Import Question JSON

Current Question (ID: 18757)

Question:

$\text{The correct statement, among the following, regarding } [\text{Mn(CN)}_6]^{3-} \text{ is:}$ $1. \text{It is } \text{sp}^3\text{d}^2 \text{ hybridised and octahedral in shape}$ $2. \text{It is } \text{sp}^3\text{d}^2 \text{ hybridised and tetrahedral in shape}$ $3. \text{It is } \text{d}^2\text{sp}^3 \text{ hybridised and octahedral in shape}$ $4. \text{It is } \text{dsp}^2 \text{ hybridised and square planar in shape}$

Options:

1. $\text{It is } \text{sp}^3\text{d}^2 \text{ hybridised and octahedral in shape}$
2. $\text{It is } \text{sp}^3\text{d}^2 \text{ hybridised and tetrahedral in shape}$
3. $\text{It is } \text{d}^2\text{sp}^3 \text{ hybridised and octahedral in shape}$
4. $\text{It is } \text{dsp}^2 \text{ hybridised and square planar in shape}$

Solution:

$\text{HINT : } [\text{Mn(CN)}_6]^{3-} \text{ is } \text{d}^2\text{sp}^3 \text{ hybridised.}$ $\text{Explanation:}$ $\text{In } [\text{Mn(CN)}_6]^{3-}, \text{ Mn is in +3 oxidation state which has electronic configuration } 3\text{d}^4 4\text{s}^0$ $\text{CN}^- \text{ is a strong field ligand. So, pairing of electrons will take place and the hybridisation of } [\text{Mn(CN)}_6]^{3-} \text{ is } \text{d}^2\text{sp}^3 \text{ and it is octahedral.}$

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
}