Import Question JSON

Current Question (ID: 18750)

Question:

$\text{The correct statement among the following about } [\text{Co(CN)}_6]^{3-} \text{ is:}$

Options:

1. $\text{It has no unpaired electrons and will be in a low-spin configuration}$
2. $\text{It has four unpaired electrons and will be in a low-spin configuration}$
3. $\text{It has four unpaired electrons and will be in a high-spin configuration}$
4. $\text{It has no unpaired electrons and will be in a high-spin configuration}$

Solution:

$\text{HINT: CN}^- \text{ is a strong field ligand.}$ $\text{Explanation:}$ $\text{STEP 1: In } [\text{Co(CN)}_6]^{3-}, \text{ the electronic configuration of Co}^{3+} \text{ is } 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$ $\text{STEP 2: CN}^- \text{ is a strong field ligand and as it approaches the metal ion, the electrons must pair up.}$ $\text{The splitting of the d-orbitals into two sets of orbitals in an octahedral } [\text{Co(CN)}_6]^{3-} \text{ may be represented as:}$ $\text{Here, for d}^6 \text{ ions, three electrons first enter orbitals with parallel spin, the remaining may pair up in t}_{2g} \text{ orbital giving rise to low spin complex (strong ligand) field.}$ $\text{STEP 3: Thus, } [\text{Co(CN)}_6]^{3-} \text{ has no unpaired electron and will be in a low spin configuration.}$

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
}