What is #Pb(IV) + NiCl_2 ->#?

#"Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl"_2(g)#

This is a "simplified" double-replacement reaction (i.e. we ignore some anion coupled with the #"Pb"^(4+)#), so we initially get---noting that #IV = 4#:

#"Pb"^(4+)(aq) + "NiCl"_2(aq) -> "PbCl"_4(s) + "Ni"^(2+)(aq)#

The #"NiCl"_2(aq)# is rather deliquescent (i.e. hygroscopic), and sops up surrounding moisture to form an aqueous solution.

#"PbCl"_4(s)# is going to be written as a solid for now, but it decomposes in aqueous solution into #"PbCl"_2(s)# and #"Cl"_2(g)#. #"Ni"^(2+)# is evidently going to form, but to balance the charge, we must rebalance the mass.

#"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_4(s) + 2"Ni"^(2+)(aq)#

And now, we incorporate the decomposition.

#"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> cancel("PbCl"_4(s)) + 2"Ni"^(2+)(aq)#

#cancel("PbCl"_4(s)) stackrel("H"_2"O"" ")(->) "PbCl"_2(s) + "Cl"_2(g)#

#"-----------------------------------------------------------------"#

#"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_2(s) + 2"Ni"^(2+)(aq) + "Cl"_2(g)#

We expect the nickel(II) chloride to dissociate, and thus we are not done:

#"Pb"^(4+)(aq) + cancel(2"Ni"^(2+)(aq)) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + cancel(2"Ni"^(2+)(aq)) + "Cl"_2(g)#

This gives us:

#color(blue)("Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl"_2(g))#