Ch. 9.1-9.4: Ionic Bonds & Lattice Energy; Covalent Bonds
•
Lewis (Electron) Dot Symbols
•
Ionic Bonding
o Electrostatic attraction between positive cation(s) and negative anion(s)
§ Electrons transferred from one atom to another to form ions
Na
Ca
+
+
+
Cl
Na
Br
Ca2+ + 2 Br
Cl
NaCl
CaBr2
Br
•
o Overall charge of compound = 0, so formula indicates quantity of each ion needed to donate and accept
equal numbers of electrons (balance positive and negative charge)
o Form lattice of alternating + & – charges
§ Each ion surrounded by oppositely-charged ion
Lattice Energy
o Energy released when 1 mole of an ionic compound forms its crystal lattice
§ Lattice formation always exothermic, lattice energy always positive
qq
qq
o Depends on charges on ions and radius (Coulomb’s Law): E = F ⋅d = k 1 2 2 ⋅r = k 1 2
r
r
§ Directly proportional to charges, inversely proportional to distance between them:
Lattice Energies and Melting Points of Some Alkali Metal and
• Increasing radius of anion
Alkaline Earth Metal Halides and Oxides
(LiF to LiI) or cation (LiCl
Compound
Lattice Energy (kJ/mol)
Melting Point (°C)
to KCl) decreases LE
o 1/r effect, approximate
LiF
1017
845
• Increasing charge
LiCl
838
610
LiBr
787
550
• of ion with similar radius
LiI
732
450
(NaCl to MgCl2) increases
LE
NaCl
788
801
o By that factor, approx.
NaBr
736
750
o More pronounced than
NaI
686
662
1/r effect
KCl
699
772
o Increasing charge of
KBr
689
735
both ions is
KI
632
680
multiplicative
MgCl2
2527
714
Na2O
2570
Sublimes @ 1275
MgO
3890
2800
Ionic Bonds-Lattice Energy-Born Haber Cycle.docx
o Born-Haber Cycle
§ Formation of Ionic Compounds
• Thermodynamic cycle like Hess’s
Law
• Starts with elements in their
standard state, shows energy for
every step of reaction to form
ionic compound
§ Large lattice energy is why O2– forms
stable ions even though O→O2– has
large endothermic EA
• Energy required to form cations
& some anions
• Formation of lattice exothermic
§ LE allows MgO to form even though
O → O2– has large endothermic EA
(why?) and Mg → Mg2+ has large IE
• Mg+O– would have too low LE
(why?), even with lower IE and
exo EA, overall rxn would be
endo.
§ NaF lattice energy is much lower, IE
is much lower and EA is exo, so rxn
is exo.
• Na2+F–2 would have more exo LE, but IE2 for Na so large that overall rxn is endo.
§ Ion stability comes from LE overcoming endo processes, not from “octet.”
p.2