Is there a difference in the electronegativity differences of tin(IV) chloride and tin(II) chloride?

Question

I'm wondering if this is why one is a covalent bond and one is an ionic bond. (Less than 1.7 and more than 1.7in electronegativity difference).

Answer 1

There are like three of you all asking the same crazy "why is SnCl2 ionic" thing.  Whose goofy assignment is this?  Look, tin (II) chloride is NOT an ionic compound.  It just isn't.

At room temperature, SnCl2 consists of polymeric chains -[-ClSn(Cl)-]-, where each pyramidal Sn is connected to two bridging Cl atoms and one terminal Cl atom, and bears a lone pair.  (See the wikipedia page for a drawing.)  It's a perfectly fine octet structure, with directional bonds and lone pairs, just like any other covalent compound.  More importantly, SnCl2 does *not* consist of a repeating lattice of distinct Sn2+ cations and Cl- anions, which is what most people's definition of an ionic compound would require.

And in the gas phase, above 623ºC, SnCl2 goes monomeric, it's a discrete molecular species with a bent geometry. So there's no way you could justify calling that an ionic compound.

A crude approach to electronegativity, where you're just reading Pauling values off the periodic table, gives Sn at 1.96 and Cl at 3.16, so that's a difference of only 1.2. There are different formulas you can use to approximate the "ionic character" of a bond based on electronegativity differences, but Pauling's was (percent ionic) = 1 – e^(–0.25*(delchi)^2), which gets you only 30 percent ionic character in those bonds.

OK, now here's the interesting part.  You're actually on the right track, looking for some reason for the electronegativity of tin to vary between the two compounds.  And it does.  A more sophisticated approach admits that elements in different geometries and different canonical oxidation states ought really to be assigned different electronegativities.  It's easier to rationalize using a Mulliken definition, where electronegativity measures the average of ionization energy and electron affinity -- harder to remove an electron and easier to add one equals greater electronegativity.  So, Sn(IV) is harder to ionize and more likely to accept electrons than Sn(II).  Which means that Sn(IV) should have higher electronegativity.  As well, consider the geometries, and the implications for the valence orbitals.  SnCl4 is perfectly tetrahedral, so you can call that sp3 hybrids.  The angles in polymeric SnCl2 are much more acute, so there's much more p character in the bonds, which means higher energy valence orbitals, which again makes it less electronegative (easier to ionize with higher E orbitals).

So, conclusion: Sn(IV) has higher e-neg than Sn(II), so Sn(II)-Cl bonds are more polar than Sn(IV)-Cl.

But it's not that big a difference.  This only gets you a Pauling value of 1.80 for Sn(II) vs the 1.96 for Sn(IV), and that's not enough for your magic >1.7 = ionic designation.  You're still only at 37 percent ionic, i.e. still more than three fifths covalent bonding. The differences are even smaller on the Mulliken scale.

Both the stannous and the stannic chlorides are primarily covalent bonding, neither consists of a lattice of discrete tin cations and chloride anions. Yes, there is an electronegativity difference between Sn(II) and Sn(IV), but not enough to make tin (II) chloride an ionic compound.