Ikatan Ion

1

7.17.1 Formation of Ionic Bonds: Donating and AcceptingFormation of Ionic Bonds: Donating and Accepting

ElectronsElectrons

7.2 7.2 Energetics of Formation of Ionic CompoundsEnergetics of Formation of Ionic Compounds

7.3 7.3 Stoichiometry of Ionic CompoundsStoichiometry of Ionic Compounds

7.47.4 Ionic CrystalsIonic Crystals

7.57.5 Ionic RadiiIonic Radii

Ionic BondingIonic Bonding77

2

Bonding & Structure

3

Lewis Model G.N. Lewis in 1916

Only the outermost (valence) electrons are involved significantly in bond formation

Successful in solving chemical problems

4

Why are some elements so reactive (e.g.Na) and others inert (e.g. noble

gases)?

Why are there compounds with chemical formulae H2O and NaCl, but not H3O and NaCl2?

Why are helium and the other noble gases monatomic, when molecules of hydrogen and chlorine are diatomic?

5

Chemical bonds are strong electrostatic forces holding atoms or ions together, which are formed by the rearrangement (transfer or sharing) of outermost electrons

Atoms tend to form chemical bonds in such a way as to achieve the

electronic configurations of the nearest noble gases (The Octet Rule )

6

Ion s,p,d,f notationIsoelectronic

noble gas

Be2+

O2-

Sc3+

Br-

Ba2+

At-

Q.1 Write the s,p,d,f notation for the ions in the table below

7

Ion s,p,d,f notation Isoelectronic

noble gas

Be2+ 1s2 (2) He

O2- [He] 2s2,2p6 (2,8) Ne

Sc3+ [Ne] 3s2,3p6 (2,8,8) Ar

Br- [Ar] 3d10,4s2,4p6 (2,8,18,8) Kr

Ba2+ [Kr] 4d10,5s2,5p6 (2,8,18,18,8) Xe

At- [Xe] 4f14,5d10,6s2,6p6 (2,8,18,32,18,8) Rn

8

Three types of chemical bondsThree types of chemical bonds

1.Ionic bond (electrovalent bond)

Formed by transfer of electrons

Introduction (SB p.186)

9

Na Cl

Sodium atom, Na1s22s22p63s1

Chlorine atom, Cl1s22s22p63s23p5

7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)


1.Ionic bond (electrovalent bond)

10


1.Ionic bond


Electrostatic attraction between positively charged particles and negatively charged particles

11


2. Covalent bond

Formed by sharing of electrons


12


2.Covalent bond


Electrostatic attraction between nuclei and shared electrons

13


3.Metallic bond

Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)


14


3. Metallic bond


Formed by sharing of a large number of delocalized electrons

15

Ionic bonds and Covalent bonds are only extreme cases of a continuum.

In real situation, most chemical bonds are intermediate between ionic and covalent

16

Ionic Bonds with incomplete transfer of electrons have covalent character.Covalent Bonds with unequal sharing of electrons have ionic character.

17

Electronegativity and Types of Chemical BondsIonic or covalent depends on the electron-attracting ability of bonding atoms in a chemical bond.

Ionic bonds are formed between atoms with great difference in their electron-attracting abilities

Covalent bonds are formed between atoms with small or no difference in their electron-attracting abilities.

18

Ways to compare the electron-attracting ability of atoms

1. Ionization Enthalpy

2. Electron affinity

3. Electronegativity

19

Electronegativity and Types of Chemical Bonds1. Ionization enthalpy

The enthalpy change when one mole of electrons are removed from one mole of atoms or positive ions in gaseous state.

X(g) X+(g) + e- H 1st

I.E.

X+(g) X2+(g) + e- H 2nd

I.E.

Ionization enthalpies are always positive.

20

X(g) + e- X-(g) H 1st

E.A.

X(g) + e- X2(g) H 2nd

E.A.

Electronegativity and Types of Chemical Bonds2. Electron affinity

The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state.

Electron affinities can be positive or negative.

21

I.E. and E.A. only show e- releasing/attracting power of

free, isolated atoms

However, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.

22

Electronegativity and Types of Chemical Bonds3. Electronegativity

The ability of an atom to attract electrons in a chemical bond.

23

Mulliken’s scale of electronegativity

Electronegativity(Mulliken) )(2

1EAIE

Nobel Laureate in Chemistry, 1966

24

Pauling’s scale of electronegativity

Nobel Laureate in Chemistry, 1954

Nobel Laureate in Peace, 1962

25

Pauling’s scale of electronegativity calculated from bond enthalpies

cannot be measured directly

having no unit

always non-zero

the most electronegative element, F, is arbitrarily assigned a value of 4.00

26

I A I I A I I I A IVA VA VIA VI IA VI I IA

H 2.1

He -

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Ne -

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ar -

K 0.8

Ca 1.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr -

Rb 0.8

Sr 1.0

I n 1.7

Sn 1.8

Sb 2.0

Te 2.1

I 2.5

Xe -

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.8

Bi 1.9

Po 2.0

At 2.2

Rn -

Pauling’s scale of electronegativity

27

What trends do you notice about the EN values in the Periodic Table? Explain. I A I I A I I I A IVA VA VIA VI IA VI I IA

H 2.1

He -

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Ne -

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ar -

K 0.8

Ca 1.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr -

Rb 0.8

Sr 1.0

I n 1.7

Sn 1.8

Sb 2.0

Te 2.1

I 2.5

Xe -

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.8

Bi 1.9

Po 2.0

At 2.2

Rn -

28


H 2.1

He -

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Ne -

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ar -

K 0.8

Ca 1.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr -

Rb 0.8

Sr 1.0

I n 1.7

Sn 1.8

Sb 2.0

Te 2.1

I 2.5

Xe -

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.8

Bi 1.9

Po 2.0

At 2.2

Rn -

The EN values increase from left to right across a Period.

29

What trends do you notice about the EN values in the Periodic Table? Explain.The EN values increase from left to right across a Period.

The atomic radius decreases from left to right across a Period.Thus,the nuclear attraction experienced by the bonding electrons increases accordingly.

