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Current Question (ID: 8043)

Question:

Among 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 in :

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 No. $= \frac{1}{2} [V + M - C + A]$ Explanation: Hybridisation No. $= \frac{1}{2} [V + M - C + A]$; where, $V = \text{no. of valence electrons}$; $M = \text{no. of monovalent atom}$; $C = \text{Positive charge}$; $A = \text{Negative Charge}$ $\text{BF}_3$: $\frac{1}{2} [3 + 3 - 0 + 0] = 3 \Rightarrow \text{sp}^2$ $\text{NO}_2^-$: $\frac{1}{2} [5 + 0 - 0 + 1] = 3 \Rightarrow \text{sp}^2$ $\text{NH}_2^-$: $\frac{1}{2} [5 + 2 - 0 + 1] = 4 \Rightarrow \text{sp}^3$ $\text{H}_2\text{O}$: $\frac{1}{2} [6 + 2 - 0 + 0] = 4 \Rightarrow \text{sp}^3$ Hence, $\text{BF}_3$ and $\text{NO}_2^-$ have $\text{sp}^2$ hybridised central atom among the given species.

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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
}