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Current Question (ID: 20873)

Question:
$\text{Using the following statements, identify the correct set of statements:}$ $\text{(i) } n \text{ (principal quantum number) can have values } 1, 2, 3, 4, \ldots$ $\text{(ii) The number of orbitals for a given value of } l \text{ is } (2l+1).$ $\text{(iii) The value of spin quantum numbers is always } \pm \frac{1}{2}.$ $\text{(iv) For } l=5, \text{ the total number of orbitals is } 9.$
Options:
  • 1. $(i), (ii), (iii)$
  • 2. $(i), (ii), (iv)$
  • 3. $(i), (ii), (iii), (iv)$
  • 4. $(i), (iii), (iv)$
Solution:
$\text{Hint: For any sub-shell (defined by } 'l' \text{ value) } 2l+1 \text{ values of } m_l \text{ are possible}$ $\text{For } l=5 \text{ total number of orbital is } (2l+1)=11$

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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
}