Periodicity - ilc.edu.hk

Report
Periodic Variation in
Physical Properties of
the Elements H to Ar
1
Elements are arranged in the increasing order of atomic
number
2
The modern Periodic Table
Horizontal rows  periods  same no. of occupied shells
7 periods
3
The modern Periodic Table
Vertical columns  groups
 same no. of outermost shell electrons
18
groups
4
The modern Periodic Table
Periodicity : Properties of elements are periodic functions
of atomic number
5
The modern Periodic Table
Periodicity : Similar properties of elements recur
periodically
6
The modern Periodic Table
Periodicity : Properties of elements vary periodically with
atomic number
7
The modern Periodic Table
Four blocks  s, p, d, f
Properties depends on electronic configuration
8
s-block : Groups 1A, 2A
Outermost orbitals : ns1  ns2
9
p-block : Groups 3A  8A(0)
Outermost orbitals : ns2 np1  ns2 np6
10
s-block & p-block elements are called
representative elements
11
d-block : Transition elements (Groups 3B  1B)
Outermost orbitals : ns2 (n-1)d1  ns2 (n-1)d10
n4
12
d-block : Inner transition elements
Outermost orbitals : ns2 (n-2)f1  ns2 (n-2)f14
n6
Lanthanides
Actinides
13
Q.1
2
14
Q.1
2
10
15
Q.1
2
10
18
16
Q.1
2
10
18
36
17
Q.1
2
10
18
36
54
18
Q.1
2
10
18
36
54
86
19
Q.1
Atomic no. = 109  Period 7
= 118-6-3  6d
2
10
18
36
54
86
118
20
Q.1
Atomic no. = 123  Period 8
= 118+2+3  5g
2
10
18
36
54
86
118
8
21
Q.1
Atomic no. = 151  Period 8
= 118+2+18+13  6f
Period 8 can hold up to 50 elements(119 to 168)
2
10
18
36
54
86
118
8
22
23
Periodic variation in physical properties of
the elements H to Ar
1.
Melting point
2.
Atomic radius
3.
First ionization enthalpy
4.
Electronegativity
24
Melting point
A measure of the ease of the change from
solid phase to liquid phase
Depends on
25
(a)
The strength of the bonds to be broken
(b)
The extent of bond breaking
(c)
The structure of the crystal lattice
Melting point  periodic variation
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
26
Closedpacked
metallic
Giant
covalent
Simple
molecular
Structure and Bonding
A summary of the variations in structure and bonding of
elements across both Periods 2
27
Structure and Bonding
A summary of the variations in structure and bonding of
elements across both Periods 3
28
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling point / oC
883
1090
2517
2900
277
444
34
186
29
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling point / oC
883
1090
2517
2900
277
444
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
30
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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  accordingly
Boron  giant covalent
31
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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)
32
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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-directional
Extent of bond breaking on melting
Covalent >> metallic
33
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling point / oC
883
1090
2517
2900
277
444
34
186
Interpretation : - * C sublimes at 1 atm
 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
34
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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)
35
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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
36
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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
37
Element
Li
Be
B
C
N
O
F
Ne
Melting point / oC
181
1287
2076
3527*
210
219
220
249
Boiling 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
189
Boiling 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
38
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)
39
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.
S
S
S
S
S
S
S8
S
P
S
P
P
P
P4
 bond formation is not favoured due to
poor side-way overlap between 3p orbitals
40
Q.2

O
O
Si
C
O
O
O
O
O

O
Si
O

O
O
C
O

Each Si atom forms four single bonds
rather than two double bonds with O atoms
(∵ poor 3pz-2pz overlap)
41

O
O
Si
C
O
O
O
O
O

O
Si
O

O
O
C
O

C=O is preferred to C-O because
1. 2pz-2pz overlap > 3pz-2pz overlap
2. Polarization of  bond (mesomeric effect)
results in stronger double bond
B.E. (kJ mol1) : C=O(749) > 2C-O(358)
42
Giant covalent

