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Based on my knowledge, lattice energy is proportional to the multiplcation of the charge of the ions, divided by the sum of the radius of ions, as follows.

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Since iodide has a larger radius than fluoride, by this equation lithium iodide would have a less exothermic lattice energy (lower lattice energy).

However, I also learned that an ionic bond with higher covalent character is stronger (i.e. more exothermic lattice energy). Since the electron cloud of iodide ion is larger and is more polarisable, lithium iodide should have greater covalent character than lithium fluoride and thus its lattice energy should be more exothermic (higher lattice energy)

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    $\begingroup$ The iod atom or ion is too large to make good orbital overlap with basically any partner. Covalent iodides are invariably rather unstable. Same for bromides, or e.g. sulfides, compared to oxides. $\endgroup$
    – Karl
    Aug 4 '20 at 12:49
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    $\begingroup$ The real issue here is that the formula quoted is for the lattice energy as calculated within the ionic model, which by definition has no covalent component whatsoever. A good answer will talk about the difference between a simple ionic model and a more complex one that includes "covalent" effects, and the need to know enough chemistry to understand which model is best to apply in a given situation. I'm sure the down voters of this good question will find time to discuss fully this once they get around to justifying their down votes. I hope to find time to give such an answer. $\endgroup$
    – Ian Bush
    Aug 4 '20 at 18:30

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