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

🧠 Lead's Amphoteric Reflex

Lead chromate ($\mathrm{PbCrO_4}$, the bright yellow paint pigment) reacts with hot NaOH to release lead as the plumbite/plumbate anion. The signature product:

$\mathrm{PbCrO_4 + NaOH (\text{hot}) \to Na_2[Pb(OH)_4]} + \ldots$

That product is $[\mathrm{Pb(OH)_4}]^{2-}$ — a dianionic complex with coordination number 4 (four hydroxide donors).

🗺️ Identify the Two Numbers

Charge of the lead complex. Pb is in $+2$, four $\mathrm{OH^-}$ contribute $-4$. Net charge $= +2 + 4(-1) = -2$. So dianionic.

Coordination number. Four hydroxide ligands bound to Pb → CN = 4.

So: dianionic, CN = 4 → option (4).

⚡ The "Amphoteric Hydroxide Cage" Pattern

Amphoteric metal oxides/hydroxides ($\mathrm{Al}$, $\mathrm{Sn}$, $\mathrm{Pb}$, $\mathrm{Zn}$, $\mathrm{Cr}$) all dissolve in excess NaOH to form tetrahydroxido anions:

• $[\mathrm{Al(OH)_4}]^-$ (aluminate)
• $[\mathrm{Sn(OH)_4}]^{2-}$ (stannite)
• $[\mathrm{Pb(OH)_4}]^{2-}$ (plumbite)
• $[\mathrm{Zn(OH)_4}]^{2-}$ (zincate)

All have CN = 4 in this stoichiometry. Memorise the family.

⚠️ CN 6 Is the Wrong Reflex

Octahedral coordination is so common in transition-metal complexes that students reach for CN = 6 by default. Lead in $[\mathrm{Pb(OH)_4}]^{2-}$ is firmly tetrahedral / square planar — CN = 4. Always verify the ligand count before guessing the geometry.

$\boxed{\text{Answer: (4) Dianionic complex with coordination number four}}$