3Lewis ModelG.N. Lewis in 1916Only the outermost (valence) electrons are involved significantly in bond formationSuccessful in solving chemical problems
4Why 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?
5Chemical 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 electronsAtoms tend to form chemical bonds in such a way as to achieve the electronic configurations of the nearest noble gases (The Octet Rule )
6Q.1 Write the s,p,d,f notation for the ions in the table below
8Three types of chemical bonds Introduction (SB p.186)Three types of chemical bonds1. Ionic bond (electrovalent bond)Formed by transfer of electrons
9Three types of chemical bonds 7.1 Formation of Ionic Bonds: Donating and Accepting Electrons (SB p.187)Three types of chemical bonds1. Ionic bond (electrovalent bond)ClNaSodium atom, Na1s22s22p63s1Chlorine atom, Cl1s22s22p63s23p5
10Three types of chemical bonds Introduction (SB p.186)Three types of chemical bonds1. Ionic bondElectrostatic attraction between positively charged particles and negatively charged particles
11Three types of chemical bonds Introduction (SB p.186)Three types of chemical bondsCovalent bondFormed by sharing of electrons
12Three types of chemical bonds Introduction (SB p.186)Three types of chemical bonds2. Covalent bondElectrostatic attraction between nuclei and shared electrons
13Three types of chemical bonds Introduction (SB p.186)Three types of chemical bonds3. Metallic bondElectrostatic attraction between metallic cations and delocalized electrons (electrons that have no fixed positions)
14Three types of chemical bonds Introduction (SB p.186)Three types of chemical bonds3. Metallic bondFormed by sharing of a large number of delocalized electronsLet's Think 1
15Ionic bonds and Covalent bonds are only extreme cases of a continuum. In real situation, most chemical bonds are intermediate between ionic and covalent
16Ionic Bonds with incomplete transfer of electrons have covalent character. Covalent Bonds with unequal sharing of electrons have ionic character.
17Electronegativity 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 abilitiesCovalent bonds are formed between atoms with small or no difference in their electron-attracting abilities.
18Ways to compare the electron-attracting ability of atoms Ionization EnthalpyElectron affinityElectronegativity
19Electronegativity and Types of Chemical Bonds Ionization enthalpyThe 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.
20Electronegativity and Types of Chemical Bonds 2. Electron affinityThe 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.
21I.E. and E.A. only show e- releasing/attracting power of free, isolated atomsHowever, whether a bond is ionic or covalent depends on the ability of atoms to attract electrons in a chemical bond.
22Electronegativity and Types of Chemical Bonds The ability of an atom to attract electrons in a chemical bond.
23Mulliken’s scale of electronegativity Electronegativity(Mulliken) Nobel Laureate in Chemistry, 1966
24Pauling’s scale of electronegativity Nobel Laureate in Chemistry, 1954Nobel Laureate in Peace, 1962
25Pauling’s scale of electronegativity calculated from bond enthalpiescannot be measured directlyhaving no unitalways non-zerothe most electronegative element, F, is arbitrarily assigned a value of 4.00
27What 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.
28The EN values increase from left to right across a Period.
29What 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.
31What 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.
33XeF2, 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
34Formation 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
35Formation 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
36The 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
38Notes 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).
39Tendency for the Formation of Ions An ion will be formed easily if1. 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 issmall for an anion, orlarge for a cation.
40For 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.
41Li+ 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+
42For anion, Smaller size of parent atom more negative E.A. or more -ve sum ofsuccessive E.A.s, easier formation of anion,atoms of Group VIA & Group VIIA elements form anions easily.
43N3 O2 F P3 S2 Cl Br I Ease of formation of anion
44Stable 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.
45Q.2 Write the s,p,d,f notation for the following cations
61(c) Cations of transition elements Fe [Ne] 3s2, 3p6, 3d6, 4s2Fe2+ [Ne] 3s2, 3p6, 3d6Fe3+ [Ne] 3s2, 3p6, 3d5Which 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)
62B. 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 anionGroup VIA and Group VIIA elements form anions easily.
