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Current Question (ID: 18817)

Question:
$\text{Which of the following ions has the most unpaired electrons?}$
Options:
  • 1. $\text{Fe}^{2+}$
  • 2. $\text{Fe}^{3+}$
  • 3. $\text{Co}^{3+}$
  • 4. $\text{Co}^{2+}$
Solution:
$\text{Paramagnetic behaviour} \propto \text{Number of unpaired electrons}$ $\text{Draw its valence orbital diagram and find the unpaired electron.}$ $\text{The diagram is as follows:}$ $\begin{array}{c|c} \text{4s} & \text{3d} \\ \hline \text{Fe}^{2+} & \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \\ \text{Fe}^{3+} & \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \uparrow \\ \text{Co}^{2+} & \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \uparrow \\ \text{Co}^{3+} & \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \uparrow \uparrow \\ \end{array}$ $\therefore \text{Fe}^{3+} \text{ has maximum number (= 5) of unpaired electrons.}$

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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
}