1. 3.0 PERIODIC TABLE 1 1

2. At the end of the lesson the students should be able to :
LEARNING OUTCOMES
At the end of the lesson the students should be able to :
(i) Indicate period, group and block (s, p, d, f).
(ii) Specify the position of metals, metalloids
and non-metals in the periodic table.
(iii) Deduce the position of elements in the
periodic table from its electronic
configuration. 2

3. 3.1 Classification of elements
The periodic table is a table that arranges all the known elements in order of increasing proton number.
This order generally coincides with increasing atomic mass. 3

4. Elements in the same group have the same number of valence electrons.
A vertical column of elements is called a group and a horizontal row is known as a period.
Elements in the same group have the same number of valence electrons.
Group number = number of valence electrons
(if the element is in block s and d)
Group number = number of valence electrons + 10
(if the element is in block p) 4

5. Transition metals
For example, oxygen and sulphur are both found in group 16 which means that they both have 6 valence electrons. 5

6. GROUP Main Groups in Periodic Table Group 1 : alkali metals (except H)
Group : alkaline earth metals
Group : transition metals
Group : halogens
Group : inert/ noble gases
Group : Zn, Cd, Hg 6

7. The periods in the Periodic Table are numbered from 1 to 7
For example, hydrogen and helium are in Row 1 or Period 1 because their principal quantum number, n, of the main electron shell is (H:1s1 ;He: 1s2)
Period number = Principle quantum number 7

8. Blocks
All the elements in the Periodic Table can be classified into 4 main blocks according to their valence electrons configuration.
These main blocks are s, p, d and f block. 8

9. s- block 20Ca : 1s2 2s2 2p6 3s2 3p6 4s2 Group 1 and 2
The filling of valence electrons involve the s orbital
Configuration of the valence electrons :
Eg:
11Na : 1s2 2s2 2p6 3s1
20Ca : 1s2 2s2 2p6 3s2 3p6 4s2
ns1 to ns2 9

10. p-block
Group 13 to 18
The filling of valence electrons involve s and p orbital.
The configuration of valence electrons:
Eg.
13Al : 1s2 2s2 2p6 3s2 3p1
52Te : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2
5p4
ns2 np1 to ns2 np6 10

11. d-block
Groups 3 to 12
The filling of valence electrons involve s and d orbitals.
Group 3 to 11 known as Transition metal.
Configuration of valence electron :
Eg.
23V : 1s2 2s2 2p6 3s2 3p6 3d3 4s2 or
[Ar] 3d3 4s2
where [Ar] = 18 electrons
(n-1) d1 ns2 to (n-1) d10 ns2 11

12. f-block
Involve the elements in the series of lanthanides (Ce to Lu) and actinides (Th to Lr).
The filling of valence electrons happen in the subshell of 4f and 5f. 12 12

13. Number of velence electron
elements
block
period
group
Number of velence electron
19K : 1s2 2s2 2p6 3s2 3p6 4s1 s 4 1
12Mg : 1s2 2s2 2p6 3s2 3 2
15P : 1s2 2s2 2p6 3s2 3p3 p 15 5
35Bv : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 17 7
28Ni : 1s2 2s2 2p6 3s2 3p6 4s2 3d8 d 10
40Zr : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 3d2

15. B …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 C …….1s2 2s2 2p6 3s2 3p6 4s2
Example
Classify the following elements into its appropriate group, period and block.
A ……1s2 2s2 2p6 3s2 3p6
B …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
C …….1s2 2s2 2p6 3s2 3p6 4s2
D …….1s2 2s2 2p6 3s2 3p6 3d3 4s2
E …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 15

16. Transition element/ block d
Group
Period
Class/block A B C D E 18 17 2 5 3 4
Inert gas / block p
Block p
Block s
Transition element/ block d 16

17. 4.2.1 Periodic trends in the size of atom (atomic radii)
3.2 Periodicity
4.2.1 Periodic trends in the size of atom
(atomic radii)
The size /radius of atom is difficult to be defined exactly because the electron cloud has no clear boundary.
Therefore, the atomic radius is taken as half of the distance between the nuclei of two adjacent identical atom. 17

