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Current Question (ID: 8452)

Question:

$\text{The tendency of } \text{BF}_3, \text{BCl}_3 \text{ and } \text{BBr}_3 \text{ to behave as Lewis acids decreases in