CHAPTER 3 PERIODIC TABLE

3.1 Classification of elements

LEARNING OUTCOMES

At the end of the lesson the students should be able to :

(a) Describe period, group and block (s, p, d, f). (b) Specify the position of metals, metalloids and non- metals in the periodic table. (c) Deduce the position of elements in the periodic table from its electronic configuration.

The elements in periodic table are arranged in ascending order of relative atomic mass PERIOD

G R O U P

4 main BLOCK

Group 1 : Alkali metals Group 2 : Alkaline earth metals

Group 18 : Inert/noble gases

Group 17 : Halogens

Group 3-12 : Transition metals

GROUPS- The groups in the periodic table are numbered from 1 to 18. - Elements in the same group have the have the same number of valence electrons . Example:22s1 Li : 1s 3 22s22p63s1 Na : 1s 11

- Both found in Group 1 - Both have 1 valence electron

Group number = Number of valence electron, block s and d if the element is in __________ OR Group number = Number of valance electron + 10, if the element is in ______ block p

Dmitri Mendeleev Russian Chemist worked out the 1st Periodic Table

PERIODS- The periods in the periodic table are numbered 1 to 7. - Period number = _____________________________ _____________________________ Example: - Sodium and chlorine are in row 3 or Period 3. - Because their principal quantum number of the valence electron shell is 3. Na: 1s2 2s2 2p6 3s1 Cl: 1s2 2s2 2p6 3s2 3p5

BLOCKS

THE POSITION OF METALS, METALLOIDS & NON-METALS IN THE PERIODIC TABLE

PHYSICAL PROPERTIESnitrogen

tellurium

THE POSITION OF ELEMENTS IN THE PERIODIC TABLE FROM ITS ELECTRONIC CONFIGURATION- The position of an element in the periodic table can be deduced from its ______________________________

Example: The electronic configuration of magnesium is shown below: Mg : 1s2 2s2 2p6 3s2 Determine its position in the periodic table.

Strategy: Step 1: Identify the outermost electronic configuration...3s2 Step 2: Identify the block to which the element belongs. Mg is an s-block element. Step 3: Identify the principal quantum number & the number of valence electrons. Principal quantum number, n = 3 denotes the period No. of valence electron = 2 denotes the group Solution: The outermost electronic configuration of Mg is 3s2. Thus, Mg is in Period 3 and Group 2 of the periodic table.

Example: Identify the positions of the elements P, Q, R, S and T in the periodic table and classify these elements. P : 1s2 2s2 2p6 3s2 3p6 Q: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

R : 1s2 2s2 2p6 3s2 3p6 4s2 S : 1s2 2s2 2p6 3s2 3p6 3d3 4s2 T : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Answer:

3.2 Periodicity

LEARNING OUTCOMESAt the end of the lesson the students should be able to : (a) Explain the variation in atomic radii: i. across a period ii. across the first row of transition elements. iii. down a group (b) Compare the atomic radius of an element and its corresponding ionic radius. (c) Define the term isoelectronic. (d) Explain the radius of isoelectronic species. (e) Explain the variation in the ionic radii across Period 3. (f) Define the first and second ionisation energies.

LEARNING OUTCOMES

At the end of the lesson the students should be able to : (g) Explain the variation in the ionisation energy i. across a period ii. down a group (h) Explain the increase in the successive ionisation energies of an element (i) Deduce the electronic configuration of an element and its position in the periodic table based on successive ionisation energy data. (j) Define electronegativity. (k) Explain the variation in electronegativity of elements i. across a period ii. down a group

LEARNING OUTCOMES

At the end of the lesson the students should be able to : (l) Describe the periodicity of elements across Period 3 and down a Group 1 and 17 for the following physical properties: i. metallic character ii. melting point iii. boiling point (m) Explain the acid-base character of oxides of elements in Period 3

VARIATION IN ATOMIC RADII

The atomic radius is taken as half of the distance between the nuclei of two adjacent identical atom.