64E.A. becomes more –ve from Gp 1 to Gp 7 across a period H-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
65The electrons added experience stronger nuclear attraction when the atoms are getting smaller across the period.H-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
66+ve E.A. for Gp 2 and Gp 0 because the electron added occupies a new shell / subshell -73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
67Goes 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-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
68More –ve E.A. for Gp 1 because the ions formed have full-filled s-subshells. E.g. Li(1s22s1) Li(1s22s2)H-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
69Less –ve E.A. for Gp 5 because the stable half-filled p-subshell no longer exists. E.g. N(2s22p3) N(2s22p4)H-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
70H -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-73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
71It is because the halide ions formed have full-filled s/p subshells (octet structure). -73He+21Li-60Be+18B-23C-122N~0O-142F-328Ne+29Na-53MgAl-44Si-134P-72S-200Cl-349Ar+35K-48Br-324Kr+39
72The electrons added are repelled strongly by the negative ions. Q.4 Why are the second E.A.s of O(+844 kJmol1) and S(+532 kJmol1) positive ?O(g) + e O2(g)S(g) + e S2(g)The electrons added are repelled strongly by the negative ions.
73Why is the E.A. of F less negative than that of Cl ? HalogenFClBrIE.A.kJ mol1328349325295The 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.
74By Hess’s law, H1 = IE1(K) + EA1(F) + H2 Energetics of Formation of Ionic CompoundsA. Formation of an ion pair in gaseous stateConsider the formation of a KF(g) ion pair from K(g) & F(g)H1K(g) F(g) KF(g)IE1(K)K+(g)H2EA1(F)+ F(g)By Hess’s law, H1 = IE1(K) + EA1(F) + H2
78I.E. and E.A. are NOT the driving forces for the formation of ionic bond. H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
79K+(g) & F(g) tend to come close together in order to become stable. H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
80Coulomb stabilization is the driving force for the formation of ionic bond. H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
81The most energetically stable state is reached. When R = 2.1701010 mThe most energetically stable state is reached.H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
82The ions cannot come any closer than Re as it will results in less stable states H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
83Repulsions between electron clouds and between nuclei > attraction between ions H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
84Repulsion : Minimum V when R = 2.170 1010 m Attraction : between opposite chargesRPotential energy (V)Minimum V when R = 1010 mAttraction :between opposite charges
85What is the significance of the lowest energy state of the neutral state of K + F ? H2 = 1.0631018 JIE1(K) + EA1(F) = 1019 J
86K F A covalent bond is formed K F KFElectrostatic attraction between positive nuclei and bond electron pair stabilizes the system
87By Hess’s law, Hf = H1 + H2 A. Formation of Ionic CrystalsConsider the formation of NaCl(s) from Na(s) & Cl2(g)HfNa(s) Cl2(g) NaCl(s)Na+(g) Cl(g)H1H2By Hess’s law, Hf = H1 + H2
88H1 is the sum of four terms of enthalpy changes 1. Standard enthalpy change of atomization of Na(s) Na(s) Na(g) H = kJ mol-12. First ionization enthalpy of Na(g) Na(g) Na+(g) + e- H= +500 kJ mol-13. Standard enthalpy change of atomization of Cl2(g) 1/2Cl2(g) Cl(g) H = kJ mol-14. First electron affinity of Cl(g) Cl(g) + e- Cl-(g) H = -349 kJ mol-1
89H2 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
90Direct determination of lattice enthalpy by experiment is very difficult, but it can be obtained from1. theoretical calculation using an ionic model, Or2. experimental results indirectly with the use of a Born-Haber cycle.
93Lattice 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 compoundLattice enthalpy is a measure of the strength of ionic bond.
94Determination of Lattice Enthalpy From Born-Haber cycle (experimental method, refers to Q.7)From theoretical calculation based on the ionic model
95Assumptions 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.
96M is the Madelung constants that depends on the crystal structure
100(r+ + r-) is the nearest distance between the nuclei of cation and anion r+ is the ionic radius of the cationr- is the ionic radius of the anion
101Stoichiometry of Ionic Compounds Stoichiometry of an ionic compound is thesimplest whole number ratio of cationsand anions involved in the formation ofthe compound.There are two ways to predict stoichiometry.
