The document discusses periodic trends in effective nuclear charge (Zeff) and how it relates to ionization energy. It provides an example calculating the Zeff of the outermost electrons as argon is sequentially ionized from Ar+ to Ar18+. With each additional ionization, an electron is removed and the Zeff of the remaining electrons increases. This leads to higher ionization energies due to the greater attraction felt by the electrons to the increasingly exposed positive nuclear charge.

1. PERIODIC TRENDS AND Zeff

2. PERIODIC TRENDS AND Zeff
The effective Nuclear Charge (Zeff) that acts on an
electron is related its ionization energy.
2p
2s Dramatic changes in
ionization energy
1s occur between
energy levels and
subshells
nucleus

3. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar (1s2) (2s22p6) (3s23p6)
7 other 3s 3p electrons = 7 x 0.35 = 2.45
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 2.45 + 6.80 + 2 = 11.25
Zeff = Z – S
= 18 – 11.25
= 6.75

4. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar1+ (1s2) (2s22p6) (3s23p5)
6 other 3s 3p electrons = 6 x 0.35 = 2.1
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 2.1 + 6.80 + 2 = 10.9
Zeff = Z – S
= 18 – 10.9
= 7.1

5. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar2+ (1s2) (2s22p6) (3s23p4)
5 other 3s 3p electrons = 5 x 0.35 = 1.75
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 1.75 + 6.80 + 2 = 10.55
Zeff = Z – S
= 18 – 10.55
= 7.45

6. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar3+ (1s2) (2s22p6) (3s23p3)
4 other 3s 3p electrons = 4 x 0.35 = 1.4
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 1.4 + 6.80 + 2 = 10.2
Zeff = Z – S
= 18 – 10.2
= 7.8

7. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar4+ (1s2) (2s22p6) (3s23p2)
3 other 3s 3p electrons = 3 x 0.35 = 1.05
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 1.05 + 6.80 + 2 = 9.85
Zeff = Z – S
= 18 – 9.85
= 8.15

8. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar5+ (1s2) (2s22p6) (3s23p1)
2 other 3s electrons = 2 x 0.35 = 0.7
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 0.7 + 6.80 + 2 = 9.5
Zeff = Z – S
= 18 – 9.5
= 8.5

9. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar6+ (1s2) (2s22p6) (3s2)
1 other 3s electron = 1 x 0.35 = 0.35
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 0.35 + 6.80 + 2 = 9.15
Zeff = Z – S
= 18 – 9.15
= 8.85

10. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar7+ (1s2) (2s22p6) (3s1)
0 other 3s electrons = 0 x 0.35 = 0
8 2s 2p electrons = 8 x 0.85 =6.80
2 1s electrons = 2 x 1.00 = 2
S = 0 + 6.80 + 2 = 8.80
Zeff = Z – S
= 18 – 8.80
= 9.2

11. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar8+ (1s2) (2s22p6)
7 other 2s 2p electrons = 7 x 0.35 = 2.45
2 1 s electrons = 2 x 0.85 = 1.7
S = 2.45 + 1.7 = 4.15
Zeff = Z – S
= 18 – 4.15
= 13.85

12. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar9+ (1s2) (2s22p5)
6 other 2s 2p electrons = 6 x 0.35 = 2.1
2 1 s electrons = 2 x 0.85 = 1.7
S = 2.1 + 1.7 = 3.8
Zeff = Z – S
= 18 – 3.8
= 14.2

13. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar10+ (1s2) (2s22p4)
5 other 2s 2p electrons = 5 x 0.35 = 1.75
2 1 s electrons = 2 x 0.85 = 1.7
S = 1.75 + 1.7 = 3.45
Zeff = Z – S
= 18 – 3.45
= 14.55

14. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar11+ (1s2) (2s22p3)
4 other 2s 2p electrons = 4 x 0.35 = 1.4
2 1 s electrons = 2 x 0.85 = 1.7
S = 1.4 + 1.7 = 3.1
Zeff = Z – S
= 18 – 3.1
= 14.9

15. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar12+ (1s2) (2s22p2)
3 other 2s 2p electrons = 3 x 0.35 = 1.05
2 1 s electrons = 2 x 0.85 = 1.7
S = 1.05 + 1.7 = 2.75
Zeff = Z – S
= 18 – 2.75
= 15.25

16. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar13+ (1s2) (2s22p1)
2 other 2s 2p electrons = 2 x 0.35 = 0.7
2 1 s electrons = 2 x 0.85 = 1.7
S = 0.7 + 1.7 = 2.4
Zeff = Z – S
= 18 – 2.4
= 15.6

17. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar14+ (1s2) (2s2)
1 other 2s 2p electron = 1 x 0.35 = 0.35
2 1 s electrons = 2 x 0.85 = 1.7
S = 0.35 + 1.7 = 2.4
Zeff = Z – S
= 18 – 2.05
= 15.95

18. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar15+ (1s2) (2s1)
0 other 2s 2p electrons = 0 x 0.35 = 0
2 1 s electrons = 2 x 0.85 = 1.7
S = 0 + 1.7 = 1.7
Zeff = Z – S
= 18 – 1.7
= 16.3

19. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar16+ (1s2)
1 other 1s electron = 1 x 0.35 = 0.35
S = 0.35
Zeff = Z – S
= 18 – 0.35
= 17.65

20. PERIODIC TRENDS AND Zeff
Example #1 - Calculate the effective nuclear charge of the
outermost electron in the following:
The electron configuration of argon is 1s2 2s2 2p6 3s2 3p6
Ar17+ (1s1)
0 other 1s electrons = 0 x 0.35 = 0
S=0
Zeff = Z – S
= 18 – 0
= 18

21. PERIODIC TRENDS AND Zeff
Zeff Vs . Ioniz ation E nerg y of the 18 E lec trons in Arg on
450000 20
The jump from the
second energy level to
400000 18
the first energy level
16
350000
14
300000
Ioniz ation E nerg y (eV )
12
250000
I.E .
Z eff
10
(eV )
200000 Zeff
8
150000
6
100000
4
50000 The jump2 from the
third energy level to
0 the second energy
0
0 2 4 6 8 10 12 14 16 level
18 20
Oute rm ost e le c tron to inne rm ost e le c tron

22. PERIODIC TRENDS AND Zeff
Zeff Vs . Ioniz ation E nerg y of the 18 E lec trons in Arg on
TREND: As the effective nuclear
450000 20
charge on an electron increases, so
18
does its ionization energy
400000
16
350000
14
300000
Ioniz ation E nerg y (eV )
12
250000
I.E .
Z eff
10
(eV )
200000 Zeff
8
150000
6
100000
4
50000 2
0 0
0 2 4 6 8 10 12 14 16 18 20
Oute rm ost e le c tron to inne rm ost e le c tron

23. PERIODIC TRENDS AND Zeff
TREND: As the effective nuclear
charge on an electron increases, so
does its ionization energy
Lower Zeff.
Attraction by
2p nucleus weaker.
Higher Zeff.
Attraction by 2s
nucleus stronger. .: less energy
1s required to
.: more energy remove this
required to O electron
remove this
electron

24. PERIODIC TRENDS AND Zeff
Example #2 - Calculate the effective nuclear charge of the
outermost electron in the first 36 elements
H (1s1)
0 other 1s electrons = 0 x 0.35 = 0
S=0
Zeff = Z – S
=1–0
=1

25. PERIODIC TRENDS AND Zeff
Example #2 - Calculate the effective nuclear charge of the
outermost electron in the first 36 elements
He (1s2)
1 other 1s electrons = 1 x 0.35 = 0.35
S = 0.35
Zeff = Z – S
= 2 – 0.35
= 1.65

26. PERIODIC TRENDS AND Zeff
Example #2 - Calculate the effective nuclear charge of the
outermost electron in the first 36 elements
Li (1s2) (2s1)
2 1s electrons = 2 x 0.85 = 1.7
S = 1.7
Zeff = Z – S
= 3 – 1.7
= 1.3

