In the previous section 2.17, we completed a discussion on energies of orbitals. We saw the chart in fig.2.51(c), which gives the increasing order of energies of orbitals. So in this section, we are ready to fill orbitals with electrons. Let us see how it is done:
• We cannot put electrons into any orbitals we like. There are certain rules to be followed
• The filling must take place according to the aufbau principle
• The aufbau principle is based on three rules:
(i) Pauli’s exclusion principle
(ii) Hund’s rule of maximum multiplicity
(iii) Relative energy of orbitals
• On hearing ‘building up’, many things come to our minds:
♦ Building a house from bottom foundation towards the top
♦ Building a machine from bottom base towards the top
♦ so on . . .
2. Our present case of ‘filling orbitals’ is also similar to the above cases
• We fill the ‘orbitals with lower energies’ first
• Then we gradually move upwards to fill ‘orbitals with higher energies’
■ The aufbau principle states that:
In ground state of the atoms, orbitals are filled in order of their increasing energies
3. So we can write:
• The electrons first occupy the lowest energy orbitals available to them
• They enter into higher energy orbitals only after filling the lower energy orbitals
During the process of filling, we will learn about the Pauli’s exclusion principle and Hund’s rule of maximum multiplicity
1. The first element is hydrogen
• It has one electron. This electron will be placed in the orbital with the lowest energy, which is the 1s orbital
2. To convey the information about ‘where the electron is located’ in the hydrogen atom, we draw a diagram
• It is called the orbital diagram
• The orbital diagram of hydrogen is shown in fig.2.52 below:
3. Let us see the features of this diagram
(i) We denote each orbital of the atom by a small rectangular box
• So, upon seeing a box, we must immediately recognize that, it is an orbital
(ii) Name of that orbital is written below the box
• In our present case, it is 's'
(iii) In front of that name, we write the ‘number of the main-shell’ in which that orbital is situated
• In our present case, it is '1'
(iv) Inside the box, we show the electrons which are present in that orbital
• The electrons are indicated using up arrow (↑) or down arrow (↓)
• In our present case, there is only one electron. It is denoted by the (↑) arrow
4. So the orbital diagram clearly conveys the information about the 'distribution of electrons into the orbitals'
■ The 'distribution of electrons into the orbitals' of an atom is called the electronic configuration of that atom
5. What we saw above, is called the orbital diagram method for writing the electronic configuration
• There is another method for writing the electronic configuration
• It is called the s, p, d, f notation
6. In the s, p, d,f notation, the electronic configuration of hydrogen is: 1s1
• Let us see the features of this method:
(i) In 1s1, there are:
♦ One letter, which is: ‘s’
♦ Two numbers, which are: ‘1’ and ‘1’
(ii) The letter ‘s’ indicates the name of the sub-shell
(iii) The number ‘1’ in front of ‘s’ indicates the main-shell in which the sub-shell s is situated
(iv) The number ‘1’ written as the superscript of ‘s’ indicates the number of electrons in the sub-shell 1s
7. Electronic configuration of hydrogen using s, p, d, f method is also added in fig.2.52
• It is shown in green color
• When the two method are shown together, we can make an easy comparison
1. Helium has two electrons
• It’s first electron can go into the 1s orbital just like in hydrogen
• We have to find a place for the second electron
2. Here we need to apply Pauli’s exclusion principle
(We have seen the basics about this rule when we learnt about spin quantum number ms)
■ It states that, no two electrons in an atom can have the same set of four quantum numbers
Let us elaborate
(i) Consider an atom of any element (other than hydrogen)
• That atom will contain two or more electrons
(ii) Consider any two electrons from them
• Their addresses should not be the same
(iii) We know that, the address of each electron is specified by a set of four quantum numbers: {n, l, ml, ms}
• Each electron has it's own set
(iv) Inside an atom, no two sets should be identical
• So according to Pauli's exclusion principle:
♦ If n, l, and ml of two electrons are the same, their ms values should be different
♦ If n, l, and ms of two electrons are the same, their ml values should be different
♦ If n, ml, and ms of two electrons are the same, their l values should be different
♦ If l, ml, and ms of two electrons are the same, their n values should be different
3. Let us apply it to the second electron of helium
(i) We will try to place it in the same 1s orbital where the first electron is residing
• Then the first three quantum numbers of both the electrons will be the same:
(n = 1); (l = 0); (ml = 0)
(ii) To avoid ms values also becoming the same, the two electrons can be given opposite spins
• Let the first electron be (↑)
• Let the second electron be (↓)
■ Then the ms values of the first and second electrons will be +1⁄2 and -1⁄2 respectively
(iii) We can write:
♦ First electron will have the quantum numbers: {1, 0, 0, +1⁄2}
♦ First electron will have the quantum numbers: {1, 0, 0, -1⁄2}
• So there will be no violation of Pauli's exclusion principle
4. The two electrons of helium can reside safely in the 1s orbital
• The electronic configuration of helium using the orbital diagram method is shown in fig.2.53 below:
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• When the two method are shown together, we can make an easy comparison
