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Current Question (ID: 8114)

Question:
$\text{O} - \text{O}$ bond length (in $\mathring{\text{A}}$) in $\text{H}_2\text{O}_2$ and $\text{O}_2\text{F}_2$ is respectively:
Options:
  • 1. $1.22, 1.48$
  • 2. $1.48, 1.22$
  • 3. $1.22, 1.22$
  • 4. $1.48, 1.48$
Solution:
$\text{HINT: Apply Bent's rule.} \text{Explanation:} \text{According to Bent, Rule in }\text{O}_2\text{F}_2 \text{ there is more p-character in }\text{O} - \text{O} \text{ bond in }\text{H}_2\text{O}_2\text{. Hence, the s-character in the }\text{O} - \text{O} \text{ bond is greater in }\text{O}_2\text{F}_2 \text{ than }\text{H}_2\text{O}_2\text{. The structures are as follows:} \text{For }\text{H}_2\text{O}_2\text{, the O-O bond length is }1.48 \mathring{\text{A}}.\n\text{For }\text{O}_2\text{F}_2\text{, the O-O bond length is }1.22 \mathring{\text{A}}.\n \text{Therefore, the bond lengths are } 1.48 \mathring{\text{A}} \text{ and } 1.22 \mathring{\text{A}} \text{ respectively, matching option 2.}$

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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
}