7 Ionic Bonding 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds 7.4 Ionic Crystals 7.5 Ionic Radii

Bonding & Structure

Lewis Model G.N. Lewis in 1916 Only the outermost (valence) electrons are involved significantly in bond formation Successful in solving chemical problems

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

Chemical bonds are strong. electrostatic forces holding atoms 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 )

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

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 1. Ionic bond (electrovalent bond) Formed by transfer of electrons

Three types of chemical bonds 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187) Three types of chemical bonds 1. Ionic bond (electrovalent bond) Cl Na Sodium atom, Na 1s22s22p63s1 Chlorine atom, Cl 1s22s22p63s23p5

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 1. Ionic bond Electrostatic attraction between positively charged particles and negatively charged particles

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds Covalent bond Formed by sharing of electrons

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 2. Covalent bond Electrostatic attraction between nuclei and shared electrons

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 3. Metallic bond Electrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)

Three types of chemical bonds Introduction (SB p.186) Three types of chemical bonds 3. Metallic bond Formed by sharing of a large number of delocalized electrons Let's Think 1

Ionic bonds and Covalent bonds are only extreme cases of a continuum. In real situation, most chemical bonds are intermediate between ionic and covalent

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

Electronegativity and Types of Chemical Bonds Ionic 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.

Ways to compare the electron-attracting ability of atoms Ionization Enthalpy Electron affinity Electronegativity

Electronegativity and Types of Chemical Bonds 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.

Electronegativity and Types of Chemical Bonds 2. Electron affinity The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state. X(g) + e-  X-(g) H 1st E.A. X(g) + e-  X2(g) H 2nd E.A. Electron affinities can be positive or negative.

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.

Electronegativity and Types of Chemical Bonds The ability of an atom to attract electrons in a chemical bond.

Mulliken’s scale of electronegativity Electronegativity(Mulliken)  Nobel Laureate in Chemistry, 1966

Pauling’s scale of electronegativity Nobel Laureate in Chemistry, 1954 Nobel Laureate in Peace, 1962

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

Pauling’s scale of electronegativity

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

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

The EN values decrease down a Group.

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

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

XeF2, XeF4 and XeF6 are present Why are the E.A of noble gases not shown ? 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. XeF2, XeF4 and XeF6 are present

Formation of Ionic Bond Atoms 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

Formation of Ionic Bond Atoms 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

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

Dots and Crosses Diagram

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).

Tendency for the Formation of Ions An 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.

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.

Li+ Be2+ Na+ Mg2+ Al3+ K+ Ca2+ Sc3+ Rb+ Sr2+ Y3+ Zr4+ Ease of formation of cation  Ease of formation of cation  Li+ Be2+ Na+ Mg2+ Al3+ K+ Ca2+ Sc3+ Rb+ Sr2+ Y3+ Zr4+ Cs+ Ba2+ La3+ Ce4+

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.

N3 O2 F P3 S2 Cl Br I Ease of formation of anion 

Stable Ionic Structures Not 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.

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

Q.2

1. 18 - electrons group e.g. Zn2+ Fully-filled s, p & d-subshells

Fully-filled s, p & d-subshells 2. (18 + 2 ) - electron group e.g. Pb2+ Fully-filled s, p & d-subshells

3. Variable arrangements for ions of. transition. metals 3. Variable arrangements for ions of transition metals. ns2,np6, nd1 to ns2,np6, nd9 Ti3+ [Ne] 3s23p63d1 Mn3+ [Ne] 3s23p63d4

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

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

(a) 18 - electrons group cations of Groups IB, IIB, IIIA and IVA

Less stable than ions with a noble gas structure.

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

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

due to presence of 4f electrons. (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)

(b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1 6s electrons are poorly shielded by the inner 4f electrons. 6s electrons experience stronger nuclear attraction.