18. Down the group, atomic radii increases.
Radius, r = a/2 (Å)
Size volume
V = 4/3 πr3 , V  r
Down the group, atomic radii increases.
Across period, atomic radii decreases.
Across the period of d-block (transition elements)
the change in atomic radii is small as valence
electrons are filled in the 3d degenerate orbitals. 18

19. Effective nuclear charge experienced by the valence electrons
Two factors that influence the changes of atomic radii in the Periodic Table are:
Effective nuclear charge experienced by the valence electrons
ii. The principal quantum number, n, of the valence electrons 19

20. i. Effective nuclear charge (Zeff )
Electrons around the nucleus experience different nucleus attraction.
Those electrons closer to the nucleus experience a greater attraction than those that are farther away.
The actual nuclear charge experienced by an electron is called the effective nuclear charge, Zeff 20

21. Effective nuclear charge increase, nucleus attraction stronger, atomic radii decrease
Across the period, the effective nuclear charge increases as proton number increase.
As a result, the attraction between the nucleus and valence electrons become stronger, causing the atomic radius to decrease. 21 21

22. The valence electrons are farther from the nucleus.
ii. The principal quantum number of the valence electrons
As we move down a group, the number of shells increases, more inner electrons are present to shield the valence electrons from the nucleus.
The valence electrons are farther from the nucleus.
Thus, the attraction between the nucleus and valence electrons decreases, therefore, the atomic radius increase. 22

23. Down a group, the atomic radius increases because of the increasing principal quantum number (n) of the valence electron. 23

24. Across
period 3
Across
Period 2 24

25. Atomic radius decreases when : * Across a period (from left to right)
The graph shows that :
Atomic radius decreases when :
* Across a period (from left to right)
* Moving up a group in the periodic table.
Atomic radius increases when
* Going down the group
The greater the nucleus attraction, the
smaller the atomic radius. 25

26. Example
Arrange the following atoms in order of increasing radius P,Si,N.
Solution
N and P are in the same group and N is above P.
Atomic radius increases as we go down the group.
Therefore, the radius of N is smaller than that of P
Both Si and P are in the third period and Si is to the left of P.
Atomic radius decreases as we move from left to right.
Therefore, the radius of P is smaller than Si.
Thus the order of increasing radius :
N<P<Si
Group 14
Group 15 N Si P 26

27. 3.2.2 Trends in the ionic radii
When electrons are added to an atom, the mutual repulsions between them increase.
This enlarge the domain of electron cloud.
Therefore, negative ions (anions) are larger than the atoms from which they are formed. 27

28. When electrons are removed from the valence shell, the electron-electron repulsions decrease but the nuclear charge remains the same.
So the remaining electrons are to be pulled closer together around the nucleus.
Therefore, cations are smaller than the atoms from which they are formed. 28

29. 29

30. Isoelectronic species
Isoelectronic species are groups of atoms and ions which have the same electronic configuration.
Within isoelectronic species:
a) the more positive the charge, the smaller the
species
E.g :
Na+, Mg2+, Al3+and Si4+ ions are isoelectronic
(10 e) with the electron configurations as
1s2 2s2 2p6. 30

31. Isoelectronic species with electronic configuration 1s2 2s2 2p6 (10 electrons)
When proton number increase, effective nuclear charge increase.
The attraction between nucleus and remaining electron increase.
Therefore, the ionic radii decrease.
The ionic radii of Na+ > Mg2+ > Al3+ > Si4+
species
Number of proton
Na+ 11
Mg2+ 12
Al3+ 13
Si4+ 14

32. Isoelectronic species with electronic 1s2 2s2 2p6 3s2 3p6 (18 electrons)
When proton number increase, effective nuclear charge increase.
The attraction between nucleus and remaining electron increase.
Therefore, the ionic radii decrease.
The ionic radii of Cl- < S2- < P3-
species
Number of proton
P3- 15
S2- 16
Cl- 17

33. N3- > O2- > F- > Na+ > Mg2+ > Al3+> Si4+
Exercise
Na+,Si4+ ,Mg2+, N3- ,O2- ,Al3+ and F- are isoelectronic with the electronic configuration as 1s2 2s2 2p6. Arrange in an descending order the size of those isoelectronic species.
Answer :
N3- > O2- > F- > Na+ > Mg2+ > Al3+> Si4+ 33

