1 Periodic Variation in Physical Properties of the 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 shells 7 periods

4 The modern Periodic Table Vertical columns  groups  same no. of outermost shell electrons 18 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 configuration Four blocks  s, p, d, f

9 Outermost orbitals : ns 1  ns 2 s-block : Groups 1A, 2A

10 Outermost orbitals : ns 2 np 1  ns 2 np 6 p-block : Groups 3A  8A(0)

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

12 Outermost orbitals : ns 2 (n-1)d 1  ns 2 (n-1)d 10 n  4 d-block : Transition elements (Groups 3B  1B)

13 Outermost orbitals : ns 2 (n-2)f 1  ns 2 (n-2)f 14 n  6 d-block : Inner transition elements Lanthanides Actinides

14 Q.1 2

15 Q

16 Q

17 Q

18 Q

19 Q

20 Q Atomic no. = 109  Period 7 =  6d

21 Q Atomic no. = 123  Period 8 =  5g 8

22 Q Atomic no. = 151  Period 8 =  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 point 2.Atomic radius 3.First ionization enthalpy 4.Electronegativity

25 Melting point A measure of the ease of the change from solid phase to liquid phase Depends 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 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 from Closed- packed metallic Giant covalent Simple molecular  periodic variation

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

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

29 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186

30 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 A.Variations in m.p. across a period Patterns : -  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 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  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  accordingly Boron  giant covalent

32 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 Interpretation : -  m.p.  from group 1A to 3A because (ii)Packing efficiency : - Gp2A/3A(hcp/fcc) > Gp1A(bcc)

33 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 Interpretation : -  Gp4A elements(C & Si)  giant covalent  Covalent bonds are highly directional Metallic bonds are non-directional Extent of bond breaking on melting Covalent >> metallic

34 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  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 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 Interpretation : -  Sharp  in m.p. from Gp4A to Gp5A because Covalent bond(Gp4A) >> van der Waals forces(Gp5A)

36 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  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 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 Interpretation : -  m.p. : N > O > F > Ne (regular) ∵ molecular size : N 2 > O 2 > F 2 > Ne Strength of v.d.w. forces : N 2 > O 2 > F 2 > Ne

38 ElementLiBeBCNOFNe Melting point / o C *  210  219  220  249 Boiling point / o C *  196  183  188  246 ElementNaMgAlSiP(white)SClAr Melting point / o C  101  189 Boiling point / o C  34  186 Interpretation : -  m.p. : S > P > Cl > Ar (irregular) ∵ molecular size : S 8 > P 4 > Cl 2 > Ar Strength of v.d.w. forces : S 8 > P 4 > Cl 2 > Ar

39 Atoms of elements in period 2 tend to form multiple bonds (double or triple) with one another. Examples : O=O (1  + 1  ), N  N (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.  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 3p z -2p z overlap)   Q.2

42 C=O is preferred to C-O because 1.2p z -2p z overlap > 3p z -2p z overlap 2.Polarization of  bond (mesomeric effect) results in stronger double bond B.E. (kJ mol  1 ) : C=O(749) > 2  C-O(358)  

43 Simple molecular   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 group 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 group Sn 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 elements m.p.  down the group F(-220) Cl(-101) Br(-7.2) I(114)

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

50 Atomic radius Atomic radius  when ENC  ENC depends on 1. Nuclear charge 2. Screening effect of electrons (repulsion among electrons)

51 For the first 2 or 3 elements, atomic radius  more significantly because nuclear charge  sharply

52 Then, atomic radius  less sharply because screening effect is getting more important

53 Refer to notes on ‘Electronic structure of atoms and the periodic table’, pp First ionization enthalpy

54 Variation in the first ionization enthalpy of the first 20 elements First ionization enthalpy

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

56 Increases when atomic size  Decreases when atomic size  Electronegativity

57 Electronegativity cannot be measured directly  Not a physical properties Electronegativity

58 Variation in electronegativity values of the first 20 elements

59 Periodic Relationship among the Oxides of the Elements Li to Cl

60 Li 2 OBeOB2O3B2O3 CO CO 2 N 2 O NO N 2 O 3 NO 2 N 2 O 4 N 2 O 5 O 2, O 3 OF 2 Na 2 O Na 2 O 2 MgOAl 2 O 3 SiO 2 P 4 O 6 P 4 O 10 SO 2 SO 3 Cl 2 O ClO 2 Cl 2 O 6 Cl 2 O 7 1. Bonding and Stoichiometric Composition Ionic and basic Covalent and mainly acidic Amphoteric

61 2. Reactions with water, acids and alkalis A.Ionic oxide (basic oxide) Li 2 O and Na 2 O react vigorously with water to give alkaline solutions Li 2 O(s) + H 2 O(l)  2Li + (aq) + 2OH  (aq) Na 2 O(s) + H 2 O(l)  2Na + (aq) + 2OH  (aq) -- ++

