Chapter 9.1 - Electronic configuration of elements from Magnesium to Titanium

In the previous section, we saw the electronic configuration upto the eleventh element. In this section we will see the next eleven.

12. Twelfth element Magnesium (Mg):

(i) Mg has an atomic number 12. It has 12 electrons 
(ii) We have seen how 11 electrons are filled when we saw Na. The filling up of 3s began in Na.

(iii) Any s subshell can accommodate 2 electrons. Out of these 2, only 1 is  used up in Na

(iv) So the twelfth electron in Mg can be easily accommodated in 3s. This gives us 3s².

■ So the S.E.C of Mg is 1s²2s²2p⁶3s².

---

Let us write the  configurations of the 12 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²

---

13. Thirteenth element Aluminium (Al):

(i) Al has an atomic number 13. It has 13 electrons

(ii) We have seen how 12 electrons are distributed when we saw Mg. 3s is already filled up. 

The thirteenth electron in Al can be accommodated only if another subshell is made available. 

(iii) The 'main shell 3' can have 3 subshells: 3s, 3p and 3d. So we bring in 3p. 

(iv) The thirteenth electron will be accommodated in the 3p, giving us 3p¹.

■ So the S.E.C of Al is 1s²2s²2p⁶3s²3p¹.

---

Let us write the  configurations of the 13 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  OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹

---

14. Fourteenth element Silicon (Si):

(i) Si has an atomic number 14. It has 14 electrons

(ii) We have seen how 13 electrons are distributed when we saw Al. Filling up of 3p has begun in Al.

(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 1 is used up in Al  

(iv) So the fourteenth electron in Si can be easily accommodated in 3p. This gives us 3p².

■ So the S.E.C of Si is 1s²2s²2p⁶3s²3p².

---

Let us write the  configurations of the 14 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s² OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹ OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p²  OR [Ne]3s²3p²

---

15. Fifteenth element Phosphorous (P):

(i) P has an atomic number 15. It has 15 electrons

(ii) We have seen how 14 electrons are distributed when we saw Si. The 3p is being filled up.

(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 2 is used up in Si  

(iv) So the fifteenth electron in P can be easily accommodated in 3p. This gives us 3p³.

■ So the S.E.C of P is 1s²2s²2p⁶3s²3p³.

---

Let us write the  configurations of the 15 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  OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹ OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p²  OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³  OR [Ne]3s²3p³ 

---

16. Sixteenth element Sulfur (S):

(i) S has an atomic number 16. It has 16 electrons

(ii) We have seen how 15 electrons are distributed when we saw P. The 3p is being filled up.

(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 3 is used up in P  

(iv) So the sixteenth electron in S can be easily accommodated in 3p. This gives us 3p⁴.

■ So the S.E.C of S is 1s²2s²2p⁶3s²3p⁴.

---

Let us write the  configurations of the 16 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴  

---

17. Seventeenth element Chlorine (Cl):

(i) Cl has an atomic number 17. It has 17 electrons

(ii) We have seen how 16 electrons are distributed when we saw S. The 3p is being filled up.

(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 4 is used up in S.  

(iv) So the seventeenth electron in Cl can be easily accommodated in 3p. This gives us 3p⁵.

■ So the S.E.C of Cl is 1s²2s²2p⁶3s²3p⁵.

---

Let us write the  configurations of the 17 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵

---

18. Eighteenth element Argon (Ar):

(i) Ar has an atomic number 18. It has 18 electrons

(ii) We have seen how 17 electrons are distributed when we saw Cl. The 3p is being filled up.

(iii) Any p subshell can accommodate 6 electrons. Out of these 6, only 5 is used up in Cl.  

(iv) So the eighteenth electron in Ar can be easily accommodated in 3p. This gives us 3p⁶.

■ So the S.E.C of Ar is 1s²2s²2p⁶3s²3p⁶.

