The values of van't Hoff factors for \(CaCl_2\), \(Na_2SO_4\) and \(KCl\) respectively are:

3, 3, 2

The correct answer is option 2. 3, 3, 2.

The van't Hoff factor is a measure of the number of particles a compound dissociates into when it dissolves in water. It is important for understanding colligative properties (properties that depend on the number of particles in a solution, not their identity), such as boiling point elevation, freezing point depression, and osmotic pressure.

If a compound does not dissociate (e.g., sugar), \(i = 1\). If a compound dissociates into multiple ions, \(i\) will be equal to the total number of ions produced.

Let us Analyze Each Compound:

\(CaCl_2\) (Calcium Chloride):

Dissociation in Water: \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^{-}\)

When \(CaCl_2\) dissolves in water, it dissociates into one \(Ca^{2+}\) ion and two \(Cl^{-}\) ions. The total number of ions produced is 1 (from \(Ca^{2+}\)) + 2 (from \(Cl^{-}\)) = 3. Therefore, the van't Hoff factor (\(i\)) for \(CaCl_2\) is 3

\(Na_2SO_4\) (Sodium Sulphate):

Dissociation in Water: \(Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}\)

When \(Na_2SO_4\) dissolves in water, it dissociates into two \(Na^{+}\) ions and one \(SO_4^{2-}\) ion. The total number of ions produced is 2 (from \(Na^{+}\)) + 1 (from \(SO_4^{2-}\)) = 3. Therefore, the van't Hoff factor (\(i\)) for \(Na_2SO_4\) is 3.

\(KCl\) (Potassium Chloride):

Dissociation in Water: \(KCl \rightarrow K^{+} + Cl^{-}\)

When \(KCl\) dissolves in water, it dissociates into one \(K^{+}\) ion and one \(Cl^{-}\) ion. The total number of ions produced is 1 (from \(K^{+}\)) + 1 (from \(Cl^{-}\)) = 2. Therefore, the van't Hoff factor (\(i\)) for \(KCl\) is 2.

Thus, the correct sequence of van't Hoff factors for the compounds \(CaCl_2\), \(Na_2SO_4\), and \(KCl\) is \(3, 3, 2\), which corresponds to option \(2\).