Naming Ionic Compounds
In order to communicate conveniently compounds need to be named
unambiguously.
Name the cation then the anion.
Monoatomic Cations:
Most cations are single elements and are given the same name as the element.
Ex). Na+ = sodium
Mg2+ = magnesium
If an element can give rise to different cations use a Roman numeral to indicate
the charge.
Ex)
Fe2+ = iron (II)
Au+ = gold (I)
Pb2+ = lead (II)
Fe3+ = iron (III)
Au3+ = gold (III)
Pb4+ = lead (IV)
This is necessary for most
transition metals as well
as tin and lead
Monatomic cations and the
Periodic Table
Naming Ionic Compounds
Monoatomic Anions
To indicate that an element has formed an anion, we change its suffix to -ide.
Ex)
F = fluorine
O = oxygen
F- = Fluoride
O2- = Oxide
Tend to form exclusively
from the non-metals and
one metalliod
Charge = group # -18
Free ions do not exist with
charges higher than 3
units.
carbides have little
true ionic character. But the
valence of carbon is still
often 4- in many of its
compounds
N = nitrogen
Se = selenium
N3- = Nitride
Se2- = Selenide
Naming Ionic Compounds
Polyatomic Ions
Not all ions consist of just one atom. Some groups of covalently bonded
atoms have overall charges. Because they are charged, they are ions (not
molecules).
Polyatomic cations:
NH4+ Ammonium Best known
H3O+ Hydronium
Polyatomic Anions:
ClO- Hypochlorite (Bleach)
HCO3- Bicarbonate(baking Soda)
Examples
CNOHCO32-
CrO42Cr2O72MnO41-
Cyanide
Hydroxide
Carbonate
Chromate
Dichromate
Permanganate
When H+ is added to a polyatomic atom the prefix “bi” is placed before the ‘old’
name.
HCO3-
Bicarbonate
HS-
Bisulfide
HSO4-
Bisulfate
Naming Ionic Compounds
Ionic Compounds
To name an ionic compound, name the cation then the anion. The
charge of each ion indicates how many are needed to make a neutral
compound so no prefixes are necessary.
Exercise
MgCl2
Magnesium Chloride
CuBr2
Copper(II) Bromide
NaNO3
Sn(CO3)2
Sodium Nitrate
Ammonium Sulfate
Tin (IV) Carbonate
Sn(HCO3)2
Tin (II) Bicarbonate
NaNO2
Sodium Nitrite
(NH4)2SO4
LiF formation
Li loses all e’s in its valence shell
Large reduction in atomic radius:
Li – 152 pm
Li+ - 78 pm Vol = 1/7th
Li .
Li
+
Ionization Energy consumed.
Li .
Li
+
F gains electron its valence shell
Expansion in the radius because effective Z*
is decreased due to extra electron.
F – 71 pm
F- - 133 pm Vol = 6x
..
:F .
..
.. :F :
..
Electron Affinity Energy released
LiF formation
Energy is released to form the ionic bond
Ionic Compounds
Recall that opposite charges attract each other.
A cation and an anion, experience an electrostatic force, given by:
 q1eq2 e 
F k 2 
 r 
e = 1.6022 x 10-19 C
unit of charge
k = 8.988 × 109 N m2/C2
Coulomb's Constant
This force pulls them together to make an ionic bond.
q1
+
Na
r
q2
r
Cl
+
Na
Cl
Ionic Compounds
The energy released to form an ionic bond:
 q1eq2 e 
E k

 r 
Exercise
Compute the force that a sodium cation and chloride anion experience when
10.00 nm apart
F = ke2q1q2/r2
q1 = +1
q2 = -1
r = 10 nm = 1.000*10-8 m
F = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1)
(1.000*10-8 m)2
F = -2.307*10-12 N
Ionic Compounds
Exercise:
Compute the energy of formation of the ionic bond in NaCl, where the bond
length is 279 pm
E = ke2q1q2/r
q1 = +1
q2 = -1
r = 279 pm = 2.79*10-10 m
E = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1)
(2.79*10-10 m)
E = -8.27*10-19 N m = J
How energy for one mole of NaCl bonds?
E(total) = E(bond) *(# of bonds)
= (-8.27*10-19 J/bonds)(6.022*1023 bonds/mol)
= 498,000 J/mol = 498 kJ/mol
The Alkali Metals
Ionic Compounds
– soluble in water. Dissociation
Ex) NaCl(aq)
– solvation of ions:
Indicated as Na(OH2)6+(aq) or Na+(aq).
– Solvated ions must be more stable (lower energy) than the initial
crystal lattice so that energy is released.