34. 34

35. 3.2.3 Trends in the ionization energies
The ionization energy (IE) is the minimum energy required to remove an electron from a gaseous atom in its ground state.
The first ionization energy (IE1) is the minimum energy required to remove the first electron from the atom in its ground state.
E.g:
energy + X(g) → X+(g) + e- ΔH = IE1 35

36. Ionization energy across a period :
The effective nuclear charge increases, the
atomic size decreases.
Electrons are held tightly to the nucleus
thus it is difficult to remove the first electron.
Therefore the first ionisation energy is high.
It can be said that the first ionization energy increases from left to right.
However, there are some irregularities in the trend. 36

37. ٭Be
♦ B
■ N ▲O ■ ▲ ٭ ♦ 37

38. Anomalous cases in Period 2
Between group 2 and 13
5B : 1s2 2s2 2p1 in group 13 has a lower IE1 than 4Be: 1s2 2s2 in group 2.
Be loses a 2s electron while B loses a 2p electron.
Less energy is needed to remove an electron from partially-filled 2p orbital in B than to remove an electron from fully/completely filled 2s orbital in Be. 38

39. O (group16) has lower IE1 than N (group 15)
b) Between group 15 and 16
O (group16) has lower IE1 than N (group 15)
7N :1s2 2s2 2p3 (the half-filled 2p orbital )
8O :1s2 2s2 2p4( the partially-filled 2p orbital)
When N loses an electron it must come from the half-filled 2p orbital which is more stable than that of electron of the partially-filled orbital in O.
As a result, the first ionization energy of N is higher than of O. 39

40. ii) Ionization energy going down the group
Going down the group, the atomic size increases as the energy level, n increases.
Therefore the outer electrons are farther from the nucleus and are held less tightly (weaker attraction) by the nucleus.
Thus, it is easy to remove the first electron.
Hence the Ionization Energy decreases down the group. 40

41. Second ionization energy (IE2) is the minimum energy required to remove an electron from a positive gaseous ion.
X+(g) → X2+(g) + e-
When an electron is removed from a neutral atom, the mutual repulsion among the remaining electrons decrease.
Since the nuclear charge remain constant, the electron are held tightly to the nucleus.
Therefore more energy is needed to remove another electron from the positively charged ion. 41

42. Thus, ionization energies always increase in the following order :
IE1< IE2< IE3< IE4<…..
Although the removal of a subsequent electron from an atom requires an increment amount of energy but it may not be consistence.
We can determine the electronic configuration of the valence electron for an element by using the ionization energy. 42

43. Example 1
4Be
The ionization energies (kJmol-1) of Beryllium are shown below. 43 43

44. The ratio between the ionization energies are:44

45. A sharp increase in ionization energy occurs when an inner-orbital electron is removed.
The sharp increase is in IE3. It means the 3rd electron occupies the inner shell.
Therefore, there are 2 valence electrons. Hence, Be is in group 2 with valence configuration ns2. 45

46. Example 2
Five successive ionization energies (kJmol-1) of atom M is shown below:
Determine
i) the electron configuration of the valence
electron.
ii) the group number in the periodic table. 46

47. Therefore, there are 4 valence electrons.
The sharp increase is in IE5, this means the 5th electron occupies the inner shell.
Therefore, there are 4 valence electrons.
Hence, the valence electronic configuration for M is ns2 np2 in group 14 of the periodic table. 47

48. Trends in the electronegativity
Electronegativity is the relative tendency of an atom to attract electrons to itself when bonded with another atom.
Electronegativity increases up a group and across a period. This follows the trends for ionization energy and electron affinity. 48

49. The nuclear charge increase The atomic size decrease
Across period
The nuclear charge increase
The atomic size decrease
Hence, the nucleus attraction stronger
Therefore, the electronegativity increase 49

50. The principle quantum number increase The atomic size increase
b) Down a group
The principle quantum number increase
The atomic size increase
Hence, weaker nuclear attraction
Therefore electronegativity decrease 50

