ionic bonding –evidence for ionic bonding, electron density maps –trends in radii –born haber...
TRANSCRIPT
• Ionic bonding– evidence for ionic
bonding, electron density maps
– trends in radii
– Born Haber cycles• explaining formulae,
why AlO is incorrect
– polarisation
• Metallic Bonding
• Covalency– electron density maps
– giant atomic structures
– dot-cross diagrams
– shapes of molecules • VSEPR
– electronegativity
– polarity of covalent bonds
– polarity of molecules
• Intermolecular forces– trends in physical properties
• Solutions and dissolving– why certain substances dissolve in particular solvents
Na
Mg Al
Si
PS
ClAr
0
500
1000
1500
2000
Tem
per
atu
re /K
Melting Points of Period 3 elements
Melting Points of Period 2 elements
Li
Be
B
C
N O F Ne0
1000
2000
3000
4000
Tem
pera
ture
/ K
Ionic bonding
• The electrostatic attraction between oppositely charged ions– Metals, hydrogen and ammonium form positive
ions (cations).– Non-metals form negative ions (anions).
Evidence for ionic compounds• High melting points
– strong electrostatic attractions between oppositely charged ions
• Electrical conductivity only in liquid state or aqueous solution because ions need to move.
• Coloured ions can be observed migrating to electrodes during electrolysis (e.g. CuCr2O7)
– green / blue Cu2+ (aq) moves to cathode
– yellow Cr2O72- (aq) moves to anode
• Electron density maps show low electron density between the oppositely charged ions.
NaCl (s)
Na (s) + ½ Cl2 (g)
Na+ (g) + e- + Cl (g)
Na+ (g) + Cl- (g)
Hformation = - 411.2 kJ mol-1
Na (g) + ½ Cl2 (g)
m1[Na] = + 496 kJ mol-1
Eaff[Cl] = - 348.8 kJ mol-1
Hlatt (=348.8-121.7-496-107.3-411.2)
= - 787.4 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for sodium chloride
Hat = + 107.3 kJ mol-1
Na+ (g) + e- + ½ Cl2 (g)
Hat [Cl]= + 121.7 kJ mol-1
KCl (s)
K (s) + ½ Cl2 (g)
K+ (g) + e- + Cl (g)
K+ (g) + Cl- (g)
Hformation = - 436.7 kJ mol-1
K (g) + ½ Cl2 (g)
m1[K] = + 419 kJ mol-1
Eaff[Cl] = - 348.8 kJ mol-1
Hlatt (=348.8-121.7-419-89.2-436.7)
= - 717.8 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for potassium chloride
Hat = + 89.2 kJ mol-1
K+ (g) + e- + ½ Cl2 (g)
Hat [Cl]= + 121.7 kJ mol-1
KBr (s)
K (s) + ½ Br2 (l)
K+ (g) + e- + Br (g)
K+ (g) + Br - (g)
Hformation = - 393.8 kJ mol-1
K (g) + ½ Br2 (l)
Em1[K] = + 419 kJ mol-1
Eaff[Br] = - 324.6 kJ mol-1
Hlatt =(324.6-111.9-419-89.2-393.8)
= - 689.3 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for potassium bromide
Hat [K]= + 89.2 kJ mol-1
K+ (g) + e- + ½ Br2 (l)
Hat [Br]= + 111.9 kJ mol-1
Li2O (s)
2 Li (s) + ½ O2 (g)
2 Li+ (g) + 2 e- + O (g)
2 Li+ (g) + e- + O- (g)
Hformation = - 597.9 kJ mol-1
2 Li (g) + ½ O2 (g)
Em1[Li] = + 1040 kJ mol-1
Hlatt (=-798+141.1-249.2-1040-318.8-597.6)
= - 2862.8 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for lithium oxide
