Lecture 7

Suggested HW:
4, 9, 16, 24, 25a., 29, 51*,
52*, 60
*Use pg 186 for electron
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
• When a metal reacts with a nonmetal, ions form and attract.
The result is an ionic compound.
• Let’s consider the formation of a very common ionic
compound, NaCl (s)
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
• Let’s look at the electron
configurations of Na and Cl
Mechanism of an Ionic Reaction
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-
Neutral Na atom
1st Ionization Energy
10 e11 p+
Na+ cation
Mechanism of an Ionic Reaction
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])
• Electron affinity describes the energy released
when an electron is added.
17 e17 p+
1st Electron
18 e-
17 p+
Cl- anion
Neutral Cl atom
Mechanism of an Ionic Reaction
• Na and Cl can simultaneously achieve a noble gas configuration
if an electron is transferred from the metal (Na) to the nonmetal
[Ne] 3s1 + [Ne] 3s2 3p5 ---> Na+ Cl[Ne]
Lewis dot structure of the product
Na+ Cl-
Predicting and Balancing 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,
• The overall charge of any complete 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
Predicting and Balancing Charge
Group Examples
• Write the chemical formulas and Lewis structures of the
following ionic compounds:
– Calcium oxide
– Magnesium Chloride
– Sodium Sulfide
– Potassium Phosphide
• 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
Dissolving Ionic Compounds in Water
• Ionic compounds completely dissociate in water, forming
individual ions. Ions become completely ‘hydrated’.
H2O (L)
Na+(aq) + Cl- (aq)
• Here, NaCl is the solute, water is the solvent
Dissolving Ionic Compounds in Water
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-
• 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
Ionic Radii
• Cations tend to be smaller than their neutral atom
counterparts, and anions seem to be larger
+ e-
- e-
Neutral X
Cation, X+
Anion, X-
• Anions have large electron clouds because the excess of
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
Ionic Radii
Energy Changes In Reactions
• 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
Energy Changes In Reactions
• 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)
Energy Changes In Reactions
(Ionization of Na)
Na(g)  Na+(g) + e*Positive sign means energy is added.
ΔEI = +0.824 aJ
(Ionization of Cl)
Cl(g) + e-  Cl-(g)
*Negative sign means energy is released.
ΔEEA = -0.580 aJ
(Coulombic energy)
Na+(g) + Cl-(g)  Na+Cl-(s)
Coulombic Energy
• The third step is to join the two ions, as shown below.
rNa = 102 pm
rCl = 181 pm
• The equation shown above is Coulomb’s Law, which gives
the energy change (Ec) that results when two ions come
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)
• Negative energy change indicates a favorable process
• Given the following data, calculate the energy of reaction
to form CsCl given that the first ionization energy of
Cesium is 0.624 aJ
Electron Configurations of Transition Metals
• When a transition metal forms an ion, electrons are first
removed from the preceding s-orbital.
[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
Electron Configurations of 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

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