Question

21. When tin(IV) chloride is treated with excess of conc.
hydrochloric acid, the complex ion (SnC12)- is formed. The
oxidation state of tin in this complex ion is:​

Answer 1

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$<b>$First, let's consider possibility, that indeed some cation is more stable than others. Let's look at ionization energies of tin (in eV)

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As you can see, the ionization energies increase monotonically, meaning that each subsequent step to higher charged cation is progressively harder. This means, that if only cation stability (in vacuum) was considered, first all available tin should produce tin (I) compound and only then move further. This is, obviously, not the case.

Still, if you are lazy, it is perfectly OK to go with notion of 'stable' valence states, that are quite easy to memorize. Standard electrode potentials are sometimes quite puzzling from uninvolved point of view and it is easier to use the tables without much thoughts about them.

Still, if we want a reasonably acceptable explanation. We can see, that difference between 2nd and 1st ionization energies (~7 eV) and between 3rd and 4th (~10) is lower than between 2nd and 3rd (~15). This suggests that (II) state should oxidize disproportionally harder. Since cations are stabilized by ion-dipole or ion-ion interactions that strengthens linearly with charge, this suggests extra stability of tin (II) ionic compounds comparing to (I) and (III).

The oxidation state (IV) of tin is a completely different matter. Cations of charge 3+ and higher very rarely truly bear said charge. Typically, they are involved in Lewis acid-base interactions reducing the charge considerably. Tin (IV) compounds have Tin bound covalently. In this case stability of compound is guarded by 8 (18 for d-elements) electron rule.

Mind you, explanation above are not very convincing from quantum mechanical POV, but it is good enough for most cases .

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Answer 2

Answer:

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