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Current Question (ID: 10929)

Question:

$\text{The Huckel's rule-based aromaticity is shown by:}$

Options:

1. $\text{A, B, D only}$
2. $\text{B, D only}$
3. $\text{B, D, E and F only}$ (Correct)
4. {A, B, D, E & F only}

Solution:

$\text{Hint: Follow Huckel's Rule for Aromaticity}\n\n\text{Step 1: Follow these points on each above molecules systematically}\n\n\text{A. In the first molecule, }\text{sp}^3 \text{ carbon is present in the ring. Hence, it will be non-aromatic.}\n\n\text{B. The second compound is aromatic. It has } 6\pi\text{e}^\text{--} \text{ delocalised electrons ( } 4\pi\text{e}^\text{--} + 2 \text{ lone pair electrons), all the four carbon atoms and the N atom are }\text{sp}^2\text{ hybridized. The molecule is planar.}\n\n\text{C. The third compound contains } 6\pi \text{ electrons but not in the ring hence it is non-aromatic.}\n\n\text{D. The fourth compound is aromatic. It has } 10\pi\text{e}^\text{--} \text{ obeying Huckel rule and the ring is planar.}\n\n\text{E. In this compound one six-membered planar ring has } 6\pi\text{e}^\text{--} \text{ although it has } 8\pi \text{ electrons in two rings. It is therefore aromatic.}\n\n\text{F. It has } 14\pi \text{ electrons in conjugation and in the planar ring, Huckel rule is verified. It will be aromatic.}\n\n\text{Step 2:}\n\n\text{Hence those molecules which follow the huckle's rule, planarity according to } 4\text{n}+2 \pi\text{e}^\text{--} \text{ rule, are aromatic.}\n\n\text{The name of the compound is 14-Annulene, and its an aromatic compound.}$

Import JSON File

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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
}