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Current Question (ID: 17967)

Question:

$\text{The ionic radii of A}^+ \text{ and B}^- \text{ ions are } 0.98 \times 10^{-10} \text{ m and } 1.81 \times 10^{-10} \text{ m. The coordination number of each ion in AB is:}$

Options:

1. $8$
2. $2$
3. $6$
4. $4$

Solution:

$\text{HINT: The radius ratio is } 0.414 - 0.732 \text{ for coordination number 6.}$ $\text{Explanation:}$ $\text{Given that, ionic radius of cation (A}^+) = 0.98 \times 10^{-10} \text{ m}$ $\text{Ionic radius of anion (B}^-) = 1.81 \times 10^{-10} \text{ m}$ $\text{Coordination number of each ion in AB=} ?$ $\text{As we know that, Radius ratio=} \frac{\text{Radius of cation}}{\text{Radius of anion}} = \frac{0.98 \times 10^{-10}}{1.81 \times 10^{-10}} = 0.541$ $\text{If value of radius ratio is in between } 0.414 - 0.732, \text{ ion would have octahedral structure with coordination number 6.}$

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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
}