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Current Question (ID: 18706)

Question:

$\text{A complex has a molecular formula } \text{MSO}_4\text{Cl}\cdot 6\text{H}_2\text{O}. \text{ Its aqueous solution gives white precipitate with barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution.}$ $\text{If the secondary valence of the metal is six, the correct representation of the complex is:}$

Options:

1. $[\text{M(H}_2\text{O})_4\text{Cl}]\text{SO}_4\cdot 2\text{H}_2\text{O}$
2. $[\text{M(H}_2\text{O})_6]\text{SO}_4$
3. $[\text{M(H}_2\text{O})_5\text{Cl}]\text{SO}_4\cdot \text{H}_2\text{O}$
4. $[\text{M(H}_2\text{O})_3\text{Cl}]\text{SO}_4\cdot 3\text{H}_2\text{O}$

Solution:

$\text{HINT: No ppt. of } \text{AgNO}_3 \text{ means no } \text{Cl}^- \text{ ion is present outside the complex.}$ $\text{STEP 1: Formation of white precipitate with Barium chloride indicates that } \text{SO}_4^{2-} \text{ ions are outside the coordination sphere, and no precipitate with } \text{AgNO}_3 \text{ solution indicates that the } \text{Cl}^- \text{ ions are inside the coordination sphere.}$ $\text{STEP 2: Since the coordination number of M is 6, } \text{Cl}^- \text{ and 5 } \text{H}_2\text{O} \text{ are ligands, remaining 1 } \text{H}_2\text{O} \text{ molecular and } \text{SO}_4^{2-} \text{ are in the outer coordination sphere.}$ $\text{The formula of complex will be: } [\text{M(H}_2\text{O})_5\text{Cl}]\cdot (\text{SO}_4)(\text{H}_2\text{O})$ $\text{STEP 3: The reaction between } [\text{M(H}_2\text{O})_5\text{Cl}]\cdot (\text{SO}_4)(\text{H}_2\text{O}) \text{ and BaCl}_2 \text{ can be represented as:}$ $[\text{M(H}_2\text{O})_5\text{Cl}]\cdot (\text{SO}_4)(\text{H}_2\text{O}) + \text{BaCl}_2 \rightarrow [\text{M(H}_2\text{O})_5\text{Cl}]\cdot \text{H}_2\text{O} + 2 \text{Cl}^- + \text{BaSO}_4 \text{(ppt)}$

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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
}