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Current Question (ID: 8038)

Question:

The $\text{BF}_3$ is a planar molecule whereas $\text{NF}_3$ is pyramidal because:

Options:

1. $\text{B} - \text{F}$ bond is more polar than $\text{N} - \text{F}$ bond.
2. Boron atom is bigger than nitrogen atom.
3. Nitrogen is more electronegative than boron.
4. $\text{BF}_3$ has no lone pair but $\text{NF}_3$ has a lone pair of electrons. (Correct)

Solution:

$\text{HINT: Both have different hybridization}$ $\text{Explanation:}$ $\text{Hybridization of given compound is as follows:}$ $\text{BF}_3 = \text{sp}^2$ $\text{NF}_3 = \text{sp}^3$ $\text{BF}_3$ has trigonal planar geometry due to $\text{sp}^2$ hybridization and no lone pairs on the central boron atom. $\text{NF}_3$ has trigonal pyramidal geometry due to $\text{sp}^3$ hybridization and one lone pair of electrons on the central nitrogen atom. The lone pair causes repulsion with bonding pairs, leading to a distorted tetrahedral geometry (pyramidal shape). $\text{This explains why BF}_3 \text{ is planar and NF}_3 \text{ is pyramidal.}$

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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
}