Import Question JSON

Current Question (ID: 20983)

Question:

$\text{Match the molecules in Column-I with their corresponding molecular shapes in Column-II.}$ $\text{Column-I (Molecule)} \quad \text{Column-II (Shape)}$ $(i) \ \text{BF}_3 \quad (a) \ \text{T-shape}$ $(ii) \ \text{SF}_4 \quad (b) \ \text{Square planar}$ $(iii) \ \text{XeF}_4 \quad (c) \ \text{See-saw}$ $(iv) \ \text{ClF}_3 \quad (d) \ \text{Trigonal planar}$ $\text{Choose the correct matching sequence:}$

Options:

1. $(i) \ d \quad (ii) \ c \quad (iii) \ b \quad (iv) \ a$
2. $(i) \ a \quad (ii) \ b \quad (iii) \ c \quad (iv) \ d$
3. $(i) \ a \quad (ii) \ c \quad (iii) \ d \quad (iv) \ b$
4. $(i) \ b \quad (ii) \ c \quad (iii) \ d \quad (iv) \ a$

Solution:

$\text{Hint: In shape, the lone pair is not included.}$ $\text{Explanation:}$ $\text{The compound and its shape are as follows:}$ $\text{Column I} \quad \text{Column II}$ $(i) \ \text{BF}_3 \quad (a) \ \text{Trigonal planar}$ $(ii) \ \text{SF}_4 \quad (b) \ \text{See-saw}$ $(iii) \ \text{XeF}_4 \quad (c) \ \text{Square planar}$ $(iv) \ \text{ClF}_3 \quad (d) \ \text{T-shape}$ $\text{The structure of the given compound is as follows:}$ $(i) \ \text{(Trigonal planar)} \quad (ii) \ \text{(See-saw)} \quad (iii) \ \text{(Square planar)} \quad (iv) \ \text{(T-shape)}$ $\text{Hence, option 1 is the correct answer.}$

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
}