Cambridge AS & A Level Chemistry 9701

Notes
19 Lattice Energy
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Defining Lattice Energy

Lattice energy

• Lattice energy (ΔH°latt) is the enthalpy change when 1 mole of an ionic compound
is formed from its gaseous ions under standard conditions.
• Lattice energy arises from the electrostatic force of attraction of oppositely charged
ions when the crystalline lattice is formed.
• Lattice energy is always exothermic: the value of ΔH°latt is always negative.
• The large exothermic value of the lattice energy shows that the ionic lattice is very
stable with respect to its gaseous ions.
• The more exothermic the lattice energy, the stronger the ionic bonding in the
lattice.
• Examples:
Na+ (g) + Cl−(g) → NaCl(s) ∆H°latt = −787kJmol−1
Mg 2+ (g) + 2Cl−(g) → Mg𝐶𝑙2 (s) ∆H°latt = −2526kJmol−1

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Enthalpy Change of Atomisation and Electron Affinity

Enthalpy change of atomisation

• The standard enthalpy change of atomization (ΔH°at) is the enthalpy change when
1 mole of gaseous atoms is formed from its element under standard conditions.
• Values of ΔH°at are always positive (endothermic) because energy must be
supplied to break the bonds holding the atoms in the element together.
• Examples:
Li(s) → Li(𝑔) ∆H°at = +161kJmol−1
1
𝐶𝑙 (g)
2 2
→ Cl(𝑔) ∆H°at = +122kJmol−1

Electron affinity

• Electron affinity (ΔH°ea): energy change occurring when a gaseous non-metal atom accepts one electron.
• The first electron affinity, ΔH°ea1, is the enthalpy change when 1 mole of electrons is added to 1 mole of
gaseous atoms to form 1 mole of gaseous 1– ions under standard conditions
• The enthalpy change for the first electron affinity, ΔH°eal1 is generally exothermic: ΔH°ea is negative.
• When an element forms an ion with more than one negative charge, successive electron affinities are used.
• The second electron affinity, ΔH°ea2, is the enthalpy change when 1 mole of electrons is added to 1 mole of
gaseous 1– ions to form 1 mole of gaseous 2– ions under standard conditions.
• 2nd electron affinities are always endothermic (ΔH°ea2 is positive), and so are 3rd electron affinities.
• The overall enthalpy change is found by adding together the 1st and 2nd electron affinities.
• For example, equations of oxygen:
1st electron affinity:
O(g) + e− → O−(g) EA1 = −141 kJ mol−1
2nd electron affinity:
O−(g) + e− → O2−(g) EA2 = +798 kJ mol−1
The overall enthalpy change in forming an oxide ion, O2−, from an oxygen atom is found by adding together
the first and second electron affinities:
O(g) + 2e− → O2−(g) EA1 + EA2
= (−141) + (+798)
= +657 kJ mol−1

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Enthalpy Change of Atomisation and Electron Affinity

Trends in electron affinities

• Electron affinities are less negative (less exothermic) down the group, apart from
the first member in the group.
• The value of the first electron affinity depends on the attraction between the added
electron and the positively charged nucleus.
• The stronger the attraction, the greater the amount of energy released.
• The greater the nuclear charge, the greater the attractive force between the
nucleus and the outer electrons. So chlorine, with a greater nuclear charge than
sulfur, will tend to attract an electron more readily. This means that more energy is
released when a chlorine atom gains an electron.
• The further away the outer shell electrons are from the positive nuclear charge, the
less the attractive force between the nucleus and the outer shell electrons is. Since
the number of electron shells (and the atomic radius) increases down Groups 16
and 17, the electron affinity decreases going from chlorine to bromine to iodine.
• The greater the number of electron shells, the greater the power of inner shell
electrons to shield the outer shell electrons from the nuclear charge. This also
helps to decrease the electron affinity as you go from chlorine to iodine.

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Born – Haber Cycles

Components of the Born – Haber cycle

• A Born–Haber cycle is a particular type of enthalpy cycle used to calculate lattice

energy. It can be represented by:

Where ΔH°1 + ΔH°latt = ΔH°f

So, ΔH°latt = ΔH°f – ΔH°1
• The lattice energy of a compound can be determined if these are known:
- its enthalpy change of formation, ΔH°f
- the enthalpy changes involved in changing the element from their standard states
to their gaseous ions, ΔH°1
• When constructing Born–Haber cycle, remember:
- atomise metal →
ionise metal →
atomise non-metal →
ionise non-metal →
- for ionising metals you use ionisation energies and for ionising non-metals you use
electron affinities.
• The enthalpy change ΔH°1 involves several steps.

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Born – Haber Cycles

Calculating lattice energy

For example, the relevant enthalpy cycle of lithium fluoride:

Step 1 Convert solid lithium to gaseous lithium atoms

Li(s) → Li(g) ΔH⦵at = +161 kJ mol−1
Step 2 Convert gaseous lithium atoms to gaseous
lithium ions
Li(s) → Li(g) + e−   IE1 = +520 kJ mol−1
Step 3 Convert fluorine molecules to fluorine atoms
1
2
F2(g) → F(g) ΔH⦵at = +79 kJ mol−1

Step 4 Convert gaseous fluorine atoms to gaseous

fluoride ions
F(g) + e− → F−(g) EA1 = −328 kJ mol−1
Step 5 By adding all these values together, the value
of ΔH°1 is –617 kJ mol–1.

