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Current Question (ID: 8059)

Question:
The maximum number of $90^\circ$ angles between bond pair-bond pair of electrons is observed in:
Options:
  • 1. $\text{sp}^3\text{d}^2\text{-hybridization.}$ (Correct)
  • 2. $\text{sp}^3\text{d-hybridization.}$
  • 3. $\text{dsp}^2\text{-hybridization.}$
  • 4. $\text{dsp}^3\text{-hybridization.}$
Solution:
$\text{HINT: sp}^3\text{d}^2\text{-hybridization has maximum number of }90^\circ\text{ angles between bond-pair-bond pair.}$ $\text{Explanation:}$ $\text{sp}^3\text{ d}^2\text{ hybridization has an octahedral structure such that four hybrid orbitals are at }90^\circ\text{ with respect to each other and the other two at }90^\circ\text{ with the first four. The structure is as follows:}$

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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
}