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Current Question (ID: 18774)

Question:

$[\text{Fe} (\text{H}_2\text{O})_6]^{2+}$ $\text{and}$ $[\text{Fe} (\text{CN})_6]^{4-}$ $\text{differ in:}$

Options:

1. $\text{Geometry and magnetic moment.}$
2. $\text{Geometry and hybridization.}$
3. $\text{Magnetic moment and colour.}$
4. $\text{Hybridization and number of d-electrons.}$

Solution:

$\text{HINT: CN}^- \text{is a strong field ligand and H}_2\text{O is weak field ligand.}$ $\text{STEP 1: Fe}^{2+} \text{in} [\text{Fe} (\text{H}_2\text{O})_6]^{2+} \text{is having d}^6 \text{electronic configuration.}$ $\text{H}_2\text{O is weak field ligand so pairing will not happen. It is a high spin complex.}$ $\text{Total number of unpaired electrons; n=4}$ $\text{So, } \mu = \sqrt{n(n+2)} = \sqrt{4 \times 6} = 4.9 \text{ B. M.}$ $\text{Colour: Pale green, } \mu=4.9 \text{ B.M.; octahedral}$ $\text{STEP 2: Fe}^{2+} \text{in} [\text{Fe} (\text{CN})_6]^{4-} \text{is having d}^6 \text{configuration.}$ $\text{CN}^- \text{being a strong field ligand will pair up the electrons. It is low spin complex.}$ $\text{Total number of unpaired electrons = 0 ; n = 0}$ $\text{So, } \mu = 0 \text{ B.M.}$ $\text{The colour of } [\text{Fe} (\text{CN})_6]^{4-} \text{is Prussian blue.}$ $\text{The geometry is octahedral and } \mu = 0 \text{ because it doesn't contain unpaired electrons.}$

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
}