Lattice Energy and Electron Affinity

A2 · 17 min

Ionic solids such as sodium chloride and magnesium oxide are held together by the electrostatic attraction between huge numbers of oppositely charged ions. Lattice energy measures how strong that attraction is, and electron affinity measures how readily an atom accepts the electrons that turn it into an anion. Together with the enthalpy change of atomisation, these are the building blocks of every Born–Haber cycle and every enthalpy of solution calculation in Paper 4. The definitions are examined word for word, so learn them exactly.

Standard conditions and sign conventions

Every enthalpy change in this topic is a standard enthalpy change, written with the symbol $\ominus$ , for example $\Delta H^{\ominus}_{\text{at}}$ . Standard conditions are a pressure of $101\ \text{kPa}$ and a stated temperature, normally $298\ \text{K}$ , with every substance in its standard state (its normal physical state under those conditions).

A negative $\Delta H$ means energy is released to the surroundings (exothermic). A positive $\Delta H$ means energy is taken in (endothermic). For each definition below, ask yourself one question: is energy needed to make this happen, or is it released? That single question fixes the sign.

Enthalpy change of atomisation

To build an ionic lattice from its elements on paper, you first have to turn each element into separate gaseous atoms.

Definition

The standard enthalpy change of atomisation, $\Delta H^{\ominus}_{\text{at}}$ , is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.

The definition is per mole of gaseous atoms produced, not per mole of element used. That matters for diatomic elements:

\ce{Na(s) -> Na(g)} \qquad \Delta H^{\ominus}_{\text{at}} = +107\ \text{kJ mol}^{-1}

\ce{1/2Cl2(g) -> Cl(g)} \qquad \Delta H^{\ominus}_{\text{at}} = +121\ \text{kJ mol}^{-1}

\ce{1/2Br2(l) -> Br(g)} \qquad \Delta H^{\ominus}_{\text{at}} = +112\ \text{kJ mol}^{-1}

Atomisation is always endothermic. For a metal you must overcome metallic bonding; for a non-metal you must break covalent bonds (and, for bromine and iodine, overcome intermolecular forces as well, because their standard states are liquid and solid).

For a gaseous diatomic element such as chlorine, oxygen or hydrogen, the enthalpy change of atomisation is exactly half the bond energy, because breaking one mole of $\ce{Cl-Cl}$ bonds gives two moles of atoms:

\Delta H^{\ominus}_{\text{at}}(\ce{Cl}) = \tfrac{1}{2} E(\ce{Cl-Cl}) = \tfrac{1}{2}(242) = +121\ \text{kJ mol}^{-1}

This shortcut does not work for bromine or iodine: their standard states are not gases, so atomisation also includes vaporisation or sublimation.

Electron affinity

Ionisation energy (from AS) measures the energy needed to remove electrons and make cations. Electron affinity is the matching quantity for adding electrons to make anions.

Definition

The first electron affinity, $EA_1$ , is the enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous 1– ions under standard conditions.

The second electron affinity, $EA_2$ , is the enthalpy change when one mole of electrons is added to one mole of gaseous 1– ions to form one mole of gaseous 2– ions under standard conditions.

\ce{Cl(g) + e- -> Cl^-(g)} \qquad EA_1 = -349\ \text{kJ mol}^{-1}

\ce{O(g) + e- -> O^-(g)} \qquad EA_1 = -141\ \text{kJ mol}^{-1}

\ce{O^-(g) + e- -> O^{2-}(g)} \qquad EA_2 = +798\ \text{kJ mol}^{-1}

Everything must be gaseous. A common error is writing $\ce{1/2Cl2(g) + e- -> Cl^-(g)}$ : that combines atomisation and electron affinity and is not the definition.

Why the first electron affinity is usually negative

The incoming electron is attracted by the nucleus. Even in a neutral atom, the outer electrons do not completely shield the nucleus, so an electron arriving in the outer shell feels a net attraction. As it moves in, potential energy falls and energy is released, so $EA_1$ is exothermic for most non-metals.

Why the second electron affinity is always positive

The second electron is being added to an ion that is already negative. The repulsion between the incoming electron and the negative ion is larger than the attraction to the nucleus, so energy must be supplied to force the electron on. Every second electron affinity is endothermic.

The obvious question is why oxide ions exist at all, if forming them costs energy. The answer is the lattice energy. The very large energy released when $\ce{O^{2-}}$ ions pack together with cations in a solid more than pays for the endothermic second electron affinity. You will see this in Born–Haber cycles.

