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Current Question (ID: 18678)

Question:

$\text{Two complexes, PtCl}_4 \cdot 2 \text{NH}_3 \text{ and PtCl}_4 \cdot 2 \text{KCl, do not give a precipitate of AgCl when treated with AgNO}_3. \text{ They give zero and three moles of ions, respectively, in solutions for one mole of the complex. The structural formulae of both complexes are:}$

Options:

1. $[\text{PtCl}_2 (\text{NH}_3)_2] \text{Cl}_2 \text{ and K}_2[\text{PtCl}_6]$
2. $[\text{PtCl}_4 (\text{NH}_3)_2] \text{ and K}_2[\text{PtCl}_2 (\text{NH}_3)_2]$
3. $[\text{PtCl}_4 (\text{NH}_3)_2] \text{ and K}_2[\text{PtCl}_6]$
4. $[\text{PtCl}_4] \cdot 2 \text{NH}_3 \text{ and [PtCl}_4] \cdot 2 \text{KCl}$

Solution:

$\text{Hint: If Cl atom is present outside the complex sphere then ppt of AgCl is formed.}$ $\text{Explanation:}$ $\text{Step 1: When complex reacts with AgNO}_3 \text{ then white precipitate is formed only when Cl atom present outside the complex sphere.}$ $\text{The PtCl}_4 \cdot 2 \text{NH}_3 \text{ and PtCl}_4 \cdot 2 \text{KCl did not produce white ppt. with AgNO}_3, \text{ thus, it indicates that all Cl atoms in both the complexes are present inside the coordination sphere.}$ $\text{Step 2: The structural formulae of both complexes are as follows:}$ $[\text{PtCl}_4 (\text{NH}_3)_2] \text{ and K}_2[\text{PtCl}_6]$ $\text{The first compound } [\text{PtCl}_4 (\text{NH}_3)_2] \text{ produce the only ion and K}_2[\text{PtCl}_6] \text{ produces three ions in the solution. The reaction is as follows:}$ $\text{K}_2[\text{PtCl}_6] \leftrightarrow 2\text{K}^+ + [\text{PtCl}_6]^{2-}$

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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
}