Import Question JSON

Current Question (ID: 17639)

Question:

$\text{The correct order of acidity among the following is:}$

Options:

1. $\text{CH} \equiv \text{CH} > \text{CH}_3 - \text{C} \equiv \text{CH} > \text{CH}_2 = \text{CH}_2 > \text{CH}_3 - \text{CH}_3$
2. $\text{CH} \equiv \text{CH} > \text{CH}_2 = \text{CH}_2 > \text{CH}_3 - \text{CH}_3$
3. $\text{CH}_3 - \text{CH}_3 > \text{CH}_2 = \text{CH}_2 > \text{CH}_3 - \text{C} \equiv \text{CH} > \text{CH} \equiv \text{CH}$
4. $\text{CH}_2 = \text{CH}_2 > \text{CH}_3 - \text{CH}_3 > \text{CH}_3 - \text{C} \equiv \text{CH} > \text{CH} \equiv \text{CH}$

Solution:

$\text{Hint: Electronegativity order: sp carbon} > \text{sp}^2 \text{carbon} > \text{sp}^3 \text{carbon}$ $\text{Since s orbital is closest to the nucleus, thus electron which is present in orbitals with more s character will be more attracted by the nucleus, and as a result electronegativity of sp is greater than the other two.}$ $\text{Therefore sp is more acidic followed by sp}^2 \text{and sp}^3.$ $\text{Acidic character order: sp} > \text{sp}^2 > \text{sp}^3$ $\text{The conjugate base of an alkyne is more stable than the conjugate base of alkene and alkane. In alkyne, a negative charge is formed in sp carbon, an alkene, it is formed in sp}^2 \text{carbon, and in alkane, it is formed in sp}^3 \text{carbon.}$ $\text{The sp carbon is highly electronegative hence alkyne is highly acidic followed by alkene and than alkane.}$

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
}