As far as I know tin likes to form complex compound with OH- ligands. The formula of the product depends on the oxidation state of the tin and kind of ligands. As far As I know Sn2+ coordination number is very often III, because of the electron configuration of the Sn2+ cation. Hence Sn(OH)(sub)3(sup-) can be formed. On the other hand, ligand complexes of tin at this oxidation state very often form more complex structure, because this kind of ions like to form dimers, threemers etc. - Sn2+ has additional free electron pair and attracts all electrophiles from both: solution and other molecules.

Sn (II) gets oxidized to Sn (IV) by the following mechanism:

2 KOH react to give (OH2) 2 Sn (OH) 2 by losing 2KCl. The 2OH (-) are then substituted by

2 Cl (-) form Sn (OH)4 which further loses 2H2O to form SnO2

(OH2)2 SnCl2-+2OH (-) -2 Cl (-) = (OH2)2 Sn (OH) 2 + 2Cl (-) -2HCl= Sn (OH)4- 2H2O =SnO2