Import Question JSON

Current Question (ID: 7543)

Question:
$\text{The number of orbitals that are possible at the L-energy level is}$
Options:
  • 1. $2$
  • 2. $4$
  • 3. $6$
  • 4. $1$
Solution:
$\text{Hint: Number of orbitals} = n^2$ $\text{L - Energy level means second shell. The value of n is 2. So, total number of orbitals is}$ $\text{No. of orbitals} = n^2 = (2)^2 = 4$ $\text{Detailed explanation:}$ $\text{The L-shell corresponds to the second principal energy level where n = 2.}$ $\text{For n = 2, the possible subshells are:}$ $\text{• 2s subshell (l = 0): 1 orbital}$ $\text{• 2p subshell (l = 1): 3 orbitals}$ $\text{Total orbitals in L-shell = 1 + 3 = 4 orbitals}$ $\text{This confirms the formula: Number of orbitals in nth shell} = n^2$ $\text{For n = 2: Number of orbitals} = 2^2 = 4$ $\text{Therefore, the correct answer is option 2.}$

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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
}