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Current Question (ID: 18723)

Question:

$\text{Number of geometrical isomer(s) possible for the complex } [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-} \text{ is/are:}$

Options:

1. $1$
2. $3$
3. $0$
4. $2$

Solution:

$\text{HINT: M}\left[\text{M(AA)}\right]_3^{n+/−m} \text{ did not show geometrical isomerism.}$ $\text{STEP 1: Geometrical isomers are two or more coordination compounds which contain the same number and types of atoms, and bonds (i.e., the connectivity between atoms is the same), but which have different spatial arrangements of the atoms.}$ $\text{STEP 2: Oxalate ligand is bidentate ligand where donor atoms are same.}$ $\text{The structure of oxalate ligand is:}$ $\text{STEP 3: For } [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-}, \text{ did not form geometrical isomer. It only shows, optical isomerism.}$ $\text{The structure of optical isomers is:}$

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
}