# Which has the highest melting point among NaF, NaCl, NaBr, NaI?

I've been trying to understand why $$\ce{NaF}$$ has the highest melting point among all of the given solids. I thought it would be $$\ce{NaI}$$ as it will have the biggest molar mass among them which would mean it has the strongest intermolecular forces. Is it electronegativity difference or molar mass that you look at when analysing intermolecular forces? Am I missing something or misunderstanding something? Thank you very much.

• NaF has better packed crystal structure, which is harder to break. – ABC Sep 6 '17 at 9:56
• Another point. There is no such thing as a NaCl "molecule" in solid NaCl. All the ionic charges interact. Lookup the "Madelung Constant." – MaxW Sep 6 '17 at 14:00
• A simple way of comparing the M.P.s of ionic solids is through calculating their lattice energies. – Tan Yong Boon Sep 7 '17 at 3:39
• Evidently, NaF has the highest L.E. anong the four as fluoride has the smallest ionic radius among the four anions. – Tan Yong Boon Sep 7 '17 at 3:40

All those are ionic compounds, so it is just Coulomb forces that hold them together. Two charges are attracted with the same force regardless of their mass. The distance, however, does matter a great deal. And the distance depends on the ionic radii, and I'm sure you know how those change down the group.

Things could have turned out otherwise if the compounds had different crystal structures. Luckily for us, this is not the case.

So if you are confronted with an unknown compound, you first look at the electronegativity difference, and if it tells you the compound is ionic, you use the same line of thought as I just did. Otherwise it is probably a molecular compound held together with dispersion forces, and they tend to grow with molar mass. Oh, and there might be a covalent solid...

See, chemistry is complicated, and it is not always possible to get by with a few simple rules of thumb. But at least I think I've answered your original question in the narrow sense, so let's leave it at that.

• The point that is crucial here is that ionic compounds are held together in a crystal lattice structure.

• What you have misunderstood is that there are no intermolecular forces between these ions, as number one the forces are not between molecules but ions, and number two if you break the bonds between a $$\ce{Na+}$$ and a $$\ce{Cl-}$$ ions, you have effectively broken an intramolecular bond resulting in a sodium ion and a chlorine ion.

• Hence mass has no effect on the strength of the bond but factors affecting coulombs law will.
• The reason why $$\ce{NaF}$$ would have the highest melting point is because, if you look at the electronegativity values fluorine has the highest hence it have the greatest pull or attraction towards the electrons in the ionic bond, resulting in a really strong sodium positive ion and a strong negative fluorine ion, this thereby results in a high melting point. You can also look up lattice enthalpy values to be sure.

As with boiling points, the melting point of a solid is dependent on the strength of intermolecular attractive forces. A strong attraction between molecules results in a higher melting point. In general, ionic compounds have higher melting points compared to covalent compounds, because the electrostatic forces connecting the ions (the ion-ion interaction) are stronger than molecular-molecular or polar interactions exist in covalent compound. Unlike covalent compounds, those interactions do not get stronger with increasing molecular weights. Actually, it is the contrary.

The high melting point of ionic compound reflects its high lattice energy. When ions are tightly packed together of an ionic substance, it has a higher melting point than another ionic substance with ions that do not pack well (recall that this packing depends on the ratio of ionic radii of positively and negatively charged ions, which changes the crystal structure).

Most alkali metal halides crystallize with the face-centered cubic lattices (FCC). In this structure both the metals and halides feature octahedral coordination geometry, in which each ion has a coordination number of six. The exception are cesium chloride, bromide, and iodide, which are crystallize in a body-centered cubic lattice (BCC) that accommodates coordination number of eight for the larger metal cation (and the anion as well).

Thus, the melting point of ionic solid is depends on lot of factors including ion packing. However, as Ivan Neretin's excellent answer pointed out, the major force is the Coulomb forces acting between oppositely charged ions $$\left(F_{attract} = \dfrac{q_+q_-}{d^2}\right)$$. Yet, the strength of these forces changes with other factors such as ionic character of the bond. This would be evident on following table:

$$\begin{array}{c|ccc} \text{Substance} & a \ (\pu{pm})^a & Z & r_+ + r_- \ (\pu{pm}) & V_m \ (\pu{cm3mol-1}) & d \ (\pu{g \! cm-3})^b & \text{m.p.} \ (\pu{^\circ C})^c \\ \hline \ce{LiF} & 401.4 & 4 & 283.8 & 9.737 & 2.66 \ (2.64) & 845 \\ \ce{LiCl} & 513.2 & 4 & 362.9 & 20.349 & 2.08 \ (2.07) & 605 \\ \ce{LiBr} & 548.9 & 4 & 388.1 & 24.898 & 3.49 \ (3.46) & 550 \\ \ce{LiI} & 600.0 & 4 & 424.3 & 32.519 & 4.12 \ (4.08) & 449 \\ \ce{NaF} & 462.0 & 4 & 326.7 & 14.846 & 2.83 \ (2.56) & 993 \\ \ce{NaCl} & 564.8 & 4 & 399.4 & 27.125 & 2.15 \ (2.17) & 801 \\ \ce{NaBr} & 593.6 & 4 & 419.7 & 31.489 & 3.27 \ (3.21) & 747 \\ \ce{NaI} & 646.2 & 4 & 456.9 & 40.624 & 3.69 \ (3.67) & 661 \\ \ce{KF} & 532.8 & 4 & 376.7 & 22.771 & 2.55 \ (2.48) & 858 \\ \ce{KCl} & 627.6 & 4 & 443.8 & 37.216 & 2.00 \ (1.98) & 770 \\ \ce{KBr} & 657.0 & 4 & 464.6 & 42.695 & 2.79 \ (2.74) & 734 \\ \ce{KI} & 706.4 & 4 & 499.5 & 53.068 & 3.13 \ (3.12) & 681 \\ \ce{RbF} & - & 4 & - & - & - \ (3.56) & 795 \\ \ce{RbCl} & 653.4 & 4 & 462.0 & 41.997 & 2.88 \ (2.80) & 718 \\ \ce{RbBr} & 693.0 & 4 & 490.0 & 50.105 & 3.30 \ (3.35) & 693 \\ \ce{RbI} & 730.8 & 4 & 516.8 & 58.759 & 3.61 \ (3.55) & 647 \\ \ce{CsF} & 600.8 & 4 & 520.3 & 32.649 & 4.65 \ (4.64) & 682 \\ \ce{CsCl} & 411.8 & 1 & 356.6 & 42.053 & 4.00 \ (3.99) & 645 \\ \ce{CsBr} & 428.7 & 1 & 371.3 & 47.446 & 4.49 \ (4.43) & 636 \\ \ce{CsI} & 455.8 & 1 & 394.7 & 57.025 & 4.56 \ (4.51) & 626 \\ \hline \end{array}$$ $$^a$$: The lattice constant $$a$$ is the physical distance of the relevant unit cell. The values are from Ref.1; $$^b$$: The density $$d$$ is calculated from the equation $$d = \text{Molar Mass}/V_m$$. The values in the parenthesis are experimental values listed in various sources; and $$^c$$: The melting points are from Ref.2.

As the table has shown, melting points of halide salts of each alkali metal decrease with the decreasing binding energies from fluorides to iodides. However, most striking feature I have noticed here is the each series of particular alkali metal halides (at least those of $$\ce{Li}, \ \ce{Na}$$, and $$\ce{K}$$) have linear relationship with their melting points and corresponding lattice constant $$a$$:

Unfortunately, $$a$$ value of $$\ce{RbF}$$ is absent from literature due to its highly hygroscopic nature (at least the time of Ref.1; nevertheless, I couldn't find that either). Therefore, whether this relationship exists with $$\ce{RbX}$$ is yet to be seen (the three remaining points are not good enough, even though it shows the linear relationship with $$R^2 = 0.967$$). All of those crystals have closely packed FCC geometry.

Strikingly different from this phenomena is melting points of the series of $$\ce{CsX}$$, which does not show the linear relationship. This behavior is expected since only $$\ce{RbF}$$ has FCC geometry while other three are Body Centered Cubic (BCC). Alike existing three $$\ce{RbX}$$ crystals, three $$\ce{CsX}$$ with BCC geometry also show the linear relationship, but three-point straight line is not so reliable.

However, although this is out of the question, I'd like to make a suggestion. Using the equation given in this three-point straight line, we can calculate approximate value of $$a$$ for $$\ce{CsF}$$, which is seldom found in literature. For the convenience, I plotted all four metal halide graphs ($$\ce{LiX}, \ \ce{NaX}, \ \ce{KX}$$, and $$\ce{RbX}$$) in one paper:

According to the equation: $$y = -0.915 x + 1319.6$$ for $$\ce{RbX}$$, when $$y=\pu{795 ^\circ C}$$ (m.p. of $$\ce{RbF}$$), $$x=a=\pu{573 pm}$$, which is a reliable number compared to the trend.

References:

1. Ralph W. G. Wyckoff, In The Structure of Crystals, Part 1; American Chemical Society Monograph Series; The Chemical Catalog Company, Inc.: New York, NY, 1924.
2. S. L. Baldochi, I. M. Ranieri, In Reference Module in Materials Science and Materials Engineering; Saleem Hashmi, Editor-in-Chief; Elsevier, Inc.: Dublin, Ireland, 2016 (The values are available online from https://www.sciencedirect.com/topics/chemical-engineering/alkali-halides).

I think it could be because the enthalpy of formation on NaF is the most negative and so NaF is the most stable compound among the alkali metal halides. Consequently the melting point would be the highest. Also the melting and boiling point always follow the trend : fluoride>chloride>bromide>iodide.