O
O
Si
C
O
O
O
O
O

O
Si
O

O
O
C
O

Simple molecular
43
B. Variation in m.p. down a group
44
For metals in Gp1A/2A/3A,
m.p.  down the group. It is because
ionic radius  down the group
charge
 the charge density,
of positive ion
size
 down the group
 the electrostatic forces of attraction
between the positive metal ions and the
delocalized electrons  down the group
 the strength of metallic bond  down the
group
45
For Gp 4A elements,
m.p.  down the group
C(3527)
Si(1414)
Ge(937)
Sn(230)
Pb(327)
46
For Gp 4A elements,
m.p.  down the group
∵ 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
47
C(3527)
Si(1414)
Ge(937)
Sn(230)
Pb(327)
For Groups 6A/7A elements
m.p.  down the group
F(-220)
∵ Size of molecules  down the group
 Extent of polarization of electron
cloud  down the group
Cl(-101)
 Strength of London dispersion forces
Br(-7.2)
 down the group
I(114)
48
Atomic radius
Refer to notes on ‘Electronic structure of atoms
and the periodic table’, pp.25-27
49
Atomic radius
Atomic radius 
when ENC 
ENC depends on
1. Nuclear charge
2. Screening effect
of electrons
(repulsion among
electrons)
50
For the first 2 or 3 elements,
atomic radius  more significantly because
nuclear
charge

sharply
51
Then, atomic radius  less sharply because
screening effect is getting more important
52
First ionization enthalpy
Refer to notes on ‘Electronic structure of atoms
and the periodic table’, pp.27-28
53
First ionization enthalpy
54
Variation in the first ionization enthalpy of
the first 20 elements
Electronegativity
Refer to notes on ‘Bonding and structure’, pp.2, 65
55
Electronegativity
Increases when atomic size 
56
Decreases when atomic size 
Electronegativity
Electronegativity cannot be measured directly
57
 Not a physical properties
Variation in electronegativity values of the first
20 elements
58
Periodic Relationship among the
Oxides of the Elements Li to Cl
59
1. Bonding and Stoichiometric Composition
Li2O
Na2O
Na2O2
BeO
MgO
B2O3
Al2O3
CO
CO2
SiO2
N2O
NO
N2O3
O2, O3
NO2
N2O4
N2O5
P4O6
P4O10
SO2
SO3
OF2
Cl2O
ClO2
Cl2O6
Cl2O7
Ionic and Amphoteric Covalent and mainly
basic
acidic
60
2. Reactions with water, acids and alkalis
A. 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)
2-
O (aq)
stronger base
61
+
H
O
H
2 OH-(aq)
weaker base
2. Reactions with water, acids and alkalis
A. 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
62
2. Reactions with water, acids and alkalis
A. Ionic oxide (basic oxide)
Na and K form more than one kind of oxides
63
Na2O
sodium oxide
Na2O2
sodium peroxide
KO2
superoxide
2. Reactions with water, acids and alkalis
A. Ionic oxide (basic oxide)
2Na2O2(s) + 2H2O(l)  4NaOH(aq) + O2(g)
4KO2(s) + 2H2O(l)  4KOH(aq) + 3O2(g)
64
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
65
Q.3(i)
Li+ has a high charge density
 It is polarizing enough to distort the
electron cloud of O22, causing it to
decompose to give Li2O.
Li+
Li+
2Li2O
[O – O]2
Li+
66
Li+
Q.3(ii)
Gp 2 ions has smaller size and greater charge
 High charge density and polarizing power
 Polarize more the electron cloud of O22
 Gp 2A metals do not form peroxides
67
B. Ionic oxide with high covalent character
(Amphoteric oxides)
BeO(s) + H2O(l) 
Al2O3(s) + H2O(l) 
68
Do not dissolve nor
react due to high lattice
enthalpies of oxides
B. Ionic oxide with high covalent character
(Amphoteric oxides)

BeO(s) + 2H+(aq)  Be2+(aq) + H2O(l)
base

BeO(s) + 2OH(aq) + H2O(l)  Be(OH)42(aq)
acid
beryllate
Al2O3(s) +
base
6H+(aq)

 2Al3+(aq) + 3H2O(l)

Al2O3(s) + 2OH(aq) + 3H2O(l)  2Al(OH)4(aq)
acid
aluminate
69
C. Covalent oxides (acidic oxides)
1. B2O3
B2O3(s) + 3H2O(l)  2H3BO3(aq)
OH
O
B
O
B
O
B
HO
OH
Boric acid or
orthoboric acid
70
H3BO3 are held by extensive intermolecular
hydrogen bonds
 A solid at room conditions
71
Thermal dehydration of H3BO3
1. Intramolecular
H
O
H
100oC
B
O
O
H
72
HO
B
O
+ H2O(l)
HBO2
metaboric acid
Thermal dehydration of H3BO3
2. Intermolecular
160C
H2B4O7 tetraboric acid
73
Q.4
Borax(硼砂), Na2B4O7, is used as a preservative
and in the making of borax glass.
Draw the structural formula of borax.
O
Na
O
B
B
O
74
O
O
B
B
O
O
Na
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)
75
CO3(aq)
2H
H2CO3(aq) + H2O(l)
+
H
O
(aq)
+
HCO
3
3 (aq)
HCO3-(aq) + H2O(l)
+
2H
O
(aq)
+
CO
3
3 (aq)
H
- O
+ C
- O
+
H
O
-O
C
H
+
O
O
H
H2CO3
carbonic acid
76
CO(g) + H2O(l)  no reaction
SiO2(s) + H2O(l)  no reaction
SiO2 is insoluble in water due to the high
lattice energy of the giant covalent structure