• It’s first two electrons can go into the 1s orbital just like in helium
• We have to find a place for the third electron
2. Can we put the third electron in the same 1s orbital ?
• To find the answer, we need to apply Pauli’s exclusion principle again
• We have already seen the details of this rule (step 2 of helium)
(i) The two electrons already present in the 1s orbital, have the following quantum numbers:
♦ First electron has the quantum numbers: {1, 0, 0, +1⁄2}
♦ Second electron has the quantum numbers: {1, 0, 0, -1⁄2}
(ii) If we put the third electron also in 1s, two possibilities arise:
♦ Third electron has the quantum numbers: {1, 0, 0, +1⁄2} OR
♦ Third electron has the quantum numbers: {1, 0, 0, -1⁄2}
(iii) Neither of the two possibilities written in (ii) can be allowed. Because:
• If the first possibility is allowed, the third electron will have the same four quantum numbers as the first electron
• If the second possibility is allowed, the third electron will have the same four quantum numbers as the second electron
(iv) So there is no place for the third electron in the 1s orbital
• We have to put it in the next higher orbital, which is: the 2s orbital
(v) From the above 4 steps (i), (ii), (iii) and (iv) we can arrive at an important conclusion:
■ No orbital can hold more than two electrons
• We proved it for the 1s orbital
• The reader may try any orbital. If a third electron is put into that orbital, two of those three will end up with the same quantum numbers
• In order to avoid such a situation, we must restrict the 'number of electrons in any orbital' to 2
(vi) So from now on, we need not write detailed steps to check 'whether any violation of Pauli's principle can occur'
• If we restrict the 'number of electrons in any orbital' to 2, no further 'checks related to Pauli's principle', need to be made
3. So we place the third electron of lithium in the 2s orbital
• The electronic configuration of lithium using the orbital diagram method is shown in fig.2.54 below:
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• When the two method are shown together, we can make an easy comparison
1. Beryllium has four electrons
• It’s first three electrons can go into the 1s and 2s orbitals just like in the previous lithium
• We have to find a place for the fourth electron
2. We saw that, any orbital can accommodate 2 and only 2 electrons
• So the fourth electron of beryllium can safely reside in the 2s orbital
• The two electrons in the 2s orbital will have opposite spins
3. The electronic configuration of beryllium using the orbital diagram method is shown in fig.2.55 below:
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
1. Boron has five electrons
• It’s first four electrons can go into the 1s and 2s orbitals just like in the previous beryllium
• We have to find a place for the fifth electron
2. With beryllium, the 1s and 2s orbitals are completely filled up
(i) From the chart in fig.2.51(c), we see that, the next higher orbital is 2p
(ii) Any p sub-shell in an atom will have 3 orbitals: px, py and pz
• Those three orbitals will be always degenerate orbitals
• That means, they will be having the same energy levels
(iii) So in our present case, we have 2px, 2py and 2pz, all with the same energy and ready to be occupied
(iv) We place our fifth electron of boron in 2px
3. The electronic configuration of boron using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
■ Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the fifth electron of boron is in the p sub-shell of the second main-shell
1. Carbon has six electrons
• It’s first five electrons can go into the 1s, 2s and 2px orbitals just like in the previous boron
• We have to find a place for the sixth electron
2. We see that, 2px can accommodate one more electron
• Also, 2py and 2pz are completely free
• Which of the three shall we choose?
