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Current Question (ID: 18748)

Question:
$\text{Which among the following complexes shows a magnetic moment of 1.73 BM?}$ $1.\ [\text{Ni(CN)}_4]^{2-}$ $2.\ \text{TiCl}_4$ $3.\ [\text{CoCl}_6]^{4-}$ $4.\ [\text{Cu(NH}_3)_4]^{2+}$
Options:
  • 1. $[\text{Ni(CN)}_4]^{2-}$
  • 2. $\text{TiCl}_4$
  • 3. $[\text{CoCl}_6]^{4-}$
  • 4. $[\text{Cu(NH}_3)_4]^{2+}$
Solution:
$\text{HINT: } 3d^9 \text{ system contains 1 unpaired electron.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{If molecule contains one unpaired electron. It's magnetic moment value is 1.73 BM.}$ $\text{Step 2:}$ $\text{The configuration of Cu}^{2+} \text{ in } [\text{Cu(NH}_3)_4]^{2+} \text{ is } 3d^9. \text{ Hence, it contains one unpaired electron. Hence, option 4th is the correct answer.}$

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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
}