In the previous section, we completed a discussion on hydrocarbons and other compounds of carbon. In this section we will learn about the Periodic table and Electronic configuration.
In a previous chapter, we saw the Bohr model electronic configuration of atoms. Details here. We will need the details that we learned there as the basis of our present discussion. Let us see some of those points that we have already learned:
• The electrons move around the nucleus in fixed orbits. These orbits are called shells
• These shells are numbered as 1, 2, 3, 4, . . . starting from the centre to the outside
♦ Besides those numbers, the shells are given names also: K, L, M, N, . . . starting from the centre to the outside. This is shown in the figure below:
• Each shell has a particular amount or level of energy. So all the electrons in a shell will have the same amount of energy, which is equal to the energy level of that shell
• When the distance of a shell from the nucleus increases, the energy level of that shell also increases. That means, outer shells have greater energy levels than inner shells
• The higher energy levels of the outer shells make the overall atom unstable. So the electrons prefer to take up positions in the inner shells. Thus in an atom, the inner shells are filled first.
• The number of electrons that can be accommodated in a shell is given by 2n2. Where n is the number of that shell
Examples:
♦ For the K shell, n = 1. So the number of electrons that can be accommodated in the K shell = 2×12 = 2 × 1 = 2
♦ For the L shell, n = 2. So the number of electrons that can be accommodated in the L shell = 2×22 = 2 × 4 = 8
♦ For the M shell, n = 3. So the number of electrons that can be accommodated in the M shell = 2×32 = 2 × 9 = 18
• The electron distribution in elements with lower atomic numbers were satisfactorily explained using the above Bohr model.
• K, L, M, N, . . . are called the 'principal energy levels' or 'main shells'
• The smaller shells within these main shells are called 'sub energy levels' or 'subshells'
3. Next we want to know how many subshells are there. For that, we need to recall the 'number' of the main shell. Based on that we have:
• The K main shell has the number 1. It has 1 subshell.
♦ The name of this subshell is 's'
• The L main shell has the number 2. It has 2 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
• The M main shell has the number 3. It has 3 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
♦ The name of the 3rd subshell is 'd'
• The N main shell has the number 4. It has 4 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
♦ The name of the 3rd subshell is 'd'
♦ The name of the 4th subshell is 'f'
■ In short, the number of subshells in a main shell is equal to the 'number' of that main shell
4. Now a problem arises:
Consider the following situation:
• A person says: "The subshell 's' is containing one electron"
But which 'subshell s' is he referring to?
♦ Is this 'subshell s' situated in the 'main shell 1'?
♦ Is this 'subshell s' situated in the 'main shell 2'?
♦ Is this 'subshell s' situated in the 'main shell 3'?
so on . . .
Similar situations can arise for other subshells also:
• A person says: "The subshell 'p' is containing one electron"
But which 'subshell p' is he referring to?
♦ Is this 'subshell p' situated in the 'main shell 2'?
♦ Is this 'subshell p' situated in the 'main shell 3'?
♦ Is this 'subshell p' situated in the 'main shell 4'?
so on . . .
• A person says: "The subshell 'd' is containing one electron"
But which 'subshell d' is he referring to?
♦ Is this 'subshell d' situated in the 'main shell 3'?
♦ Is this 'subshell d' situated in the 'main shell 4'?
■ Thus it is clear that, we need a system to indicate the correct position of a subshell. For that, the following simple rule is used:
• Write the 'number of the main shell' just in front of the subshell
Some examples:
♦ If the person is referring to the 'subshell s' in the 'main shell 2', he must write: 2s
♦ If the person is referring to the 'subshell d' in the 'main shell 3', he must write: 3d
♦ If the person is referring to the 'subshell p' in the 'main shell 3', he must write: 3p
4. Next we want to know how many electrons can be accommodated in each of these subshells.
• Based on the studies, the following numbers are discovered:
(i) The s subshell can accommodate 2 electrons
(ii) The p subshell can accommodate 6 electrons
(iii) The d subshell can accommodate 10 electrons
(iv) The f subshell can accommodate 14 electrons
5. We have reached the final step. Let us see what we have learned till now:
(i) All the electrons in an atom are arranged in subshells.
