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Current Question (ID: 8041)

Question:
The hybridization of nitrogen in $\text{NO}_2^+$, $\text{NO}_3^-$ and $\text{NH}_4^+$ respectively is :
Options:
  • 1. $\text{sp}$, $\text{sp}^3$, and $\text{sp}^2$
  • 2. $\text{sp}$, $\text{sp}^2$, and $\text{sp}^3$ (Correct)
  • 3. $\text{sp}^2$, $\text{sp}$, and $\text{sp}^3$
  • 4. $\text{sp}^2$, $\text{sp}^3$, and $\text{sp}$
Solution:
$\text{HINT: Find the sum of lone pair and bond pair.}$ $\text{Explanation:}$ $\text{The type of hybrid orbitals of nitrogen can be decided by using VSEPR theory counting bp and as lp in}$ $\text{1. } \text{NO}_2^+ = 2 \text{ bp} + 0 \text{ lp} \Rightarrow \text{sp hybridized.}$ $\text{2. } \text{NO}_3^- = 3 \text{ bp} + 0 \text{ lp} \Rightarrow \text{sp}^2 \text{ hybridized.}$ $\text{3. } \text{NH}_4^+ = 4\text{bp} + 0 \text{ lp} \Rightarrow \text{sp}^3 \text{ hybridized.}$ $\text{Therefore, the correct order of hybridization for } \text{NO}_2^+, \text{NO}_3^-, \text{ and } \text{NH}_4^+ \text{ is } \text{sp, sp}^2, \text{ and sp}^3 \text{ respectively.}$

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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
}