Given: Molality of the solution ($m$) = $0.0020 \text{ m}$ Freezing point of the solution = $-0.00732^\circ\text{C}$ Depression in freezing point ($\Delta T_f$) = $0^\circ\text{C} - (-0.00732^\circ\text{C}) = 0.00732^\circ\text{C}$ Molal depression constant ($K_f$) = $1.86^\circ\text{C m}^{-1}$

The formula for depression in freezing point is: $\Delta T_f = i \times K_f \times m$ Where $i$ is the van't Hoff factor (which represents the number of moles of particles produced per mole of the solute).

Substituting the given values into the equation: $0.00732 = i \times 1.86 \times 0.0020$ $0.00732 = i \times 0.00372$ $i = \frac{0.00732}{0.00372} \approx 1.967 \approx 2$

Thus, $1 \text{ mol}$ of the ionic compound produces $2$ moles of ions in the solution. (The complex most likely ionizes as: $[\text{Co(NH}_3)_5(\text{NO}_2)]\text{Cl} \rightleftharpoons [\text{Co(NH}_3)_5(\text{NO}_2)]^+ + \text{Cl}^-$).