62 2. Reactions with water, acids and alkalis A.Ionic oxide (basic oxide) MgO reacts slowly with water giving a slightly alkanline solution MgO(s) + H 2 O(l) Mg(OH) 2 (s) Mg(OH) 2 (s) Mg 2+ (aq) + 2OH  (aq) High Lattice enthalpies  low solubility and slight dissociation  Equilibrium position lies on the left

63 2. Reactions with water, acids and alkalis A.Ionic oxide (basic oxide) Na and K form more than one kind of oxides Na 2 Osodium oxide Na 2 O 2 sodium peroxide KO 2 superoxide

64 2. Reactions with water, acids and alkalis A.Ionic oxide (basic oxide) 2Na 2 O 2 (s) + 2H 2 O(l)  4NaOH(aq) + O 2 (g) 4KO 2 (s) + 2H 2 O(l)  4KOH(aq) + 3O 2 (g)

65 Q.3(i) Na + and K + have small charge density (small charge and large size)  They cannot polarize or distort the unstable O 2 2 , O 2  [O – O] 2 

66 Q.3(i) Li + has a high charge density  It is polarizing enough to distort the electron cloud of O 2 2 , causing it to decompose to give Li 2 O. [O – O] 2  Li + 2Li 2 O

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 O 2 2   Gp 2A metals do not form peroxides

68 B.Ionic oxide with high covalent character (Amphoteric oxides) BeO(s) + H 2 O(l)  Al 2 O 3 (s) + H 2 O(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)  Be 2+ (aq) + H 2 O(l)  base BeO(s) + 2OH  (aq) + H 2 O(l)  Be(OH) 4 2  (aq)  acid Al 2 O 3 (s) + 6H + (aq)  2Al 3+ (aq) + 3H 2 O(l)  base Al 2 O 3 (s) + 2OH  (aq) + 3H 2 O(l)  2Al(OH) 4  (aq)  acid beryllate aluminate

70 C.Covalent oxides (acidic oxides) B 2 O 3 (s) + 3H 2 O(l)  2H 3 BO 3 (aq) 1. B 2 O 3 Boric acid or orthoboric acid

71 H 3 BO 3 are held by extensive intermolecular hydrogen bonds  A solid at room conditions

o C + H 2 O(l) Thermal dehydration of H 3 BO 3 1.Intramolecular HBO 2 metaboric acid

73 Thermal dehydration of H 3 BO 3 2.Intermolecular 160  C H 2 B 4 O 7 tetraboric acid

74 Q.4 Borax( 硼砂 ), Na 2 B 4 O 7, is used as a preservative and in the making of borax glass. Draw the structural formula of borax.

75 2. CO 2, CO, SiO 2 CO 2 reacts with water to give a weak acid which ionizes in two steps to give a weakly acidic solution.

76 ++ ++ ++ -- -- -- H 2 CO 3 carbonic acid

77 CO(g) + H 2 O(l)  no reaction SiO 2 (s) + H 2 O(l)  no reaction SiO 2 is insoluble in water due to the high lattice energy of the giant covalent structure acidic SiO 2 (l) + 2NaOH(l)  Na 2 SiO 3 + H 2 O  Sodium silicate Water glass CO 2 (g) + 2NaOH(aq)  Na 2 CO 3 + H 2 O  Sodium carbonate

78 3. Oxides of nitrogen N 2 O 5 (s) + H 2 O(l)  2HNO 3 (aq) N 2 O 4 (g) + H 2 O(l)  HNO 3 (aq) + HNO 2 (aq) N 2 O 3 (g) + H 2 O(l)  2HNO 2 (aq) Acid anhydrides Nitric acid Nitrous acid Nitric oxide (NO(g)) and nitrous oxide (N 2 O(g)) are neutral and insoluble in water NO(g) + H 2 O(l)  no reaction N 2 O(g) + H 2 O(l)  no reaction

79 Q.5 N 2 O 5 (s) + H 2 O(l)  2HNO 3 (aq) N 2 O 4 (g) + H 2 O(l)  HNO 3 (aq) + HNO 2 (aq) N 2 O 3 (g) + H 2 O(l)  2HNO 2 (aq) , Not exist +2 +4

80 N 2 O 5 (l) dinitrogen pentoxide N 2 O 4 (g) dinitrogen tetroxide NO 2 (g) nitrogen dioxide

81 N 2 O 3 (g) dinitrogen trioxide NO(g) nitric oxide N 2 O(g) nitrous oxide

82 3. Oxides of phosphorus White phosphorus White phosphorus burn spontaneously to relieved the angle strain 60  ~109 

83 3. Oxides of phosphorus P 4 O H 2 O(l)  4H 3 PO 4 (aq)  P 4 O 6 + 6H 2 O(l)  3H 3 PO 4 (aq) + PH 3 (g)  Reaction with water Q

84 Thermal dehydration of H 3 PO 4 2H 3 PO 4 H 2 O + H 4 P 2 O 7 (pyrophosphoric acid) 250 o C (i)Intermolecular dehydration 250  C + H 2 O

