Import Question JSON

Current Question (ID: 7668)

Question:

The first ionisation enthalpies of $\text{Na, Mg, Al,}$ and $\text{Si}$ are in the order of:

Options:

1. $\text{Na} < \text{Al} < \text{Mg} < \text{Si}$ (Correct)
2. $\text{Na} > \text{Mg} > \text{Al} > \text{Si}$
3. $\text{Na} < \text{Mg} < \text{Al} < \text{Si}$
4. $\text{Na} > \text{Mg} > \text{Al} < \text{Si}$

Solution:

Follow the following steps to solve out such problems $\text{Step I}$: Write the electronic configuration to find position in the periodic table $\text{}_{11}\text{Na} = [\text{Ne}] \text{3s}^1$ $\text{}_{12}\text{Mg} = [\text{Ne}] \text{3s}^2$ $\text{}_{13}\text{Al} = [\text{Ne}] \text{3s}^2\text{2p}^1$ $\text{}_{14}\text{Si} = [\text{Ne}] \text{3s}^2\text{2p}^2$ $\text{Step II}$: Arrange them in the order as they are in the periodic table $\text{Na }\text{Mg }\text{Al }\text{Si}$ $\text{11 }\text{12 }\text{13 }\text{14}$ $\text{Step III}$: Follow the general trend and also keep in mind the exception The $\text{IP}$ increases along a period from left to right but $\text{IP}$ of $\text{Mg}$ is higher than that of $\text{Al}$ due to completely filled $\text{3s}$ orbital in $\text{Mg}$. $\text{Step IV}$: On the above basis find the order The order of $\text{IP}$ is $\text{Na} < \text{Al} < \text{Mg} < \text{Si}$

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
}