Import Question JSON

Current Question (ID: 7926)

Question:
$\text{Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:}$
Options:
  • 1. $\text{F} > \text{Cl} > \text{Br} > \text{I}$
  • 2. $\text{F} < \text{Cl} < \text{Br} < \text{I}$
  • 3. $\text{F} < \text{Cl} > \text{Br} > \text{I}$ (Correct)
  • 4. $\text{F} < \text{Cl} < \text{Br} > \text{I}$
Solution:
$\text{HINT: Generally down the group electron gain enthalpy decreases.}$ $\text{Explanation:}$ $\text{As we move from Cl to I, the electron gain enthalpy (i.e., energy released in electron gain) become less and less negative due to a corresponding increase in the atomic size.}$ $\text{However, the electron gain enthalpy of F is less negative than that of Cl due to its small size.}$ $\text{Thus, the negative electron gain enthalpy follows the order : F < Cl > Br > I}$ $\text{Detailed explanation:}$ $\text{Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom to form an anion.}$ $\text{Generally, electron gain enthalpy becomes less negative (decreases) down a group due to:}$ $\text{1. Increase in atomic size}$ $\text{2. Decrease in nuclear attraction for the incoming electron}$ $\text{However, fluorine is an exception because:}$ $\text{1. F atom is very small in size}$ $\text{2. The incoming electron experiences strong electron-electron repulsion in the compact 2p orbital}$ $\text{3. This makes it less favorable to add an electron to F compared to Cl}$ $\text{Therefore, Cl has the highest (most negative) electron gain enthalpy among halogens, followed by F, then Br, and finally I.}$ $\text{The correct order is: F < Cl > Br > I}$

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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
}