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 \% ionisation of this complex and coordination number of Cr is 6 , the

JEE Main 202523 JanuaryMorning Shift

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Coordination Compounds

CrCl_3 · xNH_3 can exist as a complex. 0.1 molal aqueous solution...

\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 $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$

A. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

\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 \left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \xrightarrow{\text{ aqueous solution }}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}^{2+}\right]^{2+}+2 \mathrm{Cl}^{-}

Number of ions

=3