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Current Question (ID: 18827)

Question:

$\text{Among the following, which are amphoteric oxides?}$ $1. \text{V}_2\text{O}_5, \text{Cr}_2\text{O}_3$ $2. \text{Mn}_2\text{O}_7, \text{CrO}_3$ $3. \text{CrO}, \text{V}_2\text{O}_5$ $4. \text{V}_2\text{O}_5, \text{V}_2\text{O}_4$

Options:

1. $\text{V}_2\text{O}_5, \text{Cr}_2\text{O}_3$
2. $\text{Mn}_2\text{O}_7, \text{CrO}_3$
3. $\text{CrO}, \text{V}_2\text{O}_5$
4. $\text{V}_2\text{O}_5, \text{V}_2\text{O}_4$

Solution:

$\text{HINT: } \text{V}_2\text{O}_5 \text{ and } \text{Cr}_2\text{O}_3 \text{ are amphoteric oxide because both react with alkalies as well as acids.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{An amphoteric oxide is one that can act as either an acid or a base.}$ $\text{Generally, non-metallic oxides are acidic and metallic oxides are basic.}$ $\text{Some non-metallic oxides are neutral and some metallic or semi-metallic oxides are amphoteric.}$ $\text{STEP 2:}$ $\text{There are a number of exceptions to these generalizations.}$ $\text{V}_2\text{O}_5 + 2 \text{HNO}_3 \rightarrow 2 \text{VO}_2(\text{NO}_3) + \text{H}_2\text{O}$ $\text{It also reacts with strong alkali to form polyoxovanadates, If excess aqueous sodium hydroxide is used, the product is a colourless salt, sodium orthovanadate, } \text{Na}_3\text{VO}_4.$ $\text{STEP 3:}$ $1. \text{Cr}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{CrCl}_3 + 3\text{H}_2\text{O}$ $2. \text{Cr}_2\text{O}_3 + 4\text{MO} + 3\text{O}_2 \rightarrow 4\text{MCrO}_4$ $\text{In equation 1, it reacts with acid and acts as a base and in equation 2, reacts with base and behave as acid. So it is also amphoteric in nature.}$

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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
}