In the previous section 2.16, we completed a discussion on radial and angular nodes. In this section we will see energies of orbitals
• Consider the hydrogen atom
• We know that, it has main-shells with n = 1, 2, 3 . . . ,
• Each of those main-shells will be having sub-shells
1. Consider the main-shell with n=1
2. This first main-shell will be having one sub-shell
• It is the s sub-shell
3. This s sub-shell will be having one orbital
• In s, p, d, f notation, this orbital is the 1s orbital
4. This 1s orbital will be having a particular energy
• Let us denote this energy as E1
2. This second main-shell will be having two sub-shells
• They are the s sub-shell and p sub-shell
3. The s sub-shell will be having one orbital
♦ In s, p, d, f notation, this orbital is the 2s orbital
♦ This 2s orbital will be having a particular energy
♦ Let us denote this energy as E2
4. The p sub-shell will be having three orbitals
♦ In s, p, d, f notation, they are: 2px, 2py and 2pz
♦ For our present discussion, we can call them collectively as the 2p orbital
♦ This 2p orbital will be having a particular energy
♦ This energy is the same E2 mentioned in (3)
5. That means, both the s and p orbitals in the second main-shell will be having the same energy E2
2. This third main-shell will be having three sub-shells
• They are the s sub-shell, the p sub-shell and the d sub-shell
3. The s sub-shell will be having one orbital
♦ In s, p, d, f notation, this orbital is the 3s orbital
♦ This 3s orbital will be having a particular energy
♦ Let us denote this energy as E3
4. The p sub-shell will be having three orbitals
♦ In s, p, d, f notation, they are: 3px, 3py and 3pz
♦ For our present discussion, we can call them collectively as the 3p orbital
♦ This 3p orbital will be having a particular energy
♦ This energy is the same E3 mentioned in (3)
5. The d sub-shell will be having five orbitals
♦ In s, p, d, f notation, they are: 3dxy, 3dyz . . . ,
♦ For our present discussion, we can call them collectively as the 3d orbital
♦ This 3d orbital will be having a particular energy
♦ This energy is the same E3 mentioned in (3)
6. That means, all the s, p and d orbitals in the third main-shell will be having the same energy E3
All the s, p, d and f orbitals in the fourth main-shell will be having the same energy E4
E1 < E2 < E3 < E4 < so on . . .
2. Inside the main-shells, there are orbitals having the above energy values. So we can write:
• The energy levels of orbitals are in the form:
[1s] < [2s=2p] < [3s=3p=3d] < [4s=4p=4d=4f] < so on . . .
3. Following points may be noted:
(i) The shapes of 2s and 2p orbitals are different. But they have the same energy
♦ Energy possessed by an electron in 2s orbital
♦ Will be equal to
♦ Energy possessed by an electron in 2p orbital
(ii) The shapes of 3s, 3p and 3d orbitals are different. But they have the same energy
♦ Energy possessed by an electron in 3s orbital
♦ Will be equal to
♦ Energy possessed by an electron in 3p orbital
♦ Will be equal to
♦ Energy possessed by an electron in 3d orbital
(iii) The same pattern can be written for the 4s, 4p, 4d and 4f orbitals also
4. Orbitals having same energy are called degenerate orbitals
5. The electron residing in the 1s orbital of the hydrogen atom will be having the least possible energy
• The 1s orbital of the hydrogen atom corresponds to the most stable condition
• So the 1s orbital of the hydrogen atom is called the ground state
♦ If the hydrogen electron is in the 1s orbital, that electron is said to be in the ground state
• All other orbitals of hydrogen are called excited states
♦ If the hydrogen electron is in an orbital other than the 1s, that electron is said to be in the excited state
• Next we have to see the general case of 'atoms which have more than one electron' (multi electron atoms)
• For that, we must have a greater knowledge about two items:
(i) Force between two charges
(ii) Energy possessed by the charges
We will write about them in steps:
1. The nucleus has positively charged protons. So the nucleus attracts the negatively charged electron
• This attractive force is the cause of the 'energy possessed by the electron'
[This is just like:
Potential energy possessed by 'an object at a height' is due to the gravitational force of attraction from the earth]
2. Consider any two charges q1 and q2
• The magnitude of the force of attraction/repulsion between them is given by: $\mathbf\small{\frac{k\,q_1\,q_2}{r^2}}$
• Where:
♦ k is a constant
♦ r is the distance between the two charges
