Number of Coordination Sphere Chlorides in Cobalt Complex

To find the number of chloride ions bound inside the coordination sphere of the cobalt(III) complex, we analyze the depression in freezing point of the solution.

1. Calculate the van 't Hoff factor ($i$): The depression in freezing point ($\Delta T_f$) is given by the formula: $\Delta T_f = i \cdot K_f \cdot m$

Where:

• $\Delta T_f = T_f^\circ - T_f = 0^\circ\text{C} - (-0.0558^\circ\text{C}) = 0.0558\text{ K}$
• $K_f = 1.86\text{ K kg mol}^{-1}$
• $m = 0.01\text{ mol kg}^{-1}$

Substituting the given values into the equation: $0.0558 = i \times 1.86 \times 0.01$

$i = \frac{0.0558}{0.0186} = 3$

1. Determine the number of ions produced: Since the complex behaves as a strong electrolyte ($\alpha = 1$), the van 't Hoff factor $i$ equals the total number of ions ($n$) produced per formula unit upon complete dissociation in water: $n = i = 3$

2. Determine the composition of the complex: Since Cobalt is in the $+3$ oxidation state and is paired with chloride ions ($\text{Cl}^-$), there must be a total of 3 chloride ions in the formula to maintain electrical neutrality.

Let the general formula of the cobalt(III) chloride-ammonia complex be: $[\text{Co}(\text{NH}_3)_{6-y}\text{Cl}_y]\text{Cl}_{3-y}$

Upon complete ionization in water: $[\text{Co}(\text{NH}_3)_{6-y}\text{Cl}_y]\text{Cl}_{3-y} \rightarrow [\text{Co}(\text{NH}_3)_{6-y}\text{Cl}_y]^{(3-y)+} + (3 - y)\text{Cl}^-$

The total number of ions produced per formula unit is: $n = 1 + (3 - y) = 4 - y$

Equating this to the calculated value of $n = 3$: $4 - y = 3$ $y = 1$

Thus, the number of chloride ion(s) bound inside the coordination sphere of this complex is 1.