Notes on Ionic Bonding
Electrolytes
compounds which when dissolved in water produce solutions that conduct an electric current.
Nonelectrolytes
compounds which when dissolved in water produce solutions that do not conduct an electric current.
Strong electrolytes vs weak electrolytes--some
compounds conduct a much stronger current than
others.
General rules to follow to determine if the
compound is a strong or weak electrolyte.
The following acids are considered strong acids and are also
strong electrolytes: HCl, HBr, HI, HNO3, H2SO4, and HClO4.
The rest of the acids are considered weak acids and are
also weak electrolytes.
The hydroxides of Groups I and II are considered strong bases
and are also strong electrolytes. The rest of the bases are
considered as weak bases and are weak electrolytes.
Most other ionic compounds are strong electrolytes.
Halides and cyanides of "heavy metals" are generally weak
electrolytes. An example is PbI2.
Most organic compounds are nonelectrolytes. Exceptions are
organic acids and bases.
The difference between strong and weak electrolytes is the
extent to which the ionic compounds dissociate into ions when
placed in water. The greater the amount of dissociation, the
greater the electrical conductance of the solution. The strong
electrolytes are usually considered to be 100% dissociated,
especially in dilute solutions, and weak electrolytes are usually
dissociated <10%. The quantitative values of conductance is
given for 0.10 M aqueous solutions at 25°C, and is in
seimens.cm2/mol.
Ionic Bond
bond formed between oppositely charged ions because of
electrostatic forces of attraction.
example: sodium + chlorine ----> sodium chloride
Ionic compound
a compound composed of ions.
Metal atoms lose electrons to form positive ions, called cations,
which are smaller than the original atom.
The loss of an energy level with electrons, plus the excess
positive charge draws the remaining electrons toward the
nucleus and causes the electron cloud to contract even
more.
Nonmetal atoms gain electrons to form negative ions, called
anions, which are larger than the original atom.
The addition of an extra electron increases the electronelectron repulsion and causes the electron cloud to expand.
The main Group elements (s and p orbitals) lose or gain
electrons to attain a configuration like a noble gas.
Transition elements (d orbitals) lose their s orbital electrons first
and then one or more d orbital electron(s).
H for the formation of an ionic compound
Ionic bonding can be broken down into 3 steps: ex. sodium
chloride
Formation of the sodium ions (ionization) Na(g) + energy ----> Na+(g) (H = +496 kJ/mol) Formation of the chloride ions (electron affinity) Cl(g) + e- ----> Cl-(g) + energy (H = -348 kJ/mol) Formation of the ion pair (bond energy) Na+(g) + Cl-(g) ----->
NaCl(g) + energy (H = -504 kJ/mol) According to Hess's Law, the H for the formation of one mole of
sodium chloride ion-pairs from one mole of sodium atoms and
one mole of chlorine atoms is Hrxn = H1 + H2 + H3 = -356
kJ/mol
Coulomb's Law is used to calculate the energy in
forming the ion-pair. It says that the energy of
interaction of two ions is directly proportional to
the product of their electrical charges and is
inversely proportional to the distance between
their centers.
E = kZ1Z2/deq
Z1 & Z2 are charges of the 2 ions
deq = distance between their centers
k = proportionality constant (2.31 x 10-16J pm / ion-pair )
NOTE:
All ionic bonding is exothermic.
In the real world ionic compounds don't actually exist as ion-pairs
which were formed from naturally occurring gaseous atoms.
Instead, they exist as crystalline solids consisting of many ions,
each cation being surrounded by anions and each anion being
surrounding by cations to form a crystal lattice. Each cationanion pair represents the release of energy.
Therefore, the reaction of the solid sodium atoms and the
diatomic chlorine gas molecule to form the ionic solid sodium
chloride can be broken down into 5 steps.
Vaporize the sodium atoms into gaseous sodium atoms. (use
heat of vapaorization) Na(s) -----> Na(g) (H = +93
kJ/mol) Dissociate the chlorine molecules into chlorine atoms. (use
bond energy) 1/2Cl2(g) -----> Cl(g) (H = +122 kJ/mol)
Ionize the sodium atoms. This is the oxidation step. (use
ionization energy) Na(g) -----> Na+(g) + e- (H = +496
kJ/mol) Add electron to chlorine atoms to form chloride ions. This is the
reduction step. (use electron affinity) Cl(g) + e- -----> Cl-(g)
H = -348 kJ/mol)
Formation of the crystal lattice (use lattice energy) Na+(g) +
Cl-(g) -----> NaCl(s) (H = -780 kJ/mol) The overall change in energy for the reaction, Hrxn = 417 kJ/mol
Born-Haber Cycle- a complete cycle showing all the steps
in a process.