30

The EN values decrease down a Group. I A I I A I I I A IVA VA VIA VI IA VI I IA

H 2.1

He -

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Ne -

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ar -

K 0.8

Ca 1.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr -

Rb 0.8

Sr 1.0

I n 1.7

Sn 1.8

Sb 2.0

Te 2.1

I 2.5

Xe -

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.8

Bi 1.9

Po 2.0

At 2.2

Rn -

31

The EN values decrease down a Group.The atomic radius increases down a Group, thus weakening the forces of attraction between the nucleus and the bonding electrons.

What trends do you notice about the EN values in the Periodic Table? Explain.

32


H 2.1

He -

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Ne -

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

Ar -

K 0.8

Ca 1.0

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr -

Rb 0.8

Sr 1.0

I n 1.7

Sn 1.8

Sb 2.0

Te 2.1

I 2.5

Xe -

Cs 0.7

Ba 0.9

Tl 1.8

Pb 1.8

Bi 1.9

Po 2.0

At 2.2

Rn -

Why are the E.A. of noble gases not shown ?

33

EN is a measure of the ability of an atom to attract electrons in a chemical bond.

However, noble gases are so inert that they rarely form chemical bonds with other atoms.

Why are the E.A of noble gases not shown ?

XeF2, XeF4 and XeF6 are present

34

Formation of Ionic BondAtoms of Group IA and IIA elements ‘tend’ to achieve the noble gas structures in the previous Period by losing outermost electron(s).

In fact, formations of positive ions from metals are endothermic and not spontaneous.

I.E. values are always positive

35

Formation of Ionic BondAtoms of Group VIA and VIIA elements tend to achieve the noble gas structures in the same Periods by gaining electron(s)

First E.A. of Group VIA and VIIA elements are always negative.

Spontaneous and exothermic processes

36

The oppositely charged ions are stabilized by coming close to each other to form the ionic bond.

Ionic bond is the result of electrostatic interaction between oppositely charged ions.

Interaction = attraction + repulsion

37

Dots and Crosses Diagram

38

Notes on Dots & Crosses representation Electrons in different atoms are

indistinguishable. The dots and crosses do not indicate

the exact positions of electrons. Not all stable ions have the noble

gas structures(Refer to pp.4-5).

39

Tendency for the Formation of IonsAn ion will be formed easily if 1. The electronic structure of the ion

is stable;2. The charge on the ion is small;3. The size of parent atom from which

the ion is formed is small for an anion, or

large for a cation.

40

For cation,

Larger size of parent atom

less positive I.E. or less +ve sum of successive I.E.s, easier formation of cation,

atoms of Group IA & Group IIA elements form cations easily.

41

Li+ Be2+

Na+ Mg2+ Al3+

K+ Ca2+ Sc3+

Rb+ Sr2+ Y3+ Zr4+

Cs+ Ba2+ La3+ Ce4+

Ease of formation of cation Ease of formation of cation

42

For anion,Smaller size of parent atom

more negative E.A. or more -ve sum of

successive E.A.s,

easier formation of anion, atoms of Group VIA & Group VIIA elements form anions easily.

43

N3 O2 F

P3 S2 Cl

Br

I

Ease of formation of anion

Ease

of fo

rmatio

n o

f an

ion

44

Stable Ionic StructuresNot all stable ions have the noble gas electronic structures.

Q.2 Write the s,p,d,f notation for each of the following cations.

Use your answers to identify three types of commonly occurring arrangement of outershell electrons of cations other than the stable octet structure.

45

Ion s,p,d,f notationZn2+

Pb2+

V3+

Cr3+

Fe3+

Fe2+

Co2+

Ni2+

Cu2+

Q.2 Write the s,p,d,f notation for the following cations

46

Ion s,p,d,f notationZn2+ 1s2,2s2,2p6,3s2,3p6,3d10

Pb2+ 1s2,2s2,2p6,3s23p63d10,4s2,4p6,4d10,4f14,5s2,5p6,5d10,6s2

V3+ 1s2,2s2,2p6,3s23p63d2

Cr3+ 1s2,2s2,2p6,3s23p63d3

Fe3+ 1s2,2s2,2p6,3s23p63d5

Fe2+ 1s2,2s2,2p6,3s23p63d6

Co2+ 1s2,2s2,2p6,3s23p63d7

Ni2+ 1s2,2s2,2p6,3s23p63d8

Cu2+ 1s2,2s2,2p6,3s23p63d9

Q.2

47



V3+ 1s2,2s2,2p6,3s23p63d2

Cr3+ 1s2,2s2,2p6,3s23p63d3

Fe3+ 1s2,2s2,2p6,3s23p63d5

Fe2+ 1s2,2s2,2p6,3s23p63d6

Co2+ 1s2,2s2,2p6,3s23p63d7

Ni2+ 1s2,2s2,2p6,3s23p63d8

Cu2+ 1s2,2s2,2p6,3s23p63d9

1. 18 - electrons group e.g. Zn2+

Fully-filled s, p & d-subshells

48



V3+ 1s2,2s2,2p6,3s23p63d2

Cr3+ 1s2,2s2,2p6,3s23p63d3

Fe3+ 1s2,2s2,2p6,3s23p63d5

Fe2+ 1s2,2s2,2p6,3s23p63d6

Co2+ 1s2,2s2,2p6,3s23p63d7

Ni2+ 1s2,2s2,2p6,3s23p63d8

Cu2+ 1s2,2s2,2p6,3s23p63d9

Fully-filled s, p & d-subshells

2. (18 + 2 ) - electron group e.g. Pb2+

49



V3+ 1s2,2s2,2p6,3s23p63d2

Cr3+ 1s2,2s2,2p6,3s23p63d3

Fe3+ 1s2,2s2,2p6,3s23p63d5

Fe2+ 1s2,2s2,2p6,3s23p63d6

Co2+ 1s2,2s2,2p6,3s23p63d7

Ni2+ 1s2,2s2,2p6,3s23p63d8

Cu2+ 1s2,2s2,2p6,3s23p63d9

3. Variable arrangements for ions of transition metals. ns2,np6, nd1 to ns2,np6, nd9

Ti3+ [Ne] 3s23p63d1 Mn3+ [Ne] 3s23p63d4

50

Electronic arrangements of stable cations

Ionization enthalpy determines the ease of formation of cations.

