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Current Question (ID: 21331)

Question:
$\text{A non-aromatic compound among the following is:}$
Options:
  • 1. $\text{Option 1: Non-Aromatic}$
  • 2. $\text{Option 2: Aromatic}$
  • 3. $\text{Option 3: Aromatic}$
  • 4. $\text{Option 4: Aromatic}$
Solution:
$\text{Hint: Each atom in the cyclic system must have a p orbital perpendicular to the ring.}$ $\text{The hybridization of atoms in the cyclic system is sp}^2.$ $\text{Step 1:}$ $\text{The condition of aromaticity is as follows:}$ $1) \text{Cyclic - the molecule must be a cyclic polyene;} \ 2) \text{Planar - the geometry of the cyclic part must be planar;} \ 3) \text{Each atom in the cyclic system must have a p orbital perpendicular to the ring.}$ $\text{The hybridization of atoms in the cyclic system is sp}^2.$ $4) \text{The cyclic system has } 4n+2 \ \Pi \text{ electrons.}$ $\text{Step 2:}$ $\text{Option 1: Non-Aromatic}$ $\text{Because one carbon of the cyclic system is sp}^3.$

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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
}