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Current Question (ID: 18766)

Question:

$\text{The crystal field stabilisation energy for the high spin } d^4 \text{ octahedral complex is:}$

Options:

1. $-1.8 \Delta_o$
2. $-1.6 \Delta_o$
3. $-1.2 \Delta_o$
4. $-0.6 \Delta_o$

Solution:

$\text{HINT: CFSE for high spin } d^4 \text{ octahedral complex is } -0.6 \Delta_o$ $\text{Explanation:}$ $\text{STEP 1: The crystal field stabilization energy (CFSE) is the stability that results from placing a transition metal ion in the crystal field generated by a set of ligands.}$ $\text{As a result, if there are any electrons occupying these orbitals, the metal ion is more stable in the ligand field by the amount known as the CFSE.}$ $\text{STEP 2: High spin } d^4 \text{ complexes have electronic configuration:}$ $\text{STEP 3: Thus, CFSE can be calculated as:}$ $\Delta_o(\text{CFSE}) = (0.4 \times 3) + (0.6 \times 1) \Delta_o$ $= -0.6 \Delta_o$

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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
}