In the previous section 2.15, we completed a discussion on shapes of orbitals. In this section we will see nodes
1. Consider a wave trapped inside a container
• That wave propagates in one direction and hits the wall of the container
• Then it reflects back
2. So there will be two waves: The original wave and the reflected wave
• Those two waves interfere with each other
• The result is a standing wave
• An animation can be seen in wikipedia. The link is given below:
https://commons.wikimedia.org/wiki/File:Waventerference.gif
3. In ordinary waves, all points along the wave will be in motion
• But in standing waves, there will be certain points which will be stationary
■ Such points are called nodes
4. We have seen that, electrons have wave nature
• Those 'waves associated with electrons' are trapped inside the orbitals
• So we get ‘electron standing waves’
5. There will be nodes in those ‘electron standing waves’ too
• Remember that, the electron waves are 3-dimensional waves. That is., they are spread out in a 3-dimensional space
• So the nodes in the electron standing waves are not ‘points’. They are surfaces
■ At those ‘nodal surfaces’, we will not find any electrons
Radial nodes in the orbitals of s sub-shell
1. We know that, the orbitals in the s sub-shell have a spherical shape
• So the standing wave is present in a sphere
• The resulting nodal surfaces will be inside the sphere
2. To see those surfaces, we will have to cut open the spherical orbital
• Such a diagram is shown in fig.2.44(a) below:
3. In the fig.a, the nodal surfaces are shown in white color
• Those nodal surfaces are called radial nodes
• The cyan portions are electron clouds
• The cross section of the spherical orbital will appear as shown in fig.b
• In the figs. a & b, we see two radial nodes each
Radial nodes in the orbitals of p sub-shell
1. We know that, the orbitals in the p sub-shell have a dumbbell shape
• So the standing wave is present in a dumbbell
• The resulting nodal surfaces will be inside the dumbbell
2. To see those surfaces, we will have to cut open the dumbbell orbital
• Such a diagram is shown in fig.2.45(a) below:
3. In the fig.a, the nodal surfaces are shown in white color
• Those are the radial nodes of p orbitals
• The magenta portions are electron clouds
• The cross section of the dumbbell will appear as shown in fig.b
• In the figs. a & b, we see two radial nodes each
1. In addition to the radial nodes, the orbitals have another type of nodes also. They are called angular nodes
2. Angular nodes are not present in the orbitals of s sub-shell
• That means: 1s, 2s, 3s . . . , orbitals, do not have angular nodes
■ What is the reason for s-orbitals not having angular nodes?
• We will see the answer later in this section
3. Since the s orbitals do not have angular nodes, we will take up the p orbitals for the discussion on such nodes
• We have already seen the px orbital
• It is is shown again below in fig.2.46(a)
4. Consider the plane which contains both the y and z axes
• We know the name of such a plane: The yz-plane. It is shown in fig.b
• In this plane, we will not find any 'electrons belonging to the px orbital'
5. The reason is obvious. We will write it in steps:
(i) The ‘probability for finding electrons (electrons belonging to the px orbital)’ increases as we move away from the nucleus, towards left/right of the x axis
(ii) The probability at the exact center ('center' is the origin of the axes) will be zero
(iii) The yz-plane passes through this center
(iv) So, as far as the px orbital is concerned, the yz plane is a nodal surface
6. Note that, to become qualified as the 'nodal surface', the plane must be the exact yz-plane
• If any tilt is applied to that, plane, it will no longer be the nodal surface
• This is because, if there is any tilt, there will be some probability for finding the electrons
7. This surface is a perfect plane. So we can call it a ‘nodal plane’, rather than ‘nodal surface’
• Such planes, where we will not find any electrons, are called angular nodes
■ So we can write:
• Nodes are surfaces where we will not find any electrons
♦ Radial nodes are curved in shape
♦ Angular nodes are mainly planes
(There are indeed some angular nodes which are not planes. We will see them in higher classes)
8. The px orbital have only one angular node
• Similarly, the py orbital also, has only one angular node
♦ It is the xz-plane shown in the fig.c
• Similarly, the pz orbital also, has only one angular node
♦ It is the xy-plane shown in fig.d
9. Let us see the angular nodes of d orbitals:
• We have already seen the $\mathbf\small{d_{xy}}$ orbital
• It is is shown again below in fig.2.47(a)
