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Current Question (ID: 8651)

Question:

$\text{The first ionization energy value (kJ) mol}^{-1} \text{ for H, Li, F, and Na is given below but not in a correct order}$ $1681, 520, 1312, 495.$ $\text{The ionization energy value of H would be -}$

Options:

1. $1681$
2. $1312$
3. $520$
4. $485$

Solution:

$\text{Hint: I.E}_{\text{Alkali metals}} < \text{I.E}_\text{H} < \text{I.E}_\text{F}$ $\text{In chemistry, ionization energy is the amount of energy required for all of the atoms in a mole of substance to lose one electron each.}$ $\text{Left to right across the period ionization energy increases because effective nuclear charge increases. Down the group ionization energy decreases because the number of shells increases because its effective nuclear charge decreases.}$ $\text{The ionization energy trend is Na} < \text{Li} < \text{H} < \text{F.}$ $\text{The values are } 495 < 520 < 1312 < 1681$ $\text{The ionization energy of hydrogen is more than alkali metals but less than fluorine atoms.}$ $\text{Hence, option second is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}