3. In our present case, the first charge q1 is the 'total positive charge in the nucleus'
• It will be equal to (Z×e)
♦ Where Z is the atomic number of the atom, and e is the charge of one electron/proton
• The second charge q2 is the charge of the one electron under consideration
♦ So q2 = e
♦ r is the distance between the nucleus and the electron
■ So the magnitude of the force of attraction between the nucleus and the electron will be equal to $\mathbf\small{\frac{k\,(Ze)\,e}{r^2}}$
4. The energy possessed by the electron is given by $\mathbf\small{-\frac{k\,(Ze)\,e}{r}}$
(We will learn about these expressions in physics classes. At that time we will see the reason for putting the -ve sign)
5. Due to the presence of the -ve sign, it becomes necessary to carefully analyse the results
We will write an analysis using some examples
Example 1:
• Let Z = 7 and r = 5 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(7e)\,e}{5^2}=0.28\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(7e)\,e}{5}=-1.4\,k\,e^2}$ J
Example 2:
• Let Z = 10 and r = 5 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(10e)\,e}{5^2}=0.4\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(10e)\,e}{5}=-2\,k\,e^2}$ J
(i) This example is same as example 1, except that, Z is increased
• When Z is increased, the number of protons in the nucleus increases
• This will increase the force of attraction on the 'electron under consideration'
• The 'increase in force' is due to the fact that, Z is in the numerator
(ii) We find that, the force has increased from 0.28 N to 0.4 N
(iii) Now consider the energy. It has 'increased numerically'
• The initial value of 1.4 has become 2
• But note the negative sign:
Actually, -1.4 has become -2
(iv) -2 is 'more negative'. So the energy has actually decreased
(v) We can write:
When force of attraction increases, energy of the electron decreases
• 'Increased force of attraction' means that, the electron is more strongly bound to the nucleus
■ So we can write:
When the binding force on the electron increases, it's energy decreases
■ The converse is also true:
When the binding force on the electron decreases, it's energy increases
Example 3:
• Let Z = 7 and r = 2 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(7e)\,e}{2^2}=1.75\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(7e)\,e}{2}=-3.5\,k\,e^2}$ J
(i) This example is same as example 1, except that, r is decreased
• When r is decreased, the force of attraction on the 'electron under consideration' increases
• This 'increase in force' is due to the fact that, r is in the denominator
(ii) We find that, the force has increased from 0.28 N to 1.75 N
(iii) Now consider the energy. It has 'increased numerically'
• The initial value of 1.4 has become 3.5
• But note the negative sign:
Actually, -1.4 has become -3.5
(iv) -3.5 is 'more negative'. So the energy has actually decreased
(v) So we can write the same inference as we wrote in example 2:
When force of attraction increases, energy of the electron decreases
• 'Increased force of attraction' means that, the electron is more strongly bound to the nucleus
■ So we can write:
When the binding force on the electron increases, it's energy decreases
■ The converse is also true:
When the binding force on the electron decreases, it's energy increases
• In example 3, the 'increase in force' is due to the decrease in distance
(i) When the distance from the nucleus increases, the energy decreases numerically
• This is because, the distance r is in the denominator
(ii) But the -ve sign is present
• So 'numerically decreasing' is equivalent to 'becoming less negative'
(iii) So we can write:
• When the distance from the nucleus increases, the energy of the electron increases
• Finally, when the distance from the nucleus becomes infinity, the energy of the electron becomes zero. This is because, any quantity divided by infinity is zero
(iv) So we can write:
If the electron is far away from the influence of the nucleus, that electron will have zero energy
1. First, consider the case of the single electron hydrogen atom
(i) We want to calculate the energy possessed by it's electron
(ii) The calculation will involve only one term
• That term will take care of the attraction between the electron and the nucleus
• There is only one term because, there is only one force
♦ It is the force of attraction between the single electron and the nucleus
2. Now consider the case of a multi electron atom
(i) Let this be a small atom having lesser number of electrons
• We want to calculate the energy possessed by one of it's outermost electrons
