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:
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- [Cr(NH\(_3\))\(_6\)]Cl\(_3\)
- [Cr(NH\(_3\))\(_4\)]Cl\(_2\)Cl
- [Cr(NH\(_3\))\(_5\)]Cl\(_2\)
- [Cr(NH\(_3\))\(_3\)]Cl\(_3\)
The Correct Option is C
Approach Solution - 1
The problem involves the ionization of a coordination complex and the depression in freezing point to deduce the correct complex structure.
- Given that the depression in freezing point (\(\Delta T_f\)) is 0.558°C for a 0.1 molal solution, we can use the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\)where \(i\) is the Van't Hoff factor, \(K_f\) is the cryoscopic constant (for water \(K_f = 1.86 \, \text{°C kg/mol}\)), and \(m\) is the molality.
- The value of \(i\) can be calculated since: \(i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.558}{1.86 \cdot 0.1} \approx 3\). This means the complex dissociates into 3 ions.
- According to the coordination number of Cr being 6, we evaluate each of the provided complex structures:
- [Cr(NH\(_3\))\(_6\)]Cl\(_3\) dissociates into 1 complex ion and 3 Cl\(^-\) ions, resulting in 4 ions total. This does not match \(i = 3\).
- [Cr(NH\(_3\))\(_4\)]Cl\(_2\)Cl dissociates into 1 complex ion and 3 Cl\(^-\) ions, resulting in 4 ions total. This does not match \(i = 3\).
- [Cr(NH\(_3\))\(_5\)]Cl\(_2\) dissociates into 1 complex ion and 2 Cl\(^-\) ions, resulting in 3 ions total. This matches \(i = 3\).
- [Cr(NH\(_3\))\(_3\)]Cl\(_3\) dissociates into 1 complex ion and 3 Cl\(^-\) ions, resulting in 4 ions total. This does not match \(i = 3\).
- Thus, the correct structure of the complex is [Cr(NH\(_3\))\(_5\)]Cl\(_2\), which is consistent with the value of \(i = 3\) from the empirical data.
- Therefore, the correct answer is [Cr(NH\(_3\))\(_5\)]Cl\(_2\).
Approach Solution -2
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\(_2\).
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