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Other than the X-ray difference , how co...

Other than the X-ray difference , how could be the following pairs of isomers be distinguished from one another by :
`[Cr(NH_(3))_(6)][Cr(NO_(2))_(6)] and [Cr(NH_(3))_(4)(NO_(2))_(2)][Cr(NH_(3))_(2)(NO_(2))_(4)]`

A

cryoscopic method

B

measurement of molar conductance

C

measuring magnetic moment

D

observing their colours.

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The correct Answer is:
To distinguish between the pairs of isomers `[Cr(NH₃)₆][Cr(NO₂)₆]` and `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]`, we can follow these steps: ### Step 1: Identify the Isomers The first pair consists of two different complexes: 1. `[Cr(NH₃)₆][Cr(NO₂)₆]` 2. `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]` ### Step 2: Analyze the Ionic Composition - The first complex `[Cr(NH₃)₆][Cr(NO₂)₆]` contains two chromium ions, one with six ammonia ligands and one with six nitro ligands. - The second complex `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]` contains two chromium ions, one with four ammonia and two nitro ligands, and the other with two ammonia and four nitro ligands. ### Step 3: Use Molar Conductance Molar conductance can be used to distinguish these complexes. Molar conductance depends on the number of ions and their charges in solution. - For `[Cr(NH₃)₆][Cr(NO₂)₆]`, the charge on the complex is +3 for both chromium ions (total charge = +3 + +3 = +6). - For `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]`, the charge on the first chromium ion is +1 and on the second is also +1 (total charge = +1 + +1 = +2). ### Step 4: Conclusion Since the total charge and the number of ions differ between the two pairs of isomers, measuring the molar conductance will provide a clear distinction between them. ### Final Answer The isomers can be distinguished by measuring their molar conductance due to the difference in their ionic charges. ---

To distinguish between the pairs of isomers `[Cr(NH₃)₆][Cr(NO₂)₆]` and `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]`, we can follow these steps: ### Step 1: Identify the Isomers The first pair consists of two different complexes: 1. `[Cr(NH₃)₆][Cr(NO₂)₆]` 2. `[Cr(NH₃)₄(NO₂)₂][Cr(NH₃)₂(NO₂)₄]` ### Step 2: Analyze the Ionic Composition ...
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