Import Question JSON

Current Question (ID: 10924)

Question:

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

Options:

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

Solution:

$\text{Hint: As the percentage of s character in carbon atom increases, the electronegativity of carbon atom also increases.}$ $\text{Greater the s-character of C-atom in hydrocarbons, greater the electronegativity of that carbon and thus greater the acidic nature of the H attached to electronegative carbon.}$ $\text{CH}\equiv\text{CH} \quad \text{CH}_2=\text{CH}_2 \quad \text{CH}_3-\text{CH}_3$ $\text{Hybridization:} \quad \text{sp} \quad \text{sp}^2 \quad \text{sp}^3$ $\text{s-character:} \quad 50\% \quad 33\% \quad 25\%$ $\text{Electronegativity:} \quad \leftarrow \text{Max.}$ $\text{Acidic character} \quad \leftarrow \text{Max.}$ $\text{of terminal H}$ $\text{Thus, } \text{CH}\equiv\text{CH} > \text{CH}_3\text{C}\equiv\text{CH} > \text{CH}_2=\text{CH}_2 > \text{CH}_3-\text{CH}_3$

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
}