(ii) The calculation will involve two terms
• As before, the first term will take care of the attraction between the electron and the nucleus
• The second term takes care of the repulsion between the 'electron under consideration' and other electrons
(iii) So the first term considers attraction and the second term considers repulsion
(iv) Obviously, the attraction will be greater than repulsion
• Otherwise, the electron will move away from the nucleus
(v) We know that, the energy possessed by the electron is due to the attraction from the nucleus
• Now, a repulsion (which is lesser than the attraction) has also come into play
• So the net effect is that, in a simple multi electron atom, the outermost electron will possess only a lesser energy than expected
3. Consider the case of another multi electron atom
(i) Let this be a larger atom having greater number of electrons
• We want to calculate the energy possessed by one of it's outermost electrons
(ii) The calculation will involve three terms
• As before, the first term will take care of the attraction between the electron and the nucleus
• As before, the second term takes care of the repulsion between the 'electron under consideration' and other electrons
• The third term takes care of the shielding effect
(iii) So calculations related to multi electron atoms with greater number of electrons involves more work. We first have to learn about shielding effect
• In the fig.2.50 below, the upper cloud A represents the outermost electron
• The lower cloud B represents the electrons in the inner shell
2. This lower cloud will act as a shield
• It will shield the outermost electron from the nucleus
• So the attraction on the outermost electron will be reduced
3. We know that, the force of attraction is given by: $\mathbf\small{\frac{k\,(Ze)\,e}{r^2}}$
• Due to the shielding effect, we cannot consider the full $\mathbf\small{(Ze)}$
• We get only the effective nuclear charge $\mathbf\small{(Z_{eff}e)}$
• $\mathbf\small{(Z_{eff})}$ is less than $\mathbf\small{Z}$
• So the energy will be reduced
4. Let us now see the 'factors which increase or decrease the shielding effect'
(i) Consider the lower cloud B in fig.2.50
• This lower cloud is causing the shielding effect
(ii) If this lower cloud is part of an s orbital, the shielding effect will be very high
• In other words, the s orbital is the one which causes the most shielding effect
• The p orbital cannot cause as much shielding as s
• The d orbital cannot cause as much shielding as p
• The f orbital cannot cause as much shielding as d
■ Let us write it in mathematical form:
The shielding capacity in ascending order is: f < d < p < s
(iii) s has the most shielding capacity because, it has a spherical shape
5. So, if we want a greater 'shielding effect', we must have s or p orbitals below the outer electron
• d and f orbitals cannot provide much shielding
6. If the principal quantum number n of the outermost electron is large, there will be more orbitals s, p, d and f below it
• If the principal quantum number n is small, there will be lesser orbitals s and p or even s only
■ So, to achieve greater shielding, and thus a decrease in energy, n must be as small as possible
7. Next consider the azimuthal quantum number l
(i) l = 0 corresponds to s orbital
• The electrons in the s orbitals are very strongly bound to the nucleus
• Remember what we saw in earlier sections: s orbitals become denser as we move towards the center
(ii) l = 1 corresponds to p orbital
• The electrons in the p orbitals are also strongly bound to the nucleus
• But the 'strength of binding' is less when compared to the s orbital
• Remember what we saw in earlier sections: p orbitals become denser as we move away from the center
(iii) l = 2 corresponds to d orbital
• The electrons in the d orbitals are not so strongly bound to the nucleus
• The 'strength of binding' is less when compared to the p orbital
• Remember what we saw in earlier sections: d orbitals become denser as we move away from the center
(iv) l = 3 corresponds to f orbital
• The electrons in the f orbitals are also not so strongly bound to the nucleus
• The 'strength of binding' is less when compared to the d orbital
• Remember what we saw in earlier sections: f orbitals have a diffused shape
■ Let us write it in mathematical form:
The strength of binding in the ascending order is: f < d < p < s
8. So, if we want the 'electron in the outer most shell' to be strongly bound to the nucleus, the l value must be as low as possible