3. To find the answer, we have to apply the Hund's rule of maximum multiplicity
(i) We have to apply this rule whenever we face 'orbitals belonging to the same sub-shell'
• That means, we apply this rule when we face any one of the following 3 situations:
♦ We face the three orbitals px, py, pz of the p sub-shell
(all three are degenerate orbitals)
♦ We face the five orbitals $\mathbf\small{d_{xy},\,d_{xz},\,d_{yz},\,d_{x^2-y^2},\,d_{z^2}}$ of the d sub-shell
(all five are degenerate orbitals)
♦ We face the seven orbitals of the f sub-shell
(all seven are degenerate orbitals)
(ii) The word 'multiplicity' means: More than one of 'something'
• Consider 'something'. For example, a quantity 'k'
♦ If we have more than one of 'k', we call it a 'multiplicity'
(iii) In our present case, we have a set of 3 degenerate orbitals in front of us
• Each of them have the same energy
• Initially, all of them are empty
(iv) We will get maximum multiplicity if we follow the procedure given below:
♦ Put one electron in each orbital
♦ Thus all three will become 'half occupied'
♦ It is a multiplicity of 'half occupied orbitals'
♦ To attain a maximum of this multiplicity, we try to get as many 'half filled orbitals' as possible
♦ Here we can attain a maximum of three 'half filled orbitals'. This is shown below:
(In the case of the d sub-shell, we can attain a maximum multiplicity, if we allow all of the five orbitals to remain 'half filled')
(v) After making all the three orbitals 'half occupied', put the next electron in the first orbital. This is shown below:
(vi) Thus, pairing of electrons can start only after making all the degenerate orbitals 'half occupied'
• This type of filling gives extra stability to the atoms
(vii) Hund's rule of maximum multiplicity states that:
Pairing of electrons in the orbitals belonging to the same sub-shell does not take place until each orbital belonging to that sub-shell has got one electron each
4. Applying this rule to our present case, we can write:
The sixth electron of carbon must go into the py orbital
5. The electronic configuration of carbon using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the sixth electron of carbon is in the p sub-shell of the second main-shell
1. Nitrogen has seven electrons
• It’s first six electrons can go into the 1s, 2s, 2px and 2py orbitals just like in the previous carbon
• We have to find a place for the seventh electron
2. It is obvious that, according to Hund's rule, the seventh electron must go into the pz orbital
• The electronic configuration of nitrogen using the orbital diagram method is shown in fig.2.56 below
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the 7th electron of nitrogen is in the p sub-shell of the second main-shell
• We cannot put electrons into any orbitals we like. There are certain rules to be followed
• The filling must take place according to the aufbau principle
• The aufbau principle is based on three rules:
(i) Pauli’s exclusion principle
(ii) Hund’s rule of maximum multiplicity
(iii) Relative energy of orbitals
Afbau principle
1. The word ‘aufbau’ in German means ‘building up’• On hearing ‘building up’, many things come to our minds:
♦ Building a house from bottom foundation towards the top
♦ Building a machine from bottom base towards the top
♦ so on . . .
