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):
We have reached the twenty second element. In the next section, we will see the elements from twenty three up to thirty.
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.
13. Thirteenth element Aluminium (Al):
(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
8. O - 1s22s22p4
9. F - 1s22s22p5
10. Ne - 1s22s22p6
11. Na - 1s22s22p63s1 OR [Ne]3s1
12. Mg - 1s22s22p63s2 OR [Ne]3s2
(i) Al has an atomic number 13. It has 13 electrons
14. Fourteenth element Silicon (Si):
(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
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
(i) Si has an atomic number 14. It has 14 electrons
15. Fifteenth element Phosphorous (P):
(ii) We have seen how 13 electrons are distributed when we saw Al. Filling up of 3p has begun in Al.
■ So the S.E.C of Si is 1s22s22p63s23p2.
(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.
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
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
(i) P has an atomic number 15. It has 15 electrons
16. Sixteenth element Sulfur (S):
(ii) We have seen how 14 electrons are distributed when we saw Si. The 3p is being filled up.
■ So the S.E.C of P is 1s22s22p63s23p3.
(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.
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
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
(i) S has an atomic number 16. It has 16 electrons
17. Seventeenth element Chlorine (Cl):
(ii) We have seen how 15 electrons are distributed when we saw P. The 3p is being filled up.
■ So the S.E.C of S is 1s22s22p63s23p4.
(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.
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
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
(i) Cl has an atomic number 17. It has 17 electrons
18. Eighteenth element Argon (Ar):
(ii) We have seen how 16 electrons are distributed when we saw S. The 3p is being filled up.
■ So the S.E.C of Cl is 1s22s22p63s23p5.
(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.
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
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
(i) Ar has an atomic number 18. It has 18 electrons
19. Nineteenth element Potassium (K):
(ii) We have seen how 17 electrons are distributed when we saw Cl. The 3p is being filled up.
■ So the S.E.C of Ar is 1s22s22p63s23p6.
(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.
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.
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.
(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
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.
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.
20. Twentieth element Calcium (Ca):
(i) Ca has an atomic number 20. It has 20 electrons.
• 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 |
■ 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 |
• 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.
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.
(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.
21. Twenty first element Scandium (Sc):
(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.
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.
(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.
22. Twenty second element Titanium (Ti):
(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.
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.
(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.
■ So the S.E.C of Ti is 1s22s22p63s23p64s23d2
(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.
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.
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.
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