Import Question JSON

Current Question (ID: 20977)

Question:

$\text{The pair of species that has identical shapes among the following is:}$ $1.\ \text{CF}_4,\ \text{SF}_4$ $2.\ \text{XeF}_2,\ \text{CO}_2$ $3.\ \text{BF}_3,\ \text{PCl}_3$ $4.\ \text{PF}_5,\ \text{IF}_5$

Options:

1. $\text{CF}_4,\ \text{SF}_4$
2. $\text{XeF}_2,\ \text{CO}_2$
3. $\text{BF}_3,\ \text{PCl}_3$
4. $\text{PF}_5,\ \text{IF}_5$

Solution:

$\text{Hint: In the case of shape, only bonds are included not a lone pair.}$ $1.\ \text{The hybridization of C in CF}_4\ \text{is sp}^3\ \text{and the shape is tetrahedral.}$ $\text{The hybridization of S in SF}_4\ \text{is sp}^3d.\ \text{The shape of SF}_4\ \text{is}$ $\text{a see-saw. The shape of both CF}_4\ \text{and SF}_4\ \text{are as follows:}$ $2.\ \text{The hybridization of Xe in XeF}_2\ \text{is sp}^3d\ \text{and the shape is linear.}$ $\text{The hybridization of C in CO}_2\ \text{is sp and the shape is linear.}$ $\text{The shape of both XeF}_2\ \text{and CO}_2\ \text{are as follows:}$ $3.\ \text{The hybridization of B in BF}_3\ \text{is sp}^2\ \text{and the shape is trigonal planar.}$ $\text{The hybridization of P in PCl}_3\ \text{is sp}^3\ \text{and the shape is trigonal pyramidal.}$ $\text{The shape of both BF}_3\ \text{and PCl}_3\ \text{are as follows:}$ $4.\ \text{The hybridization of P in PF}_5\ \text{is sp}^3d\ \text{and the shape is trigonal bipyramidal.}$ $\text{The hybridization of I in IF}_5\ \text{is sp}^3d^2\ \text{and the shape is square pyramidal.}$ $\text{The shape of PF}_5\ \text{and IF}_5\ \text{are as follows:}$

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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
}