Import Question JSON

Current Question (ID: 20876)

Question:

$\text{The quantum number of four electrons are given below:}$ $\begin{array}{l} \text{I. } n = 4, \ l = 2, \ m_l = -2, \ m_s = -\frac{1}{2} \\ \text{II. } n = 3, \ l = 2, \ m_l = 1, \ m_s = +\frac{1}{2} \\ \text{III. } n = 4, \ l = 1, \ m_l = 0, \ m_s = +\frac{1}{2} \\ \text{IV. } n = 3, \ l = 1, \ m_l = 1, \ m_s = -\frac{1}{2} \end{array}$ $\text{The correct order of their increasing energies will be:}$

Options:

1. $\text{I < III < II < IV}$
2. $\text{IV < II < III < I}$
3. $\text{I < II < III < IV}$
4. $\text{IV < III < II < I}$

Solution:

$\text{Hint: Find the value of (n+l)}$ $\text{Higher the value of (n + l), higher will be energy of orbital.}$ $\text{If (n + l) value are equal, then higher the value of n, higher will be the energy of the orbital.}$ $\text{The (n+l) value for the quantum number of four electrons are as follows:}$ $\text{I. } n + l = 6 \quad n = 4, \ l = 2. \ \text{The orbital is 4d.}$ $\text{II. } n + l = 5 \quad n = 3, \ l = 2. \ \text{The orbital is 3d.}$ $\text{III. } n + l = 5 \quad n = 4, \ l = 1. \ \text{The orbital is 4p.}$ $\text{IV. } n + l = 4 \quad n = 3, \ l = 1. \ \text{The orbital is 3p.}$ $\text{The energy order is } 3p < 3d < 4p < 4d.$ $\text{Hence, the correct order of quantum number of four electrons in their increasing energies is}$ $\text{IV < II < III < I}$

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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
}