To determine in which reactions $ H_2PO_4^- $ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases. According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.

Let's analyze each reaction :

1. 1. $ H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^- $

   
   
   • In this reaction, $ H_3PO_4 $ is donating a proton to $ H_2O $, forming $ H_3O^+ $ and $ H_2PO_4^- $.
   
   
   • $ H_2PO_4^- $ is the product of this reaction and is not acting as an acid here.



2. 2. $ H_2PO_4^- + H_2O \to HPO_4^{2−} + H_3O^+ $

   
   
   • In this reaction, $ H_2PO_4^- $ donates a proton to $ H_2O $, resulting in $ HPO_4^{2−} $ and $ H_3O^+ $.
   • 
     $ H_2PO_4^- $ is acting as an acid in this reaction.
3. 
   

   3. $ H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-} $

   • 
     This reaction is not correctly balanced and does not follow standard chemical reaction rules. The product $ O^{2-} $ is highly unlikely in aqueous solutions due to its reactivity.
   • 
     A more correct reaction would be $ H_2PO_4^- + OH^- \to HPO_4^{2−} + H_2O $. In this corrected reaction, $ H_2PO_4^- $ is donating a proton to $ OH^- $, which makes it an acid.

Given the information, the correct option is :

Option A : (ii) only

This is because in reaction (ii), $ H_2PO_4^- $ is clearly acting as an acid by donating a proton to water. In reaction (i), $ H_2PO_4^- $ is not acting as an acid but rather is formed as a product. Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $ H_2PO_4^- $ acting as an acid.