Step 1: Understanding the Concept:
The color of transition metal ions is generally due to $d-d$ transitions of electrons. This requires the presence of at least one unpaired electron in the $d$-subshell (i.e., a partially filled $d$-orbital, $d^{1}$ to $d^{9}$).
Step 3: Detailed Explanation:
Let's check the electronic configurations of the given ions ($Ar$ core omitted):
1. \(Cu^{2+}\): Atomic number 29 ($3d^{10}4s^{1}$). $Cu^{2+}$ is $3d^{9}$. It has one unpaired electron. Colored (Blue).
2. \(Fe^{3+}\): Atomic number 26 ($3d^{6}4s^{2}$). $Fe^{3+}$ is $3d^{5}$. It has five unpaired electrons. Colored (Yellow/Brown).
3. \(Ti^{3+}\): Atomic number 22 ($3d^{2}4s^{2}$). $Ti^{3+}$ is $3d^{1}$. It has one unpaired electron. Colored (Purple).
4. \(Zn^{2+}\): Atomic number 30 ($3d^{10}4s^{2}$). $Zn^{2+}$ is $3d^{10}$. It has a completely filled $d$-subshell (no unpaired electrons). Colorless.
Out of the four ions, three ($Cu^{2+}, Fe^{3+}, Ti^{3+}$) impart color.
Step 4: Final Answer:
The number of ions that impart color is 3.