9. If this electron is strongly bound to the nucleus, what advantage do we get?
Answer:
• We have seen at the beginning of this section (Examples 1, 2 and 3) that:
When the binding force on the electron increases, it's energy decreases
10. So we can write:
Lesser the l value of an electron, lesser will be it's energy
11. From (6) we get:
If the outermost electron is to have a lower energy, it's n value must be as low as possible
• From (10), we get:
If the outermost electron is to have a lower energy, it's l value must be as low as possible
■ Combining the results in (6) and (10), we can write:
If the outermost electron is to have a lower energy, it's (n+l) value must be as low as possible
1. First we will write the energy levels in hydrogen atom:
(i) First energy level: 1s
(ii) Next higher energy level: 2s, 2p (both have the same energy)
(iii) Next higher energy level: 3s, 3p, 3d (All three have the same energy)
(iv) Next higher energy level: 4s, 4p, 4d (All four have the same energy)
2. The perfect order seen in the case of hydrogen atom cannot be obtained for higher atoms
(i) The order will be lost when we apply the (n+l) rule
(ii) The first disorder occurs in the case of 3d and 4s
♦ For 3d, the (n+l) value is: (3+2) = 5
♦ For 4s, the (n+l) value is: (4+0) = 4
• So 4s will have a lower energy than 3d
• In the case of hydrogen, we see that, 4s has greater energy than 3d
3. The second disorder occurs in the case of 4d and 5s
♦ For 4d, the (n+l) value is: (4+2) = 6
♦ For 5s, the (n+l) value is: (5+0) = 5
• So 5s will have a lower energy than 4d
• In the case of hydrogen, we see that, 5s has greater energy than 4d
So we prepare a chart. It is a simple geometric construction. The steps are given below:
1. First write the orbitals in columns and rows as shown in fig.2.51(a) below:
• One column for s, one column for p, . . . so on
• One row for 2, one row for 3, . . . so on
2. Draw arrows through them as shown in the fig.b.
■ Arrows are drawn in a slanting position so that:
• The 'succeeding orbital' of any orbital will be the one in:
♦ column on the left and
♦ row at the bottom
3. The arrows are numbered 1, 2, 3, . . . so on. Eight arrows are shown in fig.b
• But in fact, there is only one arrow. The eight arrows are made into one, by connecting them together with a dashed line. This is shown in fig.2.51(c) below:
4. The geometrical construction is complete. Now we will see how this fig.c can be used:
• Place your finger tip at the tail end of the top most arrow. Move the finger tip along the direction of the arrow
• You will first reach the head end of the top most arrow. Then move along the dashed line
• You will reach the tail end of the second arrow. Continue in this way
• The path followed by the finger tip gives the order of the energy levels of the orbitals
♦ As the finger tip moves, the energy level increases
• We can see that, there is no problem until 3p. But after that, instead of going to 3d, the path goes to 4s
• Similarly, after 4p, instead of going to 4d, the path goes to 5s
■ Thus, the order of energy levels can be easily determined using fig.c
• If two electrons happen to have the same (n+l) value, that electron with the lower n value will have the lesser energy
• Consider the hydrogen atom
• We know that, it has main-shells with n = 1, 2, 3 . . . ,
• Each of those main-shells will be having sub-shells
1. Consider the main-shell with n=1
2. This first main-shell will be having one sub-shell
• It is the s sub-shell
3. This s sub-shell will be having one orbital
• In s, p, d, f notation, this orbital is the 1s orbital
4. This 1s orbital will be having a particular energy
• Let us denote this energy as E1
1. Consider the main-shell with n=2
• They are the s sub-shell and p sub-shell
3. The s sub-shell will be having one orbital
♦ In s, p, d, f notation, this orbital is the 2s orbital
♦ This 2s orbital will be having a particular energy
♦ Let us denote this energy as E2
4. The p sub-shell will be having three orbitals
♦ In s, p, d, f notation, they are: 2px, 2py and 2pz
♦ For our present discussion, we can call them collectively as the 2p orbital
♦ This 2p orbital will be having a particular energy
♦ This energy is the same E2 mentioned in (3)
5. That means, both the s and p orbitals in the second main-shell will be having the same energy E2
1. Consider the main-shell with n=3
• They are the s sub-shell, the p sub-shell and the d sub-shell
3. The s sub-shell will be having one orbital
♦ In s, p, d, f notation, this orbital is the 3s orbital
♦ This 3s orbital will be having a particular energy
♦ Let us denote this energy as E3
4. The p sub-shell will be having three orbitals
♦ In s, p, d, f notation, they are: 3px, 3py and 3pz
♦ For our present discussion, we can call them collectively as the 3p orbital
♦ This 3p orbital will be having a particular energy
♦ This energy is the same E3 mentioned in (3)
5. The d sub-shell will be having five orbitals
♦ In s, p, d, f notation, they are: 3dxy, 3dyz . . . ,
♦ For our present discussion, we can call them collectively as the 3d orbital
♦ This 3d orbital will be having a particular energy
♦ This energy is the same E3 mentioned in (3)
6. That means, all the s, p and d orbitals in the third main-shell will be having the same energy E3
Following this pattern, we can write the case of n = 4 also:
1. When the distance from the nucleus increases, the energy also increases. So in mathematical form, we can write:
2. Inside the main-shells, there are orbitals having the above energy values. So we can write:
• The energy levels of orbitals are in the form:
[1s] < [2s=2p] < [3s=3p=3d] < [4s=4p=4d=4f] < so on . . .
