$\mathrm{CrCl}_{3} . \mathrm{xNH}_{3}$ 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 $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$)

(C) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$

(A) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$

(B) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$

(D) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Solutions

CrCl_{3} . xNH_{3}

can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of

0.55 8^{\circ} C

. Assuming

100%

ionisation of this complex and coordination number of Cr is 6, the complex will be
(Given

K_{f} = 1.86 K kg mol^{- 1}

)

Answer: (C)

The van't Hoff factor is obtained from the observed freezing-point depression:

Δ T_{f} = i K_{f} m \Rightarrow 0.558 = i (1.86) (0.1) \Rightarrow i = 3

i

value of 3 means the complex dissociates into 3 particles in water. With coordination number 6 for chromium, the complex

[Cr (N H_{3})_{5} Cl] C l_{2}

ionises as

[Cr (N H_{3})_{5} Cl] C l_{2} \to [Cr (N H_{3})_{5} Cl]^{2 +} + 2 C l^{-}

yielding exactly 3 ions, matching

i = 3

; the other coordination formulas give

i = 4

i = 2

i = 1

and are ruled out.

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option