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Current Question (ID: 8046)

Question:

Isostructural pair among the following is:

Options:

1. $\text{BCl}_3$ and $\text{BrCl}_3$
2. $\text{NH}_3$ and $\text{NO}_3^-$
3. $\text{NF}_3$ and $\text{BF}_3$
4. $\text{BF}_4^-$ and $\text{NH}_4^+$ (Correct)

Solution:

To be isostructural, if a number of bond pairs and lone pairs are the same for the given pairs. 1. $\text{BCl}_3$: Central atom B. 3 sigma bonds, 0 lone pairs. Hybridisation $\text{sp}^2$, trigonal planar geometry. $\text{BrCl}_3$: Central atom Br. 3 sigma bonds, 2 lone pairs. Hybridisation $\text{sp}^3\text{d}$, T-shape geometry. Not isostructural. 2. $\text{NH}_3$: Central atom N. 3 sigma bonds, 1 lone pair. Hybridisation $\text{sp}^3$, pyramidal geometry. $\text{NO}_3^-$: Central atom N. 3 sigma bonds, 0 lone pairs. Hybridisation $\text{sp}^2$, trigonal planar geometry. Not isostructural. 3. $\text{NF}_3$: Central atom N. 3 sigma bonds, 1 lone pair. Hybridisation $\text{sp}^3$, pyramidal geometry. $\text{BF}_3$: Central atom B. 3 sigma bonds, 0 lone pairs. Hybridisation $\text{sp}^2$, trigonal planar geometry. Not isostructural. 4. $\text{BF}_4^-$: Central atom B. 4 sigma bonds, 0 lone pairs. Hybridisation $\text{sp}^3$, tetrahedral geometry. $\text{NH}_4^+$: Central atom N. 4 sigma bonds, 0 lone pairs. Hybridisation $\text{sp}^3$, tetrahedral geometry. Thus, $\text{BF}_4^-$ and $\text{NH}_4^+$ are isostructural. Therefore, the correct pair is $\text{BF}_4^-$ and $\text{NH}_4^+$.

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}