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Current Question (ID: 7854)

Question:
$\text{The period number and group number of "Tantalum" (Z=73) are respectively :}$
Options:
  • 1. $5, 7$
  • 2. $6, 13$
  • 3. $6, 5$ (Correct)
  • 4. $\text{None of the above.}$
Solution:
$\text{HINT: Tantalum is a d block element.}$ $\text{Explanation:}$ $\begin{array}{ccc} \text{Period number} & \text{Group number} & \text{Element} \\ 5 & 7 & \text{Tc} \\ 6 & 13 & \text{Tl} \\ 6 & 5 & \text{Ta} \end{array}$ $\text{Electronic configuration of Ta is Z = 73 [Xe]4f}^{14}\text{5d}^3\text{6s}^2$ $\text{Period no.} \rightarrow \text{Maximum quantum no.}$ $\text{Period} \rightarrow 6 \text{ and d block}$ $\text{Group = no. of e}^- \text{ in ns + no. of e}^- \text{ in (n-1)d}$ $= 2 + 3 = 5$ $\text{So, Group} \rightarrow 5$

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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
}