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Current Question (ID: 20975)

Question:

$\text{Which of the following compounds has the smallest bond angle?}$ $1.\ \text{SO}_2$ $2.\ \text{H}_2\text{O}$ $3.\ \text{H}_2\text{S}$ $4.\ \text{NH}_3$

Options:

1. $\text{SO}_2$
2. $\text{H}_2\text{O}$
3. $\text{H}_2\text{S}$
4. $\text{NH}_3$

Solution:

$\text{Hint: If central atom orbital does not form hybridized orbital then}$ $\text{bond angle is approx } 90^\circ$ $\text{The bond angle of a central atom totally depends on the hybridization}$ $\text{state of the central atom. Calculate the hybridization of } \text{SO}_2, \text{OH}_2,$ $\text{and } \text{NH}_3 \text{ as follows:}$ $1.\ \text{SO}_2$ $S.\ N = \frac{1}{2}(V + M - C + A)$ $= \frac{1}{2}(6 + 0 - 0 + 0)$ $= 3$ $\text{The hybridization is } sp^2 \text{ bond angle is } 119^\circ$ $2.\ \text{H}_2\text{O}$ $S.\ N = \frac{1}{2}(V + M - C + A)$ $= \frac{1}{2}(6 + 2 - 0 + 0)$ $= 4$ $\text{The hybridization is } sp^3 \text{ and due to lone pair-lone pair repulsion the}$ $\text{bond angle is less than } 109^\circ, \text{ it is, } 105^\circ.$ $3.\ \text{NH}_3$ $S.\ N = \frac{1}{2}(V + M - C + A)$ $= \frac{1}{2}(5 + 3 - 0 + 0)$ $= 4$ $\text{The hybridization is } sp^3 \text{ and due to lone pair-bond pair repulsion the}$ $\text{bond angle is less than } 109^\circ, \text{ it is, } 107^\circ.$ $4.\ \text{In } \text{H}_2\text{S}, \text{ hybridization does not take place and pure p orbital}$ $\text{participate in bond formation. Hence, bond angle is approx } 90^\circ.$ $\text{The correct option is third.}$

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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
}