Import Question JSON

Current Question (ID: 9149)

Question:

$\text{The correct diagonal relation of Li and Mg among the following is -}$

Options:

1. $\text{Both form solid bicarbonates.}$
2. $\text{Their oxides are less soluble in water.}$
3. $\text{Their carbonates are ionic in nature.}$
4. $\text{Both form peroxides and superoxides.}$

Solution:

$\text{HINT: Oxides of Li and Mg are less soluble in water.}$ $\text{Explanation:}$ $\text{Step 1-}$ $\text{The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose at high temperatures.}$ $2\text{LiOH} \xrightarrow{\text{Heat}} \text{Li}_2\text{O} + \text{H}_2\text{O}$ $\text{Mg(OH)}_2 \xrightarrow{\text{Heat}} \text{MgO} + \text{H}_2\text{O}$ $\text{(iii) Both Li and Mg react with } \text{N}_2 \text{ to form nitrides.}$ $6\text{Li} + \text{N}_2 \xrightarrow{\text{Heat}} 2\text{Li}_3\text{N}$ $3\text{Mg} + \text{N}_2 \xrightarrow{\text{Heat}} \text{Mg}_3\text{N}_2$ $\text{Step 2: Neither Li nor Mg form peroxides or superoxides.}$ $\text{The carbonates of both are covalent in nature. Also, these decompose on heating.}$ $\text{Li}_2\text{CO}_3 \xrightarrow{\text{Heat}} \text{Li}_2\text{O} + \text{CO}_2$ $\text{MgCO}_3 \xrightarrow{\text{Heat}} \text{MgO} + \text{CO}_2$ $\text{Li and Mg do not form solid bicarbonates.}$ $\text{Both LiCl and } \text{MgCl}_2 \text{ are soluble in ethanol owing to their covalent nature.}$

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
}