(b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1 Inert pair effect increases down a Group. Stability : Sn4+(18) > Sn2+(18+2) Pb2+(18+2) > Pb4+(18) moving down Group IV

(b) (18+2) electrons Tl [2,8,18,32], 5s2, 5p6, 5d10, 6s2, 6p1 Inert pair effect increases down a Group. Stability : Sb5+(18) > Sb3+(18+2) Bi3+(18+2) > Bi5+(18) moving down Group V

(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

(c) Cations of transition elements Fe [Ne] 3s2, 3p6, 3d6, 4s2 Fe2+ [Ne] 3s2, 3p6, 3d6 Fe3+ [Ne] 3s2, 3p6, 3d5 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)

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.

First Electron Affinity (kJ mol1) X(g) + e  X(g) H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -322 Ne +29 Na -53 Mg Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -325 Kr +39

E.A. becomes more –ve from Gp 1 to Gp 7 across a period H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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. H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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 -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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 shell H -73 He +21 Li -60 Be +18 Be(2s2)  Be(2s22p1) Ne(2p6)  Ne(2p63s1) H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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) H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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) H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39

H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na Q.3 Why are the first E.A.s of halogen atoms more negative than those of the O atom or the S atom ? H -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg 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). -73 He +21 Li -60 Be +18 B -23 C -122 N ~0 O -142 F -328 Ne +29 Na -53 Mg Al -44 Si -134 P -72 S -200 Cl -349 Ar +35 K -48 Br -324 Kr +39

The electrons added are repelled strongly by the negative ions. Q.4 Why are the second E.A.s of O(+844 kJmol1) and S(+532 kJmol1) positive ? O(g) + e  O2(g) S(g) + e  S2(g) The electrons added are repelled strongly by the negative ions.

Why is the E.A. of F less negative than that of Cl ? Halogen F Cl Br I E.A. kJ mol1 328 349 325 295 The size of Fluorine atom is so small that the addition of an extra electron results in great repulsion among the electrons. The 2nd electron shell is much smaller than the 3rd electron shell.

By Hess’s law, H1 = IE1(K) + EA1(F) + H2 Energetics of Formation of Ionic Compounds A. Formation of an ion pair in gaseous state Consider the formation of a KF(g) ion pair from K(g) & F(g) H1 K(g) + F(g) KF(g) IE1(K) K+(g) H2 EA1(F) + F(g) By Hess’s law, H1 = IE1(K) + EA1(F) + H2

Q.6 H1 = IE1(K) + EA1(F) + H2 (+)()

H1 = IE1(K) + EA1(F) + H2 = 1.5111019 J  1.0631018 J = 9.1191019 J H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J When R  Stability : K+(g) + F-(g) < K(g) + F(g) by +1.511x10-19J H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

I.E. and E.A. are NOT the driving forces for the formation of ionic bond. H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

K+(g) & F(g) tend to come close together in order to become stable. H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

Coulomb stabilization is the driving force for the formation of ionic bond. H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

The most energetically stable state is reached. When R = 2.1701010 m The most energetically stable state is reached. H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

The ions cannot come any closer than Re as it will results in less stable states H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

Repulsions between electron clouds and between nuclei > attraction between ions H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

Repulsion : Minimum V when R = 2.170  1010 m Attraction : between opposite charges R Potential energy (V) Minimum V when R = 2.170  1010 m Attraction : between opposite charges

What is the significance of the lowest energy state of the neutral state of K + F ? H2 = 1.0631018 J IE1(K) + EA1(F) = +1.5111019 J

K F A covalent bond is formed K F   K F  Electrostatic attraction between positive nuclei and bond electron pair stabilizes the system

By Hess’s law, Hf = H1 + H2 A. Formation of Ionic Crystals Consider the formation of NaCl(s) from Na(s) & Cl2(g) Hf Na(s) + Cl2(g) NaCl(s) Na+(g) + Cl(g) H1 H2 By Hess’s law, Hf = H1 + H2

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

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

Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from 1. theoretical calculation using an ionic model, Or 2. experimental results indirectly with the use of a Born-Haber cycle.