Hat = + 318.8 kJ mol-1
2 Li+ (g) + 2 e- + ½ O2 (g)
Hat [O]= + 249.2 kJ mol-1
2 Li+ (g) + O2- (g)
Eaff[O]= - 141.1 kJ mol-1
Eaff[O-] +798 kJ mol-1
Hat(Mg) = + 148 kJ mol-1
MgCl2 (s)
Mg (s) + Cl2 (g)
Mg2+ (g) + 2e- + 2 Cl (g)
Mg2+ (g) + 2 Cl- (g)
Hformation =+148+738+1451+243.4-679.6-2526
- 625.2 kJ mol-1
Mg (g) + Cl2 (g)
Em1[Mg] = + 738 kJ mol-1
aff [Cl] = - 697.6 kJ mol-1
Hlatt == - 2526 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for magnesium chloride
Mg+ (g) + e- + Cl2 (g)
Hat [Cl]= + 243.4 kJ mol-1
Mg2+ (g) + 2e- + Cl2 (g)
m2[Mg] = + 1451 kJ mol-1
MgCl (s)Mg (s) + ½ Cl2 (g)
Mg+ (g) + e- + Cl (g)
Mg+ (g) + Cl- (g)
Mg (g) + ½ Cl2 (g)
m1[Mg] = + 738 kJ mol-1
Eaff[Cl] = - 348.8 kJ mol-1
Hformation[MgCl (s)] (=+148+738+121.7-248.8-780)
= - 21.1 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for MgCl
Hat = + 148 kJ mol-1
Mg+ (g) + e- + ½ Cl2 (g)
Hat [Cl]= + 121.7 kJ mol-1
Hlatt[MgCl (s)] = - 780 kJ mol-1
MgCl3 (s)
Mg (s) + 3/2 Cl2 (g)
Mg3+ (g) + 3e- + 3 Cl (g)
Mg3+ (g) + 3 Cl- (g)
Hformation =+148+9927+366-1047-4500 = + 4894 kJ mol-1
Mg (g) + 3/2 Cl2 (g)
Em1+ Em2+Em3 [Mg]= (738+1451+7738) kJ mol-1
= + 9927 kJ mol-1
aff[Cl] = - 1047 kJ mol-1
Hlatt = - 4500 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for MgCl3
Hat = + 148 kJ mol-1
Mg3+ (g) + 3e- + 3/2 Cl2 (g)
Hat [1/2Cl2]= + 366 kJ mol-1
CaI2 (s)
Ca (s) + I2 (g)
Ca2+ (g) + 2e- + 2 I (g)
Ca2+ (g) + 2 I- (g)
Hformation = - 533.5 kJ mol-1
Ca (g) + I2 (g)
Em1[Ca] = + 590 kJ mol-1
Eaff[I]= - 590.8 kJ mol-1
Hlatt =(590.8-214-1145-590-178.2-533.5)
= - 2069.9 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for Calcium iodide
Hat [Ca] = + 178.2 kJ mol-1
Ca+ (g) + e- + I2 (g)
Hat [I]= + 214 kJ mol-1
Ca2+ (g) + 2e- + I2 (g)
Em2[Ca] = + 1145 kJ mol-1
MgO (s)
Mg (s) + ½ O2 (g)
Mg2+ (g) + 2 e- + O (g)
Mg2+ (g) + e- + O- (g)
Hformation = - 601.7 kJ mol-1
Mg (g) + ½ O2 (g)
Em1[Mg] = + 738 kJ mol-1
Eaff[O]= - 141.1 kJ mol-1
Hlatt = (-798+141.1-249.2-1451-738-147.7-601.7)
= - 3844.5 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for magnesium oxide
Mg+ (g) + e- + ½ O2 (g)
Hat [O]= + 249.2 kJ mol-1
Mg2+ (g) + 2 e- + ½ O2 (g)
m2[Mg] = + 1451 kJ mol-1
Eaff[O-] +798 kJ mol-1
Mg2+ (g) + O2- (g)
Hat(Mg) = + 147.7 kJ mol-1
BF3 (s)
B (s) + 3/2 F2 (g)
B3+ (g) + 3e- + 3 F (g)
B3+ (g) + 3 F- (g)
Hformation = - 1504.1 kJ mol-1
B (g) + 3/2 F2 (g)
Em1+ Em2+Em3 [B]= (578+1817+2745) kJ mol-1
= + 5140 kJ mol-1
aff[F] = - 984 kJ mol-1
Hlatt (984-237-5140-326.4-1504.1)
= - 6223.5 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for Boron fluoride
Hat = + 326.4 kJ mol-1
B3+ (g) + 3e- + 3/2 F2 (g)
Hat [F]= + 237 kJ mol-1
Al2O3 (s)
2 Al (s) + 3/2 O2 (g)
2 Al3+ (g) + 6 e- + 3 O (g)
2 Al3+(g)+3e-+3O- (g)
Hformation = - 1675.7 kJ mol-1
3 Eaff[O]= - 423.3 kJ mol-1
Hlatt = (-2394+423.3-747.6-10280-652.8-1675.7)
