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Given that $\ce{PtCl4.2HCl}$ when reacted with excess of $\ce{AgNO3}$ produced zero moles of $\ce{AgCl}$, I understand that all the chlorine atoms are inside the coordination sphere. But what will the formula and thus name of this complex be? And what could the coordination polyhedron look like? Would the name be $$\ce{H2[Pt(Cl)6]}$$ or $$\ce{[Pt(Cl)4(HCl)2]}$$

If someone could explain why the name is one and not the other, I'd be very grateful. Thank you in advance!

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Platinum(IV) chloride adopt an octahedral coordination geometry. This geometry is achieved by forming a polymer wherein half of the chloride ligands bridge between the platinum centers. So, whenever it is dissolved in some acids, say $\ce{HCl}$, the chloride bridging ligands breaks and the molecules gets added giving $\ce{H2PtCl6}$, also called chloroplatinic acid.

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But the story doesn't end here. Chloroplatinic acid is highly hygroscopic. It pulls water from air and forms an ionic complex consisting of hydronium ($\ce{H3O+}$) cation and hexachloroplatinate ($\ce{PtCl6^{2-}}$) anion. Its actual formula is $\ce{[H3O]2[PtCl6](H2O)_x}$ where $\pu{0 ≤ x ≤ 6}$. But for simplicity, we write the former formula.

enter image description here

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  • $\begingroup$ but which one out of $$\ce{H2[Pt(Cl)6]}$$ and $$\ce{[Pt(Cl)4(HCl)2]}$$ is the right formula for the coordination complex i mentioned? besides i dont this is platinum chloride, or is it? $\endgroup$ – sam Oct 2 '20 at 9:24
  • $\begingroup$ @sam the first one is the correct one. Its name is chloroplatinic acid. $\endgroup$ – Nilay Ghosh Oct 2 '20 at 9:27

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