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Current Question (ID: 18682)

Question:

$2.335 \text{ g of compound } X \text{ (empirical formula } \text{CoH}_{12}\text{N}_4\text{Cl}_3) \text{ upon treatment with excess } \text{AgNO}_3 \text{ solution produces } 1.435 \text{ g of a white precipitate. The primary and secondary valences of cobalt in compound } X, \text{ respectively are:}$ $[\text{Given Atomic mass: Co = 59, Cl = 35.5, Ag = 108}]$

Options:

1. $3, 6$
2. $3, 4$
3. $2, 4$
4. $4, 3$

Solution:

$\text{HINT: One mole of Cl}^- \text{ gives one mole of AgCl.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Secondary valence is the number of groups bound directly to the metal ion.}$ $\text{Secondary valence is coordination number.}$ $\text{Primary valence is the oxidation state of Co in the complex.}$ $\text{Step 2:}$ $X + \text{AgNO}_3 \rightarrow \text{AgCl} \downarrow$ $\text{mole of } X \text{ used } = 0.01$ $\text{moles of AgCl produced } = 0.01$ $\text{Compound } X \text{ is } [\text{Co }(\text{NH}_3)_4 \text{Cl}_2]\text{Cl}$ $\text{Hence, the primary and secondary valences of cobalt in compound } (X) \text{ 3 and 6.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}