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Current Question (ID: 18691)

Question:

$\text{Given the following pair of isomers:}$ $[\text{Cr} (\text{NH}_3)_6][\text{Cr} (\text{NO}_2)_6]$ $\text{and}$ $[\text{Cr} (\text{NH}_3)_4(\text{NO}_2)_2][\text{Cr} (\text{NH}_3)_2(\text{NO}_2)_4]$ $\text{Aside from X-ray diffraction, how could the following pairs of isomers be distinguished from one another?}$

Options:

1. $\text{Cryoscopic method}$
2. $\text{Measuring their molar conductance}$
3. $\text{Measuring their magnetic moments}$
4. $\text{Observing their colors}$

Solution:

$\text{HINT: Charge on complex is different.}$ $\text{Explanation:}$ $1.$ $[\text{Cr} (\text{NH}_3)_6][\text{Cr} (\text{NO}_2)_6] \text{ and } [\text{Cr} (\text{NH}_3)_4(\text{NO}_2)_2][\text{Cr} (\text{NH}_3)_2(\text{NO}_2)_4]$ $\text{have the same number of ions, so the cryoscopic method cannot be used.}$ $2. \text{ Molar conductance depends on the number of ions as well as on the charge on the complexes.}$ $[\text{Cr} (\text{NH}_3)_6][\text{Cr} (\text{NO}_2)_6] \text{ and } [\text{Cr} (\text{NH}_3)_4(\text{NO}_2)_2][\text{Cr} (\text{NH}_3)_2(\text{NO}_2)_4]$ $\text{have the same number of ions but different electrical charges.}$ $3. \text{ Both complexes have same number of unpaired electrons. So, measuring magnetic moments methods can not be used.}$ $4. \text{ They have similar color because of similar conditions.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}