Periodic Variation in Physical Properties of the Elements H to Ar

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Periodic Variation in Physical Properties of the Elements H to Ar. Elements are arranged in the increasing order of atomic number. The modern Periodic Table. Horizontal rows  periods  same no. of occupied shells 7 periods. The modern Periodic Table. Vertical columns  groups - PowerPoint PPT Presentation

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1

Periodic Variation in Periodic Variation in Physical Properties of Physical Properties of the Elements H to Arthe Elements H to Ar

2 The modern Periodic Table

Elements are arranged in the increasing order of atomic number

3 The modern Periodic Table

Horizontal rows periods same no. of occupied shells7 periods

4 The modern Periodic Table

Vertical columns groups same no. of outermost shell

electrons18

groups

5 The modern Periodic Table

Periodicity : Properties of elements are periodic functions of atomic number

6 The modern Periodic Table

Periodicity : Similar properties of elements recur periodically

7 The modern Periodic Table

Periodicity : Properties of elements vary periodically with atomic number

8

Properties depends on electronic configurationFour blocks s, p, d, f

9

Outermost orbitals : ns1 ns2

s-block : Groups 1A, 2A

10

Outermost orbitals : ns2 np1 ns2 np6

p-block : Groups 3A 8A(0)

11

s-block & p-block elements are called representative elements

12

Outermost orbitals : ns2 (n-1)d1 ns2 (n-1)d10

n4

d-block : Transition elements (Groups 3B 1B)

13

Outermost orbitals : ns2 (n-2)f1 ns2 (n-2)f14

n6

d-block : Inner transition elements

LanthanidesActinides

14

Q.1

2

15

Q.1

210

16

Q.1

21018

17

Q.1

2101836

18

Q.1

210183654

19

Q.1

21018365486

20

Q.1

21018365486118

Atomic no. = 109

Period 7 = 118-6-3

6d

21

Q.1

21018365486118

Atomic no. = 123

Period 8 = 118+2+3

5g

8

22

Q.1

21018365486118

Atomic no. = 151

Period 8 = 118+2+18+13

6f

8

Period 8 can hold up to 50 elements(119 to 168)

23

24

Periodic variation in physical properties of the elements H to Ar

1. Melting point2. Atomic radius3. First ionization enthalpy4. Electronegativity

25

Melting point Melting point A measure of the ease of the change from solid phase to liquid phaseDepends on(a) The strength of the bonds to be broken(b) The extent of bond breaking(c) The structure of the crystal lattice

26

Melting point Melting point Across a period,1. the type of bonding changes from

Strong metallic

Strong covalent

Weak van der Waals’ forces

2. the structure of elements changes fromClosed-packed metallic

Giant covalent

Simple molecular

periodic variation

27

Structure and Structure and BondingBonding

A summary of the variations in structure and bonding of elements across both Periods 2

28

Structure and Structure and BondingBonding

A summary of the variations in structure and bonding of elements across both Periods 3

29

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

30

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

A.Variations in m.p. across a periodPatterns :

- increases steadily from group 1A to 3A, reaching a maximum in group 4A

drops sharply from group 4A to 5A, and eventually reaching a minimum in group

0

31

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - m.p. from group 1A to 3A because

(i) the no. of outermost electrons involved in metallic bonds from 1 to 3

strength of bond accordinglyBoron giant covalent

32

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - m.p. from group 1A to 3A because

(ii) Packing efficiency : - Gp2A/3A(hcp/fcc) > Gp1A(bcc)

33

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - Gp4A elements(C & Si) giant covalent Covalent bonds are highly directional

Metallic bonds are non-directionalExtent of bond breaking on meltingCovalent >> metallic

34

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - For metals, the differences between m.p. and b.p. are great

∵ extent of bond breaking : boiling >> melting

Particles are completely separated on boiling For Gp4A elements, the differences between m.p. and b.p. are relatively small

∵ extent of bond breaking : boiling melting

* C sublimes at 1 atm

35

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - Sharp in m.p. from Gp4A to Gp5A because

Covalent bond(Gp4A) >> van der Waals forces(Gp5A)