2. Our present case of ‘filling orbitals’ is also similar to the above cases
• We fill the ‘orbitals with lower energies’ first
• Then we gradually move upwards to fill ‘orbitals with higher energies’
■ The aufbau principle states that:
In ground state of the atoms, orbitals are filled in order of their increasing energies
3. So we can write:
• The electrons first occupy the lowest energy orbitals available to them
• They enter into higher energy orbitals only after filling the lower energy orbitals
• From the chart that we saw in fig.2.51(c) of the previous section, we know the order in which the ‘energies of orbitals’ increases. For convenience, the chart is shown again below:
• So let us start filling:
 |
| Fig.2.51 |
1. The first element is hydrogen
• It has one electron. This electron will be placed in the orbital with the lowest energy, which is the 1s orbital
2. To convey the information about ‘where the electron is located’ in the hydrogen atom, we draw a diagram
• It is called the orbital diagram
• The orbital diagram of hydrogen is shown in fig.2.52 below:
 |
| Fig.2.52 |
(i) We denote each orbital of the atom by a small rectangular box
• So, upon seeing a box, we must immediately recognize that, it is an orbital
(ii) Name of that orbital is written below the box
• In our present case, it is 's'
(iii) In front of that name, we write the ‘number of the main-shell’ in which that orbital is situated
• In our present case, it is '1'
(iv) Inside the box, we show the electrons which are present in that orbital
• The electrons are indicated using up arrow (↑) or down arrow (↓)
• In our present case, there is only one electron. It is denoted by the (↑) arrow
4. So the orbital diagram clearly conveys the information about the 'distribution of electrons into the orbitals'
■ The 'distribution of electrons into the orbitals' of an atom is called the electronic configuration of that atom
5. What we saw above, is called the orbital diagram method for writing the electronic configuration
• There is another method for writing the electronic configuration
• It is called the s, p, d, f notation
6. In the s, p, d,f notation, the electronic configuration of hydrogen is: 1s1
• Let us see the features of this method:
(i) In 1s1, there are:
♦ One letter, which is: ‘s’
♦ Two numbers, which are: ‘1’ and ‘1’
(ii) The letter ‘s’ indicates the name of the sub-shell
(iii) The number ‘1’ in front of ‘s’ indicates the main-shell in which the sub-shell s is situated
(iv) The number ‘1’ written as the superscript of ‘s’ indicates the number of electrons in the sub-shell 1s
7. Electronic configuration of hydrogen using s, p, d, f method is also added in fig.2.52
• It is shown in green color
• When the two method are shown together, we can make an easy comparison
So we have completed the electronic configuration of hydrogen (H). Next is the element with atomic number 2, which is: helium (He)
• It’s first electron can go into the 1s orbital just like in hydrogen
• We have to find a place for the second electron
2. Here we need to apply Pauli’s exclusion principle
(We have seen the basics about this rule when we learnt about spin quantum number ms)
■ It states that, no two electrons in an atom can have the same set of four quantum numbers
Let us elaborate
(i) Consider an atom of any element (other than hydrogen)
• That atom will contain two or more electrons
(ii) Consider any two electrons from them
• Their addresses should not be the same
(iii) We know that, the address of each electron is specified by a set of four quantum numbers: {n, l, ml, ms}
• Each electron has it's own set
(iv) Inside an atom, no two sets should be identical
• So according to Pauli's exclusion principle:
♦ If n, l, and ml of two electrons are the same, their ms values should be different
♦ If n, l, and ms of two electrons are the same, their ml values should be different
♦ If n, ml, and ms of two electrons are the same, their l values should be different
♦ If l, ml, and ms of two electrons are the same, their n values should be different
3. Let us apply it to the second electron of helium
(i) We will try to place it in the same 1s orbital where the first electron is residing
• Then the first three quantum numbers of both the electrons will be the same:
(n = 1); (l = 0); (ml = 0)
(ii) To avoid ms values also becoming the same, the two electrons can be given opposite spins
• Let the first electron be (↑)
• Let the second electron be (↓)
■ Then the ms values of the first and second electrons will be +1⁄2 and -1⁄2 respectively
(iii) We can write:
♦ First electron will have the quantum numbers: {1, 0, 0, +1⁄2}
♦ First electron will have the quantum numbers: {1, 0, 0, -1⁄2}
• So there will be no violation of Pauli's exclusion principle
4. The two electrons of helium can reside safely in the 1s orbital
• The electronic configuration of helium using the orbital diagram method is shown in fig.2.53 below:
 |
| Fig.2.53 |
• When the two method are shown together, we can make an easy comparison
So we have completed the electronic configuration of helium. Next is the element with atomic number 3, which is: lithium (Li)