(ii) Consider an atom of any element. It will have main shells and subshells.
• Take any subshell in that atom, 2p or 1s, or 3d or whatever.
• We want to write the number of electrons contained in that subshell. How will we write it?
Ans: We follow a simple rule:
• Write the 'number of electrons in the subshell' as a superscript just after the 'name of the subshell'
Some examples:
• The 2p subshell in an atom contains 4 electrons.
♦ We write this as: 2p4
♦ It is read as: 'Two p Four'
• The 1s subshell in an atom contains 2 electrons.
♦ We write this as: 1s2
♦ It is read as: 'One s Two'
• The 3d subshell in an atom contains 10 electrons.
♦ We write this as: 3d10
♦ It is read as: 'Three d Ten'
We have reached the eleventh element. In the next section, we will see the next eleven.
In a previous chapter, we saw the Bohr model electronic configuration of atoms. Details here. We will need the details that we learned there as the basis of our present discussion. Let us see some of those points that we have already learned:
• The electrons move around the nucleus in fixed orbits. These orbits are called shells
• These shells are numbered as 1, 2, 3, 4, . . . starting from the centre to the outside
♦ Besides those numbers, the shells are given names also: K, L, M, N, . . . starting from the centre to the outside. This is shown in the figure below:
• Each shell has a particular amount or level of energy. So all the electrons in a shell will have the same amount of energy, which is equal to the energy level of that shell
• When the distance of a shell from the nucleus increases, the energy level of that shell also increases. That means, outer shells have greater energy levels than inner shells
• The higher energy levels of the outer shells make the overall atom unstable. So the electrons prefer to take up positions in the inner shells. Thus in an atom, the inner shells are filled first.
• The number of electrons that can be accommodated in a shell is given by 2n2. Where n is the number of that shell
Examples:
♦ For the K shell, n = 1. So the number of electrons that can be accommodated in the K shell = 2×12 = 2 × 1 = 2
♦ For the L shell, n = 2. So the number of electrons that can be accommodated in the L shell = 2×22 = 2 × 4 = 8
♦ For the M shell, n = 3. So the number of electrons that can be accommodated in the M shell = 2×32 = 2 × 9 = 18
• But when we come to elements with higher atomic numbers, the experimental results are different from the theoretical results.
■ So scientists conducted further studies and they discovered finer details about the distribution of electrons around the nucleus. Let us see those new findings:
1. Each of the shells K, L, M, N, . . . contains smaller shells with in them. The electrons are distributed among those smaller shells
2. We must be able to distinguish between the main shells and smaller shells. So, they are given special names:• K, L, M, N, . . . are called the 'principal energy levels' or 'main shells'
• The smaller shells within these main shells are called 'sub energy levels' or 'subshells'
3. Next we want to know how many subshells are there. For that, we need to recall the 'number' of the main shell. Based on that we have:
• The K main shell has the number 1. It has 1 subshell.
♦ The name of this subshell is 's'
• The L main shell has the number 2. It has 2 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
• The M main shell has the number 3. It has 3 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
♦ The name of the 3rd subshell is 'd'
• The N main shell has the number 4. It has 4 subshells.
♦ The name of the 1st subshell is 's'
♦ The name of the 2nd subshell is 'p'
♦ The name of the 3rd subshell is 'd'
♦ The name of the 4th subshell is 'f'
■ In short, the number of subshells in a main shell is equal to the 'number' of that main shell
4. Now a problem arises:
Consider the following situation:
• A person says: "The subshell 's' is containing one electron"
But which 'subshell s' is he referring to?
♦ Is this 'subshell s' situated in the 'main shell 1'?
♦ Is this 'subshell s' situated in the 'main shell 2'?
♦ Is this 'subshell s' situated in the 'main shell 3'?
so on . . .