51. Trends in the melting or boiling point
The melting or boiling point depends on the types of intermolecular forces that exist between the molecules. 51

52. Across the period Melting and boiling point of the 3rd period elements52

53. (a) Metallic structure (Na to Al) (b) Gigantic covalent structure (Si)
The variation of melting and boiling point of elements in the 3rd period can be discussed as:
(a) Metallic structure (Na to Al)
(b) Gigantic covalent structure (Si)
(c) Simple molecular structure (P to Ar) 53

54. Metallic structure (Na to Al)
Metal has positive metal ions attracted to the electrons sea which form the metallic bonding.
Strength of metallic bonding is proportional to the number of valence electrons. e e e e
The more valence electrons, the stronger the metallic bond and the higher the melting / boiling point
Na+
Na+
Na+ e e e e
Na+
Na+ e e e e
Na+
Na+ 54

55. b) Gigantic covalent structure (Si)
Silicon has a gigantic covalent structure.
Melting and boiling point of Si is very high because high energy is needed to break the infinity amount of the strong covalent bond. 55

56. c) Simple molecular structure ( P to Ar)
The non-metal that exist as molecules of P4, S8, Cl2 and Ar (monoatom).
The covalent bond between the atoms is very strong but the intermolecular force (Van der Waals), is very weak.
The strength of Van der Waals force is proportional to molecular size (relative molecular weight)
Molecular size: Ar < Cl2 < P4 < S8
therefore melting / boiling point : Ar < Cl < P < S 56

57. Therefore, less energy is needed to overcome the attraction.
Melting and boiling points down the group
Group 1
The size increase, the attraction between nucleus and electron sea become weaker.
Therefore, less energy is needed to overcome the attraction.
Thus, melting and boiling point decrease. 57

58. Group 17
The size of molecules increase, the intermolecular forces (Van der Waals) become stronger.
Therefore, more energy is needed to overcome the attraction - Thus, melting and boiling point increase. 58

59. Metallic properties are:
Metallic character
Metallic properties are:
- shiny with various colours and most are silvery.
- malleable and ductile
- good thermal and electric conductor
Generally, metallic character :
- Increases down a group
- Decreases across the period 59

60. metal Metalloids (semimetal) nonmetal
The easier to the electrons to be removed
from an atom, the more metallic the
element.
Therefore metallic character increases down
a group and decreases across a period.
metal
Metalloids
(semimetal)
nonmetal 60

61. Acid-base behavior of oxides of Period 3
For Period 3:
When react with oxygen : (a) Na & Mg form basic oxide
(b) Al form amphoteric (both acidic and basic) oxide.
(c) Si, P, S & Cl form acidic oxide
Na Mg Al Si P S Cl 61

62. MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l) base acid
Na reacts with oxygen to form a basic oxide.
4Na (s) + O2 (g)→2Na2O (s)
The oxide will produce base solution when react with water.
Na2O (s) + H2O (l) → 2NaOH (aq)
Mg burns in oxygen to form a basic oxide, MgO Mg (s) + O2 (g) →2MgO (s)
MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l) base acid 62

63. Al forms amphoteric oxide, can react either with an acid or a base.
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3 H2O (l)
base acid
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4 (aq) acid base sodium aluminate 63

64. Si, P, S & Cl burn in oxygen to form acidic oxide. Si :
Si (s) + O2 (g) → SiO2 (s)
SiO2 (s) + NaOH (aq) → Na2SiO3(aq)+H2O (l)
acid base 64

65. P4O6 (s) + 6H2O (l) → 4H3PO3 (aq) phosphorus acid
P4 (s) + 3O2 (g) → P4O6 (s)
P4O6 (s) + 6H2O (l) → 4H3PO3 (aq) phosphorus acid 65

66. Cl2O7 (g) + H2O (l) → 2HClO4 (aq) hypochloric acid
S :
S (s) + O2 (g) →SO2 (g)
SO2 (g) + H2O (l) → H2SO3 (aq)
sulfurous acid
Cl :
Cl2O7 (g) + H2O (l) → 2HClO4 (aq) hypochloric acid 66