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Current Question (ID: 17622)

Question:
$\text{Decreasing order of acidic behavior of benzene, n-hexane, and ethyne is:}$
Options:
  • 1. $\text{Hexane} > \text{Ethyne} > \text{Benzene}$
  • 2. $\text{Benzene} > \text{Hexane} > \text{Ethyne}$
  • 3. $\text{Ethyne} > \text{Benzene} > \text{Hexane}$
  • 4. $\text{Benzene} > \text{Ethyne} > \text{Hexene}$
Solution:
$\text{Hint: Acidic character of a species is defined on the basis of ease with which it can lose its H}^- \text{ atoms.}$ $\text{Step 1: The hybridization state of carbon in the given compound is:}$ $\text{Hexane } sp^3 \quad \text{Ethyne } sp \quad \text{Benzene } sp^2$ $\text{As the } s\text{-character increases, the electronegativity of carbon increases, and the electrons of C-H bond pair lie closer to the carbon atom. As a result, the partial positive charge on H}^- \text{ atom increases, and H}^+ \text{ ions are set free.}$ $\text{Step 2: The } s\text{-character increases in the order as follows:}$ $sp^3 < sp^2 < sp$ $\text{Hence, the decreasing order of acidic behavior is Ethyne} > \text{Benzene} > \text{Hexane}$

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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
}