JEE Mains · Chemistry · STD 12 - 5. Co-ordination chemistry
\(\mathrm{CrCl}_3 \cdot \mathrm{xNH}_3\) can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of \(0.558^{\circ} \mathrm{C}\). Assuming \(100 \%\) ionisation of this complex and coordination number of Cr is 6 , the complex will be _______.
(Given \(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) )
- A \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)
- B \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\)
- C \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\)
- D \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}\)
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta T_f=i K_f m \\ & 0.558=i \times 1.86 \times 0.1 \\ & i=\frac{0.558}{0.186}=3\end{aligned}\) Number of ions when 100\% ionisation takes place \(=3\)…
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