Factors affecting electron affinity

The size of the energy released when an electron is added depends on how strongly the nucleus attracts it. Three factors decide that:

Nuclear charge. More protons attract the incoming electron more strongly, so $EA_1$ becomes more exothermic.
Atomic radius (distance). The further the outer shell is from the nucleus, the weaker the attraction for the incoming electron, so $EA_1$ becomes less exothermic.
Shielding. More inner shells of electrons between the nucleus and the outer shell reduce the effective attraction, so $EA_1$ becomes less exothermic.

Across a period, nuclear charge increases while shielding stays roughly the same, so electron affinity generally becomes more exothermic (Group 17 elements have the most exothermic values in each period). Down a group, the extra shells increase the radius and the shielding, which outweighs the increase in nuclear charge.

Electron affinities of Groups 16 and 17

Group 17	$EA_1$ / kJ mol⁻¹	Group 16	$EA_1$ / kJ mol⁻¹
F	–328	O	–141
Cl	–349	S	–200
Br	–325	Se	–195
I	–295	Te	–190

From chlorine downwards (and sulfur downwards), $EA_1$ becomes less exothermic. The first member of each group (F and O) is less exothermic than the second (Cl and S).

The fluorine and oxygen anomaly

On radius alone, fluorine should have the most exothermic electron affinity in Group 17. It does not, because the fluorine atom is very small. Its outer electrons are crowded into the compact $n = 2$ shell, so an incoming electron is repelled strongly by the electrons already there. This electron–electron repulsion partly cancels the strong nuclear attraction, so less energy is released than for chlorine, whose larger $n = 3$ shell gives the incoming electron more room.

Oxygen and sulfur in Group 16 follow exactly the same pattern for the same reason.

Watch out

Do not say "fluorine has a smaller nuclear charge" or "fluorine is less electronegative". Fluorine is the most electronegative element. The correct explanation is repulsion between the incoming electron and the existing electrons in the small, crowded 2p subshell of fluorine.

Group 16 values are less exothermic than Group 17 values in the same period because a Group 16 atom has one fewer proton for the same shielding, so it attracts the incoming electron less strongly.

Lattice energy

Definition

The lattice energy, $\Delta H^{\ominus}_{\text{latt}}$ , is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

Cambridge defines lattice energy in the formation direction: from gas phase ions to solid lattice.

\ce{Na^+(g) + Cl^-(g) -> NaCl(s)} \qquad \Delta H^{\ominus}_{\text{latt}} = -787\ \text{kJ mol}^{-1}

\ce{Mg^{2+}(g) + 2Cl^-(g) -> MgCl2(s)} \qquad \Delta H^{\ominus}_{\text{latt}} = -2526\ \text{kJ mol}^{-1}

Bringing oppositely charged ions together from infinite separation releases energy, so lattice energies are always negative (exothermic). The more negative the value, the stronger the ionic bonding in the lattice.

Lattice energy cannot be measured directly: you cannot make a mole of isolated gaseous ions and let them crash together in a calorimeter. It is calculated indirectly using a Born–Haber cycle and Hess's law.

Tip

Some textbooks and websites define lattice energy as the energy needed to break the lattice (a positive value). Cambridge 9701 uses the formation definition, so lattice energies in this course are negative. If a question gives a positive "lattice dissociation enthalpy", reverse the sign before using it.

What controls the size of a lattice energy

The attraction between two ions depends on their charges and on how close their centres can get. Qualitatively,

\left|\Delta H^{\ominus}_{\text{latt}}\right| \;\propto\; \frac{(\text{charge on cation}) \times (\text{charge on anion})}{\text{sum of the ionic radii}}

You do not need to use this as an equation; you need to use the two ideas it contains.

Factors affecting the magnitude of lattice energy

Ionic charge. Higher charges on the ions give stronger electrostatic attraction, so the lattice energy is more exothermic (larger in magnitude).
Ionic radius. Smaller ions can approach more closely, so the attraction between their centres is stronger and the lattice energy is more exothermic.

Charge usually has the bigger effect: doubling both charges multiplies the product of the charges by four.

The data show both effects clearly.