10. Consider the two planes: xz-plane and yz-plane
• They are shown in fig.2.47(b)
• In those two planes, we will not find any 'electrons belonging to the $\mathbf\small{d_{xy}}$ orbital'
■ So we can write:
The $\mathbf\small{d_{xy}}$ orbital has two angular nodes: xz-plane and yz-plane
■ Similarly, from fig.c, we see that:
The $\mathbf\small{d_{xz}}$ orbital has two angular nodes: xy-plane and yz-plane
■ Similarly, from fig.d, we see that:
The $\mathbf\small{d_{yz}}$ orbital has two angular nodes: xy-plane and xz-plane
11. The two angular nodes of $\mathbf\small{d_{x^2-y^2}}$ orbital are a bit different
• They are shown in fig.e
• We can write three properties of those two planes:
(i) Both the planes are perpendicular to the xy-plane
(Note that, xy-plane is the plane in which the $\mathbf\small{d_{x^2-y^2}}$ orbital lies)
(ii) Both the planes are perpendicular to each other
(iii) Both the planes bisect the '90o angle' between the x and y axes
12. The $\mathbf\small{d_{z^2}}$ orbital also has two angular nodes
• They are conical in shape. We will see the details in higher classes
The answer can be written in four steps:
1. The s orbitals are spherical in shape
2. Consider a plane, passing through the center of the s orbital
• We can think of infinite number of orientations for that plane
3. What ever be the orientation, that plane will be cutting through the sphere
• So there is always a probability for finding the electrons in that plane
• None of the orientations will give 'zero probability'
4. So we can write:
The s orbitals do not have angular nodes
Our next task is to find the 'number of nodes possessed' by the various orbitals
This task can be elaborated in 3 steps as follows:
1. Given any random orbital: 1s, 3p, 5d etc.,
2. We want to know the number of radial nodes in that orbital
3. We want to know the number of angular nodes in that orbital
■ To get those numbers, we use the following steps:
1. Scientists discovered that, the total number of nodes in any orbital will be equal to (n-1)
Where n is the main-shell in which the orbital is situated
2. So we can write:
Total number of nodes
= Number of radial nodes + Number of angular nodes
= (n-1)
3. Scientists also discovered that, the number of angular nodes in any orbital will be equal to the azimuthal quantum (l) number of that orbital
4. So we can write:
Number of angular nodes = l
5. Substituting (4) in (2), we get:
Number of radial nodes + l = (n-1)
⇒ Number of radial nodes = (n-1-l)
6. Thus from (4) and (5), we have two important results:
(i) Number of angular nodes = l
(ii) Number of radial nodes = (n-1-l)
• Using these two results, we can find the number of nodes in any orbital
Let us see some solved examples:
Solved example 2.74
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the first main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the first main-shell (n=1)
3. So in the s, p, d, f notation, the given orbital is: 1s
• For 1s orbital:
n = 1 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (1-1-0) = 0
5. So for the 1s orbital, there are no nodes at all. This is shown in fig.2.48(a) below:
Solved example 2.75
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the second main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the second main-shell (n=2)
3. So in the s, p, d, f notation, the given orbital is: 2s
• For 2s orbital:
n = 2 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (2-1-0) = 1
5. So for the 2s orbital, there is one radial node. But there are no angular nodes. This is shown in fig.2.48(b) above
Solved example 2.76
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the third main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3s
• For 2s orbital:
n = 3 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (3-1-0) = 2
5. So for the 3s orbital, there are two radial nodes. But there are no angular nodes. This is shown in fig.2.48(c) above
Solved example 2.77
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the second main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the second main-shell (n=2)
3. So in the s, p, d, f notation, the given orbital is: 2p
• For any of the three 2p orbital:
n = 2 and l = 1
4. Consider any one from among the three: 2px, 2py, 2pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (2-1-1) = 0
5. So for the 2p orbitals, there are no radial nodes at all. This is shown in fig.2.49(a) below:
• But there is one angular node. We already saw it earlier in fig.2.46
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 2px, 2py and 2pz
• That means, each of px, py and pz, (in the second main-shell) have zero radial nodes and one angular node
Solved example 2.78