36

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - m.p. of Mg m.p. of Al

∵ only an average of TWO outermost shell electrons per atom of aluminium participate in

the formation of metallic bonds

37

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - m.p. : N > O > F > Ne (regular)

∵ molecular size : N2 > O2 > F2 > Ne Strength of v.d.w. forces : N2 > O2 > F2 >

Ne

38

Element Li Be B C N O F NeMelting point / oC 181 1287 2076 3527* 210 219 220 249Boiling point / oC 1342 2469 3927 4027* 196 183 188 246

Element Na Mg Al Si P(white) S Cl Ar

Melting point / oC 98 650 660 1414 44 115 101 189Boiling point / oC 883 1090 2517 2900 277 444 34 186

Interpretation : - m.p. : S > P > Cl > Ar (irregular)

∵ molecular size : S8 > P4 > Cl2 > Ar Strength of v.d.w. forces : S8 > P4 > Cl2 >

Ar

39

Atoms of elements in period 2 tend to form multiple bonds (double or triple) with one another. Examples : O=O (1 + 1), NN (1 + 2)

40

Atoms of elements in period 3 do not form multiple bonds with one another. Instead, they form cyclic structure in which all bonds are bonds.

SS

SS

SSS

S

S8

P

P

PP

P4

bond formation is not favoured due to poor side-way overlap between 3p orbitals

41

Each Si atom forms four single bonds rather than two double bonds with O atoms (∵ poor 3pz-2pz overlap)

Si

O

O OO

O Si O O C O

C

O

O OO

Q.2

42

C=O is preferred to C-O because1. 2pz-2pz overlap > 3pz-2pz overlap2. Polarization of bond (mesomeric effect) results in stronger double bond

B.E. (kJ mol1) : C=O(749) > 2C-O(358)

Si

O

O OO

O Si O O C O

C

O

O OO

43

Simple molecular

Si

O

O OO

O Si O O C O

C

O

O OO

Giant covalent

44

B. Variation in m.p. down a group

45

the electrostatic forces of attraction between the positive metal ions and the delocalized electrons down the group

the charge density, of positive ion down the groupsize

charge

For metals in Gp1A/2A/3A,

the strength of metallic bond down the group

m.p. down the group. It is because ionic radius down the group

46

For Gp 4A elements,m.p. down the group

C(3527)

Si(1414)

Ge(937)

Sn(230)

Pb(327)

47

∵ atomic radius down the group Extent of orbital overlap down the group Strength of covalent bond down

the groupSn and Pb are metals and thus have exceptionally low m.p. due to less extensive breaking of metallic bonds

C(3527)

Si(1414)

Ge(937)

Sn(230)

Pb(327)

For Gp 4A elements,m.p. down the group

48

∵ Size of molecules down the group Extent of polarization of electron cloud down the group Strength of London dispersion forces down the group

For Groups 6A/7A elementsm.p. down the group F(-220)

Cl(-101)

Br(-7.2)

I(114)

49

Atomic Atomic radius radius

Refer to notes on ‘Electronic structure of atoms and the periodic table’, pp.25-27

50

Atomic Atomic radius radius Atomic radius when ENC ENC depends on1. Nuclear charge2. Screening effect of electrons (repulsion among

electrons)

51

For the first 2 or 3 elements,atomic radius more significantly becausenuclear charge sharply

52

Then, atomic radius less sharply becausescreening effect is getting more important

53

Refer to notes on ‘Electronic structure of atoms and the periodic table’, pp.27-28

First ionization enthalpy First ionization enthalpy

54Variation in the first ionization enthalpy of

the first 20 elements

First ionization enthalpy First ionization enthalpy

55

Refer to notes on ‘Bonding and structure’, pp.2, 65

Electronegativity Electronegativity

56

Increases when atomic size

Decreases when atomic size

Electronegativity Electronegativity

57

Electronegativity cannot be measured directly Not a physical properties

Electronegativity Electronegativity

58

Variation in electronegativity values of the first 20 elements

59

Periodic Relationship among the Oxides of the Elements Li

to Cl

60

Li2O BeO B2O3CO CO2

N2O NON2O3 NO2N2O4 N2O5

O2, O3 OF2

Na2O Na2O2

MgO Al2O3 SiO2P4O6 P4O10

SO2 SO3

Cl2OClO2Cl2O6Cl2O7

1. Bonding and Stoichiometric Composition

Ionic and basic

Covalent and mainly acidic

Amphoteric

61

2. Reactions with water, acids and alkalisA. Ionic oxide (basic oxide)

Li2O and Na2O react vigorously with water to give alkaline solutions Li2O(s) + H2O(l) 2Li+(aq) + 2OH(aq)

