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Current Question (ID: 18764)

Question:

$\text{Spin-only magnetic moments of } [\text{Fe(NH}_3\text{)}_6]^{3+} \text{ and } [\text{FeF}_6]^{3-} \text{ in BM are, respectively:}$

Options:

1. $1.73 \text{ and } 1.73$
2. $5.92 \text{ and } 1.73$
3. $1.73 \text{ and } 5.92$
4. $5.92 \text{ and } 5.92$

Solution:

$\text{Hint:}$ $\text{The VBT diagram for}$ $[\text{Fe(NH}_3\text{)}_6]^{3+}$ $\text{Fe}^{+3} = 3d^54s^0$ $\text{In case of } \text{Fe}^{+3}, \text{ NH}_3 \text{ is strong ligand and F}^- \text{ are weak field ligands.}$ $\text{In the case of F}^-, \text{ the pairing of electrons does not take place.}$ $\text{Number of unpaired electrons} = 5$ $\text{Magnetic moment} = \sqrt{n(n+2)} \text{ BM}$ $= \sqrt{5(5+2)}$ $= \sqrt{35}$ $= 5.92$ $\text{In the case of NH}_3, \text{ the electron will be paired in 3d orbital because ammonia}$ $\text{is a strong field ligand for Fe}^{3+} \text{ ion, so 1 unpaired electron will remain, so the}$ $\text{magnetic moment will be } 1.73$

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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
}