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3.0 PERIODIC TABLE

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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.

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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.

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• 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)

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

Transition metals

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GROUP

• Main Groups in Periodic Table– Group 1 : alkali metals (except H)– Group 2 : alkaline earth metals– Group 3-11 : transition metals– Group 17 : halogens– Group 18 : inert/ noble gases– Group 12 : Zn, Cd, Hg

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• 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 1. (H:1s1 ;He: 1s2)

Period number = Principle quantum number

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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.

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s- block

• 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

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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

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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

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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.

elements block period group Number of

velence

electron

19K : 1s2 2s2

2p6 3s2 3p6 4s1

s 4 1 1

12Mg : 1s2 2s2

2p6 3s2

s 3 2 2

15P : 1s2 2s2

2p6 3s2 3p3

p 3 15 5

35Bv : 1s2 2s2

2p6 3s2 3p6 4s2 3d10 4p5

p 4 17 7

28Ni : 1s2 2s2

2p6 3s2 3p6 4s2 3d8

d 4 10 10

40Zr : 1s2 2s2

2p6 3s2 3p6 4s2 3d10 4p6 5s2 3d2

d 5 4 4

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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

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Element Group Period Class/block

A

B

C

D

E

18

17

2

5

18

3

4

4

4

4

Inert gas / block p

Block p

Block s

Transition element/ block d

Inert gas / block p

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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.

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a

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.

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Two factors that influence the changes of atomic radii in the Periodic Table are:

i. Effective nuclear charge experienced by the valence electrons

ii. The principal quantum number, n, of the valence electrons

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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, Z

eff

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• 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.

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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.

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

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Across period 3

AcrossPeriod 2

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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.

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Example

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

Arrange the following atoms in order of increasing radius P,Si,N.

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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.

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• 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.

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• 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.

Isoelectronic species

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

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

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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.

Exercise

Answer :

N3- > O2- > F- > Na+ > Mg2+ > Al3+> Si4+

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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

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i) 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.

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٭♦

Be٭♦ B

■▲

■ N▲O

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Anomalous cases in Period 2

a) 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.

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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.

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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.

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• 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.

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• 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.

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Example 1

• 4Be

• The ionization energies (kJmol-1) of Beryllium are shown below.

IE1 IE2 IE3 IE4 899 1757 14850 21005

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• The ratio between the ionization energies are:

95.1899

1757

1

2 IE

IE

45.8175714850

2

3 IE

IE

41.11485021005

3

4 IE

IE

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• 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.

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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.

IE1 IE2 IE3 IE4 IE5 800 1580 3230 4360 16000

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• 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.

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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.

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a) Across period

• The nuclear charge increase

• The atomic size decrease

• Hence, the nucleus attraction stronger

• Therefore, the electronegativity increase

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b) Down a group

• The principle quantum number increase• The atomic size increase• Hence, weaker nuclear attraction• Therefore electronegativity decrease

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Trends in the melting or boiling point

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

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Across the period

• Melting and boiling point of the 3rd period elements

Element Na Mg Al Si P S Cl Ar Melting point (oC)

98 650 660 1410 44 120 -101 -189

Boiling point (oC)

890 1120 2450 2360 280 445 -34 -186

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• 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)

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Na+ Na+

Na+ Na+

Na+ Na+

a)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.

Na+ Na+

Na+Na+

Na+ Na+

e eThe more valence

electrons, the stronger the metallic bond and

the higher the melting / boiling point

e

e ee

e

e

e

e

ee

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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.

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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

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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.

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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.

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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

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• 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.

metalMetalloids

(semimetal)nonmetal

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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

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• 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. 2Mg (s) + O2 (g) →2MgO (s)

MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l)base acid

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• 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

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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

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P :

P4 (s) + 3O2 (g) → P4O6 (s)

P4O6 (s) + 6H2O (l) → 4H3PO3 (aq) phosphorus acid

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S :

S (s) + O2 (g) →SO2 (g)

SO2 (g) + H2O (l) → H2SO3 (aq)

sulfurous acid

Cl :

Cl2O7 (g) + H2O (l) → 2HClO4 (aq) hypochloric acid