Import Question JSON

Current Question (ID: 18767)

Question:

$\Delta_{\text{oct}} \text{(CFSE in the octahedral field) will be maximum in:}$ $\text{(Atomic number Co = 27)}$

Options:

1. $[\text{Co} (\text{H}_2\text{O})_6]^{3+}$
2. $[\text{Co} (\text{NH}_3)_6]^{3+}$
3. $[\text{Co} (\text{CN})_6]^{3-}$
4. $[\text{Co} (\text{C}_2\text{O}_4)_3]^{3-}$

Solution:

$\text{STEP 1:}$ $\Delta_0 \text{ depends on a few factors. These factors are as follows:}$ $\text{C. F. S. E.} \propto \text{O. No. of central atom}$ $\propto \text{geometry, sp} > \text{Oct} > \text{tetra}$ $\propto \text{Strength of ligands}$ $\text{STEP 2: In the given complexes, Co is in +3 oxidation state and metal is also}$ $\text{the same with 6 coordination number.}$ $\text{Thus, } \Delta_0 \text{ depends on the strength of the ligand.}$ $\text{Using spectrochemical series:}$ $\text{I}^- < \text{Br}^- < \text{SCN}^- < \text{Cl}^- < \text{S}^{2-} < \text{N}_3^- < \text{F}^- < \text{ONO}^- < \text{OH}^- < \text{SO}_4^{2-} < \text{NO}_4^- < \text{C}_2\text{O}_4^{2-} < \text{H}_2\text{O} < \text{NCS}^- < \text{EDTA} < \text{NH}_3 \sim \text{Py} < \text{en} < \text{bpy} \sim \text{phen}$ $\text{From } \text{I}^- \text{ to } \text{H}_2\text{O} \text{ are weak field ligands}$ $\text{From NCS}^- \text{ to CO are strong field ligands}$ $\text{STEP 3: So,}$ $\text{H}_2\text{O} = \text{Weakest} \rightarrow \text{lowest CFSE}$ $\text{CN} = \text{Strongest} \rightarrow \text{Highest CFSE}$

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
}