1021. 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
103 NaF MgCl2 1. Considerations in terms of electronic configurations Na F [Na]+ [F]-2,8,1 2, ,8 2,8 NaFMg Cl [Cl]- [Mg]2+ [Cl]-2,8,2 2,8, ,8,8 2,8 2,8,8 MgCl2
1042. 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.
1052. 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.
106Q. 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.
107Hf[MgCl(s)] = Hat[Mg(s)] + 1st IE of Mg + Hat[1/2Cl2(g)] + 1st EA of Cl + HL[MgCl(s)]= ( ) kJ mol-1= kJ mol-1
113Since 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 isMgCl2
114Reasons 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 CaF2Short range repulsive forces between oppositely charged ions have not been considered.Check Point 7-2
115Factors affecting lattice enthalpy 7.2 Energetics of Formation of Ionic Compounds (SB p.196)Factors affecting lattice enthalpyEffect of ionic size: The greater the ionic size The lower (or less negative) is the lattice enthalpy
116Factors affect lattice enthalpy 7.2 Energetics of Formation of Ionic Compounds (SB p.196)Factors affect lattice enthalpyEffect of ionic charge: The greater the ionic charge The higher (or more negative) is the lattice enthalpyCheck Point 7-3
130Simple cubic structure Space fillingSpace latticeUnit cell View this Chemscape 3D structure
131Body-centred cubic (b.c.c.) structure Space latticeSpace fillingUnit cell View this Chemscape 3D structure
132Face-centred cubic (f.c.c.) structure View this Chemscape 3D structure Unit cellView this Chemscape 3D structure
133The f.c.c. structure can be obtained by stacking the layers of particles in the pattern abcabc...
134Interstitial sites - The empty spaces in a crystal lattice Two types of interstitial sites in f.c.c. structureOctahedral siteTetrahedral site
135Octahedral site : -It is the space between the 3 spheres in one layer and 3 other spheres in the adjacent layer.
136When 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.beforebehind45o
137Tetrahedral site : -It is the space between the 3 spheres of one layer and a fourth sphere on the upper layer.
138In 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.
140Top layer Q.9 Label a, b, and c as Oh sites or Td sites c is Oh site a and b are Td sitesc is Oh siteTop layer
14113 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.2113 Oh sites34564 Th sites in the front13569874 Th sites at the back782411101213View Octahedral sites and tetrahedral sites
142Since 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.
143The crystal structures of sodium chloride,caesium chloride andcalcium fluoride are discussed.
144Sodium Chloride (The Rock Salt Structure) f.c.c. unit cell of larger Cl- ions, with all Oh sites occupied by the smaller Na+ ionsOh sites of f.c.c. lattice of Cl- ions
145A 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
146F.C.C. Structure of Sodium Chloride View this Chemscape 3D structure
147Structure of Sodium Chloride 7.4 Ionic Crystals (SB p.201)Structure of Sodium ChlorideUnit cell of NaClCo-ordination number of Na+ = 66 : 6 co-ordinationCo-ordination number of Cl- = 6
148Only 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.
149Fraction of particles occupied by a unit cell 1/2BodyFaceEdgeCorner11/2Fraction of particles occupied by a unit cell
150Fraction of particles occupied by a unit cell 1/4BodyFaceEdgeCorner11/21/4Fraction of particles occupied by a unit cell
151Fraction of particles occupied by a unit cell 1/8BodyFaceEdgeCorner11/21/41/8Fraction of particles occupied by a unit cell
152Q. 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 corners6 facesNo. of Na+ ions =12 edges1 bodyExample 7-4
153Structure of Caesium Chloride Link 7.4 Ionic Crystals (SB p.202)Structure of Caesium Chloride LinkCs+ 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
154A cubical site is the space confined by 4 spheres in one layer and 4 other spheres in the adjacent layer.
