$\mathrm{CrCl}_3 \cdot \mathrm{xNH}_3$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $0.558^{\circ} \mathrm{C}$. Assuming $100 \%$ ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

<p>Given the problem, we have the following information:</p> <p><p>Depression in freezing point, $\Delta T_f = 0.558^\circ \text{C}$</p></p> <p><p>Freezing point depression constant, $k_f = 1.86 \, \text{K kg mol}^{-1}$</p></p> <p><p>Molal concentration of the solution, $0.1 \, \text{mol kg}^{-1}$</p></p> <p><p>Coordination number of Cr is 6.</p></p> <p>Using the formula for freezing point depression:</p> <p>$ \Delta T_f = i \times k_f \times m $</p> <p>where $i$ is the van 't Hoff factor, $k_f$ is the freezing point depression constant, and $m$ is the molality of the solution.</p> <p>Substituting the known values:</p> <p>$ 0.558 = i \times 1.86 \times 0.1 $</p> <p>Solving for $i$:</p> <p>$ i = \frac{0.558}{1.86 \times 0.1} = 3 $</p> <p>The van 't Hoff factor $i = 3$ indicates that the complex ionizes into three particles in solution. Given the coordination number of Cr is 6, we analyze the possible complexes:</p> <p><p>$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$</p></p> <p><p>$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$</p></p> <p><p>$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$</p></p> <p><p>$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$</p></p> <p>The complex that gives three ions upon dissociation could be $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$. This is because the complex will dissociate as:</p> <p>$ \left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} + 2 \, \mathrm{Cl}^- $</p> <p>This results in three ions total, matching the van 't Hoff factor $i = 3$.</p> <p>Therefore, the complex is $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.</p>