Compound	Ionic radii / nm	$\Delta H^{\ominus}_{\text{latt}}$ / kJ mol⁻¹
NaF	0.102 + 0.133	–923
NaCl	0.102 + 0.181	–787
NaBr	0.102 + 0.196	–747
NaI	0.102 + 0.220	–704
MgO	0.072 + 0.140	–3791
CaO	0.100 + 0.140	–3401
SrO	0.118 + 0.140	–3223
BaO	0.135 + 0.140	–3054

Down the sodium halides, the anion gets larger, the inter-ionic distance increases, and the lattice energy becomes less exothermic. Down the Group 2 oxides, the cation gets larger with the same result.

Now compare NaF and MgO. Their inter-ionic distances are similar (0.235 nm and 0.212 nm), yet the lattice energy of MgO is about four times larger. The product of the charges is $1 \times 1 = 1$ for NaF but $2 \times 2 = 4$ for MgO. Charge dominates.

Watch out

Lattice energy depends on ionic radius, not atomic radius. A sodium ion (0.102 nm) is much smaller than a sodium atom (0.186 nm) because it has lost its outer shell. When you explain a trend, talk about "the radius of the ions" or "the distance between the centres of the ions".

Worked examples

Writing defining equations

Write equations, with state symbols, for each of the following.

(a) The enthalpy change of atomisation of iodine. (b) The second electron affinity of sulfur. (c) The lattice energy of calcium chloride.

Solution

(a) Iodine is a solid in its standard state, and the definition is per mole of atoms formed:

\ce{1/2I2(s) -> I(g)}

(b) The second electron affinity starts from the gaseous 1– ion:

\ce{S^-(g) + e- -> S^{2-}(g)}

\ce{Ca^{2+}(g) + 2Cl^-(g) -> CaCl2(s)}

Each equation earns its mark only with every state symbol correct and exactly one mole of the thing being defined.

Atomisation from bond energy

The bond energy of $\ce{O=O}$ is $496\ \text{kJ mol}^{-1}$ and that of $\ce{Br-Br}$ is $193\ \text{kJ mol}^{-1}$ .

(a) Calculate $\Delta H^{\ominus}_{\text{at}}$ of oxygen. (b) Explain why $\Delta H^{\ominus}_{\text{at}}$ of bromine ( $+112\ \text{kJ mol}^{-1}$ ) is not equal to half the $\ce{Br-Br}$ bond energy.

Solution

(a) Breaking one mole of $\ce{O=O}$ gives two moles of oxygen atoms, and atomisation is per mole of atoms:

\Delta H^{\ominus}_{\text{at}}(\ce{O}) = \tfrac{1}{2} \times 496 = +248\ \text{kJ mol}^{-1}

(b) Bond energies refer to gaseous molecules. Bromine's standard state is a liquid, so atomisation is $\ce{1/2Br2(l) -> Br(g)}$ . This includes vaporising half a mole of liquid bromine (overcoming instantaneous dipole–induced dipole forces) as well as breaking half a mole of bonds, so it is larger than $\tfrac{1}{2} \times 193 = 96.5\ \text{kJ mol}^{-1}$ .

Explaining electron affinity trends

The first electron affinities of chlorine, bromine and iodine are $-349$ , $-325$ and $-295\ \text{kJ mol}^{-1}$ . The first electron affinity of fluorine is $-328\ \text{kJ mol}^{-1}$ .

(a) Explain the trend from chlorine to iodine. (b) Suggest why fluorine does not follow this trend.

Solution

(a) Down the group, the atoms have more electron shells, so the outer shell is further from the nucleus and there is more shielding. Although nuclear charge increases, the incoming electron is less strongly attracted to the nucleus, so less energy is released: $EA_1$ becomes less exothermic.

(b) A fluorine atom is very small. The incoming electron enters the compact 2p subshell, where it experiences strong repulsion from the electrons already present. This repulsion reduces the energy released, so $EA_1$ of fluorine is less exothermic than that of chlorine, even though the incoming electron is closer to the nucleus.

Ordering lattice energies

Place the following in order of increasingly exothermic lattice energy and explain your order: $\ce{KCl}$ , $\ce{CaO}$ , $\ce{NaCl}$ , $\ce{MgO}$ .

Solution

Order (least exothermic first): $\ce{KCl} < \ce{NaCl} < \ce{CaO} < \ce{MgO}$ .

