# Enthalpies of formation of alkali metal halides

Shriver and Atkins's Inorganic Chemistry has this graph:

How can I explain the different direction of the trends for fluorides and chlorides? I know that the decreasing lattice energy (as M goes from Li to Cs) is one factor that needs to be taken into account, but the trend in lattice energy should be similar for all the metals.

The equation for the formation of $\ce{MX(s)}$ from its constituent elements is:

$$\ce{M(s) + 1/2 X2(g) -> MX (s)}$$

In general, $\ce{X2}$ could be liquid ($\ce{X} = \ce{Br}$) or solid ($\ce{X} = \ce{I}$). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be described can be applied just as easily to the bromides and iodides without any problem.

For any question that asks about trends in thermodynamics, the best way to go about solving it is always to draw a cycle. This will allow you to identify which terms contribute to the change in the trend, and from there rationalise the observed trend. So, without further ado:

$$\large\require{AMScd}\begin{CD} \ce{M (s) + 1/2 X2 (g)} @>\color{blue}{\Delta_\mathrm{f}H(\ce{MX})}>> \ce{MX (s)} \\ @V\color{red}{\Delta_\mathrm{atm}H(\ce{M})}VV @AA\color{red}{-\Delta_\mathrm{L}H(\ce{MX})}A \\ \color{red}{\ce{M (g) + 1/2 X2 (g)}} @. \color{red}{\ce{M+ (g) + X- (g)}} \\ @V\color{red}{I_1(\ce{M})}VV @AA\color{red}{E_\mathrm{a}(\ce{X})}A \\ \color{red}{\ce{M+ (g) + e- + 1/2 X2 (g)}} @>\color{red}{\frac{1}{2}B(\ce{X-X})}>> \color{red}{\ce{M+ (g) + e- + X (g)}} \end{CD}$$

By Hess's law, the enthalpy change for the blue path must be equal to the enthalpy change for the red path. This follows from the fact that enthalpy, $H$, is a state function. So:

$$\Delta_\mathrm{f}H(\ce{MX}) = \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) + \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) - \Delta_\mathrm{L}H(\ce{MX})$$

Note the sign convention I am using for the electron affinity and lattice enthalpy (these seem to be defined differently in different places). The lattice enthalpy used here is always positive (endothermic).

Clearly, the terms $\Delta_\mathrm{atm}H(\ce{M})$ and $I_1(\ce{M})$ only depend on the identity of the metal $\ce{M}$. Therefore, it is helpful to lump them together and consider them as one term, called $\Delta H_\ce{M}$. Similarly, the next two terms only depend on the halide $\ce{X}$ and we can collect them together and call it another term, $\Delta H_\ce{X}$. It is by no means necessary to do this; however, it saves some clutter in the expressions:

$$\Delta H_\mathrm{tot} = \Delta H_\ce{M} + \Delta H_\ce{X} + \Delta H_\ce{MX}$$

with

\begin{align} \Delta H_\mathrm{tot} &= \Delta_\mathrm{f}H(\ce{MX}) \\ \Delta H_\ce{M} &= \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) \\ \Delta H_\ce{X} &= \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) \\ \Delta H_\ce{MX} &= -\Delta_\mathrm{L}H(\ce{MX}) \end{align}

# The fluorides

As shown in your graph, the trend for the metal fluorides is that going from $\ce{LiF}$ to $\ce{CsF}$, $-\Delta_\mathrm{f}H(\ce{MF})$ decreases, i.e. $\Delta_\mathrm{f}H(\ce{MF}) = \Delta H_\mathrm{tot}$ increases (it becomes less negative). Here, the halide ion $\ce{X-}$ is not changing, so $\Delta H_\ce{X}$ is a constant and we only need to consider the other two terms:

• Going from $\ce{Li}$ to $\ce{Cs}$, the atomisation enthalpy decreases and the first ionisation energy decreases. That means that $\Delta H_\ce{M}$ decreases.
• Going from $\ce{Li}$ to $\ce{Cs}$, the radius of the metal ion increases, and so the lattice enthalpy decreases. This means that $\Delta H_\ce{MX}$ (which is the negative of the lattice enthalpy) increases.

So, there are two opposing factors. The first factor says that $\Delta H_\mathrm{tot}$ should decrease, and the other says that it should increase. Since the observed trend is that $\Delta H_\mathrm{tot}$ increases, this means that the second factor is greater than the first. That is to say, the increase in $\Delta H_\ce{MX}$ outweighs the decrease in $\Delta H_\ce{M}$, leading to an overall increase in $\Delta H_\mathrm{tot}$.

# The chlorides

The analysis here - in terms of the individual trends in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ - is exactly the same. However, this time, the overall trend is reversed: $\Delta H_\mathrm{tot}$ decreases. That can only mean that, when $\ce{X} = \ce{Cl}$, the decrease in $\Delta H_\ce{M}$ now outweighs the increase in $\Delta H_\ce{MX}$. Why is that so?

First, we note that $\Delta H_\ce{M}$ is entirely dependent on the identity of the metal ion. It is independent of the identity of the halide. That means that the decrease in $\Delta H_\ce{M}$ when $\ce{X} = \ce{F}$ must be exactly the same when $\ce{X} = \ce{Cl}$ - there is absolutely no reason for it to change.

Logically, since the decrease in $\Delta H_\ce{M}$ is not changing in magnitude, it must mean that the increase in $\Delta H_\ce{MX}$ must have become smaller in the chlorides. Otherwise, there is no way for the decrease in $\Delta H_\ce{M}$ to suddenly outweigh the increase in $\Delta H_\ce{MX}$. And we can explain this very easily using what we know about lattice energies, because that's exactly what $\Delta H_\ce{MX}$ is: a lattice energy (with a minus sign).

The lattice energy of an ionic salt can be estimated using the Kapustinskii equation. For a metal halide, this reduces to the equation;

$$\Delta_\mathrm{L}H \propto \frac{1}{r(\ce{M+}) + r(\ce{X-})}$$

Clearly, going from $\ce{LiX}$ to $\ce{CsX}$, $r(\ce{M+})$ increases and $\Delta_\mathrm{L}H$ decreases. The difference is that when $\ce{X} = \ce{F}$, $r(\ce{X-})$ is smaller than when $\ce{X} = \ce{Cl}$. Mathematically, his means that in the fluorides, the lattice enthalpy is more sensitive to changes in $r(\ce{M+})$. This makes sense - if we went the other way and took the limit of $r(\ce{X-}) \to \infty$, the lattice enthalpy would not depend on $r(\ce{M+})$ at all.

That means that the decrease in $\Delta_\mathrm{L}H$ - which is equivalent to the increase in $\Delta H_\ce{MX}$ - must be larger in the fluorides, and smaller in the chlorides. That is why in the chlorides, the trend in $\Delta H_\ce{M}$ overrules the trend in $\Delta H_\ce{MX}$, and the reversal of the overall trend is observed.

# Ending words

How would you know all of this if you were not given the graph? Well, the truth is, without looking up values of thermochemical radii, ionisation enthalpies, etc., there is no real way to know. The fact that the overall trend is reversed in the chlorides means that the variations in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ must be pretty similar in magnitude. However, if you are given data and asked to rationalise it, you can definitely do so by constructing a diagram and relating the variation in the various energetic terms to variation in physical properties that you are familiar with (ionisation energy, ionic radii, and so on).