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Current Question (ID: 18811)

Question:
$\text{Which of the following has the highest spin-only magnetic moment value?}$
Options:
  • 1. $\text{Ti}^{3+}$
  • 2. $\text{Mn}^{2+}$
  • 3. $\text{Fe}^{2+}$
  • 4. $\text{Co}^{3+}$
Solution:
$\mu = \sqrt{n(n+2)} \text{ BM}$ $\text{Explanation: Ion contains the maximum number of unpaired electrons has high magnetic moment values}$ $\text{Spin magnetic moment is } \sqrt{n(n+2)}, \text{ where } n \text{ is the number of unpaired electrons.}$ $\text{STEP 1: The configuration of ions are as follows:}$ $\text{Ti}^{3+} = [\text{Ar}]3d^1 \text{ 1 unpaired electron}$ $\text{Mn}^{2+} = [\text{Ar}]3d^5 \text{ 5 unpaired electrons}$ $\text{Fe}^{2+} = [\text{Ar}]3d^6 \text{ 4 unpaired electrons}$ $\text{Co}^{3+} = [\text{Ar}]3d^6 \text{ 4 unpaired electrons}$ $\text{STEP 2:}$ $\text{The } \text{Mn}^{2+} \text{ ion contains a maximum number of unpaired electrons, hence, it has a high spin-only magnetic moment value.}$

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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
}