Q.7 Calculate the lattice enthalpy of NaCl Hf = H1 +H2 Lattice enthalpy = H2 = Hf - H1 = [(411)  (+108.3 + 500 + 121.1  349)] kJ mol1 = 791.4 kJ mol1

-349 -791.4

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

Determination of Lattice Enthalpy From Born-Haber cycle (experimental method, refers to Q.7) From theoretical calculation based on the ionic model

Assumptions made in the calculation 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.

M is the Madelung constants that depends on the crystal structure

L is the Avogadro’s constant

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

0 is the permittivity of vacuum

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

Stoichiometry 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.

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 anions

 NaF  MgCl2 1. Considerations in terms of electronic configurations Na + F  [Na]+ [F]- 2,8,1 2,7 2,8 2,8  NaF Mg + 2Cl  [Cl]- [Mg]2+ [Cl]- 2,8,2 2,8,7 2,8,8 2,8 2,8,8  MgCl2

2. Considerations in terms of enthalpy changes of formation Based on the Born-Haber cycle & the theoretically calculated lattice enthalpy, the values of Hfo of hypothetical compounds (e.g. MgCl , MgCl2 , MgCl3) can be calculated. The stoichiometry with the most negative Hfo value is the most stable one.

2. Considerations in terms of enthalpy changes of formation Or, we can determine the true stoichiometry by comparing the calculated Hfo values of hypothetical compounds with the experimentally determined Hfo value. The stoichiometry with Hfo value closest to the experimentally determined Hfo value is the answer.

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

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

Enthalpy (kJ mol) HL[MgCl] Hf[MgCl]<0 Mg(s)+ Cl2(g) Hat[1/2Cl2] Mg+(g)+Cl(g) Mg+(g)+Cl(g) 1st EA[Cl] 1st IE[Mg] Mg+(g)+ Cl2(g) MgCl(s) HL[MgCl] Hat[Mg] Mg(g)+ Cl2(g) Hf[MgCl]<0

= -752 kJ mol-1 Hf[MgCl2(s)] = Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg + 2Hat[1/2Cl2(g)] + 21st EA of Cl + HL[MgCl2(s)] = [150+736+1450+2121+2(-364)+(-2602)] kJ mol-1 = -752 kJ mol-1

Enthalpy (kJ mol) HL[MgCl2] Hf[MgCl2] <0 Mg(s)+Cl2(g) Mg2+(g)+2Cl(g) 2Hat[1/2Cl2] Mg2+(g)+2Cl(g) 21st EA[Cl] Mg2+(g)+Cl2(g) 2nd IE[Mg] MgCl2(s) HL[MgCl2] Mg+(g)+Cl2(g) 1st IE[Mg] Mg(g)+Cl2(g) Hat[Mg] Hf[MgCl2] <0

= +3907 kJ mol-1 Hf[MgCl3(s)] = Hat[Mg(s)] + 1st IE of Mg + 2nd IE of Mg + 3rd IE of Mg + 3Hat[1/2Cl2(g)] + 31st EA of Cl + HL[MgCl3(s)] = 150+736+1450+7740+3121+3(-364)+(-5440) = +3907 kJ mol-1

Enthalpy ( kJ mol) Hf[MgCl3] >0 Mg3+(g)+3Cl(g) 3Hat[1/2Cl2] 3rd IE[Mg] Mg3+(g)+ Cl2(g) Mg3+(g)+3Cl(g) 31st EA[Cl] MgCl3(s) HL[MgCl3] Enthalpy ( kJ mol) Mg2+(g)+ Cl2(g) 2nd IE[Mg] Hf[MgCl3] >0 Mg+(g)+ Cl2(g) 1st IE[Mg] Hat[Mg] Mg(g)+ Cl2(g) Mg(s)+3/2Cl2(g)

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

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 7-2

Factors affecting lattice enthalpy 7.2 Energetics of Formation of Ionic Compounds (SB p.196) Factors affecting lattice enthalpy Effect of ionic size:  The greater the ionic size  The lower (or less negative) is the lattice enthalpy

