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Current Question (ID: 7991)

Question:
Compounds that have a maximum value of lattice energy among the following, is:
Options:
  • 1. $1.\ \text{NaF}$ (Correct)
  • 2. $2.\ \text{NaCl}$
  • 3. $3.\ \text{NaBr}$
  • 4. $4.\ \text{NaI}$
Solution:
$\text{HINT: Lattice energy is inversely proportional to the size of ions.}$ $\text{Explanation:}$ $\text{The energy which is used to decompose the ionic solids into their constituent cations and anions is known as lattice energy. For example when sodium and chloride ions combine to form sodium chloride.}$ $\text{Both sodium and chloride ions are present in the gaseous state, 787.3 kJ/mol of energy release, this is known as lattice energy. So as we go down the group, the ionic radius increases which decrease the lattice energy.}$ $\text{When we move from left to right in a period or as the charge on the ions increases the lattice energy increases. Lattice energy is directly proportional to the charge on the ion and it is inversely proportional to the size of the ion.}$ $\text{lattice energy } \propto \frac{\text{charge on ion}}{\text{size of ion}}$ $\text{In the given options, the cation is common (Na}^+\text{). The anions are F}^-\text{, Cl}^-\text{, Br}^-\text{, and I}^-\text{.}$ $\text{The size of halide ions increases down the group: F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-$. $\text{Since lattice energy is inversely proportional to the size of the ions, the compound with the smallest anion will have the maximum lattice energy.}$ $\text{Therefore, NaF will have the maximum lattice energy.}$ $\text{The correct order of increasing lattice energy will be NaI < NaBr < NaCl < NaF.}$

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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
}