1. Lithium has two electrons
1. Lithium has two electrons
• We have to find a place for the third electron
2. Can we put the third electron in the same 1s orbital ?
• To find the answer, we need to apply Pauli’s exclusion principle again
• We have already seen the details of this rule (step 2 of helium)
(i) The two electrons already present in the 1s orbital, have the following quantum numbers:
♦ First electron has the quantum numbers: {1, 0, 0, +1⁄2}
♦ Second electron has the quantum numbers: {1, 0, 0, -1⁄2}
(ii) If we put the third electron also in 1s, two possibilities arise:
♦ Third electron has the quantum numbers: {1, 0, 0, +1⁄2} OR
♦ Third electron has the quantum numbers: {1, 0, 0, -1⁄2}
(iii) Neither of the two possibilities written in (ii) can be allowed. Because:
• If the first possibility is allowed, the third electron will have the same four quantum numbers as the first electron
• If the second possibility is allowed, the third electron will have the same four quantum numbers as the second electron
(iv) So there is no place for the third electron in the 1s orbital
• We have to put it in the next higher orbital, which is: the 2s orbital
(v) From the above 4 steps (i), (ii), (iii) and (iv) we can arrive at an important conclusion:
■ No orbital can hold more than two electrons
• We proved it for the 1s orbital
• The reader may try any orbital. If a third electron is put into that orbital, two of those three will end up with the same quantum numbers
• In order to avoid such a situation, we must restrict the 'number of electrons in any orbital' to 2
(vi) So from now on, we need not write detailed steps to check 'whether any violation of Pauli's principle can occur'
• If we restrict the 'number of electrons in any orbital' to 2, no further 'checks related to Pauli's principle', need to be made
3. So we place the third electron of lithium in the 2s orbital
• The electronic configuration of lithium using the orbital diagram method is shown in fig.2.54 below:
 |
| Fig.2.54 |
• When the two method are shown together, we can make an easy comparison
So we have completed the electronic configuration of lithium. Next is the element with atomic number 4, which is: beryllium (Be)
• It’s first three electrons can go into the 1s and 2s orbitals just like in the previous lithium
• We have to find a place for the fourth electron
2. We saw that, any orbital can accommodate 2 and only 2 electrons
• So the fourth electron of beryllium can safely reside in the 2s orbital
• The two electrons in the 2s orbital will have opposite spins
3. The electronic configuration of beryllium using the orbital diagram method is shown in fig.2.55 below:
 |
| Fig.2.55 |
So we have completed the electronic configuration of beryllium. Next is the element with atomic number 5, which is: boron (B)
• It’s first four electrons can go into the 1s and 2s orbitals just like in the previous beryllium
• We have to find a place for the fifth electron
2. With beryllium, the 1s and 2s orbitals are completely filled up
(i) From the chart in fig.2.51(c), we see that, the next higher orbital is 2p
(ii) Any p sub-shell in an atom will have 3 orbitals: px, py and pz
• Those three orbitals will be always degenerate orbitals
• That means, they will be having the same energy levels
(iii) So in our present case, we have 2px, 2py and 2pz, all with the same energy and ready to be occupied
(iv) We place our fifth electron of boron in 2px
3. The electronic configuration of boron using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
■ Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the fifth electron of boron is in the p sub-shell of the second main-shell
So we have completed the electronic configuration of boron. Next is the element with atomic number 6, which is: carbon (C)
• It’s first five electrons can go into the 1s, 2s and 2px orbitals just like in the previous boron
• We have to find a place for the sixth electron
2. We see that, 2px can accommodate one more electron
• Also, 2py and 2pz are completely free
• Which of the three shall we choose?
3. To find the answer, we have to apply the Hund's rule of maximum multiplicity
(i) We have to apply this rule whenever we face 'orbitals belonging to the same sub-shell'
• That means, we apply this rule when we face any one of the following 3 situations:
♦ We face the three orbitals px, py, pz of the p sub-shell
(all three are degenerate orbitals)
♦ We face the five orbitals $\mathbf\small{d_{xy},\,d_{xz},\,d_{yz},\,d_{x^2-y^2},\,d_{z^2}}$ of the d sub-shell
(all five are degenerate orbitals)
♦ We face the seven orbitals of the f sub-shell
(all seven are degenerate orbitals)
(ii) The word 'multiplicity' means: More than one of 'something'
• Consider 'something'. For example, a quantity 'k'
♦ If we have more than one of 'k', we call it a 'multiplicity'
(iii) In our present case, we have a set of 3 degenerate orbitals in front of us
• Each of them have the same energy
• Initially, all of them are empty
(iv) We will get maximum multiplicity if we follow the procedure given below:
