Import Question JSON

Current Question (ID: 7862)

Question:
$\text{Which among the following elements is the bridge element?}$
Options:
  • 1. $\text{K}$
  • 2. $\text{O}$
  • 3. $\text{Mg}$ (Correct)
  • 4. $\text{Pb}$
Solution:
$\text{HINT: Na and Mg are bridge elements.}$ $\text{Explanation:}$ $\text{Na and Mg are linked to Group IA and IB, and they are forming a bridge with}$ $\text{the elements which are placed diagonally opposite to Na and Mg, and the}$ $\text{properties of both the elements are similar due to electronegativity.}$ $\text{Bridge elements are those elements that show similarities in properties with}$ $\text{elements that are diagonally placed in the periodic table. This phenomenon}$ $\text{is known as the diagonal relationship.}$ $\text{Examples of diagonal relationships:}$ $\text{- Li (Group 1, Period 2) and Mg (Group 2, Period 3)}$ $\text{- Be (Group 2, Period 2) and Al (Group 13, Period 3)}$ $\text{- B (Group 13, Period 2) and Si (Group 14, Period 3)}$ $\text{The reason for diagonal relationship is that as we move across a period,}$ $\text{the charge on the cation increases and size decreases. As we move down}$ $\text{a group, the charge remains the same but size increases. The diagonal}$ $\text{movement combines both effects, resulting in similar charge-to-size ratios}$ $\text{and hence similar properties.}$ $\text{Among the given options, Mg is a bridge element as it shows diagonal}$ $\text{relationship with Li and forms a bridge between Group 1 and Group 2}$ $\text{elements.}$

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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
}