3. Following points may be noted:
(i) The shapes of 2s and 2p orbitals are different. But they have the same energy
♦ Energy possessed by an electron in 2s orbital
♦ Will be equal to
♦ Energy possessed by an electron in 2p orbital
(ii) The shapes of 3s, 3p and 3d orbitals are different. But they have the same energy
♦ Energy possessed by an electron in 3s orbital
♦ Will be equal to
♦ Energy possessed by an electron in 3p orbital
♦ Will be equal to
♦ Energy possessed by an electron in 3d orbital
(iii) The same pattern can be written for the 4s, 4p, 4d and 4f orbitals also
4. Orbitals having same energy are called degenerate orbitals
5. The electron residing in the 1s orbital of the hydrogen atom will be having the least possible energy
• The 1s orbital of the hydrogen atom corresponds to the most stable condition
• So the 1s orbital of the hydrogen atom is called the ground state
♦ If the hydrogen electron is in the 1s orbital, that electron is said to be in the ground state
• All other orbitals of hydrogen are called excited states
♦ If the hydrogen electron is in an orbital other than the 1s, that electron is said to be in the excited state
• The above discussion is related to hydrogen atom. A hydrogen atom has only one electron
• For that, we must have a greater knowledge about two items:
(i) Force between two charges
(ii) Energy possessed by the charges
We will write about them in steps:
1. The nucleus has positively charged protons. So the nucleus attracts the negatively charged electron
• This attractive force is the cause of the 'energy possessed by the electron'
[This is just like:
Potential energy possessed by 'an object at a height' is due to the gravitational force of attraction from the earth]
2. Consider any two charges q1 and q2
• The magnitude of the force of attraction/repulsion between them is given by: $\mathbf\small{\frac{k\,q_1\,q_2}{r^2}}$
• Where:
♦ k is a constant
♦ r is the distance between the two charges
3. In our present case, the first charge q1 is the 'total positive charge in the nucleus'
• It will be equal to (Z×e)
♦ Where Z is the atomic number of the atom, and e is the charge of one electron/proton
• The second charge q2 is the charge of the one electron under consideration
♦ So q2 = e
♦ r is the distance between the nucleus and the electron
■ So the magnitude of the force of attraction between the nucleus and the electron will be equal to $\mathbf\small{\frac{k\,(Ze)\,e}{r^2}}$
4. The energy possessed by the electron is given by $\mathbf\small{-\frac{k\,(Ze)\,e}{r}}$
(We will learn about these expressions in physics classes. At that time we will see the reason for putting the -ve sign)
5. Due to the presence of the -ve sign, it becomes necessary to carefully analyse the results
We will write an analysis using some examples
Example 1:
• Let Z = 7 and r = 5 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(7e)\,e}{5^2}=0.28\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(7e)\,e}{5}=-1.4\,k\,e^2}$ J
Example 2:
• Let Z = 10 and r = 5 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(10e)\,e}{5^2}=0.4\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(10e)\,e}{5}=-2\,k\,e^2}$ J
(i) This example is same as example 1, except that, Z is increased
• When Z is increased, the number of protons in the nucleus increases
• This will increase the force of attraction on the 'electron under consideration'
• The 'increase in force' is due to the fact that, Z is in the numerator
(ii) We find that, the force has increased from 0.28 N to 0.4 N
(iii) Now consider the energy. It has 'increased numerically'
• The initial value of 1.4 has become 2
• But note the negative sign:
Actually, -1.4 has become -2
(iv) -2 is 'more negative'. So the energy has actually decreased
(v) We can write:
When force of attraction increases, energy of the electron decreases
• 'Increased force of attraction' means that, the electron is more strongly bound to the nucleus
■ So we can write:
When the binding force on the electron increases, it's energy decreases
■ The converse is also true:
When the binding force on the electron decreases, it's energy increases
Example 3:
• Let Z = 7 and r = 2 units
• Then we get:
♦ Force of attraction = $\mathbf\small{\frac{k\,(7e)\,e}{2^2}=1.75\,k\,e^2}$ N
♦ Energy of electron = $\mathbf\small{-\frac{k\,(7e)\,e}{2}=-3.5\,k\,e^2}$ J
(i) This example is same as example 1, except that, r is decreased
• When r is decreased, the force of attraction on the 'electron under consideration' increases
• This 'increase in force' is due to the fact that, r is in the denominator
(ii) We find that, the force has increased from 0.28 N to 1.75 N
(iii) Now consider the energy. It has 'increased numerically'
• The initial value of 1.4 has become 3.5
