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Current Question (ID: 8028)

Question:

The shape of $\text{XeF}_3^+$ is:

Options:

1. Trigonal planar
2. Pyramidal
3. Bent T-shape (Correct)
4. See-saw

Solution:

$\text{HINT: XeF}_3^+$ is $\text{sp}^3\text{d}$ hybridization. $\text{Explanation:}$ $\text{Find the steric number of XeF}_3^+$ as follows: $\text{H} = \frac{\text{V+M-C+A}}{2}$ $\text{V = Number of valence electrons of central atom}$ $\text{M = Number of the monovalent atom (}\pm\text{1)}$ $\text{C = Charge on cation}$ $\text{A = Charge on anion}$ $\text{For XeF}_3^+, \text{H}= \frac{8+3-1}{2} = 5$ $\text{Steric number 5 represents sp}^3\text{d hybridization. The geometry is trigonal bipyramidal but in shape two lone pairs did not included, thus, the shape is a bent T-shape.}$

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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
}