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Current Question (ID: 7624)

Question:

$\text{The general outer electronic configuration of s, p, d, and f-block elements respectively would be:}$

Options:

1. $\text{ns}^{1-2}, \text{nd}^2\text{np}^{1-6}, (\text{n}-1)\text{d}^{1-10}\text{np}^{0-2}, (\text{n}-2)\text{f}^{1-14}(\text{n}-1)\text{d}^{0-10}\text{ns}^2$
2. $\text{ns}^{1-2}, \text{ns}^2\text{np}^{1-6}, (\text{n}-1)\text{f}^{1-10}\text{ns}^{0-2}, (\text{n}-2)\text{g}^{1-14}(\text{n}-1)\text{d}^{0-1}\text{ns}^2$
3. $\text{ns}^{1-2}, \text{ns}^2\text{np}^{1-6}, (\text{n}-1)\text{d}^{1-10}\text{ns}^{1-2}, (\text{n}-2)\text{f}^{1-14}(\text{n}-1)\text{d}^{0-1} \text{ns}^2$ (Correct)
4. $\text{np}^{1-2}, \text{nd}^2\text{np}^{1-6}, (\text{n}-1)\text{d}^{1-10}\text{ns}^{0-2}, (\text{n}-2)\text{f}^{1-14}(\text{n}-1)\text{d}^{0-10}\text{ns}^2$

Solution:

$\text{HINT: Follow the periodic table.}$\n\n$\text{Explanation:}$\n\n$\text{Element}$ $\quad\quad\quad$ $\text{General outer electronic configuration}$\n\n$\text{s–block}$ $\quad\quad\quad$ $\text{ns}^{1-2}$\n\n$\text{p–block}$ $\quad\quad\quad$ $\text{ns}^2\text{np}^{1-6}$\n\n$\text{d–block}$ $\quad\quad\quad$ $(\text{n} - 1) \text{d}^{1-10}\text{ns}^{0-2}$, $\text{where n = 3, 4, 5, 6, 7}$\n\n$\text{f–block}$ $\quad\quad\quad$ $(\text{n} - 2) \text{f}^{1-14}(\text{n} - 1)\text{d}^{0-1} \text{ns}^2$, $\text{where n = 6, 7}$

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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
}