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Current Question (ID: 8027)

Question:

Which among the following species has a triangular planar shape?

Options:

1. $N_3^{-}$
2. $NO_3^{-}$ (Correct)
3. $NO_2^{-}$
4. $CO_2$

Solution:

$\text{HINT: The trigonal planar geometry is same as plane triangular shape.}\n\n\text{Explanation:}\n\n\text{Species with } sp^2 \text{ hybridized central atom are plane triangular in shape.}\n\n\text{Among the given species } NO_3^{-} \text{ and } NO_2^{-} \text{ is } sp^2 \text{ hybridized with no lone pair and 1 lone pair of electrons on the central atom respectively.}\n\n\text{The shape of } NO_2^{-} \text{ is bent and the shape of } NO_3^{-} \text{ is trigonal planar, whereas, } N_3\text{, and } CO_2 \text{ are } sp \text{ hybridized with a linear shape.}\n\n\text{The shape of the species are as follows:}\n\n\text{For } NO_3^{-}:\n\text{Central atom: N}\n\text{Valence electrons: 5 (from N) + 3 * 6 (from O) + 1 (for charge) = 24}\n\text{Number of bonds: 3 (N-O bonds)}\n\text{Number of lone pairs on N: (24 - 6)/2 - 3 = 0}\n\text{Steric number = 3 (3 bond pairs, 0 lone pairs). Hybridization is } sp^2\text{. Shape is trigonal planar.}\n\n\text{For } N_3^{-}:\n\text{Central atom: N}\n\text{Valence electrons: 3 * 5 (from N) + 1 (for charge) = 16}\n\text{Number of bonds: 2 (N-N bonds)}\n\text{Number of lone pairs on central N: (16 - 4)/2 - 2 = 2}\n\text{Steric number = 2 (2 bond pairs, 2 lone pairs). Hybridization is } sp\text{. Shape is linear.}\n\n\text{For } NO_2^{-}:\n\text{Central atom: N}\n\text{Valence electrons: 5 (from N) + 2 * 6 (from O) + 1 (for charge) = 18}\n\text{Number of bonds: 2 (N-O bonds)}\n\text{Number of lone pairs on N: (18 - 4)/2 - 2 = 1}\n\text{Steric number = 3 (2 bond pairs, 1 lone pair). Hybridization is } sp^2\text{. Shape is bent.}\n\n\text{For } CO_2:\n\text{Central atom: C}\n\text{Valence electrons: 4 (from C) + 2 * 6 (from O) = 16}\n\text{Number of bonds: 2 (C=O bonds)}\n\text{Number of lone pairs on C: (16 - 4)/2 - 2 = 0}\n\text{Steric number = 2 (2 bond pairs, 0 lone pairs). Hybridization is } sp\text{. Shape is linear.}\n\n\text{Therefore, } NO_3^{-} \text{ has a triangular planar shape.}$

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{
  "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
}