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Current Question (ID: 8082)

Question:

The bond order of $1.5$ is shown by:

Options:

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

Solution:

$\text{HINT: B.O. } = \frac{\text{N}_\text{b} - \text{N}_\text{a}}{2}\n\n\text{Explanation:}\n\text{1. Electronic configuration of } \text{O}_2 \text{ (16 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{\ast 2}, \sigma_{2s}^2, \sigma_{2s}^{\ast 2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{\ast 1} \approx \pi_{2p_y}^{\ast 1}\n\text{Bond order} = \frac{1}{2}(\text{N}_\text{b} - \text{N}_\text{a}) = \frac{1}{2}(10 - 6) = 2\n\n\text{2. Electronic configuration of } \text{O}_2^{+} \text{ (15 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{\ast 2}, \sigma_{2s}^2, \sigma_{2s}^{\ast 2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{\ast 1} \approx \pi_{2p_y}^{\ast 0}\n\text{Bond order} = \frac{1}{2}(\text{N}_\text{b} - \text{N}_\text{a}) = \frac{1}{2}(10 - 5) = 2.5\n\n\text{3. Electronic configuration of } \text{O}_2^{-} \text{ (17 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{\ast 2}, \sigma_{2s}^2, \sigma_{2s}^{\ast 2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{\ast 2} \approx \pi_{2p_y}^{\ast 1}\n\text{Bond order} = \frac{1}{2}(\text{N}_\text{b} - \text{N}_\text{a}) = \frac{1}{2}(10 - 7) = 1.5\n\n\text{4. Electronic configuration of } \text{O}_2^{2-} \text{ (18 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{\ast 2}, \sigma_{2s}^2, \sigma_{2s}^{\ast 2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{\ast 2} \approx \pi_{2p_y}^{\ast 2}\n\text{Bond order} = \frac{1}{2}(\text{N}_\text{b} - \text{N}_\text{a}) = \frac{1}{2}(10 - 8) = 1\n\n\text{Thus, } \text{O}_2^{-} \text{ shows the bond order 1.5}$

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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
}