center is at the nucleus
• So the cloud in fig.2.42(b) is spherical in shape
• The nucleus is present at the center of that cloud
5. A ‘cloud’ is not a well defined shape
• We want to specify a ‘well defined shape’
• In that ‘well defined shape’ there must be a very high probability (90 % or 95%) for finding the electrons
6. Let us try various ‘boundaries’. They are shown in fig.2.42(c)
(i) First we try the inner most red boundary
• Though it is shown as a circle, it is in fact, a ‘hollow sphere’
♦ A ‘hollow sphere’ will have only a thin outer layer. The inside will be empty
• The red circle indicates such a hollow sphere
(ii) In fig.c, we are using the 'red hollow sphere' to enclose the central portion of the cloud
• So we have a ‘portion of the cloud’ inside the ‘red hollow sphere’
7. Is there a high probability of finding the electrons in that 'red hollow sphere' ?
• The answer can be written in steps
(i) It is true that the cloud is denser near the center
(ii) It is true that the electrons spend ‘more time’ near the center
(iii) But ‘more time’ is not equivalent to ‘always’
(iv) That means, there will be instances where we will never find the electrons inside the red hollow sphere
(v) So the red hollow sphere does not give a high probability of 90% or 95 %
8. The same reason can be written in the case of the green hollow sphere in fig.c
• There may be instances where we will never find the electrons inside the green hollow sphere
• So the green hollow sphere also does not give a high probability of 90% or 95 %
9. Now consider the cyan hollow sphere in fig.c
• It encloses most of the cloud. It is shown separately in fig.d
• It will indeed give a high probability of 90% or 95 %
10. So scientists chose the cyan hollow sphere
• The cyan hollow sphere is considered as the orbital in the s sub-shell in (n=1)
■ In other words, we can write:
The cyan sphere is the 1s orbital
• It is shown in fig.e
11. The 2s orbital will also be a sphere. But larger than the 1s sphere
• Both 1s and 2s spheres will be concentric. That is., both will have the same center
• The nucleus of the atom will be situated at this common center
• However, the 2s orbital will have a special feature called the ‘node’. We will see more about 'nodes' in the next section
• We want to know how scientists confirmed this ‘dumbbell shape’
• The 11 steps given below, will give the answer:
1. Consider the second main-shell (n=2)
• We know that, there will be an s sub-shell and a p sub-shell
• We already saw the shape of the orbital in the s sub-shell
• We want to know the shape of the orbitals in the p sub-shell
2. We know that, there will be 3 orbitals in the p sub-shell
• Consider any one of them
• We know that, there will be only two electrons in any orbital
• So where are those two electrons in that orbital situated ?
• We cannot specify the exact location of electrons
♦ We saw the reason when we discussed the 'uncertainty principle'
• We can only speak in terms of ‘probability’
■ We can say this:
There are certain regions where there is a greater probability of finding the electrons
3. Detailed calculations carried out by Schrodinger revealed the following three facts:
(i) The electrons in any one orbital of the p sub-shell spend greater time away from the nucleus
(ii) As the distance from the nucleus increases, the ‘probability of finding the electrons’ increases
(iii) This ‘increase in probability’ is not the same in ‘all directions from the nucleus’
♦ Starting from the nucleus and moving towards the left, the probability increases
♦ Starting from the nucleus and moving towards the right, the probability increases
♦ Starting from the nucleus and moving towards the top, there is only a low probability
♦ Starting from the nucleus and moving towards the bottom, there is only a low probability
♦ This variation in probabilities is shown in fig.2.43(a) below:
(Right click and choose 'open in new tab' to get an enlarged view)
4. By analyzing the three facts, scientists arrived at the following conclusions:
(i) The two electrons are continuously moving around the nucleus
• So they will appear as a cloud
(ii) Since they spend more time away from the nucleus, the cloud will appear to be denser away from the nucleus
• Since there is low probability towards top and bottom, we will not find any clouds in the top-bottom direction
• Such a cloud is shown in fig.2.43(b) above
(iii) We see that, the cloud has a shape similar to a dumbbell. It has two lobes
• The nucleus is present at the center of the two lobes
(iv) In fig.2.43(b), the nucleus is clearly visible
♦ This is because, there is no cloud near the nucleus
• In the previous fig.2.42(b), nucleus (at the center of the sphere) is not visible
♦ This is because, the cloud is very dense near the nucleus
5. A ‘cloud’ is not a well defined shape
• We want to specify a ‘well defined shape’
• In that ‘well defined shape’ there must be a very high probability (90 % or 95%) for finding the electrons
6. Let us try various ‘boundaries’. They are shown in fig.2.43(c)
(i) First we try the inner most blue boundary
• Though it is shown as a '2-dimensional closed curve', it is in fact, a ‘3-dimensional hollow space’
♦ This ‘hollow space’ will have only a thin outer layer. The inside will be empty
• The blue '2-dimensional closed curve' indicates such a 'hollow space'
(ii) In fig.c, we are using the 'blue hollow space' to enclose the left/right portion of the cloud
• So we have a ‘portion of the cloud’ inside the ‘blue hollow space’