$\ce{KCl}$ and $\ce{NaCl}$ both contain 1+ and 1– ions. $\ce{K+}$ is larger than $\ce{Na+}$ , so the distance between ion centres is greater in $\ce{KCl}$ and the attraction weaker.
$\ce{CaO}$ and $\ce{MgO}$ contain 2+ and 2– ions, so the attraction is much stronger than in either chloride; charge has the largest effect.
$\ce{Mg^{2+}}$ is smaller than $\ce{Ca^{2+}}$ , so the ions in $\ce{MgO}$ are closer together and its lattice energy is the most exothermic.

An endothermic step that still happens

The first and second electron affinities of oxygen are $-141$ and $+798\ \text{kJ mol}^{-1}$ .

(a) Calculate the enthalpy change for $\ce{O(g) + 2e- -> O^{2-}(g)}$ . (b) Explain why the second value is positive. (c) Suggest why solid magnesium oxide is nevertheless very stable relative to its elements.

Solution

(a) Add the two steps (Hess's law):

\Delta H = -141 + 798 = +657\ \text{kJ mol}^{-1}

(b) The second electron is added to a negative ion, $\ce{O^-}$ . Repulsion between the ion and the incoming electron outweighs the attraction of the nucleus, so energy must be supplied.

(c) The small, doubly charged $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions attract each other very strongly, so the lattice energy of magnesium oxide is extremely exothermic (about $-3800\ \text{kJ mol}^{-1}$ ). This more than compensates for the energy needed to form the oxide ion, so the overall enthalpy change of formation is large and negative.

Common mistakes

Mistakes examiners see every series

Wrong direction for lattice energy. In 9701 it is formation of the solid from gaseous ions and is negative.
Missing or wrong state symbols. Ions in lattice energy and electron affinity equations are all $\text{(g)}$ ; the product of a lattice energy is $\text{(s)}$ .
"Per mole of element" for atomisation. It is per mole of gaseous atoms formed, so chlorine is $\ce{1/2Cl2}$ , not $\ce{Cl2}$ .
Explaining lattice energy with atomic radius or "bigger atoms". Use ionic radius and ionic charge, and say how they affect the strength of attraction between the ions.
Saying the second electron affinity is positive "because the ion is stable". It is positive because of repulsion between the negative ion and the electron being added.

Exam technique

Exam tip

"Define" questions on these terms are usually worth 2 marks: one for "one mole of ... " with the correct species and states, one for "under standard conditions" and the direction. Learn the wording above.
When asked to "explain" a lattice energy difference, you need both ideas: the factor (charge or radius) and its effect on the strength of electrostatic attraction between ions. "MgO has a higher charge" alone earns at most one mark.
When comparing two compounds, compare like with like. Name both ions and state which is larger or more highly charged.
If a question asks you to "suggest" a value, give a number in the right direction from data you already have (for example, a lattice energy for $\ce{KBr}$ between those of $\ce{KCl}$ and $\ce{KI}$ ) and justify it in one sentence.

Summary

$\Delta H^{\ominus}_{\text{at}}$ is per mole of gaseous atoms formed from the element in its standard state; always endothermic. For gaseous diatomic elements it is half the bond energy.
$EA_1$ : one mole of electrons added to one mole of gaseous atoms giving gaseous 1– ions; usually exothermic.
$EA_2$ is always endothermic because the electron is added to a negative ion.
$EA_1$ depends on nuclear charge, distance from the nucleus and shielding; it becomes less exothermic down Groups 16 and 17 from Cl and S, but F and O are anomalously low because of electron repulsion in their small 2p subshells.
Lattice energy (9701): formation of one mole of ionic solid from gaseous ions; always exothermic; found indirectly from a Born–Haber cycle.
Higher ionic charge and smaller ionic radius give a more exothermic lattice energy; charge has the greater effect.