Na2O(s) + H2O(l) 2Na+(aq) + 2OH(aq)

HO

HO2-(aq) 2 OH-(aq)

stronger base weaker base

-

+

62

2. Reactions with water, acids and alkalisA. Ionic oxide (basic oxide)

MgO reacts slowly with water giving a slightly alkanline solution

MgO(s) + H2O(l) Mg(OH)2(s)Mg(OH)2(s) Mg2+(aq) + 2OH(aq)High Lattice enthalpies low solubility and slight dissociation Equilibrium position lies on the left

63

2. Reactions with water, acids and alkalisA. Ionic oxide (basic oxide)

Na and K form more than one kind of oxides

Na2O sodium oxideNa2O2 sodium peroxideKO2 superoxide

64

2. Reactions with water, acids and alkalisA. Ionic oxide (basic oxide)

2Na2O2(s) + 2H2O(l) 4NaOH(aq) + O2(g)4KO2(s) + 2H2O(l) 4KOH(aq) + 3O2(g)

65

Q.3(i)Na+ and K+ have small charge density (small charge and large size) They cannot polarize or distort the

unstable O22, O2

[O – O]2

66

Q.3(i)Li+ has a high charge density It is polarizing enough to distort the electron cloud of O2

2, causing it to decompose to give Li2O.

[O – O]2

Li+

Li+

Li+

Li+

2Li2O

67

Q.3(ii)Gp 2 ions has smaller size and greater charge High charge density and polarizing power Polarize more the electron cloud of O2

2

Gp 2A metals do not form peroxides

68

B. Ionic oxide with high covalent character (Amphoteric oxides)

BeO(s) + H2O(l) Al2O3(s) + H2O(l)

Do not dissolve nor react due to high lattice enthalpies of oxides

69

B. Ionic oxide with high covalent character (Amphoteric oxides)

BeO(s) + 2H+(aq) Be2+(aq) + H2O(l)

baseBeO(s) + 2OH(aq) + H2O(l) Be(OH)4

2(aq)

acid

Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)

baseAl2O3(s) + 2OH(aq) + 3H2O(l) 2Al(OH)4

(aq)

acid

beryllate

aluminate

70

C. Covalent oxides (acidic oxides)

B2O3(s) + 3H2O(l) 2H3BO3(aq)

1. B2O3

BO

BOO B

OH

HO OH

Boric acid or orthoboric acid

71

H3BO3 are held by extensive intermolecular hydrogen bonds A solid at room conditions

72

B

OH

OOH

H

B OHO100oC + H2O(l)

Thermal dehydration of H3BO31. Intramolecular

HBO2 metaboric acid

73

Thermal dehydration of H3BO32. Intermolecular

160C

H2B4O7 tetraboric acid

74

Q.4Borax(硼砂 ), Na2B4O7, is used as a preservative and in the making of borax glass.Draw the structural formula of borax.

O BO

OB O B

O

OB O NaNa

75

2. CO2, CO, SiO2

CO2 reacts with water to give a weak acid which ionizes in two steps to give a weakly acidic solution.CO2(g) + H2O(l) H2CO3(aq)

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

HCO3-(aq) + H2O(l) H3O+(aq) + CO3

2-(aq)

76

C

O

O

O

H

H

C O

O

O

H

H

+

+

+

-

-

-

H2CO3 carbonic acid

77

CO(g) + H2O(l) no reactionSiO2(s) + H2O(l) no reactionSiO2 is insoluble in water due to the high lattice energy of the giant covalent structure

acidic

SiO2(l) + 2NaOH(l) Na2SiO3 + H2O

Sodium silicateWater glass

CO2(g) + 2NaOH(aq) Na2CO3 + H2O

Sodium carbonate

78

3. Oxides of nitrogen

N2O5(s) + H2O(l) 2HNO3(aq) N2O4(g) + H2O(l) HNO3(aq) + HNO2(aq)N2O3(g) + H2O(l) 2HNO2(aq)

