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Current Question (ID: 10818)

Question:

$\text{CH}_2 = \text{C} = \text{O}$ $\text{The correct hybridization states of carbon atoms in the above compound are}$

Options:

1. $\text{C}_1 = \text{sp}, \text{C}_2 = \text{sp}^3$
2. $\text{C}_1 = \text{sp}^2, \text{C}_2 = \text{sp}$ (Correct)
3. $\text{C}_1 = \text{sp}^3, \text{C}_2 = \text{sp}$
4. $\text{C}_1 = \text{sp}, \text{C}_2 = \text{sp}^2$

Solution:

$\text{Hint: sp, sp2 and sp3 indicate the number of s and p orbitals mixed to create new, degenerate hybrid orbitals.}$ $\text{Step 1: draw the structure of the molecule}$ $\text{CH}_2 = \text{C} = \text{O}$ $\text{Step 2:}$ $\text{It is evident that the central atom (middle carbon atom) forms 2 sigma bonds and 2 pi bonds. The pi bonds are formed by the overlap of unhybridized p-orbitals. We know, that the number of hybrid orbitals formed = number of atomic orbitals taking part. Since the central atom forms two sigma bonds there are two sidewise overlapping. So, there are two hybrid orbitals which overlap with other orbitals to form two sigma bonds. Thus, there are two atomic orbitals taking part. Therefore, state of hybridization of the middle carbon atom is sp.}$ $\text{Hence,}$ $\text{C}-1 \text{ is sp}^2 \text{ hybridized.}$ $\text{C}-2 \text{ is sp hybridized.}$

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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
}