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Current Question (ID: 8080)

Question:
A pair in which both species are not likely to exist is:
Options:
  • 1. $\text{H}_2^{+}, \text{He}_2^{2-}$
  • 2. $\text{H}_2^{-}, \text{He}_2^{2+}$
  • 3. $\text{H}_2^{2+}, \text{He}_2$ (Correct)
  • 4. $\text{H}_2^{+}, \text{He}_2^{2+}$
Solution:
$\text{HINT: If diatomic molecule bond order is zero then molecule will not exist.}\n\n\text{Explanation:}\n\text{The bond order formula is :}\n\n\text{Bond order (B.O.)} = \frac{1}{2} (\text{N}_\text{b} - \text{N}_\text{a})\n\n\text{H}_2^{2+} : \text{Bond order} = \frac{0-0}{2}\n\text{(He)}_2 : \text{Bond order} = \frac{2-2}{2}\n\n\text{So, both } \text{H}_2^{2+} \text{ & } \text{He}_2 \text{ do not exist.}$

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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
}