Step 1: Understanding the Question:
The question asks for the total number of hydrogen atoms belonging to an ammonium ion ($\text{NH}_4^+$) that actively engage in bonding with surrounding solvent water molecules during the hydration/solvation process.
Step 2: Key Formula or Approach:
During aqueous solvation, a positively charged cation stabilizes itself by forming intermolecular hydrogen bonds with the partial negatively charged oxygen atoms ($\text{O}^{\delta-}$) of neighboring water molecules ($\text{H}_2\text{O}$). Any available hydrogen atom bound to a highly electronegative nitrogen atom can act as a hydrogen bond donor.
Step 3: Detailed Explanation:
Let's look at the structure of the ammonium ion ($\text{NH}_4^+$):
The central nitrogen atom is covalently bonded to four distinct hydrogen atoms in a symmetric tetrahedral geometry.
Because of the positive charge residing on the net ion and the high electronegativity of the central nitrogen, all four N-H bonds are highly polarized.
This gives each of the 4 hydrogen atoms a strong partial positive character ($\text{H}^{\delta+}$).
When dissolved in water, each of these 4 hydrogen atoms targets a lone pair on the oxygen atom of an adjacent water molecule to form an intermolecular hydrogen bond:
$$\left[\text{H}_2\text{O}\cdots\text{H-NH}_3\right]^+ \quad\rightarrow\quad \text{extending to all 4 positions}$$
As a result, all 4 hydrogen atoms of the ammonium ion are simultaneously involved in hydrogen bonding with four separate water molecules during solvation, aligning with option (B).
Step 4: Final Answer:
The number of hydrogen atoms bonded during solvation is 4, corresponding to option (B).