Import Question JSON

Current Question (ID: 8001)

Question:

Which of the following has a maximum 'Cl-O' bond order?

Options:

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

Solution:

$\text{Hint: Formula of bond order in polyatomic ion (resonance occur) = } \frac{\text{total number of bond}}{\text{total number of resonating structure}} \text{Step 1:} \text{Calculate the bond order in the given molecule as follows:} \text{1. } \text{ClO}_3^{-} \text{The structure of } \text{ClO}_3^{-} \text{ is as follows:} \text{In } \text{ClO}_3^{-}\text{, resonance takes place. Hence bond order is} \text{bond order in polyatomic ion (resonance occur) = } \frac{\text{total number of bond}}{\text{total number of resonating structure}} \text{B. O = } \frac{5}{3} \text{= 1.67} \text{2. The structure of } \text{ClO}_2 \text{ is as follows:} \text{The bond order is 1.75} \text{3. The structure of } \text{ClO}_3 \text{ is as follows:} \text{The bond order is 2} \text{4. The structure of } \text{ClO}_2^{-} \text{ is as follows:} \text{In } \text{ClO}_2^{-}\text{, resonance takes place. Hence bond order is} \text{bond order in polyatomic ion (resonance occur) = } \frac{\text{total number of bond}}{\text{total number of resonating structure}} \text{B. O = } \frac{3}{2} \text{= 1.5} \text{Step 2:} \text{Hence, the highest bond order is for } \text{ClO}_3\text{, that is, 2.}$

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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
}