Question

A complex is represented as \(\mathrm{CoCl}_{3} x \mathrm{NH}_{3}\). Its \(0.1\) molal solution in water shows \(\Delta T_{f}=0.558 \mathrm{~K}\). \(K_{f}\) for \(\mathrm{H}_{2} \mathrm{O}\) is \(1.86 \mathrm{~K}\) molality \(^{-1}\). Assuming \(100 \%\) ionisation of complex and co-ordination number of Co is six, calculate formula of complex: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) none of these

Short answer

The correct formula of the complex, considering 100% ionisation and a coordination number of six, is (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\).

Step-by-step solution

Finding the Molal Freezing Point Depression

Given the molal solution has a \(\Delta T_f = 0.558 \, K\), and the cryoscopic constant \(K_f\) for water is \(1.86 \, K\cdot{kg}/mol\), the molality \(m\) of the solution can be determined using the formula \(\Delta T_f = K_f \times m\).

Calculating the Molality of the Solution

Rearrange the equation to solve for molality: \(m = \frac{\Delta T_f}{K_f}\). Substitute the given values: \(m = \frac{0.558}{1.86}\).

Determining the Number of Particles in Solution

Given 100% ionisation, the complex will break into one \(\mathrm{Co}\) complex cation and three \(\mathrm{Cl}^-\) anions. The coordination number of \(\mathrm{Co}\) is six, which means that six \(\mathrm{NH}_3\) molecules are attached to \(\mathrm{Co}\).

Finding the Van't Hoff Factor

Since the complex ionises into four particles (one cation and three anions), the Van't Hoff factor \(i\) is 4.

Calculating the Actual Molality

The actual molality \(m\) needs to account for the Van't Hoff factor by dividing the calculated molality by \(i\): \(m' = \frac{m}{i}\).

Comparing with Given Complexes

The calculated molality that considers ionisation should match the theoretical molality based on the formula of the complex. Compare the formulas given in the options with the calculated molality to find the correct answer.

Key concepts

Molal Freezing Point Depression

The phenomenon of freezing point depression is essential for understanding how solutions behave when they freeze. Simply put, when a substance is dissolved in a liquid, the freezing point of that liquid decreases. This effect is directly proportional to the number of particles of solute in the solution.

In the context of coordination chemistry, if a complex ion dissociates fully in solution, it increases the number of particles, thereby affecting the freezing point depression more significantly. The equation to calculate this depression is given as \( \Delta T_f = K_f \times m \), where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.

Cryoscopic Constant

The cryoscopic constant (\( K_f \) characterizes the freezing point depression and is specific to the solvent. It represents the freezing point lowering of the solvent upon the addition of a one molal solution of a non-volatile solute. Crucially, it takes into account the properties of the solvent, such as its enthalpy of fusion and molal volume.

When chemists need to determine the molecular weight of a compound or understand the solution's behavior, the cryoscopic constant provides a valuable point of reference. For example, water has a cryoscopic constant of \( 1.86 \ K \cdot kg/mol \) which is utilized to calculate the molal freezing point depression in the exercise.

Van't Hoff Factor

The Van't Hoff factor, denoted as \( i \), is a key concept when investigating solutions, representing the number of particles a compound dissociates into when in solution. For example, let's consider a simple salt like NaCl. When it dissolves, it dissociates into two ions: Na\textsuperscript{+} and Cl\textsuperscript{-}, therefore having a Van’t Hoff factor of 2.

In coordination compounds, this factor can vary widely because the dissociation depends on the complex's structure. If a complex like \( [Co(NH_3)_6]Cl_3 \) dissociates completely, it results in one \( [Co(NH_3)_6]^{3+} \) ion and three \( Cl^- \) ions, leading to an \( i \) of 4. This factor is used to adjust the molal concentration to reflect the actual number of particles in solution that contribute to properties like freezing point depression.

Complex Ion Dissociation

In coordination chemistry, complex ions are composed of a central metal ion surrounded by ligands. These complexes can dissociate in solution to a varying extent based on their structure and stability.

The extent of dissociation is crucial when predicting properties such as electrical conductivity, osmotic pressure, and freezing point depression. In our example, assuming 100% ionisation, the complex \( [Co(NH_3)_6]Cl_3 \) dissociates completely into one \( [Co(NH_3)_6]^{3+} \) ion and three \( Cl^- \) ions. Understanding this dissociation helps us to accurately calculate the number of particles in solution and, in turn, solve problems like the one presented in the exercise.