Acid anhydride

s

Nitric acid

Nitrous acid

Nitric oxide (NO(g)) and nitrous oxide (N2O(g)) are neutral and insoluble in water

NO(g) + H2O(l) no reactionN2O(g) + H2O(l) no reaction

79

Q.5

N2O5(s) + H2O(l) 2HNO3(aq) N2O4(g) + H2O(l) HNO3(aq) + HNO2(aq)N2O3(g) + H2O(l) 2HNO2(aq)

+5

+5

+4

+5

+3

+3

+2, +4

O

NO

N

O+3 +3

Not exist

O

NN

O

O

+2 +4

80

NO

N

O O

OO

N2O5(l) dinitrogen pentoxide

N2O4(g) dinitrogen tetroxide

N N

OO

OO

O

NO

NO2(g) nitrogen dioxide

81

N2O3(g) dinitrogen

trioxide

NO(g) nitric oxide

N2O(g) nitrous oxide

O

NN

O

O

N O

N N O N N O0 +2 -1 +3

82

3. Oxides of phosphorus

PO O

P

O

PO

O

PO

O

O

O

O

PO O

P

O

PO

O

PO

P

P P

P

2O23O2

P4P4O6 P4O10White

phosphorus

White phosphorus burn spontaneously to relieved the angle strain

60~109 ~109

83

3. Oxides of phosphorus

P4O10 + 6H2O(l) 4H3PO4(aq)

P4O6 + 6H2O(l) 3H3PO4(aq) + PH3(g)

Reaction with water

Q.6

+5

+5

+5

-3+3

84

O

PHO OH

OH

O

POHHO

HO

Thermal dehydration of H3PO4

2H3PO4 H2O + H4P2O7 (pyrophosphoric

acid)

250oC

(i) Intermolecular dehydration

O

P

OHOH

O

POHO

HO

250C

+ H2O

85

H3PO4 H2O + HPO3 (metaphosphoric acid)

(ii) Intramolecular dehydration

900oC

Thermal dehydration of H3PO4

O

POHHO

HOP OH

O

O900C

+ H2O

86

Q.8Draw the structure of phosphorous acid, H3PO3

Dibasic acid

O

PHO H

OH

What is the oxidation number of phosphorus in phosphorous acid?

0

+4

87

O

PHO H

H

H3PO2

Hypophosphorous acidMonobasic acid

+3

88

Formation of sodium saltsH3PO4 is tribasic 3 kinds of sodium salts

H3PO4 + NaOH H2O + NaH2PO4sodium dihydrogenphosphate

NaH2PO4 + NaOH H2O + Na2HPO4disodium hydrogenphosphate

Na2HPO4 + NaOH H2O + Na3PO4sodium phosphate

89

Dehydration of sodium salts1. Inter-dehydration

2Na2HPO4 H2O + Na4P2O7

tetrasodium pyrophosphate

2. Intra-dehydrationNaH2PO4 H2O + NaPO3

sodium metaphosphate

90

O

PHO O-Na+

O-Na+

O

POH+Na-O

+Na-O

O

P

O-Na+O-Na+

O

PO+Na-O

+Na-O

Na4P2O7

O

P+Na-O OH

OHP+Na-O

O

O

NaPO3

Q.9

91

4. Oxides of oxygen and sulphurO2 is neutral and only slightly soluble in

water

SO2(g) + H2O(l) H2SO3(aq)

H2SO3(aq) + H2O(l) H3O+(aq) + HSO3-(aq)

HSO3-(aq) + H2O(l) H3O+(aq) + SO3

2-(aq)

SO2 reacts with water to give sulphurous acid which ionizes in two steps to give a weakly acidic solution.

92

SO3 reacts with water to give sulphuric acid which ionizes in two steps to give a strongly acidic solution.

