Question

CrCl\(_3\).xNH\(_3\) can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100\% ionization of this complex and coordination number of Cr is 6, the complex will be:

• A. [Cr(NH\(_3\))\(_6\)]Cl\(_3\)
• B. [Cr(NH\(_3\))\(_4\)]Cl\(_2\)Cl
• C. [Cr(NH\(_3\))\(_5\)]Cl\(_2\)
• D. [Cr(NH\(_3\))\(_3\)]Cl\(_3\)

Hint:

In colligative properties like freezing point depression, the key is the number of particles in solution. Higher ionization or dissociation results in a higher change in freezing point.

Answer 1

The Correct Option is C

Given: \[ \Delta T_f = 0.558°C \quad \text{and} \quad k_f = 1.86 \, \text{K} \, \text{kg/mol} \] We know that: \[ \Delta T_f = i \times k_f \times m \] where \(i\) is the van't Hoff factor (number of ions produced per formula unit), \(m\) is the molality, and \(k_f\) is the cryoscopic constant.
Given that the molality of the solution is 0.1 m, we have: \[ \Delta T_f = i \times 1.86 \times 0.1 \] Substituting the given value of \(\Delta T_f\): \[ 0.558 = i \times 1.86 \times 0.1 \] \[ i = \frac{0.558}{1.86 \times 0.1} = 3 \] This implies that the complex ion must dissociate into 3 ions in solution. The complex that corresponds to \(i = 3\) is [Cr(NH\(_3\))\(_5\)]Cl\(_2\), as it would dissociate into 1 Cr\(^3+\) ion and 2 Cl\(^-\) ions.
Thus, the correct complex is [Cr(NH\(_3\))\(_5\)]Cl\(_2\).