When comparing boiling points of two ionic compounds, should you look at their electronegativity difference, lattice energy, or strength from LDF forces (or all 3 factors)?

For example, between LiCL and NaCl, which one has the higher boiling point?

If we look at EN difference and LDF forces, it would be NaCl, because the EN difference here would be $2.1$ compared to $2.0$ for LiCL; also, since Na has an extra energy level, it has a larger electron cloud and thus stronger LDF's.

However, if we look at it in term of lattice energy or Coloumb's law, LiCl has a larger boiling point because high lattice energy is a combination of large charge and small ions. Therefore, even though Li and Na have the same charge, Li has a smaller atomic radius than Na.


1 Answer 1


To compare 2 ionic compounds' melting point, there are generally 3 determinants:

  1. Ionic Radii, smaller the ionic radius, closer the ions are to each other, therefore stronger the electrostatic attraction between them. (analogous to stronger attraction when closer to a magnet) A good example would be melting points of cations from the same group with the same anion (NB: LiF does not follow the trend, potentially due to its more covalent nature):

    • $\ce{NaF}\ -\ \pu{993 ^\circ C ( 1819 ^\circ F)}$
    • $\ce{KF}\ - \pu{858 ^\circ C ( 1576 ^\circ F)}$
    • $\ce{RbF}\ -\ \pu{795 ^\circ C ( 1463 ^\circ F)}$
  2. Charge of ions, larger the charge an ion has, more positive or negative the ion is, therefore there is a stronger attraction between ions. (analogous to stronger attraction with two stronger magnets in opposite poles) An example would be $\ce{NaCl}$ and $\ce{MgO}$:

    • $\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}$
    • $\ce{MgO}\ -\ \pu{2852 ^\circ C ( 5166 ^\circ F)}$
  3. Ratio of ions, in most cases, closer the ratio is to 1 : 1 between anions and cations, stronger the attraction is. larger or smaller the ratio is, the more the anions or cations will have to 'share' or 'divide' the electrostatic attraction from the oppositely-charge ion, decreasing the melting point. An example would be $\ce{NaCl}$ and $\ce{MgCl2}$:

    • $\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}\ [1 : 1]$
    • $\ce{MgCl2}\ -\ \pu{714 ^\circ C ( 1317 ^\circ F)}\ [1 : 2]$

The determinants for melting points would generally also apply to the boiling points of ionic compounds. However, it should be noted that other factors would affect the boiling point as well.

  • $\begingroup$ Hello there, thank you for your contribution! Your first two points are quite correct, but the third one doesn't seem to work so well. For example, $\ce{KCl}$ and $\ce{CaCl2}$ boil at 1420 °C and 1935 °C, respectively. Also, I would be wary of your final statement; there are factors which affect melting points but do not affect boiling points, and vice-versa (e.g. benzene and toluene have relatively close boiling points, but wildly different melting points). Therefore, the comparisons and arguments really should be made directly regarding the boiling points. $\endgroup$ Commented Nov 18, 2018 at 11:09
  • $\begingroup$ Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{ }\$ for chemical formulas/equations, \$ \$ for math term/equations, and \$\pu{ }\$ for units. More information is available in this meta post Also, it is suggested that all new users take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already. $\endgroup$
    – A.K.
    Commented Nov 18, 2018 at 15:19
  • $\begingroup$ Point 3 tries to address one of the harder parts of this, how to compare compounds with fundamentally different structures. For this one could also look at the Kapustinskii equation, see for instance en.wikipedia.org/wiki/Kapustinskii_equation $\endgroup$
    – Ian Bush
    Commented Nov 18, 2018 at 19:11

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