Similar situations can arise for other subshells also:
• A person says: "The subshell 'p' is containing one electron"
But which 'subshell p' is he referring to?
♦ Is this 'subshell p' situated in the 'main shell 2'?
♦ Is this 'subshell p' situated in the 'main shell 3'?
♦ Is this 'subshell p' situated in the 'main shell 4'?
so on . . .
• A person says: "The subshell 'd' is containing one electron"
But which 'subshell d' is he referring to?
♦ Is this 'subshell d' situated in the 'main shell 3'?
♦ Is this 'subshell d' situated in the 'main shell 4'?
■ Thus it is clear that, we need a system to indicate the correct position of a subshell. For that, the following simple rule is used:
• Write the 'number of the main shell' just in front of the subshell
Some examples:
♦ If the person is referring to the 'subshell s' in the 'main shell 2', he must write: 2s
♦ If the person is referring to the 'subshell d' in the 'main shell 3', he must write: 3d
♦ If the person is referring to the 'subshell p' in the 'main shell 3', he must write: 3p
4. Next we want to know how many electrons can be accommodated in each of these subshells.
• Based on the studies, the following numbers are discovered:
(i) The s subshell can accommodate 2 electrons
(ii) The p subshell can accommodate 6 electrons
(iii) The d subshell can accommodate 10 electrons
(iv) The f subshell can accommodate 14 electrons
5. We have reached the final step. Let us see what we have learned till now:
(i) All the electrons in an atom are arranged in subshells.
(ii) Consider an atom of any element. It will have main shells and subshells.
• Take any subshell in that atom, 2p or 1s, or 3d or whatever.
• We want to write the number of electrons contained in that subshell. How will we write it?
Ans: We follow a simple rule:
• Write the 'number of electrons in the subshell' as a superscript just after the 'name of the subshell'
Some examples:
• The 2p subshell in an atom contains 4 electrons.
♦ We write this as: 2p4
♦ It is read as: 'Two p Four'
• The 1s subshell in an atom contains 2 electrons.
♦ We write this as: 1s2
♦ It is read as: 'One s Two'
• The 3d subshell in an atom contains 10 electrons.
♦ We write this as: 3d10
♦ It is read as: 'Three d Ten'
Now we can start writing the electronic configuration of atoms of the various elements.
1. We will start with the first element Hydrogen (H):
(i) H has an atomic number 1. It has only one electron
(ii) It has only one main shell K. The number of this main shell is 1
(iii) This main shell 1 can have only one subshell [see (3) above]. That means, the subshell is s
(iv) So the subshell s is residing in the main shell 1 of the hydrogen atom. Thus the address of this subshell is 1s
(v) This s subshell can accommodate 2 electrons. But we have only one electron in H. This electron will go to the 1s subshell
■ Thus the electronic configuration of H is 1s1
2. Second element Helium (He):
(i) He has an atomic number 2. It has two electrons
(ii) It has only one main shell K. The number of this main shell is 1
(iii) This main shell 1 can have only one subshell [see (3) above]. That means, the subshell is s
(iv) So the subshell s is residing in the main shell 1 of the hydrogen atom. Thus the address of this subshell is 1s
(v) This s subshell can accommodate 2 electrons. The two electrons will go to the 1s subshell
■ Thus the electronic configuration of He is 1s2
Let us write the configurations of the 2 elements together:
1. H - 1s1
2. He - 1s2
3. Third element Lithium (Li):
(i) Li has an atomic number 3. It has three electrons
(ii) The main shell K is not sufficient. Because K can accommodate only a maximum of 2 electrons [based on the rule 2n2]. So Li has main shell L also.
• The number of the main shell K is 1
• The number of the main shell L is 2
(iii) Now we will see how the 3 electrons of Li are filled into the two main shells K and L:
Lower energy levels are filled up first. That means K main shell is filled up first.
(iv) The K main shell has only one subshell 's'. So out of the three, two electrons will fill up the 1s subshell giving 1s2. Subshell 's' can accommodate a maximum of 2 only.
