Lecture 8: Ionic Compounds
Dr Harris
9/11/12
HW: Ch 6: 4, 25, 29, 51, 58, 69
Ionic Compounds
• The nucleus of an atom is unchanged by chemical reactions (number of
protons never changes)
• However, electrons are readily added and lost and ions are formed
• When a metal reacts with a nonmetal, ions form and interact. The
result is an ionic compound.
• Let’s consider the formation of a very common ionic compound, NaCl (s)
Forming Ionic Compounds
• We know that Na(s) and Cl2(g) react together
to form NaCl (s), but how?
• The most important thing to know about
chemical reactions is that atoms
undergoing a reaction will always seek to
reach a noble gas configuration
• Let’s look at the electron configurations of Na
and Cl
Forming Ionic Compounds
Na: [Ne] 3s1
Cl: [Ne] 3s2 3p5
• For Na, the nearest noble gas is Ne. To reach the Ne configuration, it
needs to lose a single electron.
Na ( [Ne] 3s1 ) ---> Na+ ([Ne]) + e11 e-
11
p+
Na atom
1st Ionization Energy
10 e11 p+
Na+ cation
Forming Ionic Compounds
Na: [Ne] 3s1
Cl: [Ne] 3s2 3p5
• For Cl, the nearest noble gas is Ar. To reach the Ar configuration, it
needs to gain a single electron.
Cl ([Ne] 3s2 3p5) + e- ---> Cl- ([Ar])
• Unlike ionization energy, which describes the energy input needed
to extract an electron, electron affinity describes the energy
released an electron is added.
18 e17 e1st Electron Affinity
17 p+
Cl atom
17 p+
Cl- anion
Forming Ionic Compounds
• Na and Cl can simultaneously achieve a noble gas configuration if an
electron is transferred from the metal (Na) to the nonmetal (Cl)
Na ( [Ne] 3s1) + Cl ( [Ne] 3s2 3p5) ---> Na+ Cl[Ne]
[Ar]
IONIC COMPOUND
Lewis dot structure of the product
Na+ Cl-
Predicting Charge
• So now, we understand that ionic compounds form when metal and
nonmetal ions interact
• We also see why sodium chloride is NaCl, not NaCl2 or Na2Cl, etc.
• The overall charge of any molecule must be zero.
• Since the Na loses an electron to become Na+, and Cl gains an
electron to become Cl-, only one of each ion is needed to balance
the charge.
• In ionic compounds, the metal is always positively charged (cation)
and the nonmetal is always negatively charged (anion)
Predicting Charge
metals
nonmetals
1+
2+
3+
3-
2-
1-
Group Examples
• Write the chemical formulas and Lewis structures of the following ionic
compounds:
• Calcium oxide
• Magnesium Chloride
• Sodium Sulfide
• Determine the ionic product and balance:
• Mg + O2  ?
• Na + N2  ?
• Show the electron transfer process in the formation of calcium oxide
using the noble gas electron configuration, as was shown for NaCl
Dissociation
• Ionic compounds completely dissociate in water, forming individual ions.
Ions become completely ‘hydrated’.
Na+
Cl-
H2O (L)
Na+(aq) + Cl- (aq)
• Here, NaCl is the solute, water is the solvent
Dissociation of a Salt In Water
Na+
Cl• Water molecules “solvate”
ionic compounds, ripping
the ions apart.
• The negative oxygen atoms
(red) attracted to the
positive Na+, and the
positive hydrogens are
attracted to the negative
Cl-
Conducting Electric Current
• Ions in solution are capable of conducting electric current (hence, the
term electrolyte). Ions are able to transport charge across the water.
• Non-ionic solutions (covalent) do not exhibit this property because
they do not dissociate
Ion Size Depends on Charge
• As you may have noticed in my drawings in previous slides, cations
tend to be smaller than their neutral atom counterparts, and anions
seem to be larger
Neutral X
Cation, X+
Anion, X-
• Anions have larger electron clouds because the excess negative charge
causes repulsion, which leads to expansion of the electron cloud. The
excess positive charge in cations draws the electron cloud closer to the
nucleus
Ionic Radii
Energy Is Absorbed or Released When an Ionic
Compound Forms
• The electrostatic attraction, or the electrical attraction between
positive and negative ions, is what holds an ionic compound together
• When two ions form an ionic compound, there is an overall change in
energy.
• We can calculate this energy by considering:
• the ionization energy of the metal
• the electron affinity of the nonmetal
• the coulombic energy of attraction between the cation and anion
Calculating The Energy Change Due To
Formation of Ionic Compounds
• A chemical reaction is considered favorable if the energy of the product
is less than that of the reactants. In other words, the change in energy
is negative.
• Lets revisit the reaction: Na(g) + Cl(g)  NaCl(s)
• Ignore the monatomic chlorine
• To form NaCl, there are 3 steps
1. Form Na+ (ionization energy)
2. Form Cl- (electron affinity)
3. Join them together (coulombic energy)
Calculating The Energy Change Due To
Formation of Ionic Compounds
1.
(Ionization of Na)
Na(g)  Na+(g) + e*Positive sign means energy is added.
ΔEI = +0.824 aJ
2.
Cl(g) + e-  Cl-(g)
*Negative sign means energy is released.
ΔEEA = -0.580 aJ
3.
(Ionization of Cl)
(Coulombic energy)
Na+(g) + Cl-(g)  Na+Cl-(s)
Now we must calculate the coulombic energy
?
Coulombic Energy
• The third step is to join the two atoms, as shown below.
rNa = 102 pm
rCl = 181 pm
• The equation shown above is Coulomb’s Law, which gives the
coulombic energy change (Ec) that results when two ions come together.
•
•
•
Q1 and Q2 are the charges of the metal and nonmetal
d is the distance between the nuclei. This is the sum of the ionic radii.
k is a constant. (231 aJ•pm)
Solve
Negative energy change indicates a favorable process
ClNa+
Group Example
• Given the following data, calculate the energy of reaction to form CsCl
and Cs2O given that the first ionization energy of Cesium is 0.624 aJ
Electron Configurations of Transition Metal Ions
• When a transition metal forms an ion, electrons are first removed from
the preceding s-orbital.
Fe: [Ar] 4s2 3d6
Fe2+: [Ar] 3d6
Fe3+: [Ar] 3d5
• If the ionization of a transition metal results in an unpaired selectron, that electron will move into the valence d orbital
Ni: [Ar] 4s2 3d8
Ni+: [Ar] 4s1 3d8 ---> [Ar] 3d9
Roman Numeral Nomenclature for
Transition Metals
• Transition metals can have multiple positive ionic charges. To
distinguish, a roman numeral is placed in front of a transition metal
in a compound to identify its charge.
• Ex. FeCl2 ---> Here, Fe is 2+. So, we name this compound:
Iron (II) chloride
FeCl3 ---> Here, Fe is 3+.
Iron (III) chloride
• Name the following: TiO2, WCl6
Titanium (IV) oxide, Tungsten (VI) chloride`