Import Question JSON

Current Question (ID: 18715)

Question:

$\text{The ionization isomer of } [\text{Cr(H}_2\text{O})_4\text{Cl(NO}_2)]\text{Cl is:}$

Options:

1. $[\text{Cr(H}_2\text{O})_4(\text{O}_2\text{N})]\text{Cl}_2$
2. $[\text{Cr(H}_2\text{O})_4\text{Cl}_2](\text{NO}_2)$
3. $[\text{Cr(H}_2\text{O})_4\text{Cl(ONO)}]\text{Cl}$
4. $[\text{Cr(H}_2\text{O})_4\text{Cl}_2(\text{NO}_2)].\text{H}_2\text{O}$

Solution:

$\text{HINT: } [\text{Cr(H}_2\text{O})_4\text{Cl}_2](\text{NO}_2) \text{ is the ionization isomer of } [\text{Cr(H}_2\text{O})_4\text{Cl(NO}_2)]\text{Cl}$ $\text{Explanation:}$ $\text{STEP 1: Ionization isomers are identical except for a ligand that has exchanged places with an anion or neutral molecule that was originally outside the coordination complex.}$ $\text{Thus, in ionization isomerism the central ion and the other ligands are identical.}$ $\text{So, according to the definition, } \text{Cl}^- \text{ is replaced by } \text{NO}_2^- \text{ in ionization sphere.}$ $\text{STEP 2: The reaction is as follows:}$ $[\text{Cr(H}_2\text{O})_4\text{Cl(NO}_2)]\text{Cl} \rightarrow [\text{Cr(H}_2\text{O})_4\text{Cl}_2(\text{NO}_2)]^+ + \text{Cl}^-$ $[\text{Cr(H}_2\text{O})_4\text{Cl}_2](\text{NO}_2) \rightarrow [\text{Cr(H}_2\text{O})_4\text{Cl}_2]^+ + \text{NO}_2^-$

Import JSON File

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
}