= - 15 327 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for aluminium oxide
Hat [O]= + 747.6 kJ mol-1
2 Al3+ (g) + 3 e- + 3/2 O2 (g)
3 Eaff[O-] +2394 kJ mol-1
2 Al3+ (g) + 3 O2- (g)
Hat(Al) = + 652.8 kJ mol-1
2 Al (g) + 3/2 O2 (g)
2(Em1+ Em2+ Em3 )[Al]= 2(578+1817+2745) kJ mol-1
= + 10 280 kJ mol-1
B2O3 (s)
2 B (s) + 3/2 O2 (g)
2 B3+ (g) + 6 e- + 3 O (g)
2 B3+(g)+3e-+3O- (g)
Hformation = - 1273 kJ mol-1
3 Eaff[O]= - 423.3 kJ mol-1
Hlatt = (-2394+423.3-747.6-13776-1025.4-1273)
= - 18 800 kJ mol-1
Ent
halp
y (H
)
Born Haber cycle for boron oxide
Hat [½ O2 (g)]= + 747.6 kJ mol-1
2 B3+ (g) + 3 e- + 3/2 O2 (g)
3 Eaff[O-] +2394 kJ mol-1
2 B3+ (g) + 3 O2- (g)
Hat(B) = + 1025.4 kJ mol-1
2 B (g) + 3/2 O2 (g)
2(Em1+ Em2+ Em3 )[B]= 2(801+2427+3660) kJ mol-1
= + 13 776 kJ mol-1
Lattice energiesLiF -1031 LiCl -848 BeO -4443 BeCl2 -3020
NaF -918 NaCl -780 MgO -3791 MgCl2 -2526
KF -817 KCl -711 CaO -3401 CaCl2 -2258
RbF -783 RbCl -685 SrO -3223 SrCl2 -2156
CsF -747 CsCl -661 BaO -3054 BaCl2 -2056
AlO is not the formula of aluminium oxide
• Not Al2+ O2-
• Whilst successive ionisation energies of Al increase, Em3
is not especially large.• Al3+ is very much smaller than Al2+, since its 3rd
principal quantum shell is now empty.
• Consequently, the ions pack more tightly in (Al3+)2(O2-)3.
• Al3+ also carries a greater charge than Al2+,, increasing the
attraction to O2- anions.
• The lattice energy of (Al3+)2(O2-)3 is therefore much greater in magnitude than that of Al2+O2-.
AlO is not the formula of aluminium oxide
• Not Al3+ O3-
• O3- would have a greater radius than O2- since its extra electron occupies a new principal quantum shell, further from the nucleus and more shielded by inner quantum shells.
• In spite of the increased charge of the O3- anion, its large size reduces packing density of the solid.
• The electron affinity required to form O3- from O2- would be large and endothermic.
• Therefore the lattice energy of Al3+O3- does not make up for the endothermic steps in the Born Haber cycle.
Polarisation of the anion
X+ Y-
X+ Y-
Factors leading to anion polarisation
• Cation polarizing power increases with– small radius (increasing charge density)– large positive charge (increasing charge density)
• Anion polarizability increases with– large radius (outer electrons far from nucleus and shielded by
inner shells)• increasing negative charge increases its size
• Increasing anion polarisation means increasing covalent character to the bonding– indicated by large difference between theoretical and
experimental lattice energies
Metallic bonding
• The attraction between 'positive ions' and a sea of delocalised electrons.
• Why does the melting point increase across a period, Na<Mg<Al?
• Electrical and thermal conductivity due to transfer of charge and energy by the movement of delocalised electrons.
