Import Question JSON

Current Question (ID: 7912)

Question:

For the second-period elements, the correct increasing order of first ionisation enthalpy is:

Options:

1. $\text{Li} < \text{Be} < \text{B} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne}$
2. $\text{Li} < \text{Be} < \text{B} < \text{C} < \text{N} < \text{O} < \text{F} < \text{Ne}$
3. $\text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne}$ (Correct)
4. $\text{Li} < \text{B} < \text{Be} < \text{C} < \text{N} < \text{O} < \text{F} < \text{Ne}$

Solution:

HINT: $\text{IE}_1$ of $\text{Be} > \text{B}$ and $\text{N} > \text{O}$ Explanation: Ionization Enthalpy is defined as the amount of energy an isolated gaseous atom would take to lose an electron in its ground state. As we go from left to right in a period, generally the ionisation energy increases. '$\text{Be}$' and '$\text{N}$' comparatively have more stable electronic configuration than '$\text{B}$' and '$\text{O}$'. Hence, nitrogen ionisation energy is more than oxygen and beryllium ionization energy is more than boron. It is due to the half-filled configuration of nitrogen ($1s^22s^22p^3$) and fully filled configuration of $\text{Be}$ ($1s^22s^2$) So, the correct order will be : $\text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne}$

Import JSON File

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
}