♦ Put one electron in each orbital
♦ Thus all three will become 'half occupied'
♦ It is a multiplicity of 'half occupied orbitals'
♦ To attain a maximum of this multiplicity, we try to get as many 'half filled orbitals' as possible
♦ Here we can attain a maximum of three 'half filled orbitals'. This is shown below:
(In the case of the d sub-shell, we can attain a maximum multiplicity, if we allow all of the five orbitals to remain 'half filled')
(v) After making all the three orbitals 'half occupied', put the next electron in the first orbital. This is shown below:
(vi) Thus, pairing of electrons can start only after making all the degenerate orbitals 'half occupied'
• This type of filling gives extra stability to the atoms
(vii) Hund's rule of maximum multiplicity states that:
Pairing of electrons in the orbitals belonging to the same sub-shell does not take place until each orbital belonging to that sub-shell has got one electron each
4. Applying this rule to our present case, we can write:
The sixth electron of carbon must go into the py orbital
5. The electronic configuration of carbon using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the sixth electron of carbon is in the p sub-shell of the second main-shell
So we have completed the electronic configuration of carbon. Next is the element with atomic number 7, which is: nitrogen (N)
• It’s first six electrons can go into the 1s, 2s, 2px and 2py orbitals just like in the previous carbon
• We have to find a place for the seventh electron
2. It is obvious that, according to Hund's rule, the seventh electron must go into the pz orbital
• The electronic configuration of nitrogen using the orbital diagram method is shown in fig.2.56 below
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the 7th electron of nitrogen is in the p sub-shell of the second main-shell
So we have completed the electronic configuration of nitrogen. Next is the element with atomic number 8, which is: oxygen (O)
1. Oxygen has eight electrons
• It’s first seven electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous nitrogen
• We have to find a place for the eighth electron
2. It is obvious that, according to Hund's rule, the eighth electron must go back into the px orbital
• The electronic configuration of oxygen using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the 8th electron of oxygen is in the p sub-shell of the second main-shell
1. Oxygen has eight electrons
• It’s first seven electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous nitrogen
• We have to find a place for the eighth electron
2. It is obvious that, according to Hund's rule, the eighth electron must go back into the px orbital
• The electronic configuration of oxygen using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• Note that, in the s, p, d, f method, we are not able to make a distinction between px, py or pz orbitals. In this method, the only information we get is that, the 8th electron of oxygen is in the p sub-shell of the second main-shell
So we have completed the electronic configuration of oxygen. Next is the element with atomic number 9, which is: fluorine (F)
1. Fluorine has nine electrons
• It’s first eight electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous oxygen
• 2px was completely filled while 2py and 2pz were half filled
• We have to find a place for the ninth electron
2. It is obvious that, according to Hund's rule, the ninth electron must go back into the py orbital
• The electronic configuration of fluorine using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
1. Fluorine has nine electrons
• It’s first eight electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous oxygen
• 2px was completely filled while 2py and 2pz were half filled
• We have to find a place for the ninth electron
2. It is obvious that, according to Hund's rule, the ninth electron must go back into the py orbital
• The electronic configuration of fluorine using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
So we have completed the electronic configuration of fluorine. Next is the element with atomic number 10, which is: neon (Ne)
1. Neon has ten electrons
• It’s first nine electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous oxygen
• 2px and 2py were completely filled while 2pz was half filled
• We have to find a place for the tenth electron
2. It is obvious that, according to Hund's rule, the tenth electron must go back into the pz orbital
• The electronic configuration of neon using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
• It’s first nine electrons can go into the 1s, 2s, 2px, 2py and 2pz orbitals just like in the previous oxygen
• 2px and 2py were completely filled while 2pz was half filled
• We have to find a place for the tenth electron
2. It is obvious that, according to Hund's rule, the tenth electron must go back into the pz orbital
• The electronic configuration of neon using the orbital diagram method is shown in fig.2.55 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
