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Current Question (ID: 7610)

Question:
$\text{The total number of 3rd period elements with more than one electron in a 3d orbital is:}$
Options:
  • 1. $9$
  • 2. $11$
  • 3. $0$
  • 4. $8$
Solution:
$\text{HINT: Filling of 3d orbital starts from 4-period elements.}$ $\text{Explanation:}$ $\text{The filling of 3d orbital starts from 4 periods onwards. The third period}$ $\text{doesn't contain any element which has 3d electrons. Hence, the answer is}$ $\text{option third.}$ $\text{To understand this better:}$ $\text{The third period elements are: Na (11) to Ar (18).}$ $\text{According to the Aufbau principle, electrons fill orbitals in order of increasing energy:}$ $\text{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, ...}$ $\text{For third period elements, the electron configuration ends with 3s and 3p orbitals.}$ $\text{The 3d orbital is higher in energy than 4s, so it only starts filling from the fourth}$ $\text{period (starting with Scandium, Sc).}$ $\text{Therefore, no third period element has any electrons in 3d orbitals, let alone}$ $\text{more than one electron in a 3d orbital.}$

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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
}