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Current Question (ID: 7676)

Question:

$\text{1st }(\Delta H_1)\text{ and 2nd }(\Delta H_2)\text{ Ionization Enthalpies (in kJ mol}^{-1})\text{ and the }(\Delta_{eg}H)\text{ Electron Gain Enthalpy (in kJ mol}^{-1})\text{ of a few elements are given below:}$ $\begin{array}{|c|c|c|c|} \hline \text{Elements} & \Delta H_1 & \Delta H_2 & \Delta_{eg}H \\ \hline \text{I} & 520 & 7300 & -60 \\ \text{II} & 419 & 3051 & -48 \\ \text{III} & 1681 & 3374 & -328 \\ \text{IV} & 1008 & 1846 & -295 \\ \text{V} & 2372 & 5251 & +48 \\ \text{VI} & 738 & 1451 & -40 \\ \hline \end{array}$ $\text{The most reactive metal is:}$

Options:

1. $\text{VI}$
2. $\text{III}$
3. $\text{I}$
4. $\text{II}$ (Correct)

Solution:

$\text{HINT: Metals have low ionization enthalpy.}$ $\text{Explanation:}$ $\text{More reactive metal has a greater tendency to lose electrons and form positive ions. So, the metal with lowest ionization energy and highest electron gain enthalpy will be most reactive.}$ $\text{Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy }(\Delta H_1)\text{ and a low negative electron gain enthalpy }(\Delta_{eg}H)\text{.}$ $\text{Analysis of the data:}$ $\text{Looking at the first ionization enthalpies:}$ $\text{I: 520 kJ/mol}$ $\text{II: 419 kJ/mol (lowest)}$ $\text{III: 1681 kJ/mol}$ $\text{IV: 1008 kJ/mol}$ $\text{V: 2372 kJ/mol}$ $\text{VI: 738 kJ/mol}$ $\text{Element II has the lowest first ionization enthalpy (419 kJ/mol), making it the easiest to remove an electron from, which indicates high metallic reactivity.}$ $\text{The large jump from }\Delta H_1\text{ to }\Delta H_2\text{ for elements I and II (520 to 7300 and 419 to 3051 respectively) suggests these are alkali metals, with II being more reactive due to its lower first ionization energy.}$

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{
  "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
}