SiO2(l) + 2NaOH(l)  Na2SiO3 + H2O
acidic
Sodium silicate
Water glass

CO2(g) + 2NaOH(aq)  Na2CO3 + H2O
Sodium carbonate
77
3. Oxides of nitrogen
Nitric acid
N2O5(s) + H2O(l)  2HNO3(aq)
N2O4(g) + H2O(l)  HNO3(aq) + HNO2(aq)
N2O3(g) + H2O(l)  2HNO2(aq) Nitrous acid
Acid
anhydrides
NO(g) + H2O(l)  no reaction
N2O(g) + H2O(l)  no reaction
Nitric oxide (NO(g)) and nitrous oxide (N2O(g))
are neutral and insoluble in water
78
Q.5
+5
+5
N2O5(s) + H2O(l)  2HNO3(aq)
+4
+5
+3
N2O4(g) + H2O(l)  HNO3(aq) + HNO2(aq)
+3
+2, +4
N2O3(g) + H2O(l)  2HNO2(aq)
O
N
+3
O
N
+3
Not exist
79
O
O
N
+2

+4
N
O
O

N2O5(l)
dinitrogen
pentoxide
N2O4(g)
dinitrogen
tetroxide
NO2(g)
nitrogen
dioxide
80
O
O
N
N
O
O
O
O
O
N
N
O
O
N
O
O
N2O3(g)
dinitrogen
trioxide
N
O
N
O
O
NO(g)
nitric oxide
N2O(g)
nitrous oxide
81
N
0
+2
N
N
O
O
-1
+3
N
N
O
3. Oxides of phosphorus
O
~109 P
60 P
O
3O2
P
P
P
P4
White
phosphorus
P
O
~109 P
O
O
2O2
O
P
O
P
O
P4O6
O
P
O
O
P
O
P
O
O
O
P4O10
White phosphorus burn spontaneously to
relieved the angle strain
82
O
3. Oxides of phosphorus
Reaction with water
+5
P4O10

Q.6
83
+3

+ 6H2O(l) 
P4O6 + 6H2O(l)


+5
4H3PO4(aq)
+5
-3
3H3PO4(aq) + PH3(g)
Thermal dehydration of H3PO4
(i) Intermolecular
dehydration
2H3PO4
O
HO
HO
250oC
H2O + H4P2O7
(pyrophosphoric acid)
O
P
P
OH
HO
O
O
P
P
250C
OH
OH
HO
HO
O
+ H2O
84
OH
OH
Thermal dehydration of H3PO4
(ii)
Intramolecular dehydration
H3PO4
900oC
H2O + HPO3
(metaphosphoric acid)
O
O
900C
HO
HO
85
P
P
OH
O
OH + H2O
Q.8
Draw the structure of phosphorous acid, H3PO3
O
+4
HO
P
OH
Dibasic acid
H
0
What is the oxidation number of phosphorus in
phosphorous acid?
86
O
H3PO2
+3
HO
87
Hypophosphorous acid
P
H
H
Monobasic acid
Formation of sodium salts
H3PO4 is tribasic  3 kinds of sodium salts
H3PO4 + NaOH  H2O + NaH2PO4
sodium dihydrogenphosphate
NaH2PO4 + NaOH  H2O + Na2HPO4
disodium hydrogenphosphate
Na2HPO4 + NaOH  H2O + Na3PO4
sodium phosphate
88
Dehydration of sodium salts
1. Inter-dehydration
2Na2HPO4  H2O + Na4P2O7
tetrasodium pyrophosphate
2. Intra-dehydration
NaH2PO4  H2O + NaPO3
sodium metaphosphate
89
Q.9
+
-
Na O
+
Na-O
O
O
P
P
OH
HO

