Animated Solution for Chemistry - s and p-Block Elements: The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product $A$. The reaction of $1\text{ mol}$ of $A$ with excess of ${\text{AgNO}}_3$ in aqueous medium gives ...... mol(es) of $\text{Ag}$ (Round off to the nearest integer).

Imagine you are in a chemistry lab, and you drop a piece of white phosphorus into a boiling solution of sodium hydroxide. To prevent it from catching fire, you've carefully maintained an inert atmosphere. What happens next is a beautiful dance of electrons known as a disproportionation reaction.

The phosphorus atoms, initially at an oxidation state of zero, split their paths. Some gain electrons to become phosphine gas ($P{H}_3$), while others lose electrons to form sodium hypophosphite ($Na{H}_{2}P{O}_2$). This sodium hypophosphite is our mysterious Product A.

Now, the plot thickens. We take this Product A and introduce it to an excess of silver nitrate ($AgN{O}_3$). Sodium hypophosphite is notorious for being a powerful reducing agent. Why? Because it harbors two highly reactive P-H bonds. It looks at the silver ions ($A{g}^{+}$) and decides to generously donate its electrons.

Let's track the electron flow. Phosphorus in hypophosphite sits at a +1 oxidation state. It wants to reach its most stable +5 state in phosphoric acid ($H_{3}P{O}_4$). That's a leap of 4 electrons!

On the other side, each silver ion ($A{g}^{+}$) only needs 1 electron to become metallic silver ($A{g}^0$). Therefore, to balance the cosmic ledger of electrons, one molecule of sodium hypophosphite will hand out its 4 electrons to exactly 4 silver ions.

The balanced equation emerges in its full glory:

And there we have it! For every 1 mole of Product A, we precipitate exactly 4 moles of shining metallic silver. The answer is a perfect 4.