156Structure of Caesium Chloride Link 7.4 Ionic Crystals (SB p.202)Structure of Caesium Chloride LinkSimple cubic latticeCo-ordination number of Cs+ = 88 : 8 co-ordinationCo-ordination number of Cl- = 8
158Structure of Calcium Fluoride Link 7.4 Ionic Crystals (SB p.203)Structure of Calcium Fluoride LinkFluorite structureIt can be viewed as a simple cubic structure of larger fluoride ions with alternate cubical sites occupied by smaller calcium ions.
159Structure of Calcium Fluoride Link 7.4 Ionic Crystals (SB p.203)Structure of Calcium Fluoride LinkAlternately, it can be viewed as an expanded f.c.c. structure of smaller calcium ions with all tetrahedral sites occupied by larger fluoride ions.
160Face-centred cubic lattice 7.4 Ionic Crystals (SB p.203)Structure of Calcium Fluoride LinkFace-centred cubic latticeCo-ordination number of Ca2+ = 88 : 4 co-ordinationCo-ordination number of F- = 4
161CaF2 Q.13 (a) Number of F = 8 Number of Ca2+ = 7.4 Ionic Crystals (SB p.203)Q.13 (a)CaF2Number of F = 8Number of Ca2+ =
1627.4 Ionic Crystals (SB p.203)Q.13 (b)CaF2Only alternate cubical sites are occupied by Ca2+ in order to maintain electroneutrality.
163Na2O vs CaF2 Structure of Sodium oxide Link Antifluorite structure 7.4 Ionic Crystals (SB p.203)Structure of Sodium oxide LinkAntifluorite structureNa2O vs CaF2fluorite structure
164Zinc blende structure - Link Q.14(a)Zinc blende structure Link
1721. Close Packing Considerations Factors governing the structures of ionic crystals1. Close Packing ConsiderationsIons in ionic crystals tend to pack as closely as possible. (Why ?)To strengthen the ionic bonds formedTo maximize the no. of ionic bonds formed.
1731. Close Packing Considerations Factors governing the structures of ionic crystals1. Close Packing ConsiderationsThe 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
174Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystalsQ.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.
175Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystalsQ.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.
176Q.16 Is there no limit for the C.N. ? Explain your answer. Factors governing the structures of ionic crystalsQ.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.
1772. The relative size of cation and anion, Factors governing the structures of ionic crystals2. The relative size of cation and anion,In general, r > r+,
178If the cations are large, Factors governing the structures of ionic crystalsIf 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.
179Summary : - NH4+ is large Interstitial site Tetrahedral Octahedral 7.4 Ionic Crystals (SB p.203)Summary : -Interstitial siteTetrahedralOctahedralCubicalCoordination4 : 46 : 68 : 80.225 – 0.4140.414 – 0.7320.732 – 1.000ExamplesZnS, most copper(I) halidesAlkali metal halides except CsClCsCl, CsBrCsI, NH4ClNH4+ is large
182the range of ratio that favours octahedral arrangement.
183the optimal ratio for cations and anions to be in direct contact with each other in the cubical sites.Q.18
184Q.18(a) Silver chloride, AgCl octahedral site Cl ions adopt f.c.c. structureAll Oh sites are occupied by Ag+ ions to ensure electroneutrality.
185Q.18(a) Silver chloride, AgCl octahedral site Ag+ ions adopt an open f.c.c. structureAll Oh sites are occupied by Cl ions to ensure electroneutrality.
186Q.18(b) Copper(I) bromide, CuBr tetrahedral site Br ions adopt f.c.c. structureOnly alternate Td sites are occupied by Cu+ ions to ensure electroneutrality.
187Q.18(c) Copper(II) bromide, CuBr2 tetrahedral site Br ions adopt f.c.c. structureOnly 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)
188Q.18(c) Copper(II) bromide, CuBr2 Or, Cu2+ ions adopt a very open f.c.c. structureAll 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.
193Ionic 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.
194Q.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.