O-Na+
O-Na+
+
Na-O
+
Na-O
O
O
P
P
O
Na4P2O7
O

P
+
Na-O
OH
OH
O
+
Na-O
P
O
NaPO3
90
O-Na+
O-Na+
4. Oxides of oxygen and sulphur
O2 is neutral and only slightly soluble in water
SO2 reacts with water to give sulphurous acid
which ionizes in two steps to give a weakly
acidic solution.
SO 2(g) + H2O(l)
91
SO 3(aq)
2H
H2SO 3(aq) + H2O(l)
+
H
O
(aq)
+
HSO
3
3 (aq)
HSO3-(aq) + H2O(l)
+
2H
O
(aq)
+
SO
3
3 (aq)
SO3 reacts with water to give sulphuric acid
which ionizes in two steps to give a strongly
acidic solution.
SO 3(g) + H2O(l)
92
SO 4(aq)
2H
H2SO 4(aq) + H2O(l)
+
H
O
(aq)
+
HSO
3
4 (aq)
HSO4-(aq) + H2O(l)
+
2H
O
(aq)
+
SO
3
4 (aq)
O
+
S
O
O
93
O
H +
O
S
H +
O
OH
OH
O
O
S
S
O
O
SO3
S
O
H2SO4
OH
OH
H2SO3
94
O
OH
OH
S
O
O
SO2
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)
- F
+O
-F
95
H +
2H–F + O=O
- O
H +
Cl2O(g) + H2O(l)  2HOCl(aq)
+
H
O
+
Cl
Cl
+
O
-
- F
+O
-F
96
H+
2H–O–Cl
H +
2H–F + O=O
- O
H +
+1
+1
Cl2O(g) + H2O(l)  2HOCl(aq)
hypochlorous acid
chloric(I) acid
+4
+5
+3
2ClO2(g) + H2O(l)  HClO3(aq) + HClO2(aq)
chlorous acid
chloric(III) acid
+6
+7
+5
Cl2O6(l) + H2O(l)  HClO4(aq) + HClO3(aq)
chloric acid
chloric(V) acid
+7
+7
Cl2O7(l) + H2O(l)  2HClO4(aq)
perchloric acid
chloric(VII) acid
97
3-electron bond
ClCl
O
Cl
Cl
Cl2O
98
O
O
OO
ClO2
O
O
O
Cl
Cl
O
O
Cl2O6
O
O
O
Cl
O
O
O
Cl
O
Cl2O7
O
Q.10
ClCl
O
Cl
a
Cl
OO
OO
bb
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 Cl2O decreases the bond angle a
99
Q.11
O
Cl
H
O
Cl
O
HOCl
HClO2
O
O
Cl
Cl
OH
HClO3
100
HO
O
OH
O
HClO4
Q.12
HClO4 > HClO3 > HClO2 > HOCl
Explanation : -
HClOx
ClO
Average charge
on each O
Attraction for H+
Ease of leaving of H+
Acid strength
101
-1
ClOx + H+
ClO2
ClO3
-½
decreases
increases
increases
1
- 3
ClO4
-¼
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
102
The END
103
38.1 The Periodic Table (SB p.3)
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
104
38.1 The Periodic Table (SB p.5)
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
105
38.2 Periodic Variation in Physical Properties of Elements (SB p.6)
Which element would have the highest first ionization
enthalpy?
Answer
Helium
Back
106
38.2 Periodic Variation in Physical Properties of Elements (SB p.8)
Which element would have the smallest atomic radius?
Answer
Helium
Back
107
38.2 Periodic Variation in Physical Properties of Elements (SB p.12)
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
108
38.2 Periodic Variation in Physical Properties of Elements (SB p.13)
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
109
38.2 Periodic Variation in Physical Properties of Elements (SB p.13)
(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.
110
38.2 Periodic Variation in Physical Properties of Elements (SB p.13)
(b) Describe and explain the general periodic trend of
atomic radius of elements in the Periodic Table.
Answer
111
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.
112
38.2 Periodic Variation in Physical Properties of Elements (SB p.13)
(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
113
38.2 Periodic Variation in Physical Properties of Elements (SB p.13)
(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
114
39.1 Bonding of the Oxides of Periods 2 and 3 Elements (SB p.22)
(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
(a) (i) Ionic oxide
(ii) Covalent oxide
(iii) Covalent oxide
(iv) Ionic oxide with covalent character
115
Answer
39.1 Bonding of the Oxides of Periods 2 and 3 Elements (SB p.22)
(b) Carbon can form two oxides. Name the two oxides and
draw their electronic structures.
Answer
(b) Carbon monoxide (CO):
Carbon dioxide (CO2):
Back
116
39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids
and Dilute Alkalis (SB p.27)
(a) Why does silicon(IV) oxide not react with water?
(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.
117
Answer
39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids
and Dilute Alkalis (SB p.27)
(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) 
118
Answer
39.2 Behaviour of Oxides of Periods 2 and 3 Elements in Water, Dilute Acids
and Dilute Alkalis (SB p.27)
(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)
Back
119

similar documents