Applying Hess’s law to find the lattice energy of lithium fluoride:

ΔH⦵latt = ΔHf⦵ − ΔH1⦵

It is known that:

ΔH⦵1 = ΔH⦵at [Li] + IE1 [Li] + ΔH⦵at [F] + EA1 [F]

Putting in the figures:
ΔH⦵1 = (+161) + (+520) + (+79) + (−328) = +432
ΔH⦵latt = (−617) − (+432) = −1049 kJ mol−1

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Born – Haber Cycles

Born – Haber cycle as an energy level diagram

For example, the relevant enthalpy cycle of lithium fluoride:

Born–Haber cycle as an energy level diagram is the

best, and clearest, type of diagram for a Born–Haber
cycle.
To draw the cycle:
- start by putting down the elements in their
standard state on the left-hand side
- add the other enthalpy changes in the order
of steps 1 to
- complete the cycle by adding the enthalpy
change of formation and lattice energy
Note that the arrows going upwards represent an
increase in energy (ΔH° is positive) and the arrows
going downwards represent a decrease in energy
(ΔH° is negative).

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Born – Haber Cycles

Born – Haber cycle as an energy level diagram

For example, the relevant enthalpy cycle of magnesium chloride:

1. The magnesium ion is Mg2+, so the first and the

second ionisation energies need to be taken into
account:
Mg(g) → Mg+(g) + e− IE1 = +736 kJ mol−1
Mg+(g) → Mg2+(g) + e− IE2 = +1450 kJ mol−1
2. There are two chloride ions in MgCl2. 2 must
multiply the values of the enthalpy change of
atomisation and the first electron affinity of
chlorine.
Cl2(g) → 2Cl(g) 2ΔH⦵at = 2 × (+122)
= +244 kJ mol−1
2Cl(g) + 2e− → 2Cl−(g) 2EA1 = 2 × (−348)
= −696 kJ mol−1
In order to calculate the lattice energy, we need some
additional information:
ΔH⦵f [MgCl2] = −641 kJ mol−1

ΔH⦵at [Mg] = +148 kJ mol−1

According to Hess’s law:

ΔH⦵latt = ΔH⦵f − {ΔH⦵at [Mg] + IE1 [Mg] + IE2 [Mg] +

2ΔH⦵at [Cl] + 2EA1 [Cl]}

ΔH⦵latt = (−641) − {(+148) + (+736) + (+1450) + 2 ×

(+122) + 2 × (−348)}
ΔH⦵latt = (−641) − (+1882) = −2523 kJ mol−1

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Factors Affecting the Value of Lattice Energy

The factors
Factor Explanation
• As the size of the ion increases, the lattice energy becomes less exothermic. This applies
to both anions and cations.
• Figure below shows that:

- for any given anion, e.g. F–, the

lattice energy gets less exothermic
as the size of the cation increases
Ion size
from Li+ to Cs+
- for any given cation, e.g. Li+, the
lattice energy gets less exothermic
as the size of the anion increases
from F– to I–

• As the ionic charge increases, the lattice energy becomes more exothermic.
• For example, by comparing lithium fluoride, LiF, with magnesium oxide, MgO.
- These compounds have the same arrangement of ions in their lattice structure, the
cations Li+ and Mg2+ have similar sizes, and the anions F – and O2– are similar in
size.
Charge on - The major physical difference is the ionic charge: ΔH°latt [LiF] = –1049 kJ mol–1 ΔH°latt
the ions [MgO] = –3923 kJ mol–1
- Magnesium oxide has a greater lattice energy than lithium fluoride because Mg2+
and O2– ions in magnesium oxide attract each other stronger than the singly charged
ions of the same size in LiF.
• For ions of similar size, the greater the ionic charge, the higher the charge density and
results in stronger ionic bonds being formed.

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Ion Polarisation

Ion polarisation

• In some cases, the positive charge on the cation in an ionic lattice may attract the
electrons in the anion towards it, which results in a distortion of the electron cloud
of the anion and the anion is no longer spherical called ion polarisation.
• The ability of a cation to attract electrons and distort an anion is called the
polarising power of the cation.

