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Current Question (ID: 17952)

Question:

$\text{Oxygen atoms form fcc unit cells with 'A' atoms occupying all tetrahedral voids and 'B' atoms occupying all octahedral voids. If atoms are removed from two of the body diagonals then the formula of the resultant compound formed is:}$

Options:

1. $\text{A}_4\text{B}_4\text{O}_7$
2. $\text{A}_8\text{B}_6\text{O}_7$
3. $\text{A}_8\text{B}_8\text{O}_7$
4. $\text{A}_6\text{B}_8\text{O}_6$

Solution:

$\text{HINT: Number of tetrahedral voids = 2} \times \text{Number of octahedral voids.}$ $\text{STEP 1:}$ $\text{Originally, Number of oxygen atoms} = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 1 + 3 = 4 \text{ atoms}$ $\text{Number of 'A' atoms} = 8$ $\text{Number of 'B' atoms} = \frac{1}{4} \times 12 + 1 = 3 + 1 = 4 \text{ atoms}$ $\text{STEP 2:}$ $\text{After removal of atoms from two body diagonals are-}$ $\text{Number of oxygen atoms} = 4 - \frac{1}{8} \times 4 = 4 - \frac{1}{2} = \frac{7}{2}$ $\text{Number of 'A' atoms} = 8 - 4 = 4$ $\text{Number of 'B' atoms} = 4 - 1 = 3$ $\text{STEP 3:}$ $\text{Hence, the formula of the compound is}$ $\text{A}_4\text{B}_3\text{O}_{7/2} \text{ or } \text{A}_8\text{B}_6\text{O}_7$

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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
}