Import Question JSON

Current Question (ID: 8047)

Question:

Given the following hydrocarbon: $\text{CH}_3 - \text{C}(\text{CH}_3)_2 - \text{CH} = \text{CH} - \text{CH}(\text{CH}_3) - \text{C} \equiv \text{CH}$ The state of hybridisation of carbons marked 2, 3, 5 and 6 in the following hydrocarbon are, respectively :

Options:

1. $\text{sp}$, $\text{sp}^3$, $\text{sp}^2$ and $\text{sp}^3$ (Correct)
2. $\text{sp}^3$, $\text{sp}^2$, $\text{sp}^2$ and $\text{sp}$
3. $\text{sp}$, $\text{sp}^2$, $\text{sp}^2$ and $\text{sp}^3$
4. $\text{sp}$, $\text{sp}^2$, $\text{sp}^3$ and $\text{sp}^2$

Solution:

HINT: Count number of $\sigma$ bonds and then find hybridisation. Explanation: If number of $\sigma$ bonds = 2; hybridisation is $\text{sp}$. If number of $\sigma$ bonds = 3; hybridisation is $\text{sp}^2$. If number of $\sigma$ bonds = 4; hybridisation is $\text{sp}^3$. Let's analyze the hybridization of the carbons marked 2, 3, 5, and 6 in the given hydrocarbon structure: Original structure: $\text{CH}_3 - \text{C}(\text{CH}_3)_2 - \text{CH} = \text{CH} - \text{CH}(\text{CH}_3) - \text{C} \equiv \text{CH}$ Numbering the carbons from right to left as shown in the image: Carbon 1: $\text{CH}$ (at the end of triple bond) - Forms 1 sigma bond with Carbon 2 and 1 sigma bond with Hydrogen. Also forms 2 pi bonds. So, 2 sigma bonds $\Rightarrow \text{sp}$ hybridized. Carbon 2: $\text{C}$ (in triple bond) - Forms 1 sigma bond with Carbon 1 and 1 sigma bond with Carbon 3. Also forms 2 pi bonds. So, 2 sigma bonds $\Rightarrow \text{sp}$ hybridized. Carbon 3: $\text{CH}(\text{CH}_3)$ - Forms 1 sigma bond with Carbon 2, 1 sigma bond with Carbon 4, 1 sigma bond with Hydrogen, and 1 sigma bond with the methyl group ($\text{CH}_3$). So, 4 sigma bonds $\Rightarrow \text{sp}^3$ hybridized. Carbon 4: $\text{CH}$ (in double bond) - Forms 1 sigma bond with Carbon 3, 1 sigma bond with Carbon 5, and 1 sigma bond with Hydrogen. Also forms 1 pi bond. So, 3 sigma bonds $\Rightarrow \text{sp}^2$ hybridized. Carbon 5: $\text{CH}$ (in double bond) - Forms 1 sigma bond with Carbon 4, 1 sigma bond with Carbon 6, and 1 sigma bond with Hydrogen. Also forms 1 pi bond. So, 3 sigma bonds $\Rightarrow \text{sp}^2$ hybridized. Carbon 6: $\text{C}(\text{CH}_3)_2$ - Forms 1 sigma bond with Carbon 5 and 3 sigma bonds with methyl groups. So, 4 sigma bonds $\Rightarrow \text{sp}^3$ hybridized. Carbon 7: $\text{CH}_3$ - Forms 1 sigma bond with Carbon 6 and 3 sigma bonds with Hydrogens. So, 4 sigma bonds $\Rightarrow \text{sp}^3$ hybridized. We need the hybridization for carbons marked 2, 3, 5, and 6, respectively: Carbon 2: $\text{sp}$ Carbon 3: $\text{sp}^3$ Carbon 5: $\text{sp}^2$ Carbon 6: $\text{sp}^3$ Therefore, the correct sequence is $\text{sp}$, $\text{sp}^3$, $\text{sp}^2$ and $\text{sp}^3$.

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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
}