The covalent bond
• The attraction between two nuclei and a shared pair of electrons.
• One electron of the shared pair originating from each atom in a 'standard' covalent bond.
• Both electrons of the shared pair originate from the same atom in a dative bond.
• 'Standard' and dative covalent bonds are indistinguishable.
H H
Hydrogen molecule (H2)
H-H
Valence Shell Electron Pair Repulsion• Sigma bond electron pairs and lone pairs all repel each
other around the central atom.
• The electron pairs move into positions of maximum separation.– 2 pairs gives 180: 3 pairs gives 120: 4 pairs gives 109.5, 6
pairs gives 90.
• Lone pairs have a greater repulsion than sigma bond pairs.– Each lone pair reduces the expected bond pair - bond pair
angle by about 2.5.
Chlorine molecule, Cl2
Cl Cl Cl-Cl
Hydrogen chloride molecule, HCl (g){not HCl (aq) which is ionic}
H-ClClH
Why is this covalent?
Water molecule, H2O
HH O
O
HH
109.5
N N
Nitrogen molecule, N2
N N
Ammonia, NH3
HNH
H
N
H H H107
Methane, CH4
H
H
CH H
H
H HC
H
C
H
HHH
109.5
Ethane, C2H6
H C C HH H
H H
H H
HC
C
H
H
H
C
H
HHH3C
109.5
Ethene, C2H4
C
C
H
HH
H
C = CH
H H
H
121
118
O = C = O
Carbon dioxide, CO2
O
O
C
SH H
SH
H104
H2S
109.5SiH H
H
H
Si
H
HHH
SiH4
CH H
O
C
H
HO
120120
Methanal, HCHO
CH N
H C N180
HCN
H HO O
H
H104
104
O O
NH H109.5
H
H
+
N
H
HHH +
HO
-
SO O109.5
O
O
S
O
OOO
2-
2-
SS O109
O
O
2-
2-
S
O
OSO
Ethene
C
C
HHH
H • 3 -bond pairs around each C atom repel to positions of maximum separation.
• trigonal planar
C = C
H
H H
H121
118
Ethene
C CHH
HHC C
HH
H
H
C CHH
HH
bonds shown as lines and wedges
or
C CHH
H
H
Benzene, C6H6
neither
but
nor
Benzene
• bonding is delocalised over the whole ring because all 6 p orbitals are coplanar and overlap.– not 3 separate bonds
• Benzene is more stable than alkenes and tends to react by substitution rather than addition.
Pyrene, C16H10
Graphite
Flat sheet ofC atoms
Weak forces between sheets
How to draw diamond
Diamond
Group VII hydrides
HF
HClHBr
HI
-200
-100
0
100
Group VI hydrides
H2TeH2Se
H2S
H2O
-200
-100
0
100
Group V hydrides
SbH3
AsH3
PH3
NH3
-200
-100
0
100
Group IV hydrides
SnH4
GeH4SiH4
CH4
-200
-100
0
100
The ice structure
Dissolving an ionic solid
- HO H
HO H
H O
H
HO H
H
O H
+
HO H
HO H
HO H
HO H
+ (aq)
Dissolving
• Forces are broken between the particles within the solute and within the solvent
• The energy required to break these forces needs to come from new attractive interactions between solvent and solute particles.– If this energy is not supplied, the ‘solute’ does
not dissolve.
Dissolving (2)
• Non-polar molecules attract by London forces alone.
• Ionic solutes have strong electrostatic attractions between the ions.– The energy required to overcome the high lattice
energy can often be supplied through hydration of the ions by the highly polar water molecule
– Water can also hydrogen bond to polar solutes such as sugars and ethanol.
Dissolving (3)
• Water does not dissolve non-polar solutes.– Hydrogen bonds would need to be broken between solvent water
molecules.– This energy is not made up for by London forces between solvent
and solute molecules.
• Non-polar solvents dissolve non-polar solutes.– Little energy is required to break London forces between the solvent
molecules.– This energy is supplied by new London forces between solute and
solvent molecules.
Dissolving (4)
• Metals dissolve other metals
• Ionic solids (such as alumina) dissolve in ionic solvents (such as cryolite, Na3AlF6)