Factors affect lattice enthalpy 7.2 Energetics of Formation of Ionic Compounds (SB p.196) Factors affect lattice enthalpy Effect of ionic charge:  The greater the ionic charge  The higher (or more negative) is the lattice enthalpy Check Point 7-3

7.4 Ionic Crystals

Crystal Lattice(晶體格子) 7.4 Ionic Crystals (SB p.202) Crystal Lattice(晶體格子)

7.4 Ionic Crystals (SB p.202) In solid state, ions of ionic compounds are regularly packed to form 3-dimensional structures called crystal lattices(晶體格子)

7.4 Ionic Crystals (SB p.202) The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle. C.N. of Na+ = 6

7.4 Ionic Crystals (SB p.202) The coordination number (C.N.) of a given particle in a crystal lattice is the number of nearest neighbours of the particle. C.N. of Cl = 6

Identify the unit cell of NaCl crystal lattice Not a unit cell

Identify the unit cell of NaCl crystal lattice Not a unit cell

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

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

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

Unit cell Unit cell

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

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

Simple cubic structure Space filling Space lattice Unit cell  View this Chemscape 3D structure

Body-centred cubic (b.c.c.) structure Space lattice Space filling Unit cell  View this Chemscape 3D structure

Face-centred cubic (f.c.c.) structure View this Chemscape 3D structure Unit cell View this Chemscape 3D structure

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

Interstitial sites - The empty spaces in a crystal lattice Two types of interstitial sites in f.c.c. structure Octahedral site Tetrahedral site

Octahedral site : - It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.

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. before behind 45o

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

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.

c b a

Top layer Q.9 Label a, b, and c as Oh sites or Td sites c is Oh site a and b are Td sites c is Oh site Top layer

13 Oh sites 4 Th sites in the front 4 Th sites at the back Q.10 Identify the Oh sites and Td sites in the f.c.c. unit cell shown below. 2 1 13 Oh sites 3 4 5 6 4 Th sites in the front 1 3 5 6 9 8 7 4 Th sites at the back 7 8 2 4 11 10 12 13 View Octahedral sites and tetrahedral sites

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.

The crystal structures of sodium chloride, caesium chloride and calcium fluoride are discussed.

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

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

F.C.C. Structure of Sodium Chloride View this Chemscape 3D structure

Structure of Sodium Chloride 7.4 Ionic Crystals (SB p.201) Structure of Sodium Chloride Unit cell of NaCl Co-ordination number of Na+ = 6 6 : 6 co-ordination Co-ordination number of Cl- = 6

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.

Fraction of particles occupied by a unit cell 1/2 Body Face Edge Corner 1 1/2 Fraction of particles occupied by a unit cell

Fraction of particles occupied by a unit cell 1/4 Body Face Edge Corner 1 1/2 1/4 Fraction of particles occupied by a unit cell

Fraction of particles occupied by a unit cell 1/8 Body Face Edge Corner 1 1/2 1/4 1/8 Fraction of particles occupied by a unit cell

Q. 11. Calculate the net nos. of Na+ and Cl- Q.11 Calculate the net nos. of Na+ and Cl- ions in a unit cell of NaCl. No. of Cl- ions = 8 corners 6 faces No. of Na+ ions = 12 edges 1 body Example 7-4

Structure of Caesium Chloride  Link 7.4 Ionic Crystals (SB p.202) Structure of Caesium Chloride  Link Cs+ 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. Simple cubic lattice

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

Size of interstitial sites : - Cubical > octahedral > tetrahedral In f.c.c. structure In simple cubic structure

Structure of Caesium Chloride  Link 7.4 Ionic Crystals (SB p.202) Structure of Caesium Chloride  Link Simple cubic lattice Co-ordination number of Cs+ = 8 8 : 8 co-ordination Co-ordination number of Cl- = 8

Q.12 Number of Cs+ = 1 Number of Cl =

Structure of Calcium Fluoride  Link 7.4 Ionic Crystals (SB p.203) Structure of Calcium Fluoride  Link Fluorite structure It can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions.

Structure of Calcium Fluoride  Link 7.4 Ionic Crystals (SB p.203) Structure of Calcium Fluoride  Link 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.

