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Current Question (ID: 20881)

Question:

$\text{The electrons identified by quantum numbers } n \text{ and } l \text{ can be placed in order of increasing energy as:}$ $\text{a. } n = 4, l = 1$ $\text{b. } n = 4, l = 0$ $\text{c. } n = 3, l = 2$ $\text{d. } n = 3, l = 1$

Options:

1. $(d) < (b) < (c) < (a)$
2. $(b) < (d) < (a) < (c)$
3. $(a) < (c) < (b) < (d)$
4. $(c) < (d) < (b) < (a)$

Solution:

$\text{Hint: For same } (n+l) \text{ value, the orbital with lowest } n \text{ value has lowest energy.}$ $\text{Find } (n+l) \text{ value for electrons identified by quantum numbers:}$ $1. \text{ For } n = 4, l = 1: (n+l) \text{ value is } 5. \text{ The orbital is } 4p.$ $2. \text{ For } n = 4, l = 0: (n+l) \text{ value is } 4. \text{ The orbital is } 4s.$ $3. \text{ For } n = 3, l = 2: (n+l) \text{ value is } 5. \text{ The orbital is } 3d.$ $4. \text{ For } n = 3, l = 1: (n+l) \text{ value is } 4. \text{ The orbital is } 3p.$ $\text{For same } (n+l) \text{ value, the orbital with lowest } n \text{ value has lowest energy.}$ $\text{The order of orbitals according to their energy is as follows:}$ $3p < 4s < 3d < 4p$ $(d) < (b) < (c) < (a)$

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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
}