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Current Question (ID: 8031)

Question:

The bond angles of $\text{NH}_3$, $\text{NH}_4^+$ and $\text{NH}_2^-$ are in the order :

Options:

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

Solution:

$\text{HINT: As the number of lone pairs on the central atom increases, the bond angle decreases.}$ $\text{Explanation:}$ $\text{In the case of the same hybridization of the central atom, as the number of lone pairs of electrons on the central atom increases, the bond angle decreases. NH}_4^\text{+}$ has zero lone pair, $\text{NH}_3$ has 1 lp while $\text{NH}_2$ has 2 lp. $\text{So an order of bond angle is as follows: } \text{NH}_4^+ > \text{NH}_3 > \text{NH}_2^\text{-}$

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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
}