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Current Question (ID: 21303)

Question:

$\text{Which one of the following compounds exhibits geometrical isomerism?}$

Options:

1. $\text{CH}_2=\text{CHCl}$
2. $\text{CHCl}=\text{CHCH}_3$
3. $\text{CHCl}=\text{CHCl}$
4. $\text{CHCl}=\text{C(CH}_3)_2$

Solution:

$\text{Hint: If double bond carbon attached with same groups then it will not show geometrical isomerism}$ $\text{The geometrical isomerism arises when atoms or groups are arranged differently in space due to restricted rotation of a bond or bonds in a molecule.}$ $\text{The example of geometrical isomerism is as follows:}$ $\text{Cis: CH}_3\text{CH}=\text{CHCH}_3$ $\text{Trans: CH}_3\text{CH}=\text{CHCH}_3$ $\text{The compound given in option second shows geometrical isomerism. The structures are as follows:}$ $\text{HClC}=\text{CHCH}_3$ $\text{ClHC}=\text{CHCH}_3$ $\text{Both substituents (-CH}_3\text{ groups) are identical.}$ $\text{Since these groups are not different, geometrical isomerism cannot exist in this compound.}$

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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
}