• But note the negative sign:
Actually, -1.4 has become -3.5
(iv) -3.5 is 'more negative'. So the energy has actually decreased
(v) So we can write the same inference as we wrote in example 2:
When force of attraction increases, energy of the electron decreases
• 'Increased force of attraction' means that, the electron is more strongly bound to the nucleus
■ So we can write:
When the binding force on the electron increases, it's energy decreases
■ The converse is also true:
When the binding force on the electron decreases, it's energy increases
• In example 2, the 'increase in force' is due to the increase in charge of the nucleus
We can also note another important property. We will write it in steps:
• This is because, the distance r is in the denominator
(ii) But the -ve sign is present
• So 'numerically decreasing' is equivalent to 'becoming less negative'
(iii) So we can write:
• When the distance from the nucleus increases, the energy of the electron increases
• Finally, when the distance from the nucleus becomes infinity, the energy of the electron becomes zero. This is because, any quantity divided by infinity is zero
(iv) So we can write:
If the electron is far away from the influence of the nucleus, that electron will have zero energy
Now we can take up the case of multi electron atoms
(i) We want to calculate the energy possessed by it's electron
(ii) The calculation will involve only one term
• That term will take care of the attraction between the electron and the nucleus
• There is only one term because, there is only one force
♦ It is the force of attraction between the single electron and the nucleus
2. Now consider the case of a multi electron atom
(i) Let this be a small atom having lesser number of electrons
• We want to calculate the energy possessed by one of it's outermost electrons
(ii) The calculation will involve two terms
• As before, the first term will take care of the attraction between the electron and the nucleus
• The second term takes care of the repulsion between the 'electron under consideration' and other electrons
(iii) So the first term considers attraction and the second term considers repulsion
(iv) Obviously, the attraction will be greater than repulsion
• Otherwise, the electron will move away from the nucleus
(v) We know that, the energy possessed by the electron is due to the attraction from the nucleus
• Now, a repulsion (which is lesser than the attraction) has also come into play
• So the net effect is that, in a simple multi electron atom, the outermost electron will possess only a lesser energy than expected
3. Consider the case of another multi electron atom
(i) Let this be a larger atom having greater number of electrons
• We want to calculate the energy possessed by one of it's outermost electrons
(ii) The calculation will involve three terms
• As before, the first term will take care of the attraction between the electron and the nucleus
• As before, the second term takes care of the repulsion between the 'electron under consideration' and other electrons
• The third term takes care of the shielding effect
(iii) So calculations related to multi electron atoms with greater number of electrons involves more work. We first have to learn about shielding effect
1. We know that the region in which electrons are present, will appear as clouds
• The lower cloud B represents the electrons in the inner shell
 |
| Fig.2.50 |
• It will shield the outermost electron from the nucleus
• So the attraction on the outermost electron will be reduced
3. We know that, the force of attraction is given by: $\mathbf\small{\frac{k\,(Ze)\,e}{r^2}}$
• Due to the shielding effect, we cannot consider the full $\mathbf\small{(Ze)}$
• We get only the effective nuclear charge $\mathbf\small{(Z_{eff}e)}$
• $\mathbf\small{(Z_{eff})}$ is less than $\mathbf\small{Z}$
• So the energy will be reduced
4. Let us now see the 'factors which increase or decrease the shielding effect'
(i) Consider the lower cloud B in fig.2.50
• This lower cloud is causing the shielding effect
(ii) If this lower cloud is part of an s orbital, the shielding effect will be very high
• In other words, the s orbital is the one which causes the most shielding effect
• The p orbital cannot cause as much shielding as s
• The d orbital cannot cause as much shielding as p
• The f orbital cannot cause as much shielding as d
■ Let us write it in mathematical form:
The shielding capacity in ascending order is: f < d < p < s
(iii) s has the most shielding capacity because, it has a spherical shape
5. So, if we want a greater 'shielding effect', we must have s or p orbitals below the outer electron
• d and f orbitals cannot provide much shielding
6. If the principal quantum number n of the outermost electron is large, there will be more orbitals s, p, d and f below it