The size/radius of ato is difficult to be defined because the electron clo has no clear boundary

The atomic radius of an element is determined by two factors..1. Effective nuclear charge, Zeffis the charge an electron actually experiences as a result of screening effects by other electrons. Electrons moving around the nucleus do not experience the same nucleus attraction. Electrons closer to the nucleus experience greater nucleus attraction. The effective nuclear charge,Zeff which is the positive charge felt by an electron is given by: Zeff = Z S

2. Screening effectis also known as shielding effect. __________________________________ __________________________________ __________________________________ __________________________________ Also occur between electrons of the same shell but is less effective.

Z = no. of proton S = no. of electrons filled at the inner orbital/ number of inner or core electrons

EXAMPLE 1 Consider lithium atom which electron configuration is 1s2 2s1 - The inner 2 electrons (1s2), shield the outer 2s electrons from the positive charge of 3 protons. Zeff = Z S =32 =+1een=1

e- Zeff experienced by the 2s electrons is only +1

n=2

EXAMPLE 2 Consider a sodium atom. Na: (1s2 2s2 2p6 3s1) core electron

eee- e e e e- - ee e-

e-

n=1

n=2 n=3

- The 3s electron would be perfectly screened from the positively charged nucleus. - Thus, the valence electron would experience an attraction to a net positive charge of only +1 Zeff = Z S

EXAMPLE 3 i) Na : 1s2 2s2 2p6 3s1 ii) K : 1s2 2s2 2p6 3s2 3p6 4s1 - Shielding effect of valence electrons of Na is less than K. - Therefore, when n increase, shielding effect also increase.

D O W N A G R O U P

1st row of d-block elements

COMPARISON OF ATOMIC RADIUS AND ITS IONIC RADIUS

EXAMPLE 1 (a) Na Na+ : 1s2 2s2 2p6 3s1 : 1s2 2s2 2p6 : [Ne]

(b) Ca : 1s2 2s2 2p6 3s2 3p6 4s2 Ca2+ : 1s2 2s2 2p6 3s2 3p6 : [Ar]

The positive ion contains smaller number of electron than its atom and has the configuration of the noble gas.

EXAMPLE 2 (a) F F: 1s2 2s2 2p5 : 1s2 2s2 2p6 : [Ne] : 1s2 2s2 2p4 : 1s2 2s2 2p6 : [Ne]

(b) O O2-

The negative ion contains bigger number of electron than its atom and has the configuration of the noble gas.

ISOELECTRONIC- iso means same and electronic means electrons - isoelectronic series is a group of atoms or ions having the _______________________ - within isoelectronic series: The ___________ the charge, the ________ the species. The ___________ the charge, the ________ the species.

The electron configuration of the following ions is: 1s2 2s2 2p6 (isoelectronic)

3rd period

2nd period

Na+, Mg2+, Al3+, F-, O2-, and N3- are

The variation in the ionic radii across Period 3- Across the period 3, size ofcations __________ anions __________ - the proton number increase but the number of electron is the same (10e) for each ion. - thus the ____________ and causes the ____________________

- All the ______ have ionic radius smaller than the ______ in the same period.

- Because the anions have __________________________ (screening effect) for the valence electron (n=3) compared to cations (n=2).

- Therefore, the trend in ionic radius across period 3 is: P3- > S2- > Cl- > Na+ > Mg2+ > Al3+> Si4+

Cations: The ionic radius decrease as follows: _______________

Anions: The ionic radius decrease as follows: _______________

EXAMPLE: Arrange O2-, F-, Ne, Na+ and Mg2+ in order of increasing size. Explain your answer.

ANSWER: -

Ionisation Energy (IE)The ionisation energy (general definition)- Is the ______________________required to remove the

______________________from a gaseous atom or ion.

42

The first ionisation energy -Is the minimum energy required to remove one mole of electron from the outermost orbital in one mole of neutral gaseous atom. _______________________________

Example:

___________________________

The second ionisation energy -Is the energy required to remove one mole of electron from one mole of positive ion in gaseous state. __________________________________

Example: __________________________________

-The magnitude of IE correlate with:i) ________________________________________________ ii) ________________________________________________

-The lower of IE, the more easily its electrons can be removed to form cation.

45

Example:

The process is endothermic as energy must be supplied to remove the electron.

