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Current Question (ID: 8569)

Question:
$\text{The compound AgF}_2 \text{ (unstable) acts as a/an:}$
Options:
  • 1. $\text{Oxidising agent.}$
  • 2. $\text{Reducing agent.}$
  • 3. $\text{Both oxidising and reducing agent.}$
  • 4. $\text{Neither oxidising and reducing agent.}$
Solution:
$\text{Hint: Ag is unstable in +2 form.}$ $\text{EXPLANATION:}$ $\text{The oxidation state of Ag in AgF}_2 \text{ is +2. But, +2 is an unstable oxidation state of Ag.}$ $\text{Therefore, whenever AgF}_2 \text{ is formed, silver readily accepts an electron to form Ag}^+\text{.}$ $\text{This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1.}$ $\text{As a result, AgF}_2 \text{ acts as a very strong oxidizing agent.}$ $\text{Since AgF}_2 \text{ readily accepts electrons (gets reduced from +2 to +1), it causes other substances to lose electrons (get oxidized), making it an oxidizing agent.}$

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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
}