• If the principal quantum number n is small, there will be lesser orbitals s and p or even s only
■ So, to achieve greater shielding, and thus a decrease in energy, n must be as small as possible
7. Next consider the azimuthal quantum number l
(i) l = 0 corresponds to s orbital
• The electrons in the s orbitals are very strongly bound to the nucleus
• Remember what we saw in earlier sections: s orbitals become denser as we move towards the center
(ii) l = 1 corresponds to p orbital
• The electrons in the p orbitals are also strongly bound to the nucleus
• But the 'strength of binding' is less when compared to the s orbital
• Remember what we saw in earlier sections: p orbitals become denser as we move away from the center
(iii) l = 2 corresponds to d orbital
• The electrons in the d orbitals are not so strongly bound to the nucleus
• The 'strength of binding' is less when compared to the p orbital
• Remember what we saw in earlier sections: d orbitals become denser as we move away from the center
(iv) l = 3 corresponds to f orbital
• The electrons in the f orbitals are also not so strongly bound to the nucleus
• The 'strength of binding' is less when compared to the d orbital
• Remember what we saw in earlier sections: f orbitals have a diffused shape
■ Let us write it in mathematical form:
The strength of binding in the ascending order is: f < d < p < s
8. So, if we want the 'electron in the outer most shell' to be strongly bound to the nucleus, the l value must be as low as possible
9. If this electron is strongly bound to the nucleus, what advantage do we get?
Answer:
• We have seen at the beginning of this section (Examples 1, 2 and 3) that:
When the binding force on the electron increases, it's energy decreases
10. So we can write:
Lesser the l value of an electron, lesser will be it's energy
11. From (6) we get:
If the outermost electron is to have a lower energy, it's n value must be as low as possible
• From (10), we get:
If the outermost electron is to have a lower energy, it's l value must be as low as possible
■ Combining the results in (6) and (10), we can write:
If the outermost electron is to have a lower energy, it's (n+l) value must be as low as possible
Let us see the application of this (n+l) value
(i) First energy level: 1s
(ii) Next higher energy level: 2s, 2p (both have the same energy)
(iii) Next higher energy level: 3s, 3p, 3d (All three have the same energy)
(iv) Next higher energy level: 4s, 4p, 4d (All four have the same energy)
2. The perfect order seen in the case of hydrogen atom cannot be obtained for higher atoms
(i) The order will be lost when we apply the (n+l) rule
(ii) The first disorder occurs in the case of 3d and 4s
♦ For 3d, the (n+l) value is: (3+2) = 5
♦ For 4s, the (n+l) value is: (4+0) = 4
• So 4s will have a lower energy than 3d
• In the case of hydrogen, we see that, 4s has greater energy than 3d
3. The second disorder occurs in the case of 4d and 5s
♦ For 4d, the (n+l) value is: (4+2) = 6
♦ For 5s, the (n+l) value is: (5+0) = 5
• So 5s will have a lower energy than 4d
• In the case of hydrogen, we see that, 5s has greater energy than 4d
There are more disorders like this. It is not easy to remember each of them in sequential order
1. First write the orbitals in columns and rows as shown in fig.2.51(a) below:
• One column for s, one column for p, . . . so on
• One row for 2, one row for 3, . . . so on
 |
| Fig,2.51 |
■ Arrows are drawn in a slanting position so that:
• The 'succeeding orbital' of any orbital will be the one in:
♦ column on the left and
♦ row at the bottom
3. The arrows are numbered 1, 2, 3, . . . so on. Eight arrows are shown in fig.b
• But in fact, there is only one arrow. The eight arrows are made into one, by connecting them together with a dashed line. This is shown in fig.2.51(c) below:
 |
| Fig.2.51 |
• Place your finger tip at the tail end of the top most arrow. Move the finger tip along the direction of the arrow
• You will first reach the head end of the top most arrow. Then move along the dashed line
• You will reach the tail end of the second arrow. Continue in this way
• The path followed by the finger tip gives the order of the energy levels of the orbitals
♦ As the finger tip moves, the energy level increases
• We can see that, there is no problem until 3p. But after that, instead of going to 3d, the path goes to 4s
• Similarly, after 4p, instead of going to 4d, the path goes to 5s
■ Thus, the order of energy levels can be easily determined using fig.c
• We saw that, the (n+l) rule helps us to compare the energies of any two electrons
In the next section, we will see filling of orbitals in atoms
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