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Current Question (ID: 7613)

Question:
$\text{Considering the elements B, C, N, F, and Si; the correct order of their non-}$ $\text{metallic character is :}$
Options:
  • 1. $\text{B} > \text{C} > \text{Si} > \text{N} > \text{F}$
  • 2. $\text{Si} > \text{C} > \text{B} > \text{N} > \text{F}$
  • 3. $\text{F} > \text{N} > \text{C} > \text{B} > \text{Si}$ (Correct)
  • 4. $\text{F} > \text{N} > \text{C} > \text{Si} > \text{B}$
Solution:
$\text{HINT: See the general trend of metallic character along a period or down the}$ $\text{group.}$ $\text{Explanation:}$ $\text{As we go from left to right the metallic character decreases and the non}$ $\text{metallic character increases.}$ $\text{Because as we move from left to right in a period, the number of valence}$ $\text{electrons increases by one at each succeeding element but the number of}$ $\text{shells remains same. Due to this effective nuclear charge increase.}$ $\text{Thus, the decreasing order of non-metallic character is F} > \text{N} > \text{C} > \text{B.}$ $\text{The non metallic character of C} > \text{Si. However, non metallic character of B} >$ $\text{Si.}$ $\text{The correct order of non-metallic character : F} > \text{N} > \text{C} > \text{B} > \text{Si}$ $\text{To understand this better:}$ $\text{1. Across a period (left to right): Non-metallic character increases due to}$ $\text{increasing effective nuclear charge, making atoms more likely to gain electrons.}$ $\text{2. Down a group: Non-metallic character decreases due to increasing atomic size}$ $\text{and decreasing effective nuclear charge.}$ $\text{3. Positioning of elements:}$ $\text{ - F (Period 2, Group 17): Highest non-metallic character}$ $\text{ - N (Period 2, Group 15): High non-metallic character}$ $\text{ - C (Period 2, Group 14): Moderate non-metallic character}$ $\text{ - B (Period 2, Group 13): Lower non-metallic character (metalloid)}$ $\text{ - Si (Period 3, Group 14): Lowest non-metallic character (metalloid)}$

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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
}