Less positive I.E. or sum of I.E.s

easier formation of cations

51

Electronic arrangements of stable cationsOctet structure : -cations of Group IA, IIA and IIIB

I A I IA I I I B

2,8 Na+ Mg2+ Al3+(I I IA)

2,8,8 K+ Ca2+ Sc3+

2,8,18,8 Rb+ Sr2+ Y3+

52

(a) 18 - electrons group

cations of Groups IB, IIB, IIIA and IVA

I B I IB I I IA IVA

2,8,18 Cu+ Zn2+ Ga3+ - -

2,8,18,18 Ag+ Cd2+ I n3+ Sn4+

2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+

53

I B I IB I I IA IVA

2,8,18 Cu+ Zn2+ Ga3+ - -

2,8,18,18 Ag+ Cd2+ I n3+ Sn4+

2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+

Less stable than ions with a noble gas structure.

54

I B I IB I I IA IVA

2,8,18 Cu+ Zn2+ Ga3+ - -

2,8,18,18 Ag+ Cd2+ I n3+ Sn4+

2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+

Cu and Au form other ions because the nuclear charges are not high enough to hold the 18-electron group firmly.

55

I B I IB I I IA IVA

2,8,18 Cu+ Zn2+ Ga3+ - -

2,8,18,18 Ag+ Cd2+ I n3+ Sn4+

2,8,18,32,18 Au+ Hg2+ Tl3+ Pb4+

Cu2+ 2, 8, 17 Au3+ 2, 8, 18, 32, 16The d-electrons are more diffused and thus less attracted by the less positive nuclei.

56

(b) (18+2) electrons

Cations of heavier group members

due to presence of 4f electrons.

Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1

Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)

Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)

Stability : Tl+ > Tl3+ due to extra stability of

6s electrons (inert pair effect)

57


6s electrons are poorly shielded by the inner 4f electrons.

6s electrons experience stronger nuclear attraction.

Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1

Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)

Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)

58


Inert pair effect increases down a Group.

Stability :Sn4+(18) > Sn2+(18+2)

Pb2+(18+2) > Pb4+(18)

moving down Group IV

Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1

Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)

Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)

59


Inert pair effect increases down a Group.

Stability :Sb5+(18) > Sb3+(18+2)

Bi3+(18+2) > Bi5+(18)

moving down Group V

Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1

Tl3+ [2,8,18,32], 5s2, 5p6, 5d10 (18)

Tl+ [2,8,18,32], 5s2, 5p6, 5d10, 6s2 (18 + 2)

60

(c) Cations of transition elements

- variable oxidation numbers

- Electronic configurations from

ns2, np6, nd1 to ns2, np6, nd9

Fe [Ne] 3s2, 3p6, 3d6, 4s2

Fe2+ [Ne] 3s2, 3p6, 3d6

Fe3+ [Ne] 3s2, 3p6, 3d5

61

(c) Cations of transition elements

Which one is more stable, Fe2+(g) or Fe3+(g) ?

Fe2+(g) is more stable than Fe3+(g)

Energy is needed to remove electrons from Fe2+(g) to give Fe3+(g)

Fe [Ne] 3s2, 3p6, 3d6, 4s2

Fe2+ [Ne] 3s2, 3p6, 3d6

Fe3+ [Ne] 3s2, 3p6, 3d5

62

B. Anions - with noble gas structures

Electron affinity determines the ease of formation of anions.

More -ve E.A. or sum of E.A.s

more stable anion

Group VIA and Group VIIA elements form anions easily.

63

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-322

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-325

Kr+39

First Electron Affinity (kJ mol1)

X(g) + e X(g)

64

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

E.A. becomes more –ve from Gp 1 to Gp 7 across a period

65

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

The electrons added experience stronger nuclear attraction when the atoms are getting smaller across the period.

66

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

+ve E.A. for Gp 2 and Gp 0 because the electron added occupies a new shell / subshell

67

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

Goes to a new subshell Goes to a new shellBe(2s2) Be(2s22p1)

Ne(2p6) Ne(2p63s1)

68

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells.

E.g. Li(1s22s1) Li(1s22s2)

69

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

Less –ve E.A. for Gp 5 because the stable half-filled p-subshell no longer exists.

E.g. N(2s22p3) N(2s22p4)

70

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

Q.3 Why are the first E.A.s of halogen atoms more negative than those of the O atom or the S atom ?

71

H-73

He+21

Li-60

Be+18

B-23

C-122

N~0

O-142

F-328

Ne+29

Na-53

Mg+21

Al-44

Si-134

P-72

S-200

Cl-349

Ar+35

K-48

Br-324

Kr+39

It is because the halide ions formed have full-filled s/p subshells (octet structure).

72

Q.4 Why are the second E.A.s of O(+844 kJmol1) and S(+532 kJmol1) positive ?

The electrons added are repelled strongly by the negative ions.

O(g) + e O2(g)

S(g) + e S2(g)

73

Why is the E.A. of F less negative than that of Cl ?

The size of Fluorine atom is so small that the addition of an extra electron results in great repulsion among the electrons.

Halogen

F Cl Br I

E.A.kJ mol1

328 349 325 295

The 2nd electron shell is much smaller than the 3rd electron shell.

74

Energetics of Formation of Ionic CompoundsA. Formation of an ion pair in gaseous state

Consider the formation of a KF(g) ion pair from K(g) & F(g)

H1K(g) + F(g) KF(g)

IE1(K)

K+(g)

H2

By Hess’s law, H1 = IE1(K) + EA1(F) + H2

EA1(F)

+ F(g)

75

R4π

QQ

mol106.022

mol J 10328.0

mol106.022

mol J 10419.0

0

21123

13

123

13

m102.170mJC 108.8544πC)10(1.602

J101.511 1011212

21919

J19109.119

H1 = IE1(K) + EA1(F) + H2

Q.6

J101.063J101.511 1819

(+)()

76

J109.119ΔH 191

H1 = IE1(K) + EA1(F) + H2

= 1.5111019 J 1.0631018 J = 9.1191019 J

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

77

When R

Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

78

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

I.E. and E.A. are NOT the driving forces for the formation of ionic bond.

