Yellow compound of lead chromate gets dissolved on treatment with hot NaOH solution. The product of lead formed is a :

Lead chromate (PbCrO$$_4$$, yellow) dissolves in hot NaOH solution and we are to identify the lead product.

We observe that PbCrO$$_4$$ dissolves in hot NaOH because Pb$$^{2+}$$ is amphoteric:

$$PbCrO_4 + 4NaOH \to Na_2[Pb(OH)_4] + Na_2CrO_4$$

This reaction produces [Pb(OH)$$_4$$]$$^{2-}$$, which is a dianionic complex (charge = $$+2 - 4 = -2$$) with coordination number 4 (four OH$$^-$$ ligands around Pb$$^{2+}$$).

The correct answer is Option D: Dianionic complex with coordination number four.