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Current Question (ID: 17968)

Question:

$\text{In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca}^{2+}\text{) and fluoride ion (F}^-\text{) are:}$ $1.\ 4\ \text{and}\ 2$ $2.\ 6\ \text{and}\ 6$ $3.\ 8\ \text{and}\ 4$ $4.\ 4\ \text{and}\ 8$

Options:

1. $4\ \text{and}\ 2$
2. $6\ \text{and}\ 6$
3. $8\ \text{and}\ 4$
4. $4\ \text{and}\ 8$

Solution:

$\text{HINT: Coordination number of F}^-\ \text{is}\ 4\ \text{while that of Ca}^{2+}\ \text{is}\ 8.}$ $\text{Explanation:}$ $\text{In CaF}_2\ \text{i.e. Fluorite structure Ca}^{2+}\text{ions are present at all corners and at the centre of each face of the cube and F}^-\text{ions occupy all the tetrahedral sites.}$ $\text{Coordination number of F}^-\ \text{is}\ 4\ \text{while that of Ca}^{2+}\ \text{is}\ 8.}$

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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
}