Q3 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
An aqueous solution of $\mathrm{CrCl}_3 .6 \mathrm{H}_2 \mathrm{O}$ (Molar mass $=266.5 \mathrm{~g} / \mathrm{mol}$ ) containing 2.665 g of the solute after processing, when treated with excess of $\mathrm{AgNO}_3$ gave 2.87 g of AgCl (Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$ ) Choose the correct formula of the compound which give these results.
- A$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right] \cdot 3 \mathrm{H}_2 \mathrm{O}$
- Correct answerB$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$
- C$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} .2 \mathrm{H}_2 \mathrm{O}$
- D$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$
Solution
The number of moles of the complex $\mathrm{CrCl}_3 \cdot 6\mathrm{H}_2\mathrm{O}$ is calculated as:
Moles of complex $= \dfrac{2.665 \text{ g}}{266.5 \text{ g/mol}} = 0.01 \text{ mol}$.
The number of moles of $\mathrm{AgCl}$ precipitated is calculated as:
Moles of $\mathrm{AgCl} = \dfrac{2.87 \text{ g}}{143.5 \text{ g/mol}} = 0.02 \text{ mol}$.
The reaction with $\mathrm{AgNO}_3$ precipitates only the chloride ions present outside the coordination sphere. Let $x$ be the number of ionizable chloride ions per mole of the complex. The stoichiometry of the reaction is:
Moles of $\mathrm{AgCl} = x \times \text{Moles of complex}$
$0.02 = x \times 0.01$
$x = 2$.
Since there are 2 ionizable chloride ions, the formula must contain 2 chloride ions outside the coordination sphere. Among the given options, $\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}\right]\mathrm{Cl}_2 \cdot \mathrm{H}_2\mathrm{O}$ satisfies this condition.
Answer: $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$
Moles of complex $= \dfrac{2.665 \text{ g}}{266.5 \text{ g/mol}} = 0.01 \text{ mol}$.
The number of moles of $\mathrm{AgCl}$ precipitated is calculated as:
Moles of $\mathrm{AgCl} = \dfrac{2.87 \text{ g}}{143.5 \text{ g/mol}} = 0.02 \text{ mol}$.
The reaction with $\mathrm{AgNO}_3$ precipitates only the chloride ions present outside the coordination sphere. Let $x$ be the number of ionizable chloride ions per mole of the complex. The stoichiometry of the reaction is:
Moles of $\mathrm{AgCl} = x \times \text{Moles of complex}$
$0.02 = x \times 0.01$
$x = 2$.
Since there are 2 ionizable chloride ions, the formula must contain 2 chloride ions outside the coordination sphere. Among the given options, $\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_5\mathrm{Cl}\right]\mathrm{Cl}_2 \cdot \mathrm{H}_2\mathrm{O}$ satisfies this condition.
Answer: $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}$
Hence, the correct answer is option: 'B'.