Practice questions

Question

Define the term lattice energy.
Write equations, with state symbols, for: (a) the first electron affinity of bromine; (b) the enthalpy change of atomisation of magnesium; (c) the lattice energy of potassium oxide.
The $\ce{H-H}$ bond energy is $436\ \text{kJ mol}^{-1}$ . State the enthalpy change of atomisation of hydrogen.
Explain why the first electron affinity of sulfur ( $-200\ \text{kJ mol}^{-1}$ ) is more exothermic than that of oxygen ( $-141\ \text{kJ mol}^{-1}$ ).
Explain why the first electron affinity of chlorine is more exothermic than that of sulfur.
Without using data, place $\ce{LiF}$ , $\ce{LiI}$ , $\ce{MgF2}$ and $\ce{KI}$ in order of increasingly exothermic lattice energy. Explain your answer.
The lattice energies of $\ce{NaCl}$ and $\ce{NaI}$ are $-787$ and $-704\ \text{kJ mol}^{-1}$ . Suggest a value for the lattice energy of $\ce{NaBr}$ and explain your choice.
Sodium fluoride and magnesium oxide have very similar inter-ionic distances. Explain why the lattice energy of $\ce{MgO}$ is roughly four times that of $\ce{NaF}$ .
Explain why the second electron affinity of oxygen is endothermic, and why the overall process $\ce{O(g) + 2e- -> O^{2-}(g)}$ is endothermic even though $EA_1$ is exothermic.
Calcium can form $\ce{Ca+}$ or $\ce{Ca^{2+}}$ ions. Using ideas about lattice energy only, explain why the lattice energy of $\ce{CaCl2}$ is much more exothermic than that of the hypothetical compound $\ce{CaCl}$ , and state what other energy term works against forming $\ce{CaCl2}$ .

Answers

The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.
(a) $\ce{Br(g) + e- -> Br^-(g)}$ (b) $\ce{Mg(s) -> Mg(g)}$ (c) $\ce{2K^+(g) + O^{2-}(g) -> K2O(s)}$
$\tfrac{1}{2} \times 436 = +218\ \text{kJ mol}^{-1}$ , because one mole of bonds broken gives two moles of atoms.
Oxygen is very small, so the incoming electron is added to the crowded 2p subshell and is strongly repelled by the electrons already there. This repulsion outweighs the advantage of being closer to the nucleus, so less energy is released than for sulfur, whose 3p subshell is larger.
Chlorine has one more proton than sulfur but the same number of inner shells (similar shielding) and a slightly smaller radius. The incoming electron is therefore attracted more strongly by the chlorine nucleus, and more energy is released.
$\ce{KI} < \ce{LiI} < \ce{LiF} < \ce{MgF2}$ . $\ce{KI}$ , $\ce{LiI}$ and $\ce{LiF}$ all have 1+ and 1– ions. $\ce{K+}$ is larger than $\ce{Li+}$ , so $\ce{KI}$ has the greatest inter-ionic distance and weakest attraction; $\ce{F-}$ is much smaller than $\ce{I-}$ , so $\ce{LiF}$ has stronger attraction than $\ce{LiI}$ . $\ce{MgF2}$ contains the doubly charged, small $\ce{Mg^{2+}}$ ion, so its attraction is strongest of all.
About $-745\ \text{kJ mol}^{-1}$ (any value between $-704$ and $-787$ earns the mark). $\ce{Br-}$ is intermediate in size between $\ce{Cl-}$ and $\ce{I-}$ , so the inter-ionic distance, and hence the strength of attraction, is intermediate. (The data book value is $-747\ \text{kJ mol}^{-1}$ .)
The strength of attraction depends on the product of the ionic charges divided by the distance between ion centres. The distance is similar, but the product of the charges is $2 \times 2 = 4$ for $\ce{Mg^{2+}O^{2-}}$ against $1 \times 1 = 1$ for $\ce{Na+F-}$ , so the electrostatic attraction, and the energy released when the lattice forms, is roughly four times larger.
The second electron is added to a negative ion, $\ce{O-}$ , and is repelled by it more than it is attracted by the nucleus, so energy must be supplied. Overall: $-141 + 798 = +657\ \text{kJ mol}^{-1}$ ; the endothermic $EA_2$ is much larger in magnitude than the exothermic $EA_1$ .
$\ce{Ca^{2+}}$ has twice the charge of $\ce{Ca+}$ and is smaller (it has lost both 4s electrons), so it attracts chloride ions far more strongly; there are also two chloride ions per calcium ion. The lattice energy of $\ce{CaCl2}$ is therefore much more exothermic. Working against it is the large second ionisation energy of calcium ( $+1150\ \text{kJ mol}^{-1}$ ), which must be supplied to form $\ce{Ca^{2+}}$ . The extra lattice energy more than repays it, which is why $\ce{CaCl2}$ , not $\ce{CaCl}$ , is the stable compound.

Standard conditions and sign conventions

Enthalpy change of atomisation

Electron affinity

Why the first electron affinity is usually negative

Why the second electron affinity is always positive

Factors affecting electron affinity

The fluorine and oxygen anomaly

Lattice energy

What controls the size of a lattice energy

Worked examples

Common mistakes

Exam technique

Summary

Practice questions

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