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Current Question (ID: 8035)

Question:

The molecular geometry that is least likely to result from a trigonal bipyramidal electron geometry is:

Options:

1. Trigonal planar
2. See-saw
3. Linear
4. T-shaped

Solution:

HINT: $\text{Use the concept of VSEPR theory for the stable structure of any molecule.}$ Explanation: $\text{According to VSEPR theory, in trigonal bipyramidal geometry, lone pair of electrons present at an equatorial position.}$ $\text{If a molecule has 1 lone pair then trigonal bipyramidal is similar to a see-saw.}$ $\text{If it has 2 lone pairs then trigonal bipyramidal is similar to T-Shape, and if it has three lone pairs of electrons then trigonal bipyramidal is similar to linear shape.}$ $\text{It cannot be similar to trigonal planar.}$ $\text{Shapes with Five Electron Pairs}$ $\text{Trigonal bipyramidal (0 lone pairs) - AB}_5$ $\text{See-saw (1 lone pair) - AB}_4\text{E}$ $\text{T-shaped (2 lone pairs) - AB}_3\text{E}_2$ $\text{Linear (3 lone pairs) - AB}_2\text{E}_3$ $\text{Trigonal planar geometry results from an }\text{sp}^2 \text{ hybridization with three electron domains and zero lone pairs, not five electron pairs like trigonal bipyramidal electron geometry.}$

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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
}