Factor affecting ion polarisation

The degree of polarisation of an anion depends on: An anion is more likely to be polarised if:
- the charge density of the cation - the cation is small
- the ease with which the anion can be - the cation has a charge of 2+ or 3+
polarised – its polarisability. - the anion is large
- the anion has a charge of 2– or 3–

The thermal stability of Goup 2 carbonates and nitrates

Enthalpy change This trend can be explained using ideas about ion
Group 2 Decomposition
of atomization
carbonates temperature (°C)
(kJ mol-1)
polarisation:
magnesium 540 +117 - the carbonate ion has a relatively large ionic
carbonate
calcium 900 +176 radius so it is easily polarised by a small highly
carbonate charged cation
strontium 1280 +238 - the Group 2 cations increase in ionic radius
carbonate down the group: Mg2+ < Ca2+ < Sr2+ < Ba2+
barium 1360 +268 - the smaller the ionic radius of the cation, the
carbonate
better it is at polarising the carbonate ion
- the degree of polarisation of the carbonate ion
The relative stabilities of these carbonates by the Group 2 cation follows the same order
increases down the group in the order: - the greater the polarisation of the carbonate
ion, the easier it is to weaken a carbon–oxygen
BaCO3 > SrCO3 > CaCO3 > MgCO3. bond in the carbonate and form carbon dioxide
and the oxide on heating

A similar pattern is observed with the thermal decomposition of Group 2 nitrates. The order of stability with
respect to the products is in the order: Ba(NO3)2 > Sr(NO3)2 > Ca(NO3)2 > Mg(NO3)2

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Enthalpy Changes in Solution

Enthalpy change of a solution

• The enthalpy change of solution (ΔH°sol) is the energy absorbed or released when 1
mole of an ionic solid dissolves in sufficient water to form a very dilute solution.
• ΔH°sol can be positive (endothermic) or negative (exothermic).
• A compound is likely to be soluble in water only if ΔH°sol is negative or has a small
positive value; substances with large positive values of ΔH°sol are relatively insoluble.
• For example, the enthalpy changes of solution for magnesium chloride and sodium
chloride are described by the equations below:
MgCl2(s) + aq → MgCl2(aq) ΔH⦵sol = −55 kJ mol−1
or
MgCl2(s) + aq → Mg2+(aq) + 2Cl−(aq) ΔH⦵sol = −55 kJ mol−1
NaCl(s) + aq → NaCl(aq) ΔH⦵sol = +3.9 kJ mol−1
or
NaCl(s) + aq → Na+(aq) + Cl−(aq) ΔH⦵sol = +3.9 kJ mol−1

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Enthalpy Changes in Solution

Enthalpy change of hydration

• The enthalpy change of hydration, ΔH°hyd, is the enthalpy change when 1 mole of a
specified gaseous ion dissolves in sufficient water to form a very dilute solution.
• The enthalpy change of hydration is always exothermic.
• The value of ΔH°hyd is more exothermic for ions with the same charge but smaller ionic radii.
• The value of ΔH°hyd is more exothermic for ions with the same radii but a larger charge.
• For example, the enthalpy changes of hydration for calcium ions and chloride ions
are described by the equations below:
Ca2+(g) + aq → Ca2+(aq) ΔH⦵hyd = −1650 kJ mol−1
Cl−(g) + aq → Cl−(aq) ΔH⦵hyd = −364 kJ mol−1

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Enthalpy Changes in Solution

Calculating enthalpy changes in solution

The enthalpy change of solution or the enthalpy change of hydration can be calculated
by constructing an enthalpy cycle and using Hess’s law:

ΔH°latt + ΔH°sol = ΔH°hyd

Worked example:

Determine the enthalpy change of solution of ΔH°sol [NaF] = ΔH°hyd a– ΔH°latt

sodium fluoride using the following data:
= (–406) + (–506) – (–902) = –912 + 902
–1
lattice energy = –902 kJ mol
= –10kJmol–1
–1
ΔH°sol of sodium ions = –406 kJ mol
ΔH°hyd of fluoride ions = –506 kJ mol–1

ΔH°latt + ΔH°sol = ΔH°hyd

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Enthalpy Changes in Solution

The solubility of Group 2 sulfates

The solubility decreases as the radius of the metal ion increases:
Compound Solubility (mol dm-3)
magnesium sulfate 1.83
calcium sulfate 4.66 × 10−2
strontium sulfate 7.11 × 10−4
barium sulfate 9.43 × 10−6

Change in hydration enthalpy down the group Change in lattice energy down the group

• Smaller ions (with the same charge) have • Lattice energy is greater if the ions (with the
greater enthalpy changes of hydration same charge) forming the lattice are small
• The enthalpy change of hydration decreases • The lattice energy decreases in the order:
(gets less exothermic) in the order: Mg2+ > Ca2+ > Sr2+ > Ba2+
Mg2+ > Ca2+ > Sr2+ > Ba2+ • The lattice energy is inversely proportional to the
• This decrease is relatively large down the group sum of the radii of the anion and cation
and it depends entirely on the increase in the • The sulfate ion is larger than the group 2 cations
size of the cation, as the anion is unchanged (it • The decrease in lattice energy is relatively
is the sulfate ion in every case) smaller down the group and it is determined
more by the size of the large sulfate ion than the
size of the cations

Difference in enthalpy change of solution of Group 2 sulfates

• The higher the positive value of ΔH°sol the less soluble the salt
• The lattice energy of the sulfates decreases (gets less exothermic) by relatively smaller values down
the group
• The enthalpy change of hydration decreases (gets less exothermic) by relatively larger values down the
group
• By applying Hess’s law, the value of ΔH°sol gets more endothermic down the group
• The solubility of the Group 2 sulfates decreases down the group

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