Import Question JSON

Current Question (ID: 8093)

Question:
$\text{The species that does not have unpaired electrons is:}$
Options:
  • 1. $\text{N}_2^+$
  • 2. $\text{O}_2$
  • 3. $\text{O}_2^{2-}$
  • 4. $\text{B}_2$
Solution:
$\text{HINT: MOT diagram will tell about unpaired electron.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Draw the MOT diagram of the given species and find the number of unpaired electrons.}$ $\text{The electronic configuration of the given molecules are:}$ $\text{1.) } \text{N}_2^+ = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \sigma_{2p_z}^1$ $\text{It has one unpaired electron.}$ $\text{2.) } \text{O}_2 = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*1} \approx \pi_{2p_y}^{*1}$ $\text{O}_2 \text{ has two unpaired electrons.}$ $\text{3.) } \text{O}_2^{2-} = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 \approx \pi_{2p_y}^2, \pi_{2p_x}^{*2} \approx \pi_{2p_y}^{*2}$ $\text{Thus, } \text{O}_2^{2-} \text{ has no unpaired electrons.}$ $\text{4.) } \text{B}_2 = \sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^1 \approx \pi_{2p_y}^1$ $\text{Thus, } \text{B}_2 \text{ has two unpaired electrons.}$ $\text{Step 2:}$ $\text{O}_2^{2-} \text{ ion does not contain unpaired electrons.}$

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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
}