---

Let us write the  configurations of the 18 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵
18. Ar   - 1s²2s²2p⁶3s²3p⁶.

---

19. Nineteenth element Potassium (K):

(i) K has an atomic number 19. It has 19 electrons

(ii) We have seen how 18 electrons are distributed when we saw Ar. The 3p is completely filled up in Ar.
• So we have to bring in another subshell. 
• We are now in the 'main shell 3'. This main shell can have upto 3 subshell. 
• That means, we have a subshell 3d available nearby. We do not have to move on to the next 'main shell 4'.
(iii) But unfortunately, we are not allowed to bring in 3d at this time. Let us see the reason:
• The 3d subshell has a slightly greater energy level than 4s. 
• So the newly coming electrons will prefer to go to the 4s, rather than the 3d. 
• This energy difference can be seen in a graphical representation here. We can see that, 3d is situated at a slightly higher level than 4s
• So the electrons will begin to fill into the 3d, only after completely filling up 4s

---

So we have a situation:
• A subshell is present in a particular main shell
• Another subshell is present in a higher main shell
    ♦ But this second subshell, in spite of being in a higher main shell, is having a lower energy level than the first subshell.
• This is rather an unusual situation. But we have to accept it. 
■ Are there any more instances of such a situation?
Ans: There are indeed a number of instances. 
• That is., some subshells have a lower energy level than subshells in lower main shells. 
• To find such instances, we can use a simple geometrical construction. The steps are given below:
1. First write the subshells in columns and rows as shown in fig.9.1(a) below.
• One column for s, one column for p, . . . so on
• One row for 2, one row for 3, . . . so on

Fig.9.1

2. Draw arrows through them as shown in the fig.b. 
■ Arrows are drawn in a slanting position so that:
• The succeeding subshell of any subshell will be the one in: 
    ♦ column on the left and 
    ♦ row at the bottom
3. They 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.9.1(c) below. 

Fig.9.1

4. The geometrical construction is complete. Now we will see how this fig.c can be used:
• Place your finger tip at the rear end of the top most arrow. Move the finger tip along the direction of the arrow
• You will first reach the pointed end of the top most arrow. Then move along the dashed line
• You will reach the rear end of the second arrow. Continue in this way
• The path followed by the finger tip gives the order in which subshells are filled up     
• 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 filling up can be easily determined using fig.c

---

(iv) So, coming back to our main discussion, we find that, we do have to move to the next 'main shell 4'. We get 4s¹.
■ So the S.E.C of K is 1s²2s²2p⁶3s²3p⁶4s¹.

---

Let us write the  configurations of the 19 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵
18. Ar   - 1s²2s²2p⁶3s²3p⁶
19. K    - 1s²2s²2p⁶3s²3p⁶4s¹.

■ Another method for writing configuration:
• Consider the configuration of K - 1s²2s²2p⁶3s²3p⁶4s¹.
• Detach the last part 4s¹. We get: 1s²2s²2p⁶3s²3p⁶.
    ♦ But this is the configuration of Argon (Ar)
■ So we can write:
Configuration of Potassium (K) - [Ar]4s¹.

---

20. Twentieth element Calcium (Ca):
(i) Ca has an atomic number 20. It has 20 electrons.

(ii) We have seen how 19 electrons are distributed when we saw K. The filling up of 4s began in K.

(iii) Any s subshell can accommodate 2 electrons. Out of these 2, only 1 is  used up in K

(iv) So the twentieth electron in Ca can be easily accommodated in 4s. This gives us 4s².

■ So the S.E.C of Ca is 1s²2s²2p⁶3s²3p⁶4s².

---

Let us write the  configurations of the 20 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵
18. Ar   - 1s²2s²2p⁶3s²3p⁶
19. K    - 1s²2s²2p⁶3s²3p⁶4s¹   OR [Ar]4s¹.
20. Ca   - 1s²2s²2p⁶3s²3p⁶4s²  OR [Ar]4s².

