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Current Question (ID: 20962)

Question:

$\text{The correct statement about ICl}_5 \text{ and ICl}_4^- \text{ is:}$

Options:

1. $\text{Both are isostructural}$
2. $\text{ICl}_5 \text{ is square pyramidal and ICl}_4^- \text{ is square planar}$
3. $\text{ICl}_5 \text{ is trigonal bipyramidal and ICl}_4^- \text{ is tetrahedral}$
4. $\text{ICl}_5 \text{ is square pyramidal and ICl}_4^- \text{ is tetrahedral}$

Solution:

$\text{Hint: ICl}_5 \text{ and ICl}_4^- \text{ both have the different shape}$ $\text{First, find the steric number as follows:}$ $\text{Use the formula of steric number. The formula is as follows:}$ $\text{Steric number} = \frac{1}{2} \times (V + M - C + A)$ $\text{Here, V is the valence electron of the central atom, M is the monovalent atom, C is the positive charge on the compound and A is the negative charge on the compound.}$ $\text{Species} \quad \text{Calculate steric number} \quad \text{Hybrid}$ $\text{[ICl}_4^-\text{]} \quad \text{Steric number} = \frac{1}{2} \times (7 + 4 + 1) = 6 \quad \text{sp}^3\text{d}^2$ $\text{[ICl}_5\text{]} \quad \text{Steric number} = \frac{1}{2} \times (7 + 5) = 6 \quad \text{sp}^3\text{d}^2$ $\text{Both geometry is octahedral. The shape of both the compound is different. [ICl}_4^-\text{] shape is square planar, and [ICl}_5\text{] shape is square pyramidal}$

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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
}