[firzzy87]

How the product might be ?

1) Reaction of tin (IV) ion and NaOH (limited ans excess)
2) Reaction of Hg(II) ion and KI ( limited and excess)
3) Reaction of Cu(II) ion n KI (limited and excess)

thanks again ...

[Will]

How the product might be ?

1) Reaction of tin (IV) ion and NaOH (limited ans excess)
2) Reaction of Hg(II) ion and KI ( limited and excess)
3) Reaction of Cu(II) ion n KI (limited and excess)

Sn(IV)_(aq) with limited OH^- gives precipitate of Sn(OH)_4(s)
                         excess OH^- gives the stannate(IV) complex: [Sn(OH)₆]^2-_(aq)

I'm not sure about the Hg one- you probably just get HgI₂.

For the last one- I'm again unsure about the difference between the reaction with limited and excess KI- I've done the reaction before and you just get I₂ and a precipitate of CuI. See this link.

[wereworm73]

Mercury(II) ions in limited KI forms a yellow or red HgI₂ precipitate, but excess KI will dissolve it to give you a solution of K₂HgI₄.

[woelen]

Sn(4+) ions do not exist. In water you can have soluble tin(IV) compounds, e.g. the stannates, as mentioned above, but the free Sn(4+) ion simply is hydrolysed in water and cannot exist as such. It might be that in very strongly concentrated HCl, some SnCl4 may exist in solution, but then not as Sn(4+) and Cl(-) ions, but as covalent compound SnCl4. I, however, even doubt that this can exist in conc. HCl, even under such conditions I think it hydrolyses as metastannic acid, SnO(OH)4.

Hg(2+) ions form HgI2 with iodide ion, which is very bright orange/red. On addition of excess iodide, the colorless complex ion HgI4(2-) is formed.