Import Question JSON

Current Question (ID: 7633)

Question:

$\text{The incorrect match among the following options is:}$

Options:

1. $[\text{Ar}]3\text{d}^5 4\text{s}^1 \rightarrow 4^{\text{th}} \text{period}, 6^{\text{th}} \text{group}$
2. $[\text{Kr}]4\text{d}^{10} \rightarrow 5^{\text{th}} \text{period}, 12^{\text{th}} \text{group}$ (Correct)
3. $[\text{Rn}]6\text{d}^2 7\text{s}^2 \rightarrow 7^{\text{th}} \text{period}, 3^{\text{th}} \text{group}$
4. $[\text{Xe}]4\text{f}^{14} 5\text{d}^2 6\text{s}^2 \rightarrow 6^{\text{th}} \text{period}, 4^{\text{th}} \text{group}$

Solution:

$\text{HINT: Compare the electronic configuration of given elements.}$\n\n$\text{Explanation:}$\n\n$\text{1. [Ar]3d}^5\text{4s}^1 \text{electronic configuration belongs to chromium (Cr). 24- The group is 6 and period is 4.}$\n\n$\text{2. [Kr]4d}^{10} \text{electronic configuration belongs to palladium(Pd). 46 -The group is 10 and period is 5.}$\n\n$\text{3. [Rn]6d}^2\text{7s}^2 \text{electronic configuration belongs to thorium(Th). 90 -The group is 3 and period is 7.}$\n\n$\text{The actual crystal structure can only be explained when the 5f states are invoked, proving that thorium is metallurgically a true actinide}$\n\n$\text{4. [Xe]4f}^{14}\text{5d}^2\text{6s}^2 \text{electronic configuration belongs to hafnium(Hf). grp 72 -. The group is 4 and period is 6.}$\n\n$\text{So, 2nd is wrong because Pd belongs to group}$ $10^{\text{th}}$ $\text{not group}$ $12^{\text{th}}$.

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}