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Current Question (ID: 7990)

Question:
A Compound with maximum ionic character is formed from:
Options:
  • 1. $1.\ \text{Na and I}$
  • 2. $2.\ \text{Cs and F}$ (Correct)
  • 3. $3.\ \text{Cs and I}$
  • 4. $4.\ \text{Na and F}$
Solution:
$\text{HINT: The concept of the Fajan rule is used.}$ $\text{Explanation:}$ $\text{In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.}$ $\text{Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.}$ $\text{Using Fajan's rule, a larger cation and smaller anion will have maximum ionic character.}$ $\text{Hence, Cs is the most electropositive element in the first group, and F is the most electronegative element.}$ $\text{So, they will form the most ionic compound CsF.}$ $\text{To maximize the ionic character of a compound, we need the largest possible difference in electronegativity between the combining elements.}$ $\text{Electronegativity generally decreases down a group and increases across a period.}$ $\text{Among the given options:}$ $\text{Metals: Na (Sodium) and Cs (Cesium). Cesium (Cs) is below Sodium (Na) in Group 1, making it more electropositive and having lower ionization enthalpy, thus a larger cation.}$ $\text{Non-metals: F (Fluorine) and I (Iodine). Fluorine (F) is above Iodine (I) in Group 17, making it more electronegative and having higher electron gain enthalpy, thus a smaller anion.}$ $\text{Therefore, the combination of the most electropositive metal (Cs) and the most electronegative non-metal (F) will result in the compound with the maximum ionic character.}$ $\text{This corresponds to Cesium and Fluorine forming CsF.}$

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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
}