7. Is there a high probability of finding the electrons in that 'blue hollow space' ?
• The answer can be written in steps
(i) It is true that the cloud is denser near the left/right
(ii) It is true that the electrons spend ‘more time’ near the left/right
(iii) But ‘more time’ is not equivalent to ‘always’
(iv) That means, there will be instances where we will never find the electrons inside the 'blue hollow space'
(v) So the blue hollow space does not give a high probability of 90% or 95 %
8. The same reason can be written in the case of the yellow hollow space in fig.c
• There will be instances where we will never find the electrons inside the yellow hollow space
• So the yellow hollow space also does not give a high probability of 90% or 95 %
9. Now consider the magenta hollow space in fig.c
• It encloses most of the cloud. It is shown separately in fig.d
• It will indeed give a high probability of 90% or 95 %
10. So scientists chose the magenta hollow space
• The magenta hollow space is considered as the shape of orbital in the p sub-shell in (n=2)
■ In other words, we can write:
The magenta space is the 2p orbital
• It is shown in fig.e
• It resembles a dumbbell
• There will be a total of 3 such dumbbells, which are mutually perpendicular to each other
11. The 3p orbitals will also be dumbbells. But larger than the 2p dumbbells
Consider the two points
(i) The point of intersection of the three '2p dumbbells'
(ii) The point of intersection of the three '3p dumbbells'
■ Both the points are the same. And the nucleus of the atom will be present at that same point
12. Like the 2s spheres, the 2p dumbbells will also have 'nodes'. We will see them in the next section
• The cyan hollow space for s orbitals gives a probability of 90 or 95 %. The magenta hollow space for p orbitals also gives a probability of 90 or 95 %. Why don't we choose hollow spaces which gives a probability of 100%?
• The answer can be written in steps:
(i) The electron can be any where in space
• The only condition is that, it should be at a finite distance from the nucleus
♦ 'Finite distance' means 'measurable distance'
(ii) So we will have to draw a 'very large hollow space'
• Only a 'very large hollow space' can enclose 'all the possible positions of the electron'
(iii) In order to avoid such a difficulty, we do not consider 100%
(iv) We can take 99%, 98% etc.,
• But it is convenient to consider 'multiples of 5'
• That is why we take 90% or 95%
Shape of orbital in the s sub-shell
• We saw that, the orbital in the s sub-shell is spherical in shape• We want to know how scientists confirmed this ‘spherical shape’
• The 11 steps below, will give the answer:
1. Consider the first main-shell (n=1)
• We know that, the one and only sub-shell in the (n=1) is the s sub-shell
• We know that, there will be only two electrons in any orbital
2. So where are those two electrons in that orbital situated ?
• We cannot specify the exact location of electrons
♦ We saw the reason when we discussed the 'uncertainty principle'
• We can only speak in terms of ‘probability’
■ We can say this:
There are certain regions where there is a greater probability of finding the electrons
3. Detailed calculations carried out by Schrodinger revealed the following three facts:
(i) The electrons in the orbital in s sub-shell, spend greater time near the nucleus
(ii) As the distance from the nucleus increases, the ‘probability of finding the electrons’ decreases
(iii) This ‘decrease in probability’ is same in ‘all directions from the nucleus’
♦ This 'same probability' is shown in fig.2.42(a) below:
(Right click and choose 'open in new tab' to get an enlarged view)
 |
| Fig.2.42 |
(i) The two electrons are continuously moving around the nucleus
• So they will appear as a cloud
(ii) Since they spend more time at the nucleus, the cloud will appear to be denser near the nucleus
• Such a cloud is shown in fig.2.42(b) above
• A 'spherical shape' whose center is at the nucleus
• So the cloud in fig.2.42(b) is spherical in shape
• The nucleus is present at the center of that cloud
5. A ‘cloud’ is not a well defined shape
• We want to specify a ‘well defined shape’
• In that ‘well defined shape’ there must be a very high probability (90 % or 95%) for finding the electrons
6. Let us try various ‘boundaries’. They are shown in fig.2.42(c)
(i) First we try the inner most red boundary
• Though it is shown as a circle, it is in fact, a ‘hollow sphere’
♦ A ‘hollow sphere’ will have only a thin outer layer. The inside will be empty
• The red circle indicates such a hollow sphere
(ii) In fig.c, we are using the 'red hollow sphere' to enclose the central portion of the cloud
• So we have a ‘portion of the cloud’ inside the ‘red hollow sphere’
7. Is there a high probability of finding the electrons in that 'red hollow sphere' ?
• The answer can be written in steps
(i) It is true that the cloud is denser near the center
(ii) It is true that the electrons spend ‘more time’ near the center
(iii) But ‘more time’ is not equivalent to ‘always’
(iv) That means, there will be instances where we will never find the electrons inside the red hollow sphere
(v) So the red hollow sphere does not give a high probability of 90% or 95 %
8. The same reason can be written in the case of the green hollow sphere in fig.c
• There may be instances where we will never find the electrons inside the green hollow sphere
• So the green hollow sphere also does not give a high probability of 90% or 95 %
9. Now consider the cyan hollow sphere in fig.c
• It encloses most of the cloud. It is shown separately in fig.d
• It will indeed give a high probability of 90% or 95 %
10. So scientists chose the cyan hollow sphere
• The cyan hollow sphere is considered as the orbital in the s sub-shell in (n=1)
■ In other words, we can write:
The cyan sphere is the 1s orbital
• It is shown in fig.e
11. The 2s orbital will also be a sphere. But larger than the 1s sphere
• Both 1s and 2s spheres will be concentric. That is., both will have the same center
• The nucleus of the atom will be situated at this common center
• However, the 2s orbital will have a special feature called the ‘node’. We will see more about 'nodes' in the next section
Shape of orbitals in the p sub-shell
• We saw that, the orbitals in the p sub-shell have a dumbbell shape• We want to know how scientists confirmed this ‘dumbbell shape’
• The 11 steps given below, will give the answer:
1. Consider the second main-shell (n=2)
• We know that, there will be an s sub-shell and a p sub-shell
• We already saw the shape of the orbital in the s sub-shell
• We want to know the shape of the orbitals in the p sub-shell
2. We know that, there will be 3 orbitals in the p sub-shell
• Consider any one of them
• We know that, there will be only two electrons in any orbital
• So where are those two electrons in that orbital situated ?
• We cannot specify the exact location of electrons
♦ We saw the reason when we discussed the 'uncertainty principle'
• We can only speak in terms of ‘probability’
■ We can say this:
There are certain regions where there is a greater probability of finding the electrons
3. Detailed calculations carried out by Schrodinger revealed the following three facts:
(i) The electrons in any one orbital of the p sub-shell spend greater time away from the nucleus
(ii) As the distance from the nucleus increases, the ‘probability of finding the electrons’ increases
(iii) This ‘increase in probability’ is not the same in ‘all directions from the nucleus’
♦ Starting from the nucleus and moving towards the left, the probability increases
♦ Starting from the nucleus and moving towards the right, the probability increases
♦ Starting from the nucleus and moving towards the top, there is only a low probability
♦ Starting from the nucleus and moving towards the bottom, there is only a low probability
♦ This variation in probabilities is shown in fig.2.43(a) below:
(Right click and choose 'open in new tab' to get an enlarged view)
 |
| Fig.2.43 |
(i) The two electrons are continuously moving around the nucleus
• So they will appear as a cloud
(ii) Since they spend more time away from the nucleus, the cloud will appear to be denser away from the nucleus
• Since there is low probability towards top and bottom, we will not find any clouds in the top-bottom direction
• Such a cloud is shown in fig.2.43(b) above
(iii) We see that, the cloud has a shape similar to a dumbbell. It has two lobes
• The nucleus is present at the center of the two lobes
(iv) In fig.2.43(b), the nucleus is clearly visible
♦ This is because, there is no cloud near the nucleus
• In the previous fig.2.42(b), nucleus (at the center of the sphere) is not visible
♦ This is because, the cloud is very dense near the nucleus
5. A ‘cloud’ is not a well defined shape
• We want to specify a ‘well defined shape’
• In that ‘well defined shape’ there must be a very high probability (90 % or 95%) for finding the electrons
6. Let us try various ‘boundaries’. They are shown in fig.2.43(c)
(i) First we try the inner most blue boundary
• Though it is shown as a '2-dimensional closed curve', it is in fact, a ‘3-dimensional hollow space’
♦ This ‘hollow space’ will have only a thin outer layer. The inside will be empty
• The blue '2-dimensional closed curve' indicates such a 'hollow space'
(ii) In fig.c, we are using the 'blue hollow space' to enclose the left/right portion of the cloud
• So we have a ‘portion of the cloud’ inside the ‘blue hollow space’
7. Is there a high probability of finding the electrons in that 'blue hollow space' ?
• The answer can be written in steps
(i) It is true that the cloud is denser near the left/right
(ii) It is true that the electrons spend ‘more time’ near the left/right
(iii) But ‘more time’ is not equivalent to ‘always’
(iv) That means, there will be instances where we will never find the electrons inside the 'blue hollow space'
(v) So the blue hollow space does not give a high probability of 90% or 95 %
8. The same reason can be written in the case of the yellow hollow space in fig.c
• There will be instances where we will never find the electrons inside the yellow hollow space
• So the yellow hollow space also does not give a high probability of 90% or 95 %
9. Now consider the magenta hollow space in fig.c
• It encloses most of the cloud. It is shown separately in fig.d
• It will indeed give a high probability of 90% or 95 %
10. So scientists chose the magenta hollow space
• The magenta hollow space is considered as the shape of orbital in the p sub-shell in (n=2)
■ In other words, we can write:
The magenta space is the 2p orbital
• It is shown in fig.e
• It resembles a dumbbell
• There will be a total of 3 such dumbbells, which are mutually perpendicular to each other
11. The 3p orbitals will also be dumbbells. But larger than the 2p dumbbells
Consider the two points
(i) The point of intersection of the three '2p dumbbells'
(ii) The point of intersection of the three '3p dumbbells'
■ Both the points are the same. And the nucleus of the atom will be present at that same point
12. Like the 2s spheres, the 2p dumbbells will also have 'nodes'. We will see them in the next section
■ After completing the above discussion, a question arises in our minds:
• The answer can be written in steps:
(i) The electron can be any where in space
• The only condition is that, it should be at a finite distance from the nucleus
♦ 'Finite distance' means 'measurable distance'
(ii) So we will have to draw a 'very large hollow space'
• Only a 'very large hollow space' can enclose 'all the possible positions of the electron'
(iii) In order to avoid such a difficulty, we do not consider 100%
(iv) We can take 99%, 98% etc.,
• But it is convenient to consider 'multiples of 5'
• That is why we take 90% or 95%
Thus we saw how scientists confirmed the shapes of orbitals in the s and p sub-shells. Shapes of orbitals in d and f sub-shells can also be confirmed in the same way. We will see them in higher classes. In the next section, we will see 'nodes'
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