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Current Question (ID: 18751)

Question:

$\text{Complexes with halide ligands are generally:}$ $1. \text{High spin complexes.}$ $2. \text{Low spin complexes.}$ $3. \text{Both high spin and low spin complexes.}$ $4. \text{None of the above.}$

Options:

1. $\text{High spin complexes.}$
2. $\text{Low spin complexes.}$
3. $\text{Both high spin and low spin complexes.}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: Halides ion are weak ligand generally form high spin complexes.}$ $\text{Step 1: Find the nature of halide ion.}$ $\text{A chloride ion is a weak field ligand. See the spectrochemical series:}$ $\text{I}^- < \text{Br}^- < [\text{NCS}]^- < \text{Cl}^- < \text{F}^- < [\text{OH}]^- < [\text{ox}]^{2-}$ $\approx \text{H}_2\text{O} < [\text{NCS}]^- < \text{NH}_3 < \text{en} < \text{bpy} < \text{phen} < [\text{CN}]^- \approx \text{CO}$ $\text{weak field ligands} \rightarrow \text{strong field ligands}$ $\text{increasing } \Delta_{\text{oct}}$ $\text{Step 2: Find the nature of ligands and spin multiplicity}$ $\text{Weaker ligands cannot provide the energy needed to pair up the electrons.}$ $\text{Hence, chloride ions will form a high spin complex.}$

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
}