---

21. Twenty first element Scandium (Sc):

(i) Sc has an atomic number 21. It has 21 electrons.
(ii) We have seen how 20 electrons are distributed when we saw Ca. The 4s is completely filled up in Ca.
• We have to take up the next subshell. Based on sequencial order, the next subshell after 4s is 4p
• But the electrons will prefer to go to the 3d. This is because 3d has a lower energy than 4p
• This is clear from the graph and also from fig.9.1(c)
(iii) Thus the filling up of 3d begins from the twenty first element Sc. We get 3d¹.
■ So the S.E.C of Sc is 1s²2s²2p⁶3s²3p⁶4s²3d¹.

---

Let us write the  configurations of the 21 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵
18. Ar   - 1s²2s²2p⁶3s²3p⁶
19. K    - 1s²2s²2p⁶3s²3p⁶4s¹  OR [Ar]4s¹
20. Ca   - 1s²2s²2p⁶3s²3p⁶4s²  OR [Ar]4s²
21. Sc   - 1s²2s²2p⁶3s²3p⁶4s²3d¹  OR [Ar]4s²3d¹.

---

22. Twenty second element Titanium (Ti):

(i) Ti has an atomic number 22. It has 22 electrons.
(ii) We have seen how 21 electrons are distributed when we saw Sc.Filling up of 3d has begun in Sc.

(iii) Any d subshell can accommodate 10 electrons. Out of these 10, only 1 is used up in Sc  

(iv) So the Twenty second electron in Ti can be easily accommodated in 3d. This gives us 3d².

■ So the S.E.C of Ti is 1s²2s²2p⁶3s²3p⁶4s²3d²

---

Let us write the  configurations of the 22 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 OR [Ne]3s¹
12. Mg  - 1s²2s²2p⁶3s²  OR [Ne]3s²
13. Al    - 1s²2s²2p⁶3s²3p¹  OR [Ne]3s²3p¹
14. Si    - 1s²2s²2p⁶3s²3p² OR [Ne]3s²3p²
15. P    - 1s²2s²2p⁶3s²3p³ OR [Ne]3s²3p³
16. S    - 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴
17. Cl    - 1s²2s²2p⁶3s²3p⁵. OR [Ne]3s²3p⁵
18. Ar   - 1s²2s²2p⁶3s²3p⁶
19. K    - 1s²2s²2p⁶3s²3p⁶4s¹ OR [Ar]4s¹
20. Ca   - 1s²2s²2p⁶3s²3p⁶4s²  OR [Ar]4s²
21. Sc   - 1s²2s²2p⁶3s²3p⁶4s²3d¹  OR [Ar]4s²3d¹
22. Ti   - 1s²2s²2p⁶3s²3p⁶4s²3d¹  OR [Ar]4s²3d².

---

We have reached the twenty second element. In the next section, we will see the elements from twenty three up to thirty. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved

Chapter 9 - Electronic configuration and Periodic table

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. 

• 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 1^st subshell is 's'     
    ♦ The name of the 2^nd subshell is 'p'
• The M main shell has the number 3. It has 3 subshells. 
    ♦ The name of the 1^st subshell is 's'     
    ♦ The name of the 2^nd subshell is 'p'     
    ♦ The name of the 3^rd subshell is 'd'
• The N main shell has the number 4. It has 4 subshells. 
    ♦ The name of the 1^st subshell is 's'     
    ♦ The name of the 2^nd subshell is 'p'     
    ♦ The name of the 3^rd subshell is 'd'          
    ♦ The name of the 4^th 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 19^th element Potassium. Another type of adjustment has to be made for the 24^th and 29^th 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 30^th 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):

(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.

---

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

---

9. Ninth element Fluorine (F):

(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

---

10. Tenth element Neon (Ne):

(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

---

11. Eleventh element Sodium (Na):

(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.

---

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.

---

We have reached the eleventh element. In the next section, we will see the next eleven. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved