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Current Question (ID: 9171)

Question:

$\text{Li}_2\text{CO}_3 \text{ decomposes at a lower temperature because -}$

Options:

1. $\text{Lithium carbonate is not stable to heat.}$
2. $\text{Lithium carbonate is ionic in nature.}$
3. $\text{Lithium ion is unable to polarise carbonate ion.}$
4. $\text{Electropositive character of lithium is the highest.}$

Solution:

$\text{HINT: As we move down the alkali metal group, the electropositive character increases.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates. However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent.}$ $\text{Lithium-ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.}$ $\text{Step 2:}$ $\text{Li}_2\text{CO}_3 \xrightarrow{\Delta} \text{Li}_2\text{O} + \text{CO}_2$ $\text{Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.}$

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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
}