Import Question JSON

Current Question (ID: 21543)

Question:

$\text{The calculated magnetic moments (spin only value) for species}$ $[\text{FeCl}_4]^{2-}, \ [\text{Co} \ (\text{C}_2\text{O}_4)_3]^{3-} \ \text{and} \ \text{MnO}_4^{2-} \ \text{respectively are:}$

Options:

1. $5.82, \ 0 \ \text{and} \ 0 \ \text{BM}$
2. $4.90, \ 0 \ \text{and} \ 1.73 \ \text{BM}$
3. $5.92, \ 4.90 \ \text{and} \ 0 \ \text{BM}$
4. $4.90, \ 0 \ \text{and} \ 2.83 \ \text{BM}$

Solution:

$\text{Hint: Use the formula of the magnetic moment:}$ $\mu = \sqrt{n(n+2)} \ \text{BM}$ $\text{The formula of magnetic moment is}$ $\mu = \sqrt{n(n+2)} \ \text{BM, where} \ n \ \text{is the number of unpaired electrons.}$ $\text{(i) } [\text{FeCl}_4]^{2-}$ $\mu = \sqrt{4(4+2)} \ \text{BM} = \sqrt{24} \ \text{BM} \Rightarrow 4.90 \ \text{BM}$ $\text{(ii) } [\text{Co} \ (\text{C}_2\text{O}_4)_3]^{3-}$ $\text{Co}^{3+} \ \text{is} \ 3d^6 \ \text{configuration. As a strong field ligand approaches Co}^{3+} \ \text{ion, the pairing of electrons takes place.}$ $\text{In} \ [\text{Co} \ (\text{C}_2\text{O}_4)_3]^{3-}, \ \text{the number of unpaired electrons is zero.}$ $\text{Hence, the magnetic moment is zero.}$ $\text{(iii) MnO}_4^{2-}$ $\text{In MnO}_4^{2-}, \ \text{Mn oxidation state is} \ +6. \ \text{Hence, the configuration of Mn}^{6+} \ \text{is} \ 3d^1$ $\mu = \sqrt{1(1+2)} \ \text{BM} = \sqrt{3} \ \text{BM} \Rightarrow 1.73 \ \text{BM}$ $\text{Hence, option second is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}