JEE Chemistry 2025 Question with Solution

Given: Depression in freezing point is $\Delta T_f = 0.558^\circ \text{C}$, molality is $m = 0.1$ molal, and for water $K_f = 1.86 \, \text{K kg mol}^{-1}$.

Find: The correct complex of CrCl$_3$.xNH$_3$ assuming complete ionization and coordination number $6$.

Use freezing point depression:

$$\Delta T_f = iK_f m$$

Substituting the given values:

$$0.558 = i \times 1.86 \times 0.1$$

Therefore,

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

So, the complex must produce $3$ ions in aqueous solution.

Now check the options:

• [Cr(NH$_3$)$_6$]Cl$_3$ gives $4$ ions: one complex ion and $3$ chloride ions.
• [Cr(NH$_3$)$_4$]Cl$_2$Cl also gives $4$ ions according to the provided solution.
• [Cr(NH$_3$)$_5$]Cl$_2$ gives $3$ ions: one complex ion and $2$ chloride ions.
• [Cr(NH$_3$)$_3$]Cl$_3$ gives $4$ ions according to the provided solution.

Hence the structure consistent with $i = 3$ is [Cr(NH$_3$)$_5$]Cl$_2$.

Therefore, the correct option is C.

Given:

$$\Delta T_f = 0.558^\circ \text{C}, \qquad K_f = 1.86 \, \text{K kg mol}^{-1}, \qquad m = 0.1$$

Find: Which complex gives the required van't Hoff factor.

From the relation:

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

Substitute values:

$$0.558 = i \times 1.86 \times 0.1$$

So,

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

This means one formula unit must dissociate into $3$ particles.

Now compare with the options using the coordination number of chromium as $6$ and the ion count stated in the extracted solution:

1. [Cr(NH$_3$)$_6$]Cl$_3$ gives $4$ ions, so it is not correct.
2. [Cr(NH$_3$)$_4$]Cl$_2$Cl gives $4$ ions, so it is not correct.
3. [Cr(NH$_3$)$_5$]Cl$_2$ gives $3$ ions, so it matches.
4. [Cr(NH$_3$)$_3$]Cl$_3$ gives $4$ ions, so it is not correct.

Therefore, the correct complex is [Cr(NH$_3$)$_5$]Cl$_2$, so the correct option is C.