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Current Question (ID: 8073)

Question:

The bond angle between two hybrid orbitals is $105^{\circ}$. The $ \% $ s character in orbital is-

Options:

1. Between 30-31%
2. Between 9-12%
3. Between 25-26%
4. Between 22-23% (Correct)

Solution:

\text{HINT: Bond angle decreases with decrease in } \% \text{ s character.} \n\n \text{Explanation:} \n\n \text{The sp hybridization bond angle is } 180^{\circ} \text{ and percentage of s character is } 50\%. \n\n \text{The } \text{sp}^2 \text{ hybridization bond angle is } 120^{\circ} \text{ and percentage of s character is } 33.3\%. \n\n \text{The } \text{sp}^3 \text{ hybridization bond angle is } 109.7^{\circ} \text{ and percentage of s character is } 25\%. \n\n 105^{\circ} \text{ is more close to } 109.7^{\circ} \text{ hence percentage s character is in between 22-23\%}

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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
}