# Describe the formation of the [PbCl4]2- complex ion

The $$\ce{[PbCl4]^{2-}}$$ complex ion is formed when $$\ce{PbCl2}$$ is exposed to excess $$Cl^-$$ ions in solution as explained here and here through the following reversible reaction:

$$\ce{PbCl2(s) + 2Cl^-(aq) <=> [PbCl4]^{2-}(aq)}$$ What I wish to understand is the mechanism behind this reaction. I managed to find the mechanism to a similar complex ion formation of $$\ce{[CuCl4]^{2-}}$$. It is explained here that $$\ce{[CuCl4]^{2-}}$$ is formed by the hybridisation of the $$Cu^{2+}$$ 4s and 4p orbitals: $$sp^3:\boxed{\uparrow \ }\boxed{\uparrow \ }\boxed{\uparrow \ }\boxed{\uparrow \ }$$

Would the $$\ce{[PbCl4]^{2-}}$$ complex ion be obtained in a similar way by $$Pb^{2+}$$ forming the following hypridisation orbitals? $$sp^3:\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \ }\boxed{\uparrow \ }$$ or no need for the above hybridisation since the 6s orbital was already filled and there is space in the empty 6p orbital so the configuration for the complex ion would be as follows? $$6p:\boxed{\uparrow \downarrow}\boxed{\uparrow \ }\boxed{\uparrow \ }$$

• There's no hybridisation in there. – Mithoron Jan 20 at 21:39
• chemistry.stackexchange.com/questions/8717/… – Mithoron Jan 20 at 21:49
• Thank you for your input, Mithoron. The inert pair effect would explain why PbCl2 is more stable than PbCl4 and also why no hybridisation takes place in the above (the outer s orbital acting as an inner orbital rather than a valence orbital). Does this mean that when forming the complex ion, the empty p orbital is simply filled like my second suggestion in my question? – JulianS Jan 21 at 21:55
• It's more like octet rule hardly works here. Such cases can be fitted to it like SF6, though chemistry.stackexchange.com/a/5242/9961 chemistry.stackexchange.com/a/49844/9961 – Mithoron Jan 22 at 0:20
• sight best don't mention this hypervalent menace here. There was a reason I put these links specifically. – Mithoron Jan 23 at 22:59