Import Question JSON

Current Question (ID: 8045)

Question:

In which of the following molecules/ions $\text{BF}_3$, $\text{NO}_2^-$, $\text{NH}_2^-$ and $\text{H}_2\text{O}$, the central atom is $\text{sp}^2$ hybridized?

Options:

1. $\text{NO}_2^-$ and $\text{NH}_2^-$
2. $\text{NH}_2^-$ and $\text{H}_2\text{O}$
3. $\text{NO}_2^-$ and $\text{H}_2\text{O}$
4. $\text{BF}_3$ and $\text{NO}_2^-$ (Correct)

Solution:

HINT: Hybridisation = number of sigma bonds + no of lone pairs. Explanation: Central atom is $\text{sp}^2$ hybridized in $\text{BF}_3$ and $\text{NO}_2^-$. From the structure of molecules, hybridization can be known. For $\text{BF}_3$: The central atom Boron (B) forms 3 sigma bonds with 3 Fluorine (F) atoms and has no lone pairs. So, hybridisation number = $3+0=3$, which corresponds to $\text{sp}^2$ hybridisation. For $\text{NO}_2^-$: The central atom Nitrogen (N) forms 2 sigma bonds and has 1 lone pair (due to resonance, one oxygen forms a double bond, but we consider the sigma bonds). So, hybridisation number = $2+1=3$, which corresponds to $\text{sp}^2$ hybridisation. For $\text{NH}_2^-$: The central atom Nitrogen (N) forms 2 sigma bonds with 2 Hydrogen (H) atoms and has 2 lone pairs. So, hybridisation number = $2+2=4$, which corresponds to $\text{sp}^3$ hybridisation. For $\text{H}_2\text{O}$: The central atom Oxygen (O) forms 2 sigma bonds with 2 Hydrogen (H) atoms and has 2 lone pairs. So, hybridisation number = $2+2=4$, which corresponds to $\text{sp}^3$ hybridisation. Thus, $\text{BF}_3$ and $\text{NO}_2^-$ have $\text{sp}^2$ hybridized central atoms.

Import JSON File

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
}