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My chemistry teacher told us that $\ce{NaF}$ has a higher lattice energy than $\ce{CsF}$. He explained it by telling lattice energy is inversely proportional to size of ions added. He then mentioned that $\ce{Cs2SO4}$ has a higher lattice energy compared to $\ce{Na2SO4}$ (which contradicts the above relation of lattice energy and size of ions). Is this true or is he wrong? I'm not able to find an answer anywhere.

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    $\begingroup$ Lattice energies are directly comparable, regarding ion sizes , only for the same lattice structures. If ions are arranged differently, ion sizes are not the only parameters to evaluate. E.g. for 1:1 stoichiometry with spherical ions, the radius ratio <0.414 leads to coordination number 4, <0.732 to 6, >0.732 to 8 $\endgroup$
    – Poutnik
    Commented Mar 26 at 7:31
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    $\begingroup$ You will get in trouble with non-standard element symbols; stick to the convention! Because e.g., Sn in Hill formulae is about one atom of tin, SN would be about an atom sulfur and an other about nitrogen etc. In SMILES convention, uppercase C1CCCCC1 would be about cyclohexane, lowercase c1ccccc1 about benzene -- two different compounds. You surely want to be called by your name proper, too -- don't you? $\endgroup$
    – Buttonwood
    Commented Mar 26 at 8:37
  • $\begingroup$ For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and report numerical values (\pu{}) including a non-breakable space. $\endgroup$
    – Buttonwood
    Commented Mar 26 at 8:40
  • $\begingroup$ Petrusevski, Vladimir & Stojanovska, Marina & Šoptrajanov, Bojan. (2007). DEMONSTRATIONS AS A TOOL FOR IRONING-OUT PRECONCEPTIONS: 1. ON THE REACTIONS OF ALKALI METAL SULFATES WITH CONCENTRATED SULFURIC ACID. The Chemical Educator. 12. 71-74. contradicts the above assertion where it gives the lattice energy for Na2SO4 as 1827 kJmol-1 and for Cs2SO4 as 1596 kJmol-1 . This also briefly discusses the different structure of the two compounds. $\endgroup$
    – Ian Bush
    Commented Mar 26 at 10:06
  • $\begingroup$ What may be happening (and I stress this is a complete guess) is that the lattice energies of the primitive cells are being compared. This causes problems as the primitive cell for the Na salt contains half the number of formula units of the Cs salt. What you should compare is the conventional cells - the Na salt crystallizes in a higher symmetry, cubic, system and the conventional cell has twice the number of atoms of the primitive, so the same number as the Cs salt. If this mistake is being made you can half the above number for Na, and you get the claimed ordering. $\endgroup$
    – Ian Bush
    Commented Mar 26 at 10:33

2 Answers 2

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Generally speaking, large cations and large anions ($\ce{Cs2SO4}$) form more stable lattices as compared to small cations and large anions($\ce{Na2SO4}$). One reason could be that the ionic character decreases as the size of cation decreases and thus, the factors that influence the stability of a lattice are diminished with covalent character of the bond increasing. Otherwise, your teacher is correct regarding the factors that influence the stability of a lattice. It's just that in this case those factors are diminished in their effect, due to the characteristics of bond.

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    $\begingroup$ This is more applicable on spherical big anions like I-, less so on multiatomic oxo-anions, that are much less polarizable. $\endgroup$
    – Poutnik
    Commented Mar 26 at 16:17
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While what you said holds true for simpler lattices, things get complex for ionic compounds with different anions. Here, the overall charge of the cation plays a more significant role than just its size. In Na2SO4, there are two Na+ ions for every SO4²⁻ ion. This creates a stronger electrostatic attraction compared to Cs2SO4, which has only one Cs+ ion per SO4²⁻ ion. Even though Cs+ is larger than Na+, the double positive charge in Na2SO4 outweighs the size difference, leading to a higher lattice energy for Na2SO4.

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