85 H 3 PO 4 H 2 O + HPO 3 (metaphosphoric acid) (ii)Intramolecular dehydration 900 o C Thermal dehydration of H 3 PO  C + H 2 O

86 Q.8 Draw the structure of phosphorous acid, H 3 PO 3 Dibasic acid What is the oxidation number of phosphorus in phosphorous acid? 0 +4

87 H 3 PO 2 Hypophosphorous acid Monobasic acid +3

88 Formation of sodium salts H 3 PO 4 is tribasic  3 kinds of sodium salts H 3 PO 4 + NaOH  H 2 O + NaH 2 PO 4 sodium dihydrogenphosphate NaH 2 PO 4 + NaOH  H 2 O + Na 2 HPO 4 disodium hydrogenphosphate Na 2 HPO 4 + NaOH  H 2 O + Na 3 PO 4 sodium phosphate

89 Dehydration of sodium salts 1.Inter-dehydration 2Na 2 HPO 4  H 2 O + Na 4 P 2 O 7 tetrasodium pyrophosphate 2.Intra-dehydration NaH 2 PO 4  H 2 O + NaPO 3 sodium metaphosphate

90  Na 4 P 2 O 7  NaPO 3 Q.9

91 4. Oxides of oxygen and sulphur O 2 is neutral and only slightly soluble in water SO 2 reacts with water to give sulphurous acid which ionizes in two steps to give a weakly acidic solution.

92 SO 3 reacts with water to give sulphuric acid which ionizes in two steps to give a strongly acidic solution.

93 ++ ++ ++

94 SO 3 H 2 SO 4 SO 2 H 2 SO 3

95 ++ ++ ++ -- -- -- 5. Oxides of chlorine and fluorine* *F 2 O is oxygen difluoride rather than difluorine monoxide F 2 O may react slowly with water, giving oxygen gas and a slightly acidic solution. F 2 O(g) + H 2 O(l)  2HF(aq) + O 2 (g) 2H–F + O=O

96 ++ ++ ++ ++ -- -- Cl 2 O(g) + H 2 O(l)  2HOCl(aq) 2H–O–Cl ++ ++ ++ -- -- -- 2H–F + O=O

97 Cl 2 O(g) + H 2 O(l)  2HOCl(aq) 2ClO 2 (g) + H 2 O(l)  HClO 3 (aq) + HClO 2 (aq) Cl 2 O 6 (l) + H 2 O(l)  HClO 4 (aq) + HClO 3 (aq) Cl 2 O 7 (l) + H 2 O(l)  2HClO 4 (aq) hypochlorous acid chloric(I) acid chlorous acid chloric(III) acid chloric acid chloric(V) acid perchloric acid chloric(VII) acid

98 3-electron bond

99 Q.10 b > a Repulsion between a double bond and a 5-electron centre Repulsion between two single bonds > The strong repulsion between two lone pairs in Cl 2 O decreases the bond angle a

100 Q.11 HOClHClO 2 HClO 4 HClO 3

101 Q.12 HClO 4 > HClO 3 > HClO 2 > HOCl ClO  ClO 2  ClO 3  ClO 4  Average charge on each O Attraction for H + Ease of leaving of H + Acid strength -1 -½ - -¼ decreases increases Explanation : -HClO x ClO x  + H +

102 Q.12 HClO 4 > HClO 3 > HClO 2 > 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? Answer Atomic 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? Answer Rubidium: s-block Gold: d-block Astatine: p-block Uranium: f-block Back 38.1 The Periodic Table (SB p.5)

106 Which element would have the highest first ionization enthalpy? Answer Helium Back 38.2 Periodic Variation in Physical Properties of Elements (SB p.6)

107 Which element would have the smallest atomic radius? Answer Helium Back 38.2 Periodic Variation in Physical Properties of Elements (SB p.8)

108 Why is the melting point of chlorine higher than argon? Answer Chlorine 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. Answer In 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 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)

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

115 (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)

116 (b)Carbon can form two oxides. Name the two oxides and draw their electronic structures. Answer Back 39.1 Bonding of the Oxides of Periods 2 and 3 Elements (SB p.22) (b)Carbon monoxide (CO): Carbon dioxide (CO 2 ):

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

118 (b)Complete and balance the following equations: (i)K 2 O(s) + H 2 O(l)  (ii)Na 2 O 2 (s) + HCl(aq)  (iii)Al 2 O 3 (s) + H 2 SO 4 (aq)  (iv)P 4 O 10 (s) + NaOH(aq)  (v)SO 3 (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)

119 (b)(i)K 2 O(s) + H 2 O(l)  2KOH(aq) (ii)Na 2 O 2 (s) + 2HCl(aq)  2NaCl(aq) + H 2 O 2 (aq) (iii)Al 2 O 3 (s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 3H 2 O(l ) (iv)P 4 O 10 (s) + 12NaOH(aq)  4Na 3 PO 4 (aq) + 6H 2 O(l) (v)SO 3 (g) + 2NaOH(aq)  Na 2 SO4(aq) + H 2 O(l) Back 39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids and Dilute Alkalis (SB p.27)