Factors Affecting the Ionization Energy

Atomic radius Effective nuclear charge Shielding effect (screening effect)

47

Atomic radius The valence electrons of an atom with a a

______________ ionization energy.

experience

________________________________hence possesses a low

Effective nuclear charge The _____________________ the stronger the

attraction forces between _____________________. This causes the ionization energy to increase.

Shielding effect (screening effect) The shielding effect of the electrons of the inner orbitals

causes the outer electrons to be _____________________ and thus decrease the magnitude of ionization energy.

Variation in the First Ionisation Energy(i) When moving across a period, Effective nuclear charge _________ causes the size of the atoms to _________ ___________________ are closer to the nucleus . Attractions between the nucleus and the valence electrons are __________

50

More ________ to remove an electrons from each atom. ________________ is needed to remove the outermost electron from an atom.

Thus, ionisation energy (IE) increase. In general, IE increases from left to right.

What is meant by ANOMALOUS?

different abnormal weird

Anomalous CasesThe increase of IE with proton number is not uniform. -In Period 2, there are two cases: - Group 2 with group 13 (Be and B) - Group 15 with group 16 (N and O) -In Period 3, there are three cases: - Group 2 with group 13 (Mg and Al) - Group 15 with group 16 (P and S)

53

Variation Of First IE (kJ/mol) Across Period 2 & 3

54

Variation Of First IE Across Period 2 & Period 3kJ/mol 2000 1500 1000 500 Li Be B Na Al N O C Mg Si F P S

Period 2Ne

Period 3Ar Cl

56

Proton number

Period 2 : Anomalous Cases for Be and B Between group 2 and 13

Ionization energy of Be > B Electronic configuration of Be : 1s2 2s24Be:

(_________________________)1s 2s

B : 1s2 2s2 2p1

5B:

(____________________ )1s 2s 2p57

Completely filled orbital is _____________ than partially filled orbital.

Removing an electron from Be makes the atom less stable (less favourable) More energy is needed to remove the electron.

Therefore ionization energy of Be is higher than B.

In B, the electron in 2p orbital is well shielded by the electrons in 1s and 2s orbitals. Attraction between the_____________________________ Removing an electron from 2p orbital makes the atom more stable (more favourable) Less energy is needed to remove the electron. Therefore ionization energy of B is lower than Be.

59

Period 2: Anomalous Cases for N and OBetween group 15 (N) and 16 (O). Ionization energy of N > O Electronic Configuration of: N : 1s2 2s2 2p3 (half filled orbital) O : 1s2 2s2 2p4 (partially filled orbital) Half filled orbital is more stable than partially filled orbital.60

Nitrogen and Oxygen

Removing an electron from N, half filled 2p orbital (more stable) required more energy, makes the atom less stable (less favourable) More energy is needed to remove the electron. Therefore ionization energy of N is higher than O.61

Removing an electron of O from 2p partially filled orbital Makes the atom more stable (more favourable) Less energy is needed to remove the electron. Therefore ionization energy of O is lower than N.

62

Period 3: Anomalous Cases For Mg And Al12Mg:

1s

2s

2p

3s

13Al:

1s

2s

2p

3s

3p

63

IE of Mg > Al because: The electrons in the 3s orbital of Al are _____________ at shielding the electron in the 3p orbital. Attraction of nucleus towards the 3p electron become ___________ ____________ is needed to remove a single 3p electron than to remove a paired 3s orbital in Mg.

Period 3: Anomalous Cases For P and S15P:

1s

2s

2p

3s

3p

16S:

1s

2s

2p

3s

3p

65

IE of P > S because: The first electron removed from S comes from a 3p orbital that is occupied by two electrons which experience ____________________ This makes the electron ___________________ Less energy is needed to remove an electron from the 3p4 orbital of S than from the half-filled 3p orbital of P which is ____________________

Variation in the First Ionisation Energy(i)When going down a group:

Number of shell increase, the ______________ increase, _____________ also increase. Valence electrons are farther away from the nucleus and held _______________________Attractions between the nucleus and the valence electrons are___________67

_________ to remove an electron from each atom.

So, __________ is needed to remove the first electron.

Thus, ionization energy decrease.