Face-centred cubic lattice 7.4 Ionic Crystals (SB p.203) Structure of Calcium Fluoride  Link Face-centred cubic lattice Co-ordination number of Ca2+ = 8 8 : 4 co-ordination Co-ordination number of F- = 4

CaF2 Q.13 (a) Number of F = 8 Number of Ca2+ = 7.4 Ionic Crystals (SB p.203) Q.13 (a) CaF2 Number of F = 8 Number of Ca2+ =

7.4 Ionic Crystals (SB p.203) Q.13 (b) CaF2 Only alternate cubical sites are occupied by Ca2+ in order to maintain electroneutrality.

Na2O vs CaF2 Structure of Sodium oxide  Link Antifluorite structure 7.4 Ionic Crystals (SB p.203) Structure of Sodium oxide  Link Antifluorite structure Na2O vs CaF2 fluorite structure

Zinc blende structure - Link Q.14(a) Zinc blende structure - Link

Q.14(a) ZnS Number of Zn2+ = 4 Number of S2 =

Q.14(b) Zinc sulphide, ZnS  tetrahedral site S2 ions adopt f.c.c. structure Alternate Td sites are occupied by Zn2+ ions to ensure electroneutrality.

The unit cell is not a cube Q.15 Rutile structure - Link The unit cell is not a cube

Ti4+ Q.15(a) O2 Number of Ti4+ = TiO2 Number of O2 =

Q.15(a) C.N. of Ti4+ = 6 C.N. of O2 = 3  6 : 3 coordination

Q.15(b) Titanium(IV) oxide, TiO2  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.

hexagonal close-packed b a

1. Close Packing Considerations 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.

1. Close Packing Considerations Factors governing the structures of ionic crystals 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

Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystals 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.

Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystals Q.16 Is there no limit for the C.N. ? Explain your answer. 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.

Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystals Q.16 Is there no limit for the C.N. ? Explain your answer. 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.

2. The relative size of cation and anion, Factors governing the structures of ionic crystals 2. The relative size of cation and anion, In general, r > r+,

If the cations are large, Factors governing the structures of ionic crystals If the cations are large, 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.

Summary : - NH4+ is large Interstitial site Tetrahedral Octahedral 7.4 Ionic Crystals (SB p.203) Summary : - Interstitial site Tetrahedral Octahedral Cubical Coordination 4 : 4 6 : 6 8 : 8 0.225 – 0.414 0.414 – 0.732 0.732 – 1.000 Examples ZnS, most copper(I) halides Alkali metal halides except CsCl CsCl, CsBr CsI, NH4Cl NH4+ is large

Summary : - *CaF2, BaF2, BaCl2, SrF2 Examples 0.732 – 1.000 8 : 4 7.4 Ionic Crystals (SB p.203) Summary : - *CaF2, BaF2, BaCl2, SrF2 Examples 0.732 – 1.000 8 : 4 Coordination Cubical Interstitial site

Q.17 r A B 45° r r+ C

the range of ratio that favours octahedral arrangement.

the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites. Q.18

Q.18(a) Silver chloride, AgCl  octahedral site Cl ions adopt f.c.c. structure All Oh sites are occupied by Ag+ ions to ensure electroneutrality.

Q.18(a) Silver chloride, AgCl  octahedral site Ag+ ions adopt an open f.c.c. structure All Oh sites are occupied by Cl ions to ensure electroneutrality.

Q.18(b) Copper(I) bromide, CuBr  tetrahedral site Br ions adopt f.c.c. structure Only alternate Td sites are occupied by Cu+ ions to ensure electroneutrality.

Q.18(c) Copper(II) bromide, CuBr2  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)

Q.18(c) Copper(II) bromide, CuBr2 Or, Cu2+ ions adopt a very open f.c.c. structure All the Td sites are occupied by Br ions to ensure electroneutrality. However, this structure is too open to form strong ionic bonds.  non-cubic structure is preferred.