79

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

K+(g) & F(g) tend to come close together in order to become stable.

80

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

Coulomb stabilization is the driving force for the formation of ionic bond.

81

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

When R = 2.1701010 m

The most energetically stable state is reached.

82

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

The ions cannot come any closer than Re as it will results in less stable states

83

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

Repulsions between electron clouds and between nuclei > attraction between ions

84

R

Pote

ntia

l en

erg

y

(V)

Minimum V when R = 2.170 1010 m

Repulsion :

12R

1V

between opposite charges

Attraction :R

1V

between opposite charges

85

J109.119ΔH 191

IE1(K) + EA1(F) = +1.5111019 J

H2 = 1.0631018 J

What is the significance of the lowest energy state of the neutral state of K + F ?

86

K F

Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system

K FA covalent bond is formed

87

A. Formation of Ionic Crystals

Consider the formation of NaCl(s) from Na(s) & Cl2(g)

H2

By Hess’s law, Hf = H1 + H2

Na+(g) + Cl(g)

H1

HfNa(s) + Cl2(g) NaCl(s) 21

88

H1 is the sum of four terms of enthalpy changes

1. Standard enthalpy change of atomization of Na(s)

Na(s) Na(g) H = +108.3 kJ mol-1

2. First ionization enthalpy of Na(g)

Na(g) Na+(g) + e- H= +500 kJ mol-1

3. Standard enthalpy change of atomization of Cl2(g)

1/2Cl2(g) Cl(g) H = +121.1 kJ mol-1

4. First electron affinity of Cl(g)

Cl(g) + e- Cl-(g) H = -349 kJ mol-1

89

H2 is the lattice enthalpy of NaCl.

It is the enthalpy change for the formation of 1 mole of NaCl(s) from its constituent ions in the gaseous state.

Na+(g) + Cl-(g) NaCl(s) HLo

90

1.theoretical calculation using an ionic model, Or

2.experimental results indirectly with the use of a Born-Haber cycle.

Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from

91

Q.7 Calculate the lattice enthalpy of NaClHf = H1 +H2

Lattice enthalpy

= H2

= Hf - H1

= [(411) (+108.3 + 500 + 121.1 349)] kJ mol1

= 791.4 kJ mol1

92

-349

-791.4

93

Lattice enthalpy is the dominant enthalpy term responsible for the -ve Hf of an ionic compound.

More ve HLo More stable ionic compound

Lattice enthalpy is a measure of the strength of ionic bond.

94

Determination of Lattice Determination of Lattice EnthalpyEnthalpy

• From Born-Haber cycle (experimental method, refers to Q.7)

• From theoretical calculation based on the ionic model

95

Assumptions made in the Assumptions made in the calculationcalculation

• Ions are spherical and have no distortion of the charge cloud, I.e. 100% ionic.

• The crystal has certain assumed lattice structure.

• Repulsive forces between oppositely charged ions at short distances are ignored.

96

)(4 0

rr

QMLQH lattice

M is the Madelung constants that depends on the crystal structure

97

)(4 0

rr

QMLQH lattice

L is the Avogadro’s constant

98

)(4 0

rr

QMLQH lattice

Q+ and Q- are the charges on the cation and the anion respectively

99

)(4 0

rr

QMLQH lattice

0 is the permittivity of vacuum

100

)(4 0

rr

QMLQH lattice

(r+ + r-) is the nearest distance between the nuclei of cation and anionr+ is the ionic radius of the cation r- is the ionic radius of the anion

101

Stoichiometry of Ionic CompoundsStoichiometry of Ionic Compounds

Stoichiometry of an ionic compound is the simplest whole number ratio of cations and anions involved in the formation of the compound. There are two ways to predict stoichiometry.

102

1. Considerations in terms of electronic configurations

Atoms tend to attain noble gas electronic structures by losing or gaining electron(s).

The ionic compound is electrically neutral.

total charges on cation = total charges on anionstotal charges on cation = total charges on anions

103

1. Considerations in terms of electronic configurations

Na + F [Na]+ [F]-

2,8,1 2,7 2,8 2,8

Mg + 2Cl [Cl]- [Mg]2+ [Cl]-

2,8,2 2,8,7 2,8,8 2,8 2,8,8

NaF

MgCl2

104

2. Considerations in terms of enthalpy changes of formation

Based on the Born-Haber cycle & the ththeoretically calculated lattice enthalpyeoretically calculated lattice enthalpy, the values of Hf

o of hypothetical compounds (e.g. MgCl , MgCl2 , MgCl3) can be calculated.

The stoichiometry with the most negative The stoichiometry with the most negative HHff

o o value is the most stable onevalue is the most stable one.

105

2. Considerations in terms of enthalpy changes of formation

Or, we can determine the true stoichiometry by comparing the calculated Hf

o values of hypothetical compounds with the experimentally determined Hf

o value.

The stoichiometry with Hfo value closest to th

e experimentally determined Hfo value is the ans

wer.

106

Q.8 Three hypothetical formulae of magnesium chloride are proposed and their estimated lattice enthalpies are shown in the table below.

Hypothetical stoichiometry

MgCl MgCl2 MgCl3

Assumed structure

NaCl CaF2 AlCl3

Estimated HL

o(kJ mol- 1) - 771 - 2602 - 5440

107

Hf[MgCl(s)] = Hat[Mg(s)] + 1st IE of Mg + Hat[1/2Cl2(g)]

+ 1st EA of Cl + HL[MgCl(s)]

= (+150 + 736 + 121 - 364- 771) kJ mol-1

= -128 kJ mol-1

108

Hat[1/2Cl2]

Mg+(g)+Cl(g)

Mg+(g)+Cl(g)1st EA[Cl]

MgCl(s)

HL[MgCl]

Hf[MgCl]<0

1st IE[Mg]

Mg+(g)+ Cl2(g)21

Enthalpy (kJ m

ol

)

Mg(s)+ Cl2(g)21

Hat[Mg]

Mg(g)+ Cl2(g)21

109

Hf[MgCl2(s)]

= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg

+ 2Hat[1/2Cl2(g)] + 21st EA of Cl + HL[MgCl2(s)]

= [150+736+1450+2121+2(-364)+(-2602)] kJ mol-1

= -752 kJ mol-1

110

Enthalpy (kJ

mol

)

Mg(s)+Cl2(g)

Mg(g)+Cl2(g)

Hat[Mg]

Mg+(g)+Cl2(g)

1st IE[Mg]

Mg2+(g)+Cl2(g)

2nd IE[Mg]

Mg2+(g)+2Cl(g)

2Hat[1/2Cl2]

Mg2+(g)+2Cl(g)

21st EA[Cl]

MgCl2(s)

HL[MgCl2]

Hf[MgCl2] <0

111

Hf[MgCl3(s)]

= Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg + 3rd IE of Mg

+ 3Hat[1/2Cl2(g)] + 31st EA of Cl + HL[MgCl3(s)]

= 150+736+1450+7740+3121+3(-364)+(-5440)

= +3907 kJ mol-1

112

Enthalpy ( kJ m

ol

)

Mg(s)+3/2Cl2(g)

Mg3+(g)+3Cl(g)31st EA[Cl]

MgCl3(s)

HL[MgCl3]

Hf[MgCl3] >0

Mg3+(g)+3Cl(g)

3Hat[1/2Cl2]

3rd IE[Mg]

Mg3+(g)+ Cl2(g)23

Mg2+(g)+ Cl2(g)23

2nd IE[Mg]Mg+(g)+ Cl2(g)

23

1st IE[Mg]

Hat[Mg]

Mg(g)+ Cl2(g)23

113

Since the hypothetical compound MgCl2 has the most negative Hf value, and

this value is closest to the experimentally determined one,

the most probable formula of magnesium chloride is

MgCl2

114

Reasons for the discrepancy between calculated and experimental results of Hf

The ionic bond of MgCl2 is not 100% pure. I,e. the ions are not perfectly spherical.

MgCl2 has a different crystal structure from CaF2

Short range repulsive forces between oppositely charged ions have not been considered.

Check Point 7-2Check Point 7-2

115

7.2 Energetics of Formation of Ionic Compounds (SB p.196)

Factors affecting lattice enthalpyFactors affecting lattice enthalpy

• Effect of ionic size:

The greater the ionic size The lower (or less negative) is the lattice enthalpy

)(4 0

rr

QMLQH lattice

116


Factors affect lattice enthalpyFactors affect lattice enthalpy

• Effect of ionic charge:

The greater the ionic charge The higher (or more negative) is the

lattice enthalpy

Check Point 7-3Check Point 7-3

)(4 0

rr

QMLQH lattice

117

7.7.44Ionic CrystalsIonic Crystals

118

7.4 Ionic Crystals (SB p.202)

Crystal LatticeCrystal Lattice(( 晶體格子晶體格子 ))

119


In solid state, ions of ionic compounds are In solid state, ions of ionic compounds are regularly packedregularly packed to form 3-dimensional str to form 3-dimensional structures called uctures called crystal latticescrystal lattices(( 晶體格子晶體格子 ))

120


The The coordination numbercoordination number (C.N.) of a given partic (C.N.) of a given particle in a crystal lattice is the le in a crystal lattice is the number of nearest neinumber of nearest neighbours of the particleghbours of the particle..

C.N. of Na+

= 6

121


The The coordination numbercoordination number (C.N.) of a given partic (C.N.) of a given particle in a crystal lattice is the le in a crystal lattice is the number of nearest neinumber of nearest neighbours of the particleghbours of the particle..

C.N. of Cl

= 6

122

Identify the unit cell of NaCl crystal latticeIdentify the unit cell of NaCl crystal lattice

Not a unit Not a unit cellcell

123

Identify the unit cell of NaCl crystal latticeIdentify the unit cell of NaCl crystal lattice

Not a unit Not a unit cellcell

124

The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionallyrepeated 3-dimensionally in space, will in space, will generate the whole crystal lattice.generate the whole crystal lattice.

125

The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionally in space, will repeated 3-dimensionally in space, will generate the whole crystal lattice.generate the whole crystal lattice.

126

The The unit cellunit cell of a crystal lattice is the simplest of a crystal lattice is the simplest 3-D arrangement of particles which, when 3-D arrangement of particles which, when repeated 3-dimensionally in space, will repeated 3-dimensionally in space, will generate the whole crystal lattice.generate the whole crystal lattice.

127

Unit Unit cellcell

Unit Unit cell cell

128

A crystal lattice with a A crystal lattice with a cubic unit cubic unit cellcell is known as a is known as a cubic structurecubic structure..

129

Three Kinds of Cubic Three Kinds of Cubic StructureStructure• Simple cubic (primitive) structure• Body-centred cubic (b.c.c.) structure• Face-centred cubic (f.c.c.) structure

130

Simple cubic structure

View this Chemscape 3D structure

Unit cell

Space filling

Space lattice

131

Body-centred cubic (b.c.c.) structureBody-centred cubic (b.c.c.) structure


Unit cell

Space filling

Space lattice

132

Face-centred cubic (f.c.c.) structureFace-centred cubic (f.c.c.) structure


Unit cell

133

The f.c.c. structure can be obtained by stacking the layers of particles in the pattern abcabc...

134

Interstitial sitesInterstitial sites

- The empty spaces in a crystal - The empty spaces in a crystal latticelatticeTwo types of interstitial sites in f.c.c. structureOctahedral siteTetrahedral site

135

Octahedral site : -

It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.

136

When the octahedron is rotated by 45o , the octahedral site can also be viewed as the space confined by 4 spheres in one layer and 1 other sphere each in the upper and lower layers respectively.

45o

beforebehind

137

Tetrahedral site : -It is the space between the 3 spheres of one layer and a fourth sphere on the upper layer.

138

In an f.c.c. unit cell, the tetrahedral site is the space bounded by a corner atom and the three face-centred atoms nearest to it.

139

c

b a

a

140

Q.9 Label a, b, and c as Oh sites or Td sites

a and b are Td sites

c is Oh site

Top layer

141

Q.10 Identify the Oh sites and Td sites in the f.c.c. unit cell shown below.

