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Current Question (ID: 7536)

Question:

$\text{Which of the following sets of quantum numbers is possible:}$

Options:

1. $\text{n = 0, l = 0, m}_l = 0, \text{m}_s = +\frac{1}{2}$
2. $\text{n = 1, l = 0, m}_l = 0, \text{m}_s = -\frac{1}{2}$
3. $\text{n = 1, l = 1, m}_l = 0, \text{m}_s = +\frac{1}{2}$
4. $\text{n = 3, l = 3, m}_l = -3, \text{m}_s = +\frac{1}{2}$

Solution:

$\text{HINT: Use the basics of quantum numbers.}$ $\text{Analysis of each set:}$ $\text{(1) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.}$ $\text{The principal quantum number n must be a positive integer (n = 1, 2, 3, ...). Since n = 0 is given, this set is invalid.}$ $\text{(2) The given set of quantum numbers is possible.}$ $\text{For n = 1: l can be 0 (since l ranges from 0 to n-1)}$ $\text{For l = 0: m}_l \text{ can be 0 (since m}_l \text{ ranges from -l to +l)}$ $\text{m}_s = -\frac{1}{2} \text{ is always valid for electron spin}$ $\text{(3) The given set of quantum numbers is not possible.}$ $\text{For a given value of n, 'l' can have values from zero to (n - 1).}$ $\text{For n = 1, l = 0 and not 1. Since l = 1 when n = 1, this set is invalid.}$ $\text{(4) The given set of quantum numbers is not possible.}$ $\text{For n = 3, l = 0 to (3 - 1) = 0 to 2, i.e., 0, 1, 2}$ $\text{Since l = 3 when n = 3, this violates the rule that l < n, making this set invalid.}$ $\text{Therefore, only option 2 represents a possible set of quantum numbers.}$

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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
}