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Current Question (ID: 21202)

Question:

$\text{The correct statements among I to III regarding group 13 element oxides are:}$ $\text{(I) Boron trioxide or diboron trioxide is acidic.}$ $\text{(II) Oxides of aluminium and gallium are amphoteric.}$ $\text{(III) Oxides of indium and thallium are basic.}$

Options:

1. $(\text{I}) \text{ and } (\text{II}) \text{ only}$
2. $(\text{I}) \text{ and } (\text{III}) \text{ only}$
3. $(\text{II}) \text{ and } (\text{III}) \text{ only}$
4. $(\text{I}), (\text{II}), \text{ and } (\text{III}) \text{ only}$

Solution:

$\text{Hint: Down the group acidity of an oxide decreases.}$ $\text{B}_2\text{O}_3 \text{ is acidic in nature.}$ $\text{Al}_2\text{O}_3 \text{ and Ga}_2\text{O}_3 \text{ are amphoteric oxides.}$ $\text{In and Tl oxides are basic in nature.}$ $\text{The reaction is as follows:}$ $\text{B}_2\text{O}_3 + 2\text{NaOH} \rightarrow 2\text{NaBO}_2 + \text{H}_2\text{O}$ $\text{Al}_2\text{O}_3 + 6\text{NaOH} \rightarrow 2\text{Al(OH)}_3 + 3\text{Na}_2\text{O}$ $\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$ $\text{Ga}_2\text{O}_3 + 6\text{NaOH} \rightarrow 2\text{Ga(OH)}_3 + 3\text{Na}_2\text{O}$ $\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$

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
}