13 Oh sites

3

5 6

89

1011

1213

4 Th sites in the front

1

2

35 6

7 8

12

4

4

7 4 Th sites at the back

View Octahedral sites and tetrahedral sites

142

Since ionic crystal lattice is made of two kinds of particles, cations and anions,

the crystal structure of an ionic compound can be considered as two lattices of cations and anions interpenetrating with each other.

143

The crystal structures of

sodium chloride,

caesium chloride and

calcium fluoride are discussed.

144

Sodium Chloride (The Rock Salt Structure)

f.c.c. unit cell of larger Cl- ions, with all Oh sites occupied by the smaller Na+ ions

Oh sites of f.c.c. lattice of Cl- ions

145

A more open f.c.c. unit cell of smaller Na+ ions with all Oh sites occupied by the larger Cl- ions.

Oh sites of f.c.c. lattice of Na+ ions

146

F.C.C. Structure of Sodium Chloride


147

Co-ordination number of Na+ = 6Co-ordination number of Cl- = 6

6 : 6 co-ordination

Unit cell of NaClStructure of Sodium ChlorideStructure of Sodium Chloride


148

Only the particles in the centre (or body) of the unit cell belong to the unit cell entirely. Particles locating on the faces, along the edges or at the corner of a unit cell are shared with the neighboring unit cells of the crystal lattice.

149

Fraction of particles occupied by a unit cell

1/21

CornerEdgeFaceBody

1/2

150


1/41/21

CornerEdgeFaceBody

1/4

151


1/81/41/21

CornerEdgeFaceBody

1/8

152

Q.11 Calculate the net nos. of Na+ and Cl- ions in a unit cell of NaCl.

No. of Cl- ions = 42

16

8

18

No. of Na+ ions = 414

112

8 corners 6 faces

12 edges

1 body

Example 7-4Example 7-4

153

Structure of Caesium Chloride Structure of Caesium Chloride Link Link


Simple cubic latticeCs+ ions are too large to fit in the octahedral sites.

Thus Cl ions adopt the more open simple cubic structure with the cubical sites occupied by Cs+ ions.

154

A cubical site is the space confined by 4 spheres in one layer and 4 other spheres in the adjacent layer.

155

Size of interstitial sites : -

Cubical > octahedral > tetrahedral

In f.c.c. structure

In simple cubic structure

156

Co-ordination number of Cs+ = 8

Co-ordination number of Cl- = 88 : 8 co-ordination


Simple cubic lattice

Structure of Caesium Chloride Structure of Caesium Chloride Link Link

157

Q.12

Number of Cs+ = 1 1

8

18 Number of Cl =

158

Structure of Calcium Fluoride Structure of Calcium Fluoride Link Link


It can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions.

Fluorite structure

159


Alternately, it can be viewed as an expanded f.c.c. structure of smaller calcium ions with all tetrahedral sites occupied by larger fluoride ions.


160

Co-ordination number of Ca2+ = 8

Co-ordination number of F- = 48 : 4 co-ordination


Face-centred cubic lattice


161


Q.13 (a)

Number of F = 8

Number of Ca2+ = 42

16

8

18

CaF2

162


Q.13 (b) CaF2

Only alternate cubical sites are occupied by Ca2+ in order to maintain electroneutrality.

163

Structure of Sodium oxide Structure of Sodium oxide Link Link


Na2O vs CaF2

Antifluorite structure

fluorite structure

164

Q.14(a)

Zinc blende structure - Link

165

Q.14(a)

Number of Zn2+ = 4

Number of S2 = 42

16

8

18

ZnS

166

414.0402.0184.0

074.0

r

r

tetrahedral site

S2 ions adopt f.c.c. structure

Alternate Td sites are occupied by Zn2+ ions to ensure electroneutrality.

Zinc sulphide, ZnS

Q.14(b)

167

Q.15

The unit cell is not a cube

Rutile structure - Link

168

Q.15(a)

Number of O2 = 42

142

TiO2

Ti4+O2

Number of Ti4+ = 28

181

169

Q.15(a)

C.N. of Ti4+ = 6

C.N. of O2 = 3

6 : 3 coordination

170

414.0486.0140.0

068.0

r

r

octahedral site

O2 ions adopt distorted h.c.p.(not f.c.c.) structure with alternate Oh sites occupied by Ti4+ ions to ensure electroneutrality.

Titanium(IV) oxide, TiO2

Q.15(b)

171

a

b

a

hexagonal close-packed

172

Factors governing the structures of ionic crystals

1. Close Packing Considerations

Ions in ionic crystals tend to pack as closely as possible. (Why ?)

To strengthen the ionic bonds formed

To maximize the no. of ionic bonds formed.

173


1. Close Packing Considerations

The larger anions tend to adopt face-centred cubic structure (cubic closest packed, c.c.p.)

The smaller cations tend to fill the interstitial sites as efficiently as possible

174


Q.16 Is there no limit for the C.N. ? Explain your answer.

The anions tend to repel one another when they approach a given cation.

The balance between attractive forces and repulsive forces among ions limits the C.N. of the system.

175



If the cations are small, less anions can approach them without significant repulsions.

Or, if the cations are small, they choose to fit in the smaller tetrahedral sites with smaller C.N. to strengthen the ionic bonds formed.

176



If the cations are large, they choose to fit in the larger octahedral sites or even cubical sites with greater C.N. to maximize the no. of ionic bonds formed.

177


r

r2. The relative size of cation and anion,

In general, r > r+,

10

r

r

..NCr

r

178

Factors governing the structures of ionic crystalsIf the cations are large, 1

r

r

more anions can approach the cations without significant repulsions.

Or, the large cations can fit in the larger interstitial sites with greater C.N. to maximize the no. of bonds formed.