– Alkali metals have large hydration enthalpies.
ex)The beaker gets hot when NaOH is dissolved in H2O
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Ion-Dipole Interactions
Solvation of ions by polar solvents illustrate ion-dipole forces.
For water, solvation = hydration
Definite enthalpies of hydration have been established for the reaction:
MZ   nH2O  [M(OH2 )n ]z 
H hydration
The size of this energy is directly related to the size of the ion & the charge Z on the ion.
Mg2+ has an enthalpy of -1922 kJ/mol. Why?
Fion  dipole 
q
r
3
Ionic Materials
Lattice - 3-D pattern of ions
- Minimize repulsive forces
F
-Maximize attractive forces
- Charge Balance
Na
F
Na
Na
- Large lattice energy released
F
F
Na
F
Na
Na
F
- High melting point
F
- Brittle??
Na
Na
F
F
Na
Formation of
ionic materials
Whether element from ionic
compounds depends on the
balance between:
1) Ionization energy
2) Electron affinity
3) Lattice energy
4) Phase transition
energies
5) Bond energies
Electron Affinity
Dissociation
energy
Ionization
energy
vaporization
Formation energy
Lattice
energy
Exercise
Given the following data, calculate ΔLFH for NaCl(s)
Enthalpy of sublimation of Na = +108 kJ/mol
First ionization energy for Na = +496 kJ/mol
Enthalpy of bond dissociation for Cl2(g) = +243 kJ/mol
Enthalpy of electronic attraction for Cl = -349 kJ/mol
Enthalpy of formation for NaCl = -411 kJ/mol
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Enthalpy of Lattice Formation
ΔLFH is determines properties such as solubility, thermal stability and hydration:
If similar in charge and in size, they pack closely, resulting in a large negative ΔLFH .
Thus it takes a lot of energy to break up the lattice, hence and the compounds are
not very soluble.
Ionic compounds with mismatched ions tend to have less negative ΔLFH and be more
soluble.
Carbonates of alkaline earth metals decompose to the metal oxide and carbon dioxide:
MgCO3 undergoes this reaction at 300 °C while CaCO3 requires heating to 840 °C.
Because Mg2+ is smaller than Ca2+, it is a better match for the small spherical O2than the larger nonspherical CO32-
Compounds such as Mg(ClO4)2 and CuSO4 have small cations relative to their anions.
As such, they can literally pull water out of the air (and are therefore useful as drying
agents) to surround the small cation with the relatively small water molecules. This
gives a hydrated solid:
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Ionic Lattices (NaCl)
Count the number of anions (green) and cations (silver) in a unit cell.
Consider that the anions form a lattice and the cations fill holes in the
lattice.
Three types of “holes” can be found:
Octahedral - surrounded by six atoms.
Cubic - surrounded by eight atoms.
Tetrahedral - surrounded by four atoms.
What type of holes are the cations filling?
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Rock Salt NaCl Lattice Structure
This is based on the face-centred cubic unit cell, as the Cl– ions
occupy all the sites of the fcc lattice
Smaller sodium cations fit into remaining gaps to maximize
Coulombic attraction
This is the so-called lattice energy
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Structure of CsCl Lattice
Count the number of anions (teal)
and cations (gold) in a unit cell.
What kind of lattice is formed by the anions
in this picture?
What type of “hole” is found in this lattice?
Note that this lattice is not body-centered cubic (bcc)! The lattice is named for
the anions only!
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Structures of ZnS Lattice
ZnS adopts 2 structures:
i) zinc blende (ccp for S2-)
ii) wurtzite (hcp for S2-).
What kind of holes are the cations filling?
Determine thier co-ordination number ?
Note that the two structures are very similar at the level of one ion?
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Structures of ZnS Lattice
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Ionic Lattices
ZnS
Anion radius (r -)
Cation radius (r +)
184 pm
75 pm
NaCl
CsCl
181 pm
99 pm
181 pm
169 pm
6
6
8
8
Co-ordination #
anions
cations
4
4
Anion lattice type
FCC
FCC
SC
Holes for cations
Tetra.
Octa.
Cubic
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Ionic Lattices
ZnS
Anion radius (r -)
Cation radius (r +)
r+/r-
NaCl
184 pm
75 pm
181 pm
99 pm
0.408
0.547
CsCl
181 pm
169 pm
0.934
The radius ratio indicates which packing is most stable:
If r+/r – is between 0.225 and 0.414, we get a structure like ZnS
If r+/r – is between 0.414 and 0.732, we get a structure like NaCl
If r+/r – is above 0.732, we get a structure like CsCl
These threshold number can be derived geometrically.
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