Identify the compound and reaction

Yellow lead chromate is: $\mathrm{PbCrO_4}$

On treatment with hot $\mathrm{NaOH}$, lead(II) salts show amphoteric behavior and dissolve by forming a hydroxo complex.

Nature of lead(II) in alkaline medium

Since $\mathrm{Pb(OH)_2}$ is amphoteric, in excess alkali it forms plumbite-type complex ions.

The relevant dissolution can be written as: $\mathrm{PbCrO_4 + 4OH^- \rightarrow [Pb(OH)_4]^{2-} + CrO_4^{2-}}$

So, the lead-containing product is: $\mathrm{[Pb(OH)_4]^{2-}}$

Determine charge and coordination number

For $\mathrm{[Pb(OH)_4]^{2-}}$:

• Charge on complex = $-2$ → dianionic
• Number of ligands attached = $4$ → coordination number 4

Check options

• A: Tetraanionic, coordination number 6 ❌
• B: Neutral, coordination number 4 ❌
• C: Dianionic, coordination number 6 ❌
• D: Dianionic, coordination number 4 ✅

Final answer

The product of lead formed is a dianionic complex with coordination number four.