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CrCl3·xNH3 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 complex will be (Given K_f = 1.86 K kg mol^-1) (1) [Cr(NH3)6]Cl3 (2) [Cr(NH3)4Cl2]Cl (3) [Cr(NH3)3Cl]Cl2 (4) [Cr(NH3)3Cl3]

1. [Cr(NH3)6]Cl3
2. [Cr(NH3)4Cl2]Cl
3. [Cr(NH3)3Cl]Cl2
4. [Cr(NH3)3Cl3]

Correct answer: [Cr(NH3)4Cl2]Cl

Solution

The correct option, [Cr(NH3)4Cl2]Cl, accounts for the observed depression in freezing point, which indicates the total number of particles in solution. With a coordination number of 6 for Cr and assuming complete ionization, this complex would yield 3 particles in solution (1 complex ion and 2 chloride ions), resulting in a calculated freezing point depression that matches the given data.

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