By Fajans' rules, we can easily find out which compound shows covalent character.

Example: Among $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{AlCl3}$ which one is more covalent?

Answer: There is a point in Fajans' rules that the compound which has more charge number that compound will show covalent character. In $\ce{AlCl3}$, Al has the highest oxidation state (+3) so it will show more covalent character than $\ce{NaCl}$ and $\ce{MgCl2}$.

The point is that since all three compounds I have mentioned have chlorine in common, it was easy to catch the solution but if the compounds were given as $\ce{NaF}$ and $\ce{LiCl}$, which one shows more covalent character, and how can we apply Fajans' rules?


1 Answer 1


You will have to consider two factors here: polarising power of the cation and polarisability of the anion. There is more than one part of Fajan’s rule(Source-J. D. Lee, Concise Inorganic Chemistry):

  1. A small positive ion favours covalency.
  2. A large negative ion favours covalency.
  3. Large charges on either ion, or both ions, favours covalency.
  4. Polarisation and hence covalency is favoured if the positive ion does not have noble has configuration.

According to these, the $\ce{Li+}$ ion is more polarising than $\ce{Na+}$ ion, and $\ce{Cl-}$ ion is more polarisable than $\ce{F-}$ ion.

Since both factors support it, $\ce{LiCl}$ is more covalent than $\ce{NaF}$.

Somtimes, like in your question, both the factors support each other. Sometimes, when they are opposite, like in the case of $\ce{LiF}$ and $\ce{NaCl}$, it is tough to determine relative covalency using Fajan’s rule.

  • 1
    $\begingroup$ does that mean that d block elements (Transition elements) are more ionic than alkali metals and alkaline earth metal compounds ? $\endgroup$
    – Tanuj
    Feb 27, 2018 at 3:02
  • 1
    $\begingroup$ @Tanuj actually it means the opposite; the d-block elements are generally smaller than the corresponding alkali metals/ alkaline earth metals. $\endgroup$
    – Mr_Pea
    Feb 27, 2018 at 3:58
  • $\begingroup$ Okay , got it . $\endgroup$
    – Tanuj
    Feb 27, 2018 at 4:07

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