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Current Question (ID: 18768)

Question:

$\text{The species having highest magnetic moment is:}$

Options:

1. $\text{K}_4[\text{Fe(CN)}_6]$
2. $[\text{Fe(H}_2\text{O)}_6]\text{SO}_4$
3. $\text{K}_3[\text{Fe(CN)}_6]$
4. $[\text{Co(NH}_3)_6]\text{SO}_4$

Solution:

$\text{HINT: More the number of unpaired electrons, more will be magnetic moment.}$ $\text{Explanation:}$ $\text{STEP 1: In } \text{K}_4[\text{Fe(CN)}_6], \text{ Fe oxidation state is } +2. \text{ The CN is a strong field ligand hence, it will pair up the electrons present in } \text{Fe}^{2+} \text{ ion. Hence, } \text{K}_4[\text{Fe(CN)}_6] \text{ has zero unpaired electrons.}$ $\text{In } [\text{Fe(H}_2\text{O)}_6]\text{SO}_4, \text{ Fe oxidation state is } +2. \text{ The electronic configuration is } 3d^6. \text{ It contains 4 unpaired electrons.}$ $\text{STEP 2: In } \text{K}_3[\text{Fe(CN)}_6], \text{ Fe oxidation state is } +3. \text{ The CN is a strong field ligand hence, it will pair up the electrons present in } \text{Fe}^{3+} \text{ ion. Hence, it has one unpaired electron.}$ $\text{In } [\text{Co(NH}_3)_6]\text{SO}_4, \text{ Co oxidation state is } +2. \text{ The electronic configuration is } 3d^7. \text{ It contains 3 unpaired electrons.}$ $\text{STEP 3: So, } [\text{Fe(H}_2\text{O)}_6]\text{SO}_4 \text{ will have highest magnetic moment.}$

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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
}