On reaction of Lead Sulphide with dilute nitric acid which of the following is not formed?

Step 1: Understand the reagents involved

The reaction involves Lead sulphide ($\ce{PbS}$), which is an insoluble black precipitate, uniquely dissolving upon boiling aggressively with roughly 50% dilute nitric acid ($\ce{HNO3}$).

Step 2: Recall the general reaction equation

Dilute nitric acid serves dually here. It acts as an acid (providing $\ce{H+}$) and importantly as a formidable oxidizing agent. It precisely oxidizes the extremely low $-2$ state sulphide fundamentally into element zero-state Sulphur ($\ce{S}$), getting heavily reduced itself essentially into Nitric oxide ($\ce{NO}$ gas) in the balancing process.

$\ce{3PbS + 8HNO3(dil) -> 3Pb(NO3)2 + 3S v + 2NO ^ + 4H2O}$

Step 3: Analyze the products

From the balanced chemical reaction, the products structurally are:

1. $\ce{Pb(NO3)2}$ (Lead nitrate)
2. $\ce{S}$ (Elemental Sulphur, pale yellow sponge)
3. $\ce{NO}$ (Nitric oxide gas, colorless but turns brown entering air as $\ce{NO2}$)
4. $\ce{H2O}$ (Water)

Step 4: Check against given options

(a) Lead nitrate is formed. (b) Sulphur is formed. (c) Nitric oxide is formed. (d) Nitrous oxide ($\ce{N2O}$) is absolutely not formed. $\ce{N2O}$ normally arises from reacting excessively dilute $\ce{HNO3}$ actively with metals like Zinc.

Step 5: Conclusion

Nitrous oxide (d) is not formed.

Key Points to Remember:

• Any black metal sulphide belonging to Group II (except notably $\ce{HgS}$) robustly dissolves aggressively in warm dilute $\ce{HNO3}$.
• The byproduct heavily features elemental Sulphur ($\ce{S}$) and colorless Nitric oxide ($\ce{NO}$) gas.
• Concentrated $\ce{HNO3}$ conversely oxidizes sulphides terminally into fully oxidized soluble sulphates ($\ce{SO4^{2-}}$), generating brown $\ce{NO2}$ gas.