Enthalpies of formation of alkali metal halides

Shriver and Atkins's *Inorganic Chemistry* has this graph:
![enter image description here](https://i.stack.imgur.com/1Tlix.jpg)


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 


$$\begin{align}
\Delta H\_\mathrm{tot} &= \Delta\_\mathrm{f}H(\ce{MX}) \\
\Delta H\_\ce{M} &= \Delta\_\mathrm{atm}H(\ce{M}) + I\_1(\ce{M}) \\
\Delta H\_\ce{X} &= \frac{1}{2}B(\ce{X-X}) + E\_\mathrm{a}(\ce{X}) \\
\Delta H\_\ce{MX} &= -\Delta\_\mathrm{L}H(\ce{MX})
\end{align}$$




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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 $\Delta H\_\mathrm{tot}$.




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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.




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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).


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