179

Interstitial site

Tetrahedral Octahedral Cubical

Coordination 4 : 4 6 : 6 8 : 8

0.225 – 0.414 0.414 – 0.732 0.732 – 1.000

ExamplesZnS, most copper(I) halides

Alkali metal halides except C

sCl

CsCl, CsBr

CsI, NH4Cl


Summary : -Summary : -

r

r

NH4+ is large

180


Summary : -Summary : -

r

r

*CaF2, BaF2, BaCl2, SrF2Examples

0.732 – 1.000

8 : 4Coordinatio

n

CubicalInterstitial

site

181

Q.17

A B

C

r

r+

r

45coscos

rr

r

BC

BAABC

414.0

r

r

45°

182

414.0

r

r

the range of ratio that favours octahedral arrangement.

414.0732.0

r

r

183

732.0

r

r

the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites.

Q.18

184

414.0696.0181.0

126.0

r

r

octahedral site

Cl ions adopt f.c.c. structure

All Oh sites are occupied by Ag+ ions to ensure electroneutrality.

Silver chloride, AgCl

Q.18(a)

185

414.0696.0181.0

126.0

r

r

octahedral site

Ag+ ions adopt an open f.c.c. structure

All Oh sites are occupied by Cl ions to ensure electroneutrality.

Silver chloride, AgCl

Q.18(a)

186

Copper(I) bromide, CuBr

414.0379.0195.0

074.0

r

r

tetrahedral site

Br ions adopt f.c.c. structure

Only alternate Td sites are occupied by Cu+ ions to ensure electroneutrality.

Q.18(b)

187

Copper(II) bromide, CuBr2

414.0364.0195.0

071.0

r

r

tetrahedral site

Br ions adopt f.c.c. structure

Only 1/4 Td sites are occupied by Cu+ ions to ensure electroneutrality.

Some Br may form less bonds than others

Not favourable. Q.18(c)

188

Copper(II) bromide, CuBr2

All the Td sites are occupied by Br ions to ensure electroneutrality.

Q.18(c)

Or, Cu2+ ions adopt a very open f.c.c. structure

However, this structure is too open to form strong ionic bonds.

non-cubic structure is preferred.

189

Q.18(c)

4 : 2 coordinationCu2+

Br

190

Copper(II) chloride, CuCl2

Layer structure

0.295nm0.230 nm

191

Copper(II) chloride, CuCl2

6 : 3 coordination

192

7.7.55 Ionic RadiiIonic Radii

193

Ionic RadiiIonic RadiiIonic radius is the approximate radius of the spherical space occupied by the electron cloud of an ion in all directions in the ionic crystal.

194

Q.19

Why is the electron cloud of an ion always spherical in shape ?

Stable ions always have fully-filled electron shells or subshells. The symmetrical distribution of electrons accounts for the spherical shapes of ions.

195

Q.19

Li+ 1s2

Na+ [He] 2s2, 2px2, 2py

2, 2pz2

Cl [Ne] 3s2, 3px2, 3py

2, 3pz2

Zn2+ [Ar] 3dxy2, 3dxz

2, 3dyz2, ,

222 yx

3d

22z

3d

spherical Symmetrical distribution along x, y and z axes almost spherical

Symmetrical distribution along xy, xz, yz planes and along x, y and z axes almost spherical

196

Determination of Ionic Radii Pauling Scale

Interionic distance (r+ + r) can be determined by

X-ray diffraction crystallography

197

7.5 Ionic Radii (SB p.205)

Electron density plot for sodium chloride crystal

Cl

Cl

Contour having same electron density

Na+

Na+

r++ r

Electron density map for NaCl

r++ r

198

By additivity rule,

Interionic distance = r+ + r

For K+Cl,

r+ + r = 0.314 nm (determined by X-ray)

Since K+(2,8,8) and Cl(2,8,8) are isoelectronic, their ionic radii can be calculated.

r+(K+) = 0.133 nm, r(Cl) = 0.181 nm

199

For Na+Cl,

r+ + r = 0.275 nm (determined by X-ray)

Since r(Cl) = 0.181 nm(calculated)

r+(Na+) = (0.275 - 0.181) nm = 0.094 nm

200

Limitation of Additivity rule

In vacuum, the size of a single ion has no limit according to quantum mechanics.

The size of the ion is restricted by other ions in the crystal lattice.

Interionic distance < r+ + r

201

Evidence :Electron distribution is not perfectly spherical at the boundary due to repulsion between electron clouds of neighbouring ions.

202

Ionic radius depends on the bonding environment.

For example, the ionic radius of Cl ions in NaCl (6 : 6 coordination) is different from that in CsCl (8 : 8 coordination).

203

Ionic radius vs atomic radiusIonic radius vs atomic radius


Radii of cations < Radii of corresponding parent atoms cations have one less electron shell than the parent

atoms

204



p/e of cation > p/e of parent atom Less shielding effect stronger nuclear attraction for outermost electrons

smaller size

205



Radii of anions > Radii of corresponding parent atoms

206



p/e of anion < p/e of parent atom more shielding effect

weaker nuclear attraction for outermost electrons

larger size

207

Periodic trends of ionic radius

1. Ionic radius increases down a Group

Ionic radius depends on the size of the outermost electron cloud.

On moving down a Group, the size of the outermost electron cloud increases as the number of occupied electron shells increases.

208

Periodic trends of ionic radius

2. Ionic radius decreases along a series of isoelectronic ions of increasing nuclear charge

The total shielding effects of isoelectronic ions are approximately the same.

Zeff nuclear charge (Z)

Ionic radius decreases as nuclear charge increases.

209


Isoelectronic to He(2)

Isoelectronic to Ne(2,8)

Isoelectronic to Ar(2,8,8)

210


H is larger than most ions, why ?

211

Q.20 H > N3

The nuclear charge (+1) of H is too small to hold the two electrons which repel each other strongly within the small 1s orbital.

Or,

p/e of H (1/2) < p/e of N3 (7/10)

212

The END

Check Point 7-5Check Point 7-5Example 7-5Example 7-5

213

Why do two atoms bond together?

The two atoms tend to achieve an octet configuration which brings stability.

Answer


214

How does covalent bond strength compare with ionic bond strength?

Back

They are similar in strength.

Both are electrostatic attractions between charged particles.