General Trend In First Ionization EnergiesIncreasing First Ionization Energy

69

Trends you should know and be able to explain:

Across a periodZeff increases, Atomic radius decreases, I.E.increases

Down a groupShielding effect increases, Atomic radius increases, I.E.decreases70

Determination the electronic configuration of an atom using successive ionization energiesIn general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion, and that requires more work. Example: Mg (g) Mg+ (g) Mg+ (g) + e(H1:+736 kJmol-1)

Mg2+ (g) + e- (H2:+1800 kJmol-)

A large increase in successive ionization energies occurs between the removal of the last valence electron and the removal of the first core electron.

Successive Ionization Energies For Element

Methods to determine electronic configurationThe electronic configuration of the valence electron for an element can be determined by using the following methods: Method 1:By determining the IE ratios Method 2:By determining the differences in ionization energy

EXAMPLE 1Four successive ionization energies (kJmol-1) for atom X is shown below:

E -1 ol )

1 899

2 1757

3 14845

4 21000

(i) i. Predict the group in the periodic table. ii. Write the complete electronic configuration of X if it belongs to the second period.

SOLUTIONMethod 1: By determining the IE ratios: IE2 = 1757 = ________ IE1 899 IE3 = 14845 = ________ IE2 1757 IE4 = 21000 = ________ IE3 14845

-The first and second electron are removed from the same energy subshell (2s). -The third electron is removed from an inner subshell that is 1s, hence it requires a higher IE3 (8.45 times) than IE2. -Since IE3 / IE2 have the highest ratio, 2 valence electrons are present. -This element is in Group 2. - Electronic configuration: 1s2 2s2

SOLUTIONMethod 2: By determining the differences in ionization energy: IE2-IE1 = 1757 899 = ________________________

IE3-IE2 = 14845 1757 = ________________________ IE4-IE3 = 21000 14845 = ______________

- The first and second electron are removed from the same energy subshell (2s). - The third electron is removed from an inner subshell that is 1s, hence it requires a higher IE3 than IE2 (a difference of 13088 kJmol-1). - Since IE3 - IE2 have the highest difference, 2 valence electrons are present. -This element is in Group 2 - Electronic configuration: 1s2 2s2

EXAMPLE 2Five successive ionization energies (kJmol-1) for atom M is shown below:

Determine: (i)Electronic configuration of the valence electron for M. (ii) Group number of M in the periodic table.

SOLUTION Method 1: By determining the IE ratios:

The first, second, third and fourth electron are removed from the same energy subshell. The fifth electron is removed from an inner subshell, hence it requires a higher IE5 (3.67 times) than IE4. Since IE5 / IE4 have the highest ratio, 4 valence electrons are present.

Valence electronic configuration: ns2 np2 This element is in Group 14

SOLUTION Method 2: By determining the differences in ionization energy:

Advice!!!!!

Ratio method is preferable in determining the valence electronic configuration of an element.

Electronegativity-Electronegativity is the relative tendency of an atom

to________________________________________. when chemically combined with another atom.-Atoms with______________________of the nucleus

and the outer electron have the __________________

Variation in Electronegativity(a) Across A Period Going across period from left to right atomic size become _________ because of Zeff. Attraction between nucleus and outer electron become __________ Hence, the atom has _________relative tendency to attract electron to itself. Therefore when across a period, the electronegativity _______

Variation in Electronegativity(b) Down a group Going down a group, the atomic size _________ because the shielding effect __________ The attraction of the nucleus and outer electron become _______ Hence, the atom has ________ relative tendency to attract electron to itself.Therefore, the electronegativity __________

Example: Na < B < S < F

Electronegativities Of Elements

Exercise1. Arrange the elements in order of their increasing Electronegativity. a) Na, Li, Cs, K b) B, F, Li, C c) Cl, S, Si, Na

PHYSICAL PROPERTIESACROSS PERIOD 3

Group 1

Metallic characterMetallic character is depends on :

- atomic radius - ionization energyElectrons are easy to remove if :

- large atomic radius - low ionization energy

92

The element more metallic if it is easier to remove electrons from an atom. Thus, down a group 1, metallic character increases.

