# How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules?

It is easy to compare two ionic compounds when one of the ions is same. However, how do we compare two compounds if one of the ions is the same element but just has different charge and the other ion is completely different?

• Here the change of anion and the increase of oxidation, both act towards producing a similar change in the ionic character of the compound. Can you tell why aren't you able to apply Fajan's rules to solve this problem? It appears a straightforward application to me. – Satwik Pasani Mar 11 '14 at 13:41
• @SatwikPasani I am having difficulty applying the rules. Could you please apply the rules and solve this problem? Thanks! – Papul Mar 26 '14 at 17:51
• @SatwikPasani SnF2 is apparently not a straightforward case. Studies employing Mossbauer spectroscopy have shown that SnF2 is covalently bonded. – gannex Apr 15 '16 at 3:00

Fajan's rules for the prediction of bond character can be summarized in two points:

• Large cation with low positive charge + small anion $\rightarrow$ ionic

• Small cation with high positive charge + large anion $\rightarrow$ covalent

In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.

On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.

However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: 1, 2).

• Complementary the HSAB Theory can be applied. – Martin - マーチン Apr 15 '14 at 5:27
• SnF2 is a bent molecule with a a stereo-active lone pair in a hybrid orbital, so it must be covalent. I have this one professor who has spent his whole life studying tin chemistry and tin crystallography and he assures me that SnF2 is covalent. "With small and highly electronegative non-metals (F, Cl, O, S), bonding is alwayscovalent, and as a result, the tin orbitals are hybridized, usually sp3, sp3d or sp3d2,to give an electron pair distribution that is pseudotetrahedral SnX3E, pseudotrigonal bipyramidal SnX4E or pseudooctahedral SnX5E, respectively" -Georges Denes & Abdualhafed Muntasar – gannex Apr 15 '16 at 2:56