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Current Question (ID: 8058)

Question:

$\text{CH}_4\text{ does not exhibit square planar geometry because:}$

Options:

1. $\text{For square planar geometry, 5 bonds are required.}$
2. $\text{Carbon does not have d-orbitals to undergo dsp}^2\text{ hybridization.}$ (Correct)
3. $\text{Due to steric hindrance CH}_4\text{ does not exhibit square planar geometry.}$
4. $\text{Carbon does not have d-orbitals to undergo d}^2\text{sp}^3\text{ hybridization.}$

Solution:

$\text{HINT: Carbon doesn't have d- orbitals.}$ $\text{Explanation:}$ $\text{Electronic configuration of carbon atom:}$ $^6\text{C: } 1\text{s}^2 2\text{s}^2 2\text{p}^2$ $\text{In the excited state, the orbital picture of carbon can be represented as:}$ $\text{Hence, carbon atom undergoes sp}^3\text{ hybridization in CH}_4\text{ molecule and takes a tetrahedral shape.}$ $\text{For a square planar shape, the hybridization of the central atom has to be dsp}^2\text{. However, carbon atom does not have d-orbitals to undergo dsp}^2\text{ hybridization.}$ $\text{Hence, the structure of CH}_4\text{ cannot be square planar.}$

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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
}