Import Question JSON

Current Question (ID: 18728)

Question:

$\text{The total number of possible co-ordination isomers for the following compounds are:}$ $[\text{Co (en)}_3][\text{Cr(C}_2\text{O}_4)_3]$ $[\text{Cu(NH}_3)_4][\text{CuCl}_4]$ $[\text{Ni(en)}_3][\text{Co(NO}_2)_6]$

Options:

1. $4, 4, 4$
2. $2, 2, 2$
3. $2, 2, 4$
4. $4, 2, 4$

Solution:

$\text{The complex } [\text{Co(en)}_3][\text{Cr(C}_2\text{O}_4)_3] \text{ has four possible coordination isomers.}$ $\text{The structure is as follows:}$ $\text{They are } [\text{Co(en)}_3][\text{Cr(C}_2\text{O}_4)_3], [\text{Co(C}_2\text{O}_4)(en)_2][\text{Cr(C}_2\text{O}_4)_2(en)], [\text{Cr(C}_2\text{O}_4)(en)_2][\text{Co(C}_2\text{O}_4)_2(en)], \text{ and } [\text{Cr(en)}_3][\text{Co(C}_2\text{O}_4)_3].$ $\text{The complex } [\text{Cu(NH}_3)_4][\text{CuCl}_4] \text{ has two possible coordination isomers.}$ $\text{They are } [\text{Cu(NH}_3)_4][\text{CuCl}_4], \text{ and } [\text{CuCl(NH}_3)_3][\text{CuCl}_3(\text{NH}_3)].$ $\text{The complex } [\text{Ni(en)}_3][\text{Co(NO}_2)_6] \text{ has four possible coordination isomers.}$ $\text{They are } [\text{Ni(en)}_3][\text{Co(NO}_2)_6], [\text{Ni(NO}_2)_2(en)_2][\text{Co(NO}_2)_4(en)], [\text{Co(NO}_2)_2(en)_2][\text{Ni(NO}_2)_4(en)], \text{ and } [\text{Co(en)}_3][\text{Ni(NO}_2)_6].$

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
}