I read that $\ce{NaOH}$ precipitates lead hydroxide which dissolves with excess hydroxide.
In some sources, I read that lead chloride dissolves due to the formation of complex $\ce{[Pb(OH)4]^2-}$ but my teacher taught that it is because of the formation of $\ce{Na2PbO2}$
I don't think that both are same. I cannot find much information about $\ce{Na2PbO2}$ either.
Which of these two is actually formed?
Lead hydroxide do dissolve in excess of alkali solution forming complex (the solution becomes clear):
$$\ce{Pb(OH)2(s) + 2OH-(aq) -> [Pb(OH)4]^2-(aq)}$$
There can be confusion because some articles(e.g. here) says it lead to formation of sodium plumbate(II) but plumbate is poorly defined as it can considered lead oxyanion or oxides of lead.
Wikipedia further mentions:
In solution, lead(II) hydroxide is a somewhat weak base, forming lead(II) ion, $\ce{Pb^2+}$ under weakly acidic conditions. This cation hydrolyzes and under progressively increasing alkaline conditions, forms $\ce{Pb(OH)+}$, $\ce{Pb(OH)2(aq)}$, $\ce{Pb(OH)3−}$, and other polynuclear species like $\ce{Pb4(OH)4^4+}$, $\ce{Pb3(OH)4^2+}$, $\ce{Pb6O(OH)6^4+}$
