Import Question JSON

Current Question (ID: 20984)

Question:

$\text{Which of the following statement is incorrect regarding PCl}_5\text{?}$ $1.\ \text{In PCl}_5,\ \text{axial bonds are longer than equatorial bonds.}$ $2.\ \text{It has a trigonal bipyramidal structure.}$ $3.\ \text{3 chlorine atoms are in the same plane.}$ $4.\ \text{Hybridisation of 'P' is sp}^3d^2.$

Options:

1. $\text{In PCl}_5,\ \text{axial bonds are longer than equatorial bonds.}$
2. $\text{It has a trigonal bipyramidal structure.}$
3. $\text{3 chlorine atoms are in the same plane.}$
4. $\text{Hybridisation of 'P' is sp}^3d^2.$

Solution:

$\text{Hint: Hybridization = sp}^3d$ $\text{Step 1: The central atom in PCl}_5\text{ is phosphorus, which has five valence electrons.}$ $\text{In its ground state, phosphorus has two valence electrons in the s orbital and three in the p orbitals.}$ $\text{Hybridization is the process of mixing atomic orbitals of similar energies to form new orbitals with different energies and shapes.}$ $\text{In PCl}_5,\ \text{the phosphorus atom undergoes sp}^3d\ \text{ hybridization, which results in five sp}^3d\ \text{ hybrid orbitals.}$ $\text{Step 2: The five sp}^3d\ \text{ orbitals of phosphorus are directed to the five corners of a trigonal bipyramidal geometry.}$ $\text{Three equatorial bonds and two axial bonds are present in the molecule.}$ $\text{The bond angle between two equatorial P-Cl bonds is 120}^\circ,\ \text{ while the bond angle between one axial and one equatorial P-Cl bond is 90}^\circ.$ $\text{Similarly, the axial P-Cl bond length is larger than the equatorial P-Cl bond.}$ $\text{Hence, option 4 is the correct answer.}$

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
}