Download IONIC BONDS CHAPTER 6

IONIC BONDS
CHAPTER 6
(Reminder: These are only brief notes. You must read the Chapter)
The FIRST IONIZATION ENERGY is defined as the energy required to remove the least tightly held
electron from from one mole of gaseous atoms as defined by the reaction (where M represents an element).
M(g) → M+(g) + e¯
The SECOND IONIZATION ENERGY is defined as the energy required to remove the least tightly held
electron from from one mole of gaseous monopositive ions as defined by the reaction (where M represents
an element).
M+(g) → M2+(g) + e¯
The THIRD IONIZATION ENERGY is defined as the energy required to remove the least tightly held
electron from from one mole of gaseous dipositive ions as defined by the reaction (where M represents an
element).
M2+(g) → M3+(g) + e¯
First Ionization Energies for the Elements Z= 1 to 20
First Ionization Energies
6.1
The Successive Ionization Energies for the Third Row Elements
The successive ionization energies for sodium
Na
Sucessive Ionization Energies
Lg(kJ per mole)
5.5
5
4.5
4
3.5
3
2.5
1
2
3
4
5
6
7
8
9
10
11
Number of electrons removed
Electron Affinity
We know that chlorine forms the ion Cl– in simple ionic compounds like NaCl(s). The first
electron affinity, Ea, is defined by the equation:
Cl(g) + e− → Cl−(g) + 348.6 kJ mol−1 of energy released
Only the a few elements form the anion with the release of a significant amount of energy.
It requires a great deal of energy to remove one electron from a mole of gaseous Na atoms and
the removal of two looks very unlikely IN THE GAS PHASE! Combining with Cl does not seem to
be very favourable as energy would be absorbed.
Cl(g) + e− → Cl−(g) + 348.6 kJ mol−1 of energy released
Na(g) + 496 kJ mol-1 of energy → Na+(g) + e−
(Na+(g) + 4600 kJ mol-1 of energy → Na2+(g) + e−)
6.2
The process that we have imagined above is not much like what happens when NaCl(s) forms
from chlorine, Cl2(g) and sodium, Na(s), since neither the reactants or products are gaseous
atoms.
So why does NaCl form?!
The chlorine that we nomally use is dichlorine, Cl2, however we can try to use these indications to
explain the formation of NaCl(s)
Ionic Bonding - the process
2 Na(s) + Cl2(g) → 2 NaCl(s) + heat
Na(s) + ½ Cl2(g) → NaCl(s) + heat
In the reaction of the gaseous Na atom with gaseous Cl this is what happens:
Na
2
1s 2s22p6 3s1
Cl
1s2 2s22p6 3s13p5
+ e−
→
Na+
1s2 2s22p6
The same as neon
+ e−
Cl−
1s2 2s22p6 3s13p6
The same as argon
Both the Na and the Cl atom attain the same electron configuration as (they are
ISOELECTRONIC with) a nearby noble gas (Why?). (NB. The ions are very reactive, charged and
are not noble gases!). In the real solid NaCl the ions are stabilized by their strong interaction to
give what we call IONIC BONDS. The stabilization energy is called the Lattice Energy.
6.3
Now we can see what the requirements are for the formation of ionic compounds and perhaps
why they are solids at room temperature.
6.4
The Octet Rule (Law)
The s- and p-block elements undergo reactions in which they gain or lose electrons such that they
form ions with the same number of electron as the nearest noble gas.
We have already seen why this happens in some cases. Like all good rules it is frequently
broken. Just look at the fluorides of the second row - count the electrons:
LiF
NaF
MgF2 AlF3
BeF2 BF3
SiF4
CF4
PF3 & PF5
NF3
OF2
F2
SF2, SF4 & SF6
ClF & ClF3
But what about PF5, it breaks the law as it seems to have too many electrons.
Expansion of the valence shell
The third row and beyond elements all have d-levels that are close to the filling s and p levels and
these can be used for bonding.
If we look at phosphorus in PCl3 and PCl5
Ground state atom P 1s2 2s22p6 3s23p33d0
Ground state in PCl3 1s2 2s22p6 3s23p63d0 Octet rule satisfied
Ground state in PCl5 1s2 2s22p6 3s23p63d2 Expansion of the valence shell (10 e) and the octet rule
is violated. This is not possible for the second period elements since there are no 2 d orbitals.
Nitrogen does form NCl3 but not NCl5.
d-Block Cations
The ions that are formed by the d-block elements are more difficult to predict. The valence
s-electrons are lost first and then d-electrons. Scandium 3d1 4s2 forms the Sc3+ ion very much
like aluminum forms the Al3+ ion and with similar properties. Titanium gives both +2 and +4
oxidation states. As you know, chromium gives +2, +3, +6 in common compounds. The
compound/ions with the lowest oxidation numbers are the most ionic and the high oxidation
numbers are in covalent molecules or ions like CrO42-.
The elements form ions only when the energy required is low and compensated for by the Lattice
Energy. When the energy required is high bonds are formed by electron sharing rather than by
ionization to give Covalent Bonds.
Paramagnetism and Diamagnetism
How can we tell whether an atom has unpaired electrons?
We have seen that electrons have “spin”. The spin of each electron is associated with a magnetic
field. Where electrons are spin-paired the fields cancel (the spins are in opposite directions).
However, unpaired electrons will present a magnetic effect and such atoms are PARAMAGNETIC
and pulled into a magnetic field. The more unpaired electrons the greater the effect to the extent
that the number of unpaired electrons can be often be determined. Ferromagnetism (as in iron) is
an extreme and specialized example.
If all of the electrons are paired then the atoms are DIAMAGNETIC and repelled from a magnetic
field. These properties can be used to study electron configuration.
6.5