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Current Question (ID: 7682)

Question:

$\text{The electron gain enthalpies of halogens in kJ mol}^{-1}\text{ are given below.}$ $\text{F} = -332, \text{Cl} = -349, \text{Br} = -325, \text{I} = -295.$ $\text{The lesser negative value for F as compared to that of Cl is due to:}$

Options:

1. $\text{Strong electron-electron repulsions in the compact 2p-subshell of F.}$
2. $\text{Weak electron-electron repulsions in the bigger 3p-subshell of Cl.}$
3. $\text{Smaller electronegativity value of F than Cl.}$
4. $\text{1 & 2 both}$ (Correct)

Solution:

$\text{HINT: Interelectronic repulsion in fluorine.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{As we know that fluorine, chlorine, bromine, and iodine belong to the same group (17th) therefore, the size of fluorine is very small but as we go down the group the size increases.}$ $\text{The electron gain enthalpy is the tendency of elements to accept electrons in their outermost shell.}$ $\text{The halogens are present at the extreme right in the periodic table so the size of the first element of halogen(fluorine) is very small so fluorine atoms cannot add an electron due to electron crowding in its outermost shell although it is the most electronegative element. So, the electron gain enthalpy is less negative.}$ $\text{As we go down the group in halogens, the chlorine atom is having large negative electron gain enthalpy due to more electronegativity as well as increased size.}$ $\text{STEP 2:}$ $\text{As we further go down the group, the size of halogens increases and electronegativity decreases so the tendency to accept electrons in the outermost shell decreases and the electron gain enthalpy becomes less negative.}$ $\text{Therefore, the correct order of the electron gain enthalpy is fluorine (-332kJ/mol), chlorine(-349kJ/mol), bromine (-325kJ/mol) and iodine (-295kJ/mol).}$ $\text{Analysis of the options:}$ $\text{Option 1: Strong electron-electron repulsions in the compact 2p-subshell of F - This is CORRECT.}$ $\text{The small size of F leads to high electron density in the 2p orbital, causing strong repulsion when an additional electron is added.}$ $\text{Option 2: Weak electron-electron repulsions in the bigger 3p-subshell of Cl - This is also CORRECT.}$ $\text{The larger 3p orbital in Cl has more space, reducing electron-electron repulsion when an electron is added.}$ $\text{Option 3: Smaller electronegativity value of F than Cl - This is INCORRECT.}$ $\text{F is more electronegative than Cl, not less.}$ $\text{Since both statements 1 and 2 are correct and together explain why F has a less negative electron gain enthalpy than Cl, option 4 is the correct answer.}$

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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
}