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Current Question (ID: 8089)

Question:
$\text{The decreasing order of stability of } \text{O}_2, \text{O}_2^-, \text{O}_2^+ \text{ and } \text{O}_2^{2-} \text{ is:}$
Options:
  • 1. $\text{O}_2^+ > \text{O}_2 > \text{O}_2^- > \text{O}_2^{2-}$
  • 2. $\text{O}_2^{2-} > \text{O}_2^- > \text{O}_2 > \text{O}_2^+$
  • 3. $\text{O}_2 > \text{O}_2^+ > \text{O}_2^{2-} > \text{O}_2^-$
  • 4. $\text{O}_2^- > \text{O}_2^{2-} > \text{O}_2^+ > \text{O}_2$
Solution:
$\text{HINT: Bond order } \propto \text{ Stability}$ $\text{Explanation:}$ $\text{The electronic configuration of a given molecule/ion according to MOT is as follows:}$ $\text{1. Electronic configuration of } \text{O}_2 \text{ (16 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*1} \approx \pi_{2p_y}^{*1}$ $\text{Bond order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 6) = 2$ $\text{2. Electronic configuration of } \text{O}_2^+ \text{ (15 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*1} \approx \pi_{2p_y}^{*0}$ $\text{Bond order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 5) = 2.5$ $\text{3. Electronic configuration of } \text{O}_2^- \text{ (17 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*2} \approx \pi_{2p_y}^{*1}$ $\text{Bond order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 7) = 1.5$ $\text{4. Electronic configuration of } \text{O}_2^{2-} \text{ (18 electrons)} = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*2} \approx \pi_{2p_y}^{*2}$ $\text{Bond order} = \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 8) = 1$ $\text{Thus, the order of bond order is directly proportional to the relative stability.}$ $\text{The order is: } \text{O}_2^+ > \text{O}_2 > \text{O}_2^- > \text{O}_2^{2-}$

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{
  "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
}