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Current Question (ID: 7576)

Question:

$\text{The magnetic moment of 2.84 BM can be shown by:}$\n\n$\text{(At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27)}$

Options:

1. $\text{Ni}^{2+}$
2. $\text{Ti}^{3+}$
3. $\text{Cr}^{3+}$
4. $\text{Co}^{2+}$

Solution:

$\text{Hint: Use magnetic moment formula.}$\n\n$\text{Explanation:}$\n\n$\text{Magnetic moment, } \mu = \sqrt{n(n + 2)} \text{ BM where,}$\n\n$n = \text{number of unpaired electrons}$\n\n$\mu = 2.84 \text{ (given)}$\n\n$\therefore \quad 2.84 = \sqrt{n(n + 2)} \text{ BM}$\n\n$(2.84)^2 = n(n + 2)$\n\n$8 = n^2 + 2n$\n\n$n^2 + 2n - 8 = 0$\n\n$n^2 + 4n - 2n - 8 = 0$\n\n$n = 2$\n\n$\text{Ni}^{2+} = [\text{Ar}]3\text{d}^8 4\text{s}^0 \text{ (two unpaired electrons)}$\n\n$\text{Ti}^{3+} = [\text{Ar}]3\text{d}^1 4\text{s}^0 \text{ (one unpaired electrons)}$\n\n$\text{Cr}^{3+} = [\text{Ar}]3\text{d}^3 \text{ (three unpaired electrons)}$\n\n$\text{Co}^{2+} = [\text{Ar}]3\text{d}^7, 4\text{s}^0 \text{ (three unpaired electrons)}$\n\n$\text{So, only Ni}^{2+} \text{ has 2 unpaired electrons.}$

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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
}