Metallic character increasing

Boiling points and Melting pointsAll elements in group 1 are reactive metals Data for melting and boiling points of group 1:

94

Boiling points and Melting pointsDown the GROUP 1, melting and boiling points of the elements decreases It is because : - When going down a group, the _________________________ - It caused the _________________________ to the nucleus. - The __________________________________ down a group. - Therefore, ________________________________________

95

Group 17

Boiling points and Melting points- All the halogens exist as diatomic molecule, X2 - They have valence electrons with the configuration of ns2np5 - This allows them to form covalent bond between its atoms in

order to achieve noble gas configuration- Example : flourine, chlorine, bromine, iodine

96

Data for melting and boiling points of Group 17: Data for Melting and boiling points of group 17:

97

Down the group 17, the melting and boiling points increases. It is because :

- Down the group, the _____________________________ or relative molecular mass increases - The______________________________________ between molecules _______________ - Therefore, __________________________________

98

Boiling points and Melting points for ACROSS PERIOD 3The melting / boiling point of a substance depends greatly on

the type of intermolecular forces or bondings involved.Variation of melting and boiling point of elements in period 3

can be discussed in 3 parts: i) metallic bonding ii) giant molecular structure iii) simple molecular structure99

Data for melting and boiling points of period 3:

Giant molecular structure

100

i) Metallic character (Na to Al)Na to Al in period 3 is metallic character. They have metallic bonding.

METALLIC BONDING Electrostatic forces between positively charged metal ions and sea of electrons.

101

- Strength of metallic bonding is proportional to the ____________________________ -Valence electrons of Na, Mg, Al are 1,2, and 3 respectively. - Thus, melting and boiling points ________ from Na to Al.

ii) Giant covalent structure (Si)Si has giant molecular structure. Each Si is ____________________________ to four other Si

atoms infinitely.

Si Si Si Si103

Tetrahedral structure

- Si has strongest attraction between its atoms to form covalent bond. - High energy is needed to break the strong covalent bonding in melting or boiling point. - Thus, Si has the highest melting and boiling points in period 3.

iii) Simple molecular structure ( P to Ar)- P, S, Cl and Ar are non metal. - P, S and Cl exist as simple molecular structurewhich are P4, S8 and Cl2. While Ar exist as monoatom. - P, S, Cl and Ar have low melting and boiling points.

Chlorine, Cl2 Sulphur , S8105

Acid-base character of oxides (Period 3)- For Period 3, when react with oxygen, it will form :

Nabasic oxide

Mg

Al

Si S

P

Sacidic oxide

Cl

Ar

amphoteric oxide

106

ACID BASE BEHAVIOUR OF OXIDES PERIOD 3

107

1. Na reacts with oxygen to form basic oxide. The oxide will produce base when react with water.4Na (s) + O2 (g) Na2O (s) + H2O (l) 2Na2O (s) 2NaOH (aq)

2. Mg burns in oxygen to form basic oxide, MgO.2Mg (s) + O2 (g) MgO (s) + 2HCl (aq) 2MgO (s) MgCl2 (aq) + H2O (l)

108

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

Al2O3 (s) + 2NaOH (aq) + 3H2O (l)

4. Silicon burns in oxygen to form acidic oxide.Si (s) + O2 (g) SiO2 (s) + 2OH- (aq) acidic SiO2 (s) SiO32- (aq) + H2O (l)

109

5. Phosphorus burns in oxygen to form acidic oxide.P4 (s) +3O2 (g) P4O6 (g) + 6H2O (l) P4O10 (g) + 6H2O (l) P4O6 (s) 4H3PO4 (aq) phosphorus acid 4H3PO4 (aq) phosphoric acid

6. Sulphur burns in oxygen to form acidic oxide.S (s) + O2 (g) SO2 (g) + H2O (l) SO2 (g) + 2NaOH (aq) acid base SO2 (g) H2SO4 (aq) Na2SO3(aq)+H2O(l)

110

7. Halogens react indirectly with O2 to form oxidex. These oxides react with water to form acids. Cl2O7 (g) + H2O (l) 2HClO4 (aq) hypochloric acid

111

Ionisation Energy (IE)The ionisation energy (general definition)- Is the ______________________required to remove the

______________________from a gaseous atom or ion.

112

7. Halogens react indirectly with O2 to form oxidex. These oxides react with water to form acids. Cl2O7 (g) + H2O (l) 2HClO4 (aq) hypochloric acid

113