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Current Question (ID: 7618)

Question:
$\text{The period and group number of the element with Z = 114 are:}$
Options:
  • 1. $\text{8th period and 16th group}$
  • 2. $\text{7th period and 14th group}$ (Correct)
  • 3. $\text{14th period and 7th group}$
  • 4. $\text{9th group and 14th period}$
Solution:
$\text{HINT: The last element of actinoid is 103, so you need to find 11th element from Actinoid.}$\n\n$\text{Explanation:}$\n\n$\text{[Rn] 5f}^{14}\text{6d}^{10}\text{7s}^2\text{7p}^2$\n\n$\text{Those with Z = 90 – 103 are f – block elements.}$\n\n$\text{10 elements with Z = 89 and Z = 104 – 112 are d – block elements, and the elements with Z = 113 – 118 are p – block elements.}$\n\n$\text{Therefore, the element with Z = 114 is the second p – block element in the 7th period.}$

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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
}