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the third main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3p
• For any of the three 3p orbital:
n = 3 and l = 1
4. Consider any one from among the three: 3px, 3py, 3pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (3-1-1) = 1
5. So for the 3p orbitals, there is one radial node. This is shown in fig.2.49(b) above
• There is one angular node also. We already saw it earlier in fig.2.46
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 3px, 3py and 3pz
• That means, each of px, py and pz, (in the third main-shell) have one radial nod and one angular node
Solved example 2.79
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the fourth main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the fourth main-shell (n=4)
3. So in the s, p, d, f notation, the given orbital is: 4p
• For any of the three 4p orbital:
n = 4 and l = 1
4. Consider any one from among the three: 3px, 3py, 3pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (4-1-1) = 2
5. So for the 4p orbitals, there are two radial nodes. This is shown in fig.2.49(c) above
• There is one angular node also. We already saw it earlier in fig.2.46
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 4px, 4py and 4pz
• That means, each of px, py and pz, (in the fourth main-shell) have two radial nodes and one angular node
Solved example 2.80
Find the number of nodes in the orbital of the d sub-shell. Assume that, this d sub-shell is situated in the third main-shell
Solution:
1. The given orbital is present in the d sub-shell
• There are five orbitals in the d sub-shell
2. Given that, this d sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3d
• For any of the five 3d orbitals:
n = 3 and l = 2
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 2
(ii) Number of radial nodes = (n-1-l) = (3-1-2) = 0
5. So for the 3d orbitals, there are no radial nodes at all
• But there are two angular nodes. We already saw it earlier in fig.2.48(a)
6. Note that the calculated numbers in (4) is applicable to each of the five d orbitals (in the third main-shell)
1. Consider a wave trapped inside a container
• That wave propagates in one direction and hits the wall of the container
• Then it reflects back
2. So there will be two waves: The original wave and the reflected wave
• Those two waves interfere with each other
• The result is a standing wave
• An animation can be seen in wikipedia. The link is given below:
https://commons.wikimedia.org/wiki/File:Waventerference.gif
3. In ordinary waves, all points along the wave will be in motion
• But in standing waves, there will be certain points which will be stationary
■ Such points are called nodes
4. We have seen that, electrons have wave nature
• Those 'waves associated with electrons' are trapped inside the orbitals
• So we get ‘electron standing waves’
5. There will be nodes in those ‘electron standing waves’ too
• Remember that, the electron waves are 3-dimensional waves. That is., they are spread out in a 3-dimensional space
• So the nodes in the electron standing waves are not ‘points’. They are surfaces
■ At those ‘nodal surfaces’, we will not find any electrons
Now we will see the nodes in various orbitals
Radial nodes in the orbitals of s sub-shell
1. We know that, the orbitals in the s sub-shell have a spherical shape
• So the standing wave is present in a sphere
• The resulting nodal surfaces will be inside the sphere
2. To see those surfaces, we will have to cut open the spherical orbital
• Such a diagram is shown in fig.2.44(a) below:
 |
| Fig.2.44 |
• Those nodal surfaces are called radial nodes
• The cyan portions are electron clouds
• The cross section of the spherical orbital will appear as shown in fig.b
• In the figs. a & b, we see two radial nodes each
Radial nodes in the orbitals of p sub-shell
1. We know that, the orbitals in the p sub-shell have a dumbbell shape
• So the standing wave is present in a dumbbell
• The resulting nodal surfaces will be inside the dumbbell
2. To see those surfaces, we will have to cut open the dumbbell orbital
• Such a diagram is shown in fig.2.45(a) below:
 |
| Fig.2.45 |
• Those are the radial nodes of p orbitals
• The magenta portions are electron clouds
• The cross section of the dumbbell will appear as shown in fig.b
• In the figs. a & b, we see two radial nodes each
Angular nodes
1. In addition to the radial nodes, the orbitals have another type of nodes also. They are called angular nodes
2. Angular nodes are not present in the orbitals of s sub-shell
• That means: 1s, 2s, 3s . . . , orbitals, do not have angular nodes
■ What is the reason for s-orbitals not having angular nodes?
