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Current Question (ID: 7539)

Question:

$\text{The total number of electrons in an atom with the following quantum numbers would be}$ $\text{(a) n = 4, m}_s = -\frac{1}{2}$ $\text{(b) n = 3, l = 0$

Options:

1. $16, 2$
2. $11, 8$
3. $16, 8$
4. $12, 7$

Solution:

$\text{Hint: Use basics of quantum numbers.}$ $\text{(a) Total number of electrons in an atom for a value of n = 2n}^2$ $\text{For n = 4,}$ $\text{Total number of electrons} = 2 \times 4^2 = 32$ $\text{For n = 4, possible subshells are 4s, 4p, 4d and 4f. Total number of orbitals are 16. That means it can have maximum 32 electrons but only 16 will have anticlockwise spin and other 16 will have clockwise spin.}$ $\text{Since we need electrons with m}_s = -\frac{1}{2}\text{, we count only half of the total electrons.}$ $\text{So for n = 4 and m}_s = -\frac{1}{2}\text{, total electrons will be 16.}$ $\text{(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital.}$ $\text{For n = 3, l = 0 (3s orbital):}$ $\text{The 3s orbital can hold a maximum of 2 electrons.}$ $\text{Therefore, the number of electrons having n = 3 and l = 0 is 2.}$ $\text{Final Answer: For (a) n = 4, m}_s = -\frac{1}{2}\text{: 16 electrons}$ $\text{For (b) n = 3, l = 0: 2 electrons}$ $\text{Therefore, the answer is 16, 2.}$

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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
}