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Current Question (ID: 20968)

Question:
$\text{In which of the following pairs the two species are not isostructural?}$ $1.\ \text{PCl}_4^+\ \text{and}\ \text{SiCl}_4$ $2.\ \text{PF}_5\ \text{and}\ \text{BrF}_5$ $3.\ \text{AlF}_6^{3-}\ \text{and}\ \text{SF}_6$ $4.\ \text{CO}_3^{2-}\ \text{and}\ \text{NO}_3^-$
Options:
  • 1. $\text{PCl}_4^+\ \text{and}\ \text{SiCl}_4$
  • 2. $\text{PF}_5\ \text{and}\ \text{BrF}_5$
  • 3. $\text{AlF}_6^{3-}\ \text{and}\ \text{SF}_6$
  • 4. $\text{CO}_3^{2-}\ \text{and}\ \text{NO}_3^-$
Solution:
$\text{Hint: Isostructural species have the same hybridization and shape.}$ $\text{Step 1:}$ $\text{Isostructural species are those which have the same shape and hybridization.}$ $\text{Step 2:}$ $1.\ \text{PCl}_4^+\ \text{and}\ \text{SiCl}_4$ $\text{The hybridization of P in}\ \text{PCl}_4^+\ \text{and S in}\ \text{SiCl}_4\ \text{is}\ sp^3.\ \text{The shape of both compounds is tetrahedral. Thus, they are isostructural species.}$ $2.\ \text{PF}_5\ \text{and}\ \text{BrF}_5$ $\text{The hybridization of P in}\ \text{PF}_5\ \text{is}\ sp^3d\ \text{and the hybridization of Br in}\ \text{BrF}_5\ \text{is}\ sp^3d^2.\ \text{Hence, they are not an example of isostructural species.}$ $3.\ \text{AlF}_6^{3-}\ \text{and}\ \text{SF}_6$ $\text{The hybridization of Al in}\ \text{AlF}_6^{3-}\ \text{is}\ sp^3d^2\ \text{and S in}\ \text{SF}_6\ \text{is}\ sp^3d^2.\ \text{The shape of both species is the same that is octahedral. Hence, isostructural to each other.}$ $4.\ \text{CO}_3^{2-}\ \text{and}\ \text{NO}_3^-$ $\text{The hybridization of C in}\ \text{CO}_3^{2-}\ \text{is}\ sp^2\ \text{and N in}\ \text{NO}_3^-\ \text{is}\ sp^2.\ \text{The shape of both species is the same that is trigonal planar. Hence, isostructural to each other.}$ $\text{The correct answer is option second.}$

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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
}