Answer


215

Given the following data:

ΔH (kJ mol–1)

First electron affinity of oxygen –142

Second electron affinity of oxygen +844

Standard enthalpy change of atomization of oxygen +248

Standard enthalpy change of atomization ofaluminium +314

Standard enthalpy change of formation ofaluminium oxide –1669


216

ΔH (kJ mol–1)

First ionization enthalpy of aluminium +577

Second ionization enthalpy of aluminium +1820

Third ionization enthalpy of aluminium +2740

(a) (i) Construct a labelled Born-Haber cycle for the formation of aluminium oxide.

(Hint: Assume that aluminium oxide is a purely ionic compound.)

(ii) State the law in which the enthalpy cycle in (i) is based on.

(b) Calculate the lattice enthalpy of aluminium oxide.


Answer

217


(a) (i)

(ii) The enthalpy cycle in (i) is based on Hess’s law which states that the total enthalpy change accompanying a chemical reaction is independent of the route by means of which the chemical

reaction is brought about.

218


(b) ΔHf [Al2O3(s)] = 2 × ΔHatom[Al(s)] + 2 × (ΔHI.E.1 [Al(g)]

+ ΔHI.E.2 [Al(g)] + ΔHI.E.3 [Al(g)])

+ 3 × ΔHatom [O2(g)] + 3 × (ΔHE.A.1 [O(g)]

+ ΔHE.A.2 [O(g)]) + ΔHlattice[Al2O3(s)]

ΔHf [Al2O3(s)] = 2 × (+314) + 2 × (+577 + 1 820 + 2 740)

+ 3 × (+248) + 3 × (–142 + 844)

+ ΔHlattice [Al2O3(s)]

ΔHf [Al2O3(s)] = + 628 + 10 274 + 744 + 2 106 + ΔHlattice[Al2O3(s)]

ΔHlattice[Al2O3(s)] = ΔHf [Al2O3(s)] – (628 + 10 274 + 744 + 2 106)

= –1 669 – (628 + 10 274 + 744 + 2 106)

= –15 421 kJ mol–1

ø øø

ø

ø

ø

øø ø

Back

219

(a) Draw a Born-Haber cycle for the formation of magnesium oxide.

(a) The Born-Haber cycle for the formation of MgO:

Answer


220

(b)Calculate the lattice enthalpy of magnesium oxide by means of the Born-Haber cycle drawn in (a).

Given: ΔHatom [Mg(s)] = +150 kJ mol–1

ΔHI.E. [Mg(g)] = +736 kJ mol–1

ΔHI.E. [Mg+(g)] = +1 450 kJ mol–1

ΔHatom [O2(g)] = +248 kJ mol–1

ΔHE.A. [O(g)] = –142 kJ mol–1

ΔHE.A. [O–(g)] = +844 kJ mol–1

ΔHf [MgO(s)] = –602 kJ mol–1


øø

ø

Answer

221

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(b) ΔHlattice [MgO(s)]

= ΔHf [MgO(s)] – ΔHatom [Mg(s)]

– ΔHI.E. [Mg(g)] – ΔHI.E. [Mg+(g)] – ΔHatom [O2(g)]

– ΔHE.A. [O(g)] – ΔHE.A. [O–(g)]

= [–602 – 150 – 736 – 1 450 – 248 –(–142) – 844] kJ mol–1

= –3 888 kJ mol–1

ø

ø

ø

ø

222

Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound.The charges and sizes of ions will affect the value of the lattice enthalpy.

The smaller the sizes and the higher the charges of ions, the higher (i.e. more negative) is the lattice enthalpy.

Answer

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7.3 Stoichiometry of Ionic Compounds (SB p.201)

223

Write down the formula of the compound that possesses the lattice structure shown on the right:

To calculate the number of each type of particle present in the unit cell:

Number of atom A = 1

(1 at the centre of the unit cell)

Number of atom B = 8 × = 2

(shared along each edge)

Number of atom C = 8 × = 1

(shared at each corner)

∴ The formula of the compound is AB2C.

41

41

81 8

1

Answer

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224

The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of

(a) titanium; and

(b) oxygen?(a) The coordination number of titanium is 6 as there are six oxide ions surrounding each titanium ion.

(b) The coordination number of oxygen is 3.

Answer

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225

The following table gives the atomic and ionic radii of some Group IIA elements.


Element Atomic radius (nm) Ionic radius

Be 0.112 0.031

Mg 0.160 0.065

Ca 0.190 0.099

Sr 0.215 0.133

Ba 0.217 0.135

226

Explain briefly the following:

(a) The ionic radius is smaller than the atomic radius in each element.

(b) The ratio of ionic radius to atomic radius of Be is the lowest.

(c) The ionic radius of Ca is smaller than that of K(0.133 nm).


Answer

227


(a) One reason is that the cation has one electron shell less than the corresponding atom. Another reason is that in the cation, the number of protons is greater than the number of electrons. The electron cloud of the cation therefore experiences a greater nuclear attraction. Hence, the ionic radius is smaller than the atomic radius in each element.

(b) In the other cations, although there are more protons in the nucleus, the outer most shell electrons are further away from the nucleus, and electrons in the inner shells exhibit a screening effect. Be has the smallest atomic size. In Be2+ ion, the electrons experience the greatest nuclear attraction. Therefore, the contraction in size of the electron cloud is the greatest when Be2+ ion is formed, and the ratio of ionic radius to atomic radius of Be is the lowest.

228


(c) The electronic configurations of both K+ and Ca2+ ions are 1s22s22p63s23p6. Hence they have the same number and arrangement of electrons. However, Ca2+ ion is doubly charged while K+ ion is singly charged, so the outermost shell electrons of Ca2+ ion experience a greater nuclear attraction. Hence, the ionic radius of Ca2+ ion is smaller than that of K+ ion.

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229

Arrange the following atoms or ions in an ascending order of their sizes:

(a) Be, Ca, Sr, Ba, Ra, Mg

(b) Si, Ge, Sn, Pb, C

(c) F–, Cl–, Br–, I–

(d) Mg2+, Na+, Al3+, O2–, F–, N3–

(a) Be < Mg < Ca < Sr < Ba < Ra

(b) C < Si < Ge < Sn < Pb

(c) F– < Cl– < Br– < I–

(d) Al3+ < Mg2+ < Na+ < F– < O2– < N3–

Answer

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Ikatan Ion

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