SO3(g) + H2O(l) H2SO4(aq)

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)

HSO4-(aq) + H2O(l) H3O+(aq) + SO4

2-(aq)

93

O S

O

O

O

H

H

++

+

O

SO OH

OH

94

O

S

O

S

O OHOH

O

SO O

O

SO OH

OH

SO3 H2SO4

SO2H2SO3

95

O

F

F

H

O

H

+

+

+

-

-

-

5. Oxides of chlorine and fluorine*

* F2O is oxygen difluoride rather than difluorine monoxide

F2O may react slowly with water, giving oxygen gas and a slightly acidic solution.

F2O(g) + H2O(l) 2HF(aq) + O2(g)

2H–F + O=O

96

OCl

Cl

H

OH

+

+

+

+-

-

Cl2O(g) + H2O(l) 2HOCl(aq)

2H–O–Cl

O

F

F

H

O

H

+

+

+

-

-

-2H–F + O=O

97

Cl2O(g) + H2O(l) 2HOCl(aq)

2ClO2(g) + H2O(l) HClO3(aq) + HClO2(aq)

Cl2O6(l) + H2O(l) HClO4(aq) + HClO3(aq)

Cl2O7(l) + H2O(l) 2HClO4(aq)

hypochlorous acid chloric(I)

acid

chlorous acid chloric(III) acid

chloric acid chloric(V) acid

perchloric acid chloric(VII) acid

+4 +5 +3

+6 +7 +5

+1 +1

+7+7

98

ClO

Cl O

Cl

OCl

O

O

O

Cl

O

O

O

Cl

O

O

O

O Cl

O

O

O

Cl2O ClO2 Cl2O6 Cl2O7

O

ClO

3-electron bond

99

Q.10

ClO

Cl O

Cl

Oa b

O

ClO

b

b > aRepulsion between a double bond and a 5-electron centre

Repulsion between two single bonds

>

The strong repulsion between two lone pairs in Cl2O decreases the bond angle a

100

Q.11

HO

Cl HOCl

O

Cl

O

O OHO

HOCl HClO2

HClO4HClO3

O

ClO OH

101

Q.12 HClO4 > HClO3 > HClO2

> HOCl

ClO ClO2 ClO3

ClO4

Average charge on

each OAttraction for H+

Ease of leaving of H+

Acid strength

-1 -½ - -¼

31

decreases

increases

increases

Explanation : -

HClOx ClOx +

H+

102

Q.12 HClO4 > HClO3 > HClO2 > HOCl

No. of O atoms bonded to Cl atom Cl atom becomes more positively charged better electron pair acceptor stronger Lewis acid

103

The END

104

The atomic numbers of tellurium and iodine are 52 and 53 respectively. Why is tellurium heavier than

iodine?AnswerAtomic number of an element is not related to the mass of an atom

of the element. The atomic number of an element is the number of protons in an atom of the element. It is unique for each element. The mass of an atom of the element is mainly determined by the number of protons and neutrons in the nucleus. Therefore, tellurium is heavier than iodine though the atomic number of tellurium is smaller than that of iodine. Back

38.1 The Periodic Table (SB p.3)

105

To which block (s-, p-, d- or f-) in the Periodic Table do rubidium, gold, astatine and uranium belong respectively? AnswerRubidium: s-blockGold: d-blockAstatine: p-blockUranium: f-block

Back

38.1 The Periodic Table (SB p.5)

106

Which element would have the highest first ionization enthalpy?

AnswerHelium

Back

38.2 Periodic Variation in Physical Properties of Elements (SB p.6)

107

Which element would have the smallest atomic radius?

AnswerHelium

Back

38.2 Periodic Variation in Physical Properties of Elements (SB p.8)

108

Why is the melting point of chlorine higher than argon?

AnswerChlorine atom has a higher effective nuclear charge than argon atom, so the atomic radius of chlorine is smaller than that of argon. Therefore, the van der Waals’ forces between chlorine molecules are stronger than those between argon molecules. Since a higher amount of energy is needed to overcome the stronger van der Waals’ forces, the melting point of chlorine is higher than that of argon.