195Q.19 Cl [Ne] 3s2, 3px2, 3py2, 3pz2 Li+ 1s2 Symmetrical distribution along x, y and z axes almost sphericalsphericalLi+ 1s2Na+ [He] 2s2, 2px2, 2py2, 2pz2Cl [Ne] 3s2, 3px2, 3py2, 3pz2Zn2+ [Ar] 3dxy2, 3dxz2, 3dyz2, ,Symmetrical distribution along xy, xz, yz planes and along x, y and z axes almost spherical
196Determination of Ionic Radii Pauling Scale Interionic distance (r+ + r) can be determined byX-ray diffraction crystallography
197Electron density plot for sodium chloride crystal 7.5 Ionic Radii (SB p.205)ClNa+Electron density map for NaClContour having same electron densityr++ rr++ rElectron density plot for sodium chloride crystal
198By additivity rule,Interionic distance = r+ + rFor K+Cl,r+ + r = 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+) = nm, r(Cl) = nm
200Limitation 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
201Evidence :Electron distribution is not perfectly spherical at the boundary due to repulsion between electron clouds of neighbouring ions.
202Ionic 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).
203Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206)Radii of cations < Radii of corresponding parent atomscations have one less electron shell than the parent atomsIonic radius vs atomic radius
204Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206)p/e of cation > p/e of parent atom Less shielding effectstronger nuclear attraction for outermost electrons smaller sizeIonic radius vs atomic radius
205Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206)Radii of anions > Radii of corresponding parent atomsIonic radius vs atomic radius
206Ionic radius vs atomic radius 7.5 Ionic Radii (SB p.206)p/e of anion < p/e of parent atom more shielding effectweaker nuclear attraction for outermost electrons larger sizeIonic radius vs atomic radius
207Periodic trends of ionic radius Ionic radius increases down a GroupIonic 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.
208Periodic trends of ionic radius 2. Ionic radius decreases along a series of isoelectronic ions of increasing nuclear chargeThe total shielding effects of isoelectronic ions are approximately the same. Zeff nuclear charge (Z) Ionic radius decreases as nuclear charge increases.
209Isoelectronic 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)
210H is larger than most ions, why ? 7.5 Ionic Radii (SB p.206)H is larger than most ions, why ?
211Q.20H > N3The 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)
213Why do two atoms bond together? Introduction (SB p.186)Let's Think 1Why do two atoms bond together?AnswerThe two atoms tend to achieve an octet configuration which brings stability.
214How does covalent bond strength compare with ionic bond strength? Introduction (SB p.186)Let's Think 1How does covalent bond strength compare with ionic bond strength?They are similar in strength.Both are electrostatic attractions between charged particles.AnswerBack
215Example 7-2 Given the following data: ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195)Example 7-2Given the following data:ΔH (kJ mol–1)First electron affinity of oxygen –142Second electron affinity of oxygen +844Standard enthalpy change of atomization of oxygen +248Standard enthalpy change of atomization of aluminium +314Standard enthalpy change of formation of aluminium oxide –1669
216Example 7-2 Answer ΔH (kJ mol–1) 7.2 Energetics of Formation of Ionic Compounds (SB p.195)AnswerExample 7-2ΔH (kJ mol–1)First ionization enthalpy of aluminium +577Second ionization enthalpy of aluminiumThird ionization enthalpy of aluminium(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.
217Example 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.
2227.3 Stoichiometry of Ionic Compounds (SB p.201) Check Point 7-3Give two properties of ions that will affect the value of the lattice enthalpy of an ionic compound.AnswerThe 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
2237.4 Ionic Crystals (SB p.204)BackExample 7-4Write 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
224Check Point 7-4 Answer Back 7.4 Ionic Crystals (SB p.205)BackCheck Point 7-4The 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.
2257.5 Ionic Radii (SB p.208)Example 7-5The following table gives the atomic and ionic radii of some Group IIA elements.ElementAtomic radius (nm)Ionic radiusBe0.1120.031Mg0.1600.065Ca0.1900.099Sr0.2150.133Ba0.2170.135
226Example 7-5 Answer Explain briefly the following: 7.5 Ionic Radii (SB p.208)Example 7-5Explain 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
227Example 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.
228Example 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
2297.5 Ionic Radii (SB p.208)Check Point 7-5Arrange 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–AnswerBack