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Current Question (ID: 10828)

Question:

$\text{The compound that contains only sp}^3 \text{ hybridized carbon atoms is:}$

Options:

1. $\text{HCOOH}$
2. $(\text{NH}_2)_2\text{CO}$
3. $(\text{CH}_3)_3\text{COH}$ (Correct)
4. $\text{CH}_3\text{CHO}$

Solution:

$\text{Explanation:}$ $\text{Step 1: In sp}^3 \text{ hybridization, the carbon atom is bonded to four other atoms. In this case, 1 s orbital and 3 p orbitals in the same shell of an atom combine to form four new equivalent orbitals.}$ $\text{In sp}^2 \text{ hybridization, bonding takes place between 1 s-orbital and two p-orbitals. Two single bonds and one double bond between three atoms form and the hybrid orbitals come together in a triangular arrangement.}$ $\text{When a carbon atom is bound to two other atoms with the help of two double bonds or one single and one triple bond, it can be in sp hybridization state.}$ $\text{Step 2: The structures of given molecules are as follows:}$ $\text{In the third compound, all four are sigma bonds hence, carbon marked by an arrow in the third compound is sp}^3\text{.}$ $\text{Thus, option 3 is the correct choice.}$

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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
}