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Current Question (ID: 7518)

Question:

$\text{An ion among the following that has the same radius as a hydrogen atom, having n=1, is}$

Options:

1. $\text{He}^{+}, n = 2$
2. $\text{Li}^{2+}, n = 2$
3. $\text{Be}^{3+}, n = 2$
4. $\text{Li}^{2+}, n = 3$

Solution:

$\text{Hint: } r_n = 52.9 \times \frac{n^2}{Z}$\n\n$\text{Explanation:}$\n\n$\text{Calculation of the radius of an atom is the application of Bohr's model, which is applicable for hydrogen and hydrogen-like single electron-containing species such as } \text{Li}^{2+}, \text{Be}^{3+} \text{ ion.}$\n\n$\text{The radius of a hydrogen atom with n=1 is 52.9 pm}$\n\n$\text{Now calculate radius for given species as follows:}$\n\n$\text{1. He}^{+}, n = 2$\n\n$r_n = 52.9 \times \frac{2^2}{2} = 52.9 \times \frac{4}{2} = 105.8 \text{ pm}$\n\n$\text{2. Li}^{2+}, n = 2$\n\n$r_n = 52.9 \times \frac{2^2}{3} = 52.9 \times \frac{4}{3} = 70.53 \text{ pm}$\n\n$\text{3. Be}^{3+}, n = 2$\n\n$r_n = 52.9 \times \frac{2^2}{4} = 52.9 \times \frac{4}{4} = 52.9 \text{ pm}$\n\n$\text{4. Li}^{2+}, n = 3$\n\n$r_n = 52.9 \times \frac{3^2}{3} = 52.9 \times \frac{9}{3} = 158.7 \text{ pm}$\n\n$\text{Comparing with the radius of hydrogen atom with n=1 (52.9 pm), we can see that Be}^{3+} \text{ with n=2 has the same radius.}$\n\n$\text{Hence, option 3 is the correct answer.}$

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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
}