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Current Question (ID: 18711)

Question:
$\text{The complex that can exhibit linkage isomerism is:}$
Options:
  • 1. $[\text{Co} (\text{NH}_3)_5 \text{H}_2\text{O}] \text{Cl}_3$
  • 2. $[\text{Co} (\text{NH}_3)_5 (\text{NO}_2)] \text{Cl}_2$
  • 3. $[\text{Co} (\text{NH}_3)_5 \text{NO}_3] (\text{NO}_3)_2$
  • 4. $[\text{Co} (\text{NH}_3)_5 \text{Cl}] \text{SO}_4$
Solution:
$\text{HINT: } -\text{NO}_2 \text{ is an ambidentate ligand.}$ $\text{Explanation:}$ $\text{STEP 1: Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity of the metal to a ligand.}$ $\text{STEP 2: } -\text{NO}_2 \text{ is an ambidentate ligand as it connect with metal by N or O atom. Ambidentate ligand show linkage isomerism.}$ $\text{The isomers are } [\text{Co} (\text{NH}_3)_5 (\text{NO}_2)] \text{Cl}_2 \text{ and } [\text{Co} (\text{NH}_3)_5 (\text{ONO})] \text{Cl}_2$ $\text{STEP 3: All other complexes have monodentate ligands, so no linkage isomers are possible.}$

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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
}