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Current Question (ID: 21267)

Question:

$\text{Which of the following anions forms a pale yellow precipitate when reacted with aqueous } \text{AgNO}_3, \text{ and the precipitate is partially soluble in aqueous } \text{NH}_4\text{OH solution?}$

Options:

1. $\text{I}^-$
2. $\text{Cl}^-$
3. $\text{Br}^-$
4. $\text{NO}_2^-$

Solution:

$\text{Hint: } \text{Br}^- + \text{Ag}^+ \rightarrow \text{AgBr} \quad (\text{Pale yellow ppt.})$ $1. \text{ Reaction of Bromide with Silver Nitrate:}$ $\text{AgNO}_3 + \text{Br}^- \rightarrow \text{AgBr} \downarrow + \text{NO}_3^-$ $\bullet \text{ AgBr forms a pale yellow precipitate.}$ $2. \text{ Solubility in Aqueous Ammonia (NH}_4\text{OH):}$ $\text{AgBr} + 2\text{NH}_3 \rightleftharpoons [\text{Ag(NH}_3)_2]^+ + \text{Br}^-$ $\bullet \text{ AgBr is partially soluble in NH}_4\text{OH, forming a complex ion.}$ $\text{I}^- + \text{Ag}^+ \rightarrow \text{AgI} \quad (\text{Yellow ppt.})$ $\text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl} \quad (\text{White ppt.})$ $\text{Br}^- + \text{Ag}^+ \rightarrow \text{AgBr} \quad (\text{Pale yellow ppt.})$ $\text{NO}_2^- + \text{Ag}^+ \rightarrow \text{AgNO}_2 \quad (\text{White ppt.})$ $\text{AgBr is partially soluble in aq. NH}_4\text{OH solution whereas AgI is insoluble in aq. NH}_4\text{OH solution.}$ $\bullet \text{ AgI (from I}^-) \text{ forms a yellow precipitate, insoluble in NH}_4\text{OH.}$ $\bullet \text{ AgCl (from Cl}^-) \text{ forms a white precipitate, soluble in NH}_4\text{OH.}$ $\bullet \text{ NO}_2^- \text{ does not form a precipitate with AgNO}_3.$

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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
}