Back

38.2 Periodic Variation in Physical Properties of Elements (SB p.12)

109

Considering the trend of atomic radius in the Periodic Table, arrange the elements Si, N and P in the order of increasing atomic radius. Explain your answer briefly.

AnswerIn the Periodic Table, N is above P in Group VA. As the atomic radius increases down a group, the atomic radius of N is smaller than that of P.

Si and P belong to the same period. Since the atomic radius decreases across a period, the atomic radius of P is smaller than that of Si.

Therefore, the atomic radius increases in the order: N < P < Si.

Back

38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

110

(a) With the help of the Periodic Table only, arrange the elements selenium, sulphur and argon in the order of increasing first ionization enthalpies. Answer

(a) The first ionization enthalpy increases in the order: Se < S < Ar.

38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

111

(b) Describe and explain the general periodic trend of atomic radius of elements in the Periodic Table. Answer

38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

112

38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

(b) Within a given period, the atomic radii decrease progressively with increasing atomic numbers. This is because an increase in atomic number by one means that one more electron and one more proton are added in the atom. The additional electron would cause an increase in repulsion between the electrons in the outermost shell and results in an increase in atomic radius. The additional proton in the nucleus would cause the electrons to experience greater attractive forces from the nucleus. Due to the fact that the newly added electron goes to the outermost shell and is at approximately the same distance from the nucleus, the repulsion between the electrons is relatively ineffective to cause an increase in atomic radius. Therefore, the effect of increasing nuclear charge outweighs the effect of repulsion between the electrons. That means, there is an increase in effective nuclear charge. As a result, the atomic radii of elements decrease across a period.

113

(c) With reference to Fig. 38-9 on p.11 (variation in electronegativity value of the first 20 elements), explain why the alkali metals are almost at the bottom of the troughs, whereas the halogens are at the peaks of the plot. Answer

38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

114

(c) The alkali metals are almost at the bottom of troughs, indicating that they have low electronegativity values. It is because their nuclear charge is effectively shielded by the fully-filled inner electron shells of electrons, and the bonding electrons are attracted less strongly. On the other hand, the halogens appear at the peaks. This indicates that they have high electronegativity values. It is because they have one electron less than the octet electronic configuration. They tend to attract an electron to complete the octet, and the bonding electrons are attracted strongly.

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38.2 Periodic Variation in Physical Properties of Elements (SB p.13)

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(a) To which type of oxide does each of the following oxides belong?(i) Magnesium oxide(ii) Nitrogen monoxide(iii) Silicon dioxide(iv) Aluminium oxide Answer

(a) (i) Ionic oxide(ii) Covalent oxide(iii) Covalent oxide(iv) Ionic oxide with covalent character

39.1 Bonding of the Oxides of Periods 2 and 3 Elements (SB p.22)

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(b) Carbon can form two oxides. Name the two oxides and draw their electronic structures.

Answer

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39.1 Bonding of the Oxides of Periods 2 and 3 Elements (SB p.22)

(b) Carbon monoxide (CO):Carbon dioxide (CO2):

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(a) Why does silicon(IV) oxide not react with water?

Answer(a) Silicon(IV) oxide does not react with water because the electronegativity values of silicon and oxygen are very similar. The Si — O bond can be considered as nonpolar, so there is no positive centre for the lone pair electrons of the water molecule to attack.

39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids and Dilute Alkalis (SB p.27)

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(b) Complete and balance the following equations:(i) K2O(s) + H2O(l)

(ii) Na2O2(s) + HCl(aq) (iii) Al2O3(s) + H2SO4(aq)

(iv) P4O10(s) + NaOH(aq)

(v) SO3(g) + NaOH(aq) Answer

39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids and Dilute Alkalis (SB p.27)

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(b) (i) K2O(s) + H2O(l) 2KOH(aq)

(ii) Na2O2(s) + 2HCl(aq) 2NaCl(aq) + H2O2(aq)

(iii) Al2O3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2O(l )

(iv) P4O10(s) + 12NaOH(aq) 4Na3PO4(aq) + 6H2O(l)

(v) SO3(g) + 2NaOH(aq) Na2SO4(aq) + H2O(l)

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39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids and Dilute Alkalis (SB p.27)