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Current Question (ID: 7673)

Question:

$\text{If the ionization enthalpy and negative electron gain enthalpy of an element are 275 and 86 kcal mol}^{-1} \text{ respectively, then the electronegativity of the element on the Pauling scale is:}$

Options:

1. $2.8$ (Correct)
2. $0.0$
3. $4.0$
4. $2.6$

Solution:

$\text{HINT: Electronegativity = } \frac{\text{I.E. (meV) + E.A.(meV)}}{5.6}$\n\n$\text{Explanation: According to the Paulings' Formula:}$\n\n$\text{I.E. + E.A. = 275 + 86 = 361 kcal mol}^{-1}$\n\n$= 361 \times 4.184 = 1510.42 \text{ kJ mol}^{-1}$\n\n$\therefore \text{Electronegativity} = \frac{1510.42}{540} = 2.797 = 2.8$\n\n$\text{Step-by-step calculation:}$\n\n$\text{1. Given data:}$\n\n$\text{ - Ionization Enthalpy (I.E.) = 275 kcal mol}^{-1}$\n\n$\text{ - Electron Affinity (E.A.) = 86 kcal mol}^{-1}$\n\n$\text{2. Sum of I.E. and E.A.:}$\n\n$\text{ I.E. + E.A. = 275 + 86 = 361 kcal mol}^{-1}$\n\n$\text{3. Convert to kJ mol}^{-1}\text{:}$\n\n$\text{ 361 kcal mol}^{-1} \times 4.184 \text{ kJ/kcal} = 1510.42 \text{ kJ mol}^{-1}$\n\n$\text{4. Apply Pauling's formula:}$\n\n$\text{ Electronegativity} = \frac{1510.42}{540} = 2.797 \approx 2.8$\n\n$\text{The Pauling electronegativity scale uses the constant 540 kJ/mol as the denominator in the formula.}$

Import JSON File

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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
}