Import Question JSON

Current Question (ID: 18830)

Question:

$\text{The correct trend for I.E}_1 \text{ (1st Ionization Enthalpy) in the 3d series is:}$

Options:

1. $\text{Ionisation energy regularly decreases on moving along the period.}$
2. $\text{Ionisation energy regularly increases on moving along the period.}$
3. $\text{An irregular trend is observed for ionization energy because of electronic configuration.}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: Ionisation energy trend is irregular for first series transition elements.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals.}$ $\text{The irregular variations of ionization enthalpies can be attributed to the extra stability of configuration such as } d^0, d^5 \text{ and } d^{10}. \text{ Since these states are exceptionally stable, their ionization enthalpies are very high.}$ $\text{STEP 2:}$ $\text{In the case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration } (3d^5). \text{ On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals } (3d^{10}4s^2).$

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
}