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Current Question (ID: 7913)

Question:

Amongst the following electronic configurations, the highest ionization energy is represented by:

Options:

1. $[\text{Ne}]3s^23p^3$ (Correct)
2. $[\text{Ne}]3s^23p^2$
3. $[\text{Ar}]3d^{10}4s^24p^3$
4. $[\text{Ne}]3s^23p^1$

Solution:

HINT: Ionization energy is inversely proportional to the size of atom. Explanation: Ionization energies have high values in the atoms having completely filled or exactly half filled subshell values. Also, IE values increase from left to right in a period and decrease from top to bottom in a group generally . Electronic configuration $[\text{Ne}]3s^23p^3$ and $[\text{Ar}]3d^{10}4s^24p^3$ Since these $2$ options have exactly half filled configurations, hence they will have high IE values but since they belong to same group (grp-$15$) and down the group, IE decreases. Hence, $[\text{Ne}]3s^23p^3$ has highest IE value.

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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
}