JEE Main 2025 — Coordination Compounds Question with Solution

Given the problem, we have the following information:

Depression in freezing point, $\Delta T_f=0.558^{\circ }\text{C}$

Freezing point depression constant, $k_f=1.86{\text{K kg mol}}^{-1}$

Molal concentration of the solution, $0.1{\text{mol kg}}^{-1}$

Coordination number of Cr is 6.

Using the formula for freezing point depression:

$\Delta T_f=i\times k_f\times m$

where $i$ is the van 't Hoff factor, $k_f$ is the freezing point depression constant, and $m$ is the molality of the solution.

Substituting the known values:

$0.558=i\times 1.86\times 0.1$

Solving for $i$:

$i=\frac{0.558}{1.86\times 0.1}=3$

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:

$\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2$

$\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$

$\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$

$\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_3{\mathrm{Cl}}_3\right]$

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:

${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]}^{2+}+2{\mathrm{Cl}}^{-}$

This results in three ions total, matching the van 't Hoff factor $i=3$.

Therefore, the complex is $\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2$.