Q.18(c) 4 : 2 coordination Cu2+ Br

Copper(II) chloride, CuCl2 Layer structure 0.295nm 0.230 nm

Copper(II) chloride, CuCl2 6 : 3 coordination

7.5 Ionic Radii

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

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.

Q.19 Cl [Ne] 3s2, 3px2, 3py2, 3pz2 Li+ 1s2 Symmetrical distribution along x, y and z axes   almost spherical spherical Li+ 1s2 Na+ [He] 2s2, 2px2, 2py2, 2pz2 Cl [Ne] 3s2, 3px2, 3py2, 3pz2 Zn2+ [Ar] 3dxy2, 3dxz2, 3dyz2, , Symmetrical distribution along xy, xz, yz planes and along x, y and z axes  almost spherical

Determination of Ionic Radii Pauling Scale Interionic distance (r+ + r) can be determined by X-ray diffraction crystallography

Electron density plot for sodium chloride crystal 7.5 Ionic Radii (SB p.205) Cl Na+ Electron density map for NaCl Contour having same electron density r++ r r++ r Electron density plot for sodium chloride crystal

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

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

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

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

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).

Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) Radii of cations < Radii of corresponding parent atoms cations have one less electron shell than the parent atoms Ionic radius vs atomic radius

Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) p/e of cation > p/e of parent atom  Less shielding effect stronger nuclear attraction for outermost electrons  smaller size Ionic radius vs atomic radius

Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) Radii of anions > Radii of corresponding parent atoms Ionic radius vs atomic radius

Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206) p/e of anion < p/e of parent atom  more shielding effect weaker nuclear attraction for outermost electrons  larger size Ionic radius vs atomic radius

Periodic trends of ionic radius 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.

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.

Isoelectronic to He(2) Isoelectronic to Ar(2,8,8) 7.5 Ionic Radii (SB p.206) Isoelectronic to He(2) Isoelectronic to Ar(2,8,8) Isoelectronic to Ne(2,8)

H is larger than most ions, why ? 7.5 Ionic Radii (SB p.206) H is larger than most ions, why ?

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)

The END Example 7-5 Check Point 7-5

Why do two atoms bond together? Introduction (SB p.186) Let's Think 1 Why do two atoms bond together? Answer The two atoms tend to achieve an octet configuration which brings stability.

How does covalent bond strength compare with ionic bond strength? Introduction (SB p.186) Let's Think 1 How does covalent bond strength compare with ionic bond strength? They are similar in strength. Both are electrostatic attractions between charged particles. Answer Back

Example 7-2 Given the following data: ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195) Example 7-2 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 of aluminium +314 Standard enthalpy change of formation of aluminium oxide –1669

Example 7-2 Answer ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195) Answer Example 7-2 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.

Example 7-2 7.2 Energetics of Formation of Ionic Compounds (SB p.195) (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.

7.2 Energetics of Formation of Ionic Compounds (SB p.195) Back Example 7-2 (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

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (a) Draw a Born-Haber cycle for the formation of magnesium oxide. The Born-Haber cycle for the formation of MgO: Answer

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (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 ø ø Answer ø

7.2 Energetics of Formation of Ionic Compounds (SB p.197) Check Point 7-2 (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 Back

7.3 Stoichiometry of Ionic Compounds (SB p.201) Check Point 7-3 Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound. Answer 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. Back

7.4 Ionic Crystals (SB p.204) Back Example 7-4 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. Answer

Check Point 7-4 Answer Back 7.4 Ionic Crystals (SB p.205) Back Check Point 7-4 The diagram on the right shows a unit cell of titanium oxide. What is the coordination number of (a) titanium; and (b) oxygen? Answer (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.

7.5 Ionic Radii (SB p.208) Example 7-5 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

Example 7-5 Answer Explain briefly the following: 7.5 Ionic Radii (SB p.208) Example 7-5 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

Example 7-5 7.5 Ionic Radii (SB p.208) (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.

Example 7-5 Back 7.5 Ionic Radii (SB p.208) (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. Back

7.5 Ionic Radii (SB p.208) Check Point 7-5 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 Back