27. PERIODIC TRENDS AND Zeff
Example #2 - Calculate the effective nuclear charge of the
outermost electron in the first 36 elements
Be (1s2) (2s2)
1 2s electron = 1 x 0.35 = 0.35
2 1s electrons = 2 x 0.85 = 1.7
S = 0.35 + 1.7
Zeff = Z – S
= 4 – 2.05
= 1.95

28. PERIODIC TRENDS AND Zeff
Example #2 - ANSWERS
Atomic Element
Number Symbol Zeff
1 H 1 19 K 2.2
2 He 1.65 20 Ca 2.85
3 Li 1.3 21 Sc 3
4 Be 1.95 22 Ti 3.15
5 B 2.6 23 V 3.3
6 C 3.25 24 Cr 2.95
7 N 3.9 25 Mn 3.6
8 O 4.55 26 Fe 3.75
9 F 5.2 27 Co 3.9
10 Ne 5.85 28 Ni 4.05
11 Na 2.2 29 Cu 3.7
12 Mg 2.85 30 Zn 2.85
13 Al 3.5 31 Ga 5
14 Si 4.15 32 Ge 5.65
15 P 4.8 33 As 6.3
16 S 5.45 34 Se 6.95
17 Cl 6.1 35 Br 7.6
18 Ar 6.75 36 Kr 8.95

29. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2 This shows that it is
5
difficult to remove
1
the last electron of
0
an0 energy level (it
will destabilize it)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

30. PERIODIC TRENDS AND Zeff
It requires much more energy to remove the
outermost electron of helium than lithium
2s
1s 1s
He Li
This will destabilize a This will not destabilize
full orbital a full orbital
.: greater ionization .: less ionization energy
energy is required is required

31. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
5
1 2s The trend repeats
for 2s
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

32. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
2p The third electron in
the p subshell
5
1 requires more
energy to remove
0 than the fourth
0
electron
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

33. PERIODIC TRENDS AND Zeff
The third electron in the p subshell requires more energy
to remove than the fourth electron
vs
Removing the third electron produces this:
Having all three half-full orbitals is more stable than having one
orbital completely empty
vs
More stable

34. PERIODIC TRENDS AND Zeff
Stability: Rank the following from most to least stable
Most
stable
Increasing
stability
Least
stable

35. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
5
1 3s The same trend repeats for 3s
and 3p
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

36. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
5
1 The same trend repeats for 3s
3p
and 3p
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

37. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9 4s and 3d are close in energy, so
there is no great change in 25
8 ionization energy in between
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
3d 5
1
4s
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

38. PERIODIC TRENDS AND Zeff
Zeff vs 1st I.E Elements 1-36
10 30
9
4p repeats the ionization energy
8 trend of 2p and 3p
25
7
20
6
1st I.E (eV)
Zeff v1
Zeff
5 15
1st ionization energy (eV)
4
10
3
2
5
1
4p
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Atomic Number

39. PERIODIC TRENDS AND Zeff
4s and 3d are close in energy, so there is no great change in
ionization energy in between.
RECALL:
Zeff vs 1st I.E Elements 1-36
10 30
9
25
8
7
20
6
Zeff
5 15
4
10
3
2
3d
5
1
0
4s 0
1
2
3
4
5
6
7
8
9
10
17
20
29
36
11
12
13
14
15
16
18
19
21
22
23
24
25
26
27
28
30
31
32
33
34
35
Atomic Number

40. PERIODIC TRENDS AND Zeff
Zeff Vs. Atomic Radius Elements 1-36
10 TREND: As Zeff increases, the nuclear attraction 250 the
for
outermost electron increases, and thus the electrons are
9
pulled closer to the nucleus. This results in a smaller
atomic radius
8 200
7
Atomic Radium (pm)
6 150
Zeff v1
Zeff
5
Atomic Radius (pm)
4 100
3
2 50
1
0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Atomic Number

41. PERIODIC TRENDS AND Zeff
TREND: As Zeff increases, the nuclear attraction for the outermost
electron increases, and thus the electrons are pulled closer to the
nucleus. This results in a smaller atomic radius
O F
Weaker Zeff Stronger Zeff
.: weaker attraction to nucleus .: stronger attraction to nucleus