• We will see the answer later in this section
3. Since the s orbitals do not have angular nodes, we will take up the p orbitals for the discussion on such nodes
• We have already seen the px orbital
• It is is shown again below in fig.2.46(a)
 |
| Fig.2.46 |
• We know the name of such a plane: The yz-plane. It is shown in fig.b
• In this plane, we will not find any 'electrons belonging to the px orbital'
5. The reason is obvious. We will write it in steps:
(i) The ‘probability for finding electrons (electrons belonging to the px orbital)’ increases as we move away from the nucleus, towards left/right of the x axis
(ii) The probability at the exact center ('center' is the origin of the axes) will be zero
(iii) The yz-plane passes through this center
(iv) So, as far as the px orbital is concerned, the yz plane is a nodal surface
6. Note that, to become qualified as the 'nodal surface', the plane must be the exact yz-plane
• If any tilt is applied to that, plane, it will no longer be the nodal surface
• This is because, if there is any tilt, there will be some probability for finding the electrons
7. This surface is a perfect plane. So we can call it a ‘nodal plane’, rather than ‘nodal surface’
• Such planes, where we will not find any electrons, are called angular nodes
■ So we can write:
• Nodes are surfaces where we will not find any electrons
♦ Radial nodes are curved in shape
♦ Angular nodes are mainly planes
(There are indeed some angular nodes which are not planes. We will see them in higher classes)
8. The px orbital have only one angular node
• Similarly, the py orbital also, has only one angular node
♦ It is the xz-plane shown in the fig.c
• Similarly, the pz orbital also, has only one angular node
♦ It is the xy-plane shown in fig.d
9. Let us see the angular nodes of d orbitals:
• We have already seen the $\mathbf\small{d_{xy}}$ orbital
• It is is shown again below in fig.2.47(a)
 |
| Fig.2.47 |
• They are shown in fig.2.47(b)
• In those two planes, we will not find any 'electrons belonging to the $\mathbf\small{d_{xy}}$ orbital'
■ So we can write:
The $\mathbf\small{d_{xy}}$ orbital has two angular nodes: xz-plane and yz-plane
■ Similarly, from fig.c, we see that:
The $\mathbf\small{d_{xz}}$ orbital has two angular nodes: xy-plane and yz-plane
■ Similarly, from fig.d, we see that:
The $\mathbf\small{d_{yz}}$ orbital has two angular nodes: xy-plane and xz-plane
11. The two angular nodes of $\mathbf\small{d_{x^2-y^2}}$ orbital are a bit different
• They are shown in fig.e
• We can write three properties of those two planes:
(i) Both the planes are perpendicular to the xy-plane
(Note that, xy-plane is the plane in which the $\mathbf\small{d_{x^2-y^2}}$ orbital lies)
(ii) Both the planes are perpendicular to each other
(iii) Both the planes bisect the '90o angle' between the x and y axes
12. The $\mathbf\small{d_{z^2}}$ orbital also has two angular nodes
• They are conical in shape. We will see the details in higher classes
After the above discussion on angular nodes, we can now answer the question mentioned in (2) above
1. The s orbitals are spherical in shape
2. Consider a plane, passing through the center of the s orbital
• We can think of infinite number of orientations for that plane
3. What ever be the orientation, that plane will be cutting through the sphere
• So there is always a probability for finding the electrons in that plane
• None of the orientations will give 'zero probability'
4. So we can write:
The s orbitals do not have angular nodes
This task can be elaborated in 3 steps as follows:
1. Given any random orbital: 1s, 3p, 5d etc.,
2. We want to know the number of radial nodes in that orbital
3. We want to know the number of angular nodes in that orbital
■ To get those numbers, we use the following steps:
1. Scientists discovered that, the total number of nodes in any orbital will be equal to (n-1)
Where n is the main-shell in which the orbital is situated
2. So we can write:
Total number of nodes
= Number of radial nodes + Number of angular nodes
= (n-1)
3. Scientists also discovered that, the number of angular nodes in any orbital will be equal to the azimuthal quantum (l) number of that orbital
4. So we can write:
Number of angular nodes = l
5. Substituting (4) in (2), we get:
Number of radial nodes + l = (n-1)
⇒ Number of radial nodes = (n-1-l)
6. Thus from (4) and (5), we have two important results:
(i) Number of angular nodes = l
(ii) Number of radial nodes = (n-1-l)
• Using these two results, we can find the number of nodes in any orbital
Let us see some solved examples:
Solved example 2.74
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the first main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the first main-shell (n=1)
3. So in the s, p, d, f notation, the given orbital is: 1s
• For 1s orbital:
n = 1 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (1-1-0) = 0
5. So for the 1s orbital, there are no nodes at all. This is shown in fig.2.48(a) below:
 |
| Fig.2.48 |
Solved example 2.75
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the second main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the second main-shell (n=2)
3. So in the s, p, d, f notation, the given orbital is: 2s
• For 2s orbital:
n = 2 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (2-1-0) = 1
5. So for the 2s orbital, there is one radial node. But there are no angular nodes. This is shown in fig.2.48(b) above
Solved example 2.76
Find the number of nodes in the orbital of the s sub-shell. Assume that, this s sub-shell is situated in the third main-shell
Solution:
1. The given orbital is present in the s sub-shell
• There is only one orbital in the s sub-shell. It is spherical in shape
2. Given that, this s sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3s
• For 2s orbital:
n = 3 and l = 0
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 0
(ii) Number of radial nodes = (n-1-l) = (3-1-0) = 2
5. So for the 3s orbital, there are two radial nodes. But there are no angular nodes. This is shown in fig.2.48(c) above
Solved example 2.77
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the second main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the second main-shell (n=2)
3. So in the s, p, d, f notation, the given orbital is: 2p
• For any of the three 2p orbital:
n = 2 and l = 1
4. Consider any one from among the three: 2px, 2py, 2pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (2-1-1) = 0
5. So for the 2p orbitals, there are no radial nodes at all. This is shown in fig.2.49(a) below:
 |
| Fig.2.49 |
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 2px, 2py and 2pz
• That means, each of px, py and pz, (in the second main-shell) have zero radial nodes and one angular node
Solved example 2.78
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the third main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3p
• For any of the three 3p orbital:
n = 3 and l = 1
4. Consider any one from among the three: 3px, 3py, 3pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (3-1-1) = 1
5. So for the 3p orbitals, there is one radial node. This is shown in fig.2.49(b) above
• There is one angular node also. We already saw it earlier in fig.2.46
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 3px, 3py and 3pz
• That means, each of px, py and pz, (in the third main-shell) have one radial nod and one angular node
Solved example 2.79
Find the number of nodes in the orbital of the p sub-shell. Assume that, this p sub-shell is situated in the fourth main-shell
Solution:
1. The given orbital is present in the p sub-shell
• There are three orbitals in the p sub-shell: px, py and pz
• They all have the shape of a dumb-bell
2. Given that, this p sub-shell is present in the fourth main-shell (n=4)
3. So in the s, p, d, f notation, the given orbital is: 4p
• For any of the three 4p orbital:
n = 4 and l = 1
4. Consider any one from among the three: 3px, 3py, 3pz
Applying the two equations, we get:
(i) Number of angular nodes = l = 1
(ii) Number of radial nodes = (n-1-l) = (4-1-1) = 2
5. So for the 4p orbitals, there are two radial nodes. This is shown in fig.2.49(c) above
• There is one angular node also. We already saw it earlier in fig.2.46
6. Note that the calculated numbers in (4) is applicable to each of the three p orbitals: 4px, 4py and 4pz
• That means, each of px, py and pz, (in the fourth main-shell) have two radial nodes and one angular node
Solved example 2.80
Solution:
1. The given orbital is present in the d sub-shell
• There are five orbitals in the d sub-shell
2. Given that, this d sub-shell is present in the third main-shell (n=3)
3. So in the s, p, d, f notation, the given orbital is: 3d
• For any of the five 3d orbitals:
n = 3 and l = 2
4. Applying the two equations, we get:
(i) Number of angular nodes = l = 2
(ii) Number of radial nodes = (n-1-l) = (3-1-2) = 0
5. So for the 3d orbitals, there are no radial nodes at all
• But there are two angular nodes. We already saw it earlier in fig.2.48(a)
6. Note that the calculated numbers in (4) is applicable to each of the five d orbitals (in the third main-shell)
This completes our present discussion on nodes. In the next section, we will see energies of various orbitals
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