Import Question JSON

Current Question (ID: 7973)

Question:

The correct order of $\text{C} - \text{O}$ bond length among $\text{CO}$, $\text{CO}_3^{2-}$, and $\text{CO}_2$ is :

Options:

1. $\text{CO}_2 < \text{CO}_3^{2-} < \text{CO}$
2. $\text{CO} < \text{CO}_3^{2-} < \text{CO}_2$
3. $\text{CO}_3^{2-} < \text{CO}_2 < \text{CO}$
4. $\text{CO} < \text{CO}_2 < \text{CO}_3^{2-}$ (Correct)

Solution:

HINT: Bond length inversely proportional to bond order. Explanation: The multiple bond (double or triple bond) is shorter than the corresponding single bond. 1.) The $\text{C}$-atom in $\text{CO}_3^{2-}$ is $sp^2$ hybridized as shown below : $\text{O}=\text{C}-\text{O}^- \text{O}^-=\text{C}-\text{O} \text{O}^-=\text{C}-\text{O}^-$ The $\text{CO}_3^{2-}$ bond order is $1.33$. 2.) The $\text{C}$-atom in $\text{CO}_2$ is $sp$ hybridized and the bond length of carbon-oxygen bond as $122 \text{ pm}$. The bond order is $2$. $\text{O}=\text{C}=\text{O} \leftrightarrow \text{O}^- - \text{C}\equiv\text{O}^+ \leftrightarrow \text{O}^+ \equiv \text{C} - \text{O}^-$ 3.) The $\text{C}$-atom in $\text{CO}$ is $sp$ hybridized with $\text{C}-\text{O}$ bond length as $110 \text{ pm}$. The bond order in $\text{CO}$ is $3$. $:\text{C}\equiv\text{O}^{+}:$ So the correct order of bond length is $\text{CO} < \text{CO}_2 < \text{CO}_3^{2-}$

Import JSON File

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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
}