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Current Question (ID: 7687)

Question:

$\text{Elements}\quad\Delta H_1\quad\Delta H_2\quad\Delta_{eg}H$\n\n$\text{I}\quad\quad\quad 520\quad 7300\quad -60$\n\n$\text{II}\quad\quad\quad 419\quad 3051\quad -48$\n\n$\text{III}\quad\quad 1681\quad 3374\quad -328$\n\n$\text{IV}\quad\quad 1008\quad 1846\quad -295$\n\n$\text{V}\quad\quad 2372\quad 5251\quad +48$\n\n$\text{VI}\quad\quad\quad 738\quad 1451\quad -40$\n\n$\text{The least reactive element based on the above data is:}$

Options:

1. $\text{III}$
2. $\text{IV}$
3. $\text{II}$
4. $\text{V}$ (Correct)

Solution:

$\text{HINT: Energy needed to lose electrons is ionization energy and the energy that comes out to gain electrons is called electron gain enthalpy.}$\n\n$\text{Explanation:}$\n\n$\text{a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy }(\Delta H_1)\text{ and a positive electron gain enthalpy }(\Delta H_{eg}).\text{}$\n\n$\text{b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy }(\Delta H_1)\text{ and a low negative electron gain enthalpy }(\Delta H_{eg}).\text{}$\n\n$\text{c) Element III is likely to be the most reactive non-metal as it has a high first ionization enthalpy}(\Delta H_1)\text{ and the highest negative electron gain enthalpy }(\Delta H_{eg}).\text{}$

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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
}