So we have completed the electronic configuration of neon. Next is the element with atomic number 11, which is: sodium (Na)
1. Sodium has 11 electrons
• It’s first 10 electrons can go into the 1s, 2s and 2p orbitals just like in the previous neon
• We have to find a place for the 11th electron
2. With neon, all the orbitals of the 2p sub-shell are filled up
• From the chart, we see that, 3s is the next higher energy level
3. In the 3s, there is only one orbital
• That orbital can hold two electrons
• So the 11th electron of sodium will go into 3s
4. The electronic configuration of neon using the orbital diagram method is shown in fig.2.56 below:
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
5. In the orbital diagram, consider the portion to the left of the 3s box
• We see that, this left side is exactly same as neon
• So we can modify the orbital diagram into a shortened form
• In fig.2.57 below, this 'shortened form' is shown on the right side of the double arrow:
6. In the s, p, d, f notation, consider the portion to the left of the 3s1
• We see that, this left side is exactly same as neon
• So we can modify the s, p, d, f notation also into a shortened form: [Ne] 3s1
• It is also shown on the right side of the double arrow
So we have completed the electronic configuration of sodium (Na). Next is the element with atomic number 12, which is: magnesium (Mg)
1. Sodium has 11 electrons
• It’s first 10 electrons can go into the 1s, 2s and 2p orbitals just like in the previous neon
• We have to find a place for the 11th electron
2. With neon, all the orbitals of the 2p sub-shell are filled up
• From the chart, we see that, 3s is the next higher energy level
3. In the 3s, there is only one orbital
• That orbital can hold two electrons
• So the 11th electron of sodium will go into 3s
4. The electronic configuration of neon using the orbital diagram method is shown in fig.2.56 below:
 |
| Fig.2.56 |
5. In the orbital diagram, consider the portion to the left of the 3s box
• We see that, this left side is exactly same as neon
• So we can modify the orbital diagram into a shortened form
• In fig.2.57 below, this 'shortened form' is shown on the right side of the double arrow:
![In the shortened form electronic configuration for sodium and magnesium, neon [Ne] is written first](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtLT6j2jS0IQLvfUjC3hb2QfncCRWrcNBW3x-KLnCw2CGB7_BhTV-RJR2ETj62_Ls4BTc9vW1_EAQvCaLaWUonAM2fB5BE3fwa5dGMiJV75yQ1SMnwinq_fjFwe7_OxZMZwOWqa0UrRZN-/s1600/h_c_fig_2_57_n1.png "Shortened form electronic configuration for sodium and magnesium") |
| Fig.2.57 |
• We see that, this left side is exactly same as neon
• So we can modify the s, p, d, f notation also into a shortened form: [Ne] 3s1
• It is also shown on the right side of the double arrow
So we have completed the electronic configuration of sodium (Na). Next is the element with atomic number 12, which is: magnesium (Mg)
1. Mg has 12 electrons
• It’s first 11 electrons can go into the 1s, 2s, 2p and 3s orbitals just like in the previous Na
• We have to find a place for the 12th electron
2. The 3s orbital can hold 2 electrons
• So the 12th electron will go into the 3s orbital
3. The electronic configuration of Mg using the orbital diagram method is shown in fig.2.56 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
4. In the orbital diagram, consider the portion to the left of the 3s box
• We see that, this left side is exactly same as neon
• So, just as we did for sodium, we can modify the orbital diagram into a shortened form
• It is shown on the right side of the double arrow in fig.2.57 above
5. In the s, p, d, f notation, consider the portion to the left of the 3s2
• We see that, this left side is exactly same as neon
• So, just as we did for sodium, we can modify the s, p, d, f notation also into a shortened form: [Ne] 3s2
• It is shown on the right side of the double arrow in fig.2.57 above
• It’s first 11 electrons can go into the 1s, 2s, 2p and 3s orbitals just like in the previous Na
• We have to find a place for the 12th electron
2. The 3s orbital can hold 2 electrons
• So the 12th electron will go into the 3s orbital
3. The electronic configuration of Mg using the orbital diagram method is shown in fig.2.56 above
• The electronic configuration using s, p, d, f method is also shown. It is shown in green color
4. In the orbital diagram, consider the portion to the left of the 3s box
• We see that, this left side is exactly same as neon
• So, just as we did for sodium, we can modify the orbital diagram into a shortened form
• It is shown on the right side of the double arrow in fig.2.57 above
5. In the s, p, d, f notation, consider the portion to the left of the 3s2
• We see that, this left side is exactly same as neon
• So, just as we did for sodium, we can modify the s, p, d, f notation also into a shortened form: [Ne] 3s2
• It is shown on the right side of the double arrow in fig.2.57 above
So we have completed up to no.12 magnesium. In the next section, we will see from no.13 aluminium (Al) on wards
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