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Current Question (ID: 9136)

Question:

$\text{A blue coloured solution of an alkali metal in liq NH}_3 \text{ is due to the presence of -}$

Options:

1. $\text{M}^- \text{ atoms}$
2. $\text{M}^+ \text{ ions}$
3. $\text{Solvated anions}$
4. $\text{Solvated electrons}$

Solution:

$\text{HINT: The blue colour of the solution is due to the excitation of ammoniated electrons to higher energy levels.}$ $\text{Explanation:}$ $\text{The blue coloured solution of an alkali metal in ammonia is explained on the basis of the formation of ammoniated (solvated) metal cations and ammoniated (solvated) electrons in the metal-ammonia solution in the following way:}$ $\text{M} \rightarrow \text{M}^+ + \text{e}^-$ $\text{M}^+ + x\text{NH}_3 \rightarrow \{\text{M}(\text{NH}_3)_x\}^+$ $\text{e}^- + y\text{NH}_3 \rightarrow [\text{e}(\text{NH}_3)_y]^-$ $\text{M} + (x + y)\text{NH}_3 \text{ metal cation} \rightarrow [\text{M}(\text{NH}_3)_x]^+ + [\text{e}(\text{NH}_3)_y]^-$ $\text{Solvated electron}$ $\text{The blue colour of the solution is due to the excitation of free electrons to higher energy levels.}$ $\text{The absorption of photons takes place in the red region of the spectrum and hence, the solution appears blue in the transmitted light.}$ $\text{As the concentration of the alkali metal increases, the metal ion cluster formation takes place and at a very high concentration, the solution becomes coloured like that of metallic copper.}$

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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
}