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Current Question (ID: 8070)

Question:

Which species lacks tetrahedral geometry?

Options:

1. $\text{BH}_4^{-}$
2. $\text{NH}_2^{-}$
3. $\text{CO}_3^{2-}$ (Correct)
4. $\text{H}_3\text{O}^{+}$

Solution:

$\text{HINT: Molecule having } \text{sp}^3 \text{ hybridization has tetrahedral geometry.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{The sum of lone pair and sigma bond represents hybridization. If sum is 4 then hybridization is } \text{sp}^3 \text{, if sum is three than hybridization is } \text{sp}^2 \text{ and if sum is two then hybridization is sp.}$ $\text{STEP 2:}$ $\text{In } \text{BH}_4^{-} \text{, boron form four sigma bonds and doesnot contains any lone pair . Hence its hybridization is } \text{sp}^3 \text{ and geometry is tetrahedral.}$ $\text{In } \text{NH}_2^{-} \text{, nitrogen form two sigma bonds and contains two lone pairs . Hence, sum is 4 and hybridization is } \text{sp}^3 \text{. The geometry is tetrahedral.}$ $\text{In } \text{CO}_3^{2-} \text{, carbon form three sigma bonds and doesnot any lone pairs . Hence, sum is 3 and hybridization is } \text{sp}^2 \text{. The geometry is trigonal planar.}$ $\text{In } \text{H}_3\text{O}^{+} \text{, oxygen form three sigma bonds and contains one lone pair. Hence, sum is 4 and hybridization is } \text{sp}^3 \text{. The geometry is tetrahedral.}$ $\text{STEP 3:}$ $\text{Hence, } \text{CO}_3^{2-} \text{ does not has tetrahedral geometry.}$

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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
}