(v) The third electron will go to the first subshell 's' of the second main shell L. That is, subshell 2s. This will give 2s1.
■ So the final configuration of Li is 1s22s1
Let us write the configurations of the 3 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1.
In this way, we can follow a pattern and write the configuration of all elements. However, an adjustment has to be made when we reach the 19th element Potassium. Another type of adjustment has to be made for the 24th and 29th elements Chromium and Copper. We will see all such adjustments when we reach them. For now, we will continue our present task of writing the configurations. We will write upto the 30th element.
Note: The electronic configuration of atoms written in this way is called the Subshell Electronic Configuration. We will write it in short form as S.E.C.
4. Fourth element Beryllium (Be):
Note: The electronic configuration of atoms written in this way is called the Subshell Electronic Configuration. We will write it in short form as S.E.C.
4. Fourth element Beryllium (Be):
(i) Be has an atomic number 4. It has 4 electrons
(ii) The main shell K is not sufficient. Because K can accommodate only a maximum of 2 electrons [based on the rule 2n2]. So Be has main shell L also.
• The number of the main shell K is 1
• The number of the main shell L is 2
(iii) Now we will see how the 4 electrons of Be are filled into the two main shells K and L:
Lower energy levels are filled up first. That means K main shell is filled up first.
(iv) The K main shell has only one subshell 's'. So out of the four, two electrons will fill up the 1s subshell giving 1s2. Subshell 's' can accommodate a maximum of 2 only.
(v) The remaining two electrons will go to the first subshell 's' of the second main shell L. That is, subshell 2s. This will give 2s2.
■ So the S.E.C of Be is 1s22s2.
Let us write the configurations of the 4 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. Fifth element Boron (B):
(i) B has an atomic number 5. It has 5 electrons
(ii) We have seen how 4 electrons are distributed when we saw Li. 1s and 2s are already filled up.
The fifth electron in B can be accommodated only if another subshell is made available.
(iii) The main shell '2' can have 2 subshells: 2s and 2p. So we bring in 2p.
(iv) The fifth electron will be accommodated in the 2p, giving us 2p1.
■ So the S.E.C of B is 1s22s22p1.
Let us write the configurations of the 5 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. Sixth element Carbon (C):
(i) C has an atomic number 6. It has 6 electrons
(ii) We have seen how 5 electrons are distributed when we saw B. 1s and 2s are already filled up in Li. In B, 2p was also made available.
(iii) Any p subshell can accommodate 6 electrons.
(iv) So the sixth electron in C can be easily accommodated in 2p. This gives us 2p2.
■ So the S.E.C of C is 1s22s22p2
Let us write the configurations of the 6 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. Seventh element Nitrogen (N):
(i) N has an atomic number 7. It has 7 electrons
(ii) We have seen how 6 electrons are distributed when we saw C. 1s and 2s are already filled up in Li. In B and C, 2p was also made available.
(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 2 are used up in C
(iv) So the seventh electron in N can be easily accommodated in 2p. This gives us 2p3.
■ So the S.E.C of N is 1s22s22p3.
Let us write the configurations of the 7 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. N - 1s22s22p3
8. Eighth element Oxygen (O):
(i) O has an atomic number 8. It has 8 electrons
(ii) We have seen how 7 electrons are distributed when we saw N. 2p is being filled up.
(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 3 are used up in N
(iv) So the eighth electron in O can be easily accommodated in 2p. This gives us 2p4.
■ So the S.E.C of O is 1s22s22p4.
9. Ninth element Fluorine (F):
10. Tenth element Neon (Ne):
11. Eleventh element Sodium (Na):
Let us write the configurations of the 8 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. N - 1s22s22p3
8. O - 1s22s22p4
8. O - 1s22s22p4
(i) F has an atomic number 9. It has 9 electrons
(ii) We have seen how 8 electrons are distributed when we saw O. 2p is being filled up.
(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 4 are used up in O
(iv) So the ninth electron in F can be easily accommodated in 2p. This gives us 2p5.
■ So the S.E.C of F is 1s22s22p5.
Let us write the configurations of the 9 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. N - 1s22s22p3
8. O - 1s22s22p4
9. F - 1s22s22p5
8. O - 1s22s22p4
9. F - 1s22s22p5
(i) Ne has an atomic number 10. It has 10 electrons
(ii) We have seen how 9 electrons are distributed when we saw F. 2p is being filled up.
(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 5 are used up in F
(iv) So the tenth electron in Ne can be easily accommodated in 2p. This gives us 2p6.
■ So the S.E.C of Ne is 1s22s22p6.
Let us write the configurations of the 10 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. N - 1s22s22p3
8. O - 1s22s22p4
9. F - 1s22s22p5
10. Ne - 1s22s22p6
8. O - 1s22s22p4
9. F - 1s22s22p5
10. Ne - 1s22s22p6
(i) Na has an atomic number 11. It has 11 electrons
(ii) We have seen how 10 electrons are distributed when we saw Ne. 2p is completely filled up.
• 2p is the second subshell in the 'main shell 2'. It is filled up in the previous element Neon.
• 2s is the first subshell in the 'main shell 2'. It was earlier filled up No.4 Beryllium.
• So all the subshells in the 'main shell 2' are exhausted. Because, the 'main shell 2' can have only 2 subshells
• We have to move on to the next main shell, which is 'main shell 3', which is also called 'main shell M'.
(iii) Once we reach the 'main shell 3', the electrons will begin to fill in it's first subshell. The first subshell of 'main shell 3' is 3s
(iv) So the eleventh electron will be accommodated in 3s. This gives us 3s1
■ So the S.E.C of Na is 1s22s22p63s1.
• 2p is the second subshell in the 'main shell 2'. It is filled up in the previous element Neon.
• 2s is the first subshell in the 'main shell 2'. It was earlier filled up No.4 Beryllium.
• So all the subshells in the 'main shell 2' are exhausted. Because, the 'main shell 2' can have only 2 subshells
• We have to move on to the next main shell, which is 'main shell 3', which is also called 'main shell M'.
(iii) Once we reach the 'main shell 3', the electrons will begin to fill in it's first subshell. The first subshell of 'main shell 3' is 3s
(iv) So the eleventh electron will be accommodated in 3s. This gives us 3s1
■ So the S.E.C of Na is 1s22s22p63s1.
Let us write the configurations of the 11 elements together:
1. H - 1s1
2. He - 1s2
3. Li - 1s22s1
4. Be - 1s22s2
5. B - 1s22s22p1
6. C - 1s22s22p2
7. N - 1s22s22p3
8. O - 1s22s22p4
9. F - 1s22s22p5
10. Ne - 1s22s22p6
11. Na - 1s22s22p63s1.
■ Another method for writing configuration:
• Consider the configuration of Na - 1s22s22p63s1.
• Detach the last part 3s1. We get: 1s22s22p6.
♦ But this is the configuration of Neon (Ne)
■ So we can write:
Configuration of Sodium (Na) as: [Ar]3s1
■ Neon is chosen because it is the next higher element after helium, to have an octet
■ After Neon, the next element to have octet is Argon. When we reach Argon, we will see a similar modification there also.
8. O - 1s22s22p4
9. F - 1s22s22p5
10. Ne - 1s22s22p6
11. Na - 1s22s22p63s1.
■ Another method for writing configuration:
• Consider the configuration of Na - 1s22s22p63s1.
• Detach the last part 3s1. We get: 1s22s22p6.
♦ But this is the configuration of Neon (Ne)
■ So we can write:
Configuration of Sodium (Na) as: [Ar]3s1
■ Neon is chosen because it is the next higher element after helium, to have an octet
■ After Neon, the next element to have octet is Argon. When we reach Argon, we will see a similar modification there also.
We have reached the eleventh element. In the next section, we will see the next eleven.
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