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

Solution

Chemistry ma’am · 4 steps

Excess hot alkali turns Pb²⁺ into [Pb(OH)₄]²⁻: charge minus two, four ligands.

1. Why it dissolves. Lead is amphoteric. Excess hot NaOH pulls Pb²⁺ away from chromate as a soluble hydroxo complex, and chromate stays in solution as Na₂CrO₄.

   PbCrO₄ + 4NaOH → Na₂[Pb(OH)₄] + Na₂CrO₄

2. Charge on the complex. No redox happens, so lead stays +2. Four OH⁻ ligands give +2 + 4(-1) = -2, so [Pb(OH)₄]²⁻ is dianionic.
3. Coordination number. Four OH⁻ groups are bonded to Pb, so the coordination number is four. This tetrahydroxoplumbate(II) ion is also written as plumbite, PbO₂²⁻.
4. So. The lead product Na₂[Pb(OH)₄] contains a dianionic complex with coordination number four, which is option (d).

The idea to keep. Amphoteric Pb²⁺, Zn²⁺ and Sn²⁺ dissolve in excess alkali as [M(OH)₄]²⁻. Fix the oxidation state first, then the charge follows.

Answer from the NTA final answer key. Solution by PrepWiser.