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Current Question (ID: 7994)

Question:

$\text{On the left are given decreasing orders of elements for a certain property and on right side is given the property. The incorrect order for the given properties is:}$

Options:

1. $\text{AlF}_3 > \text{MgO} > \text{MgF}_2 \text{ : Lattice energy}$ (Correct)
2. $\text{Li} > \text{Na} > \text{Al} > \text{Mg} \text{ : Electron affinity}$
3. $\text{PF}_5 > \text{SF}_6 > \text{SiF}_4 \text{ : Lewis acidic character}$
4. $\text{SiCl}_4 > \text{SiBr}_4 > \text{SiI}_4 \text{ : Decreasing ionic character}$

Solution:

$\text{HINT: Lattice energy depends on the product of ionic charges.} \text{Explanation:} \text{1. In the first option, the q}_1\text{ and q}_2 \text{ factor increases, thus, lattice energy increases.} \text{2. The 2nd option is correct because, due to completely filled 3s subshell, Mg does not have tendency to accept an electron.} \text{Electron affinity values of Li = 0.618 eV, Na = 0.547 eV , Mg = -0.4 eV and Al = 0.432 eV} \text{3. The correct order of lewis acidic character: PF}_5 > \text{SiF}_4 > \text{SF}_6 \text{Although, S has vacant 3d-orbitals it cannot accept co-ordinate bonds from Lewis base due to the steric crowding factor as S-atom is already bonded to six F-atoms.} \text{4. According to Fajan's rule : Anion should be small for more ionic character.} \text{So , the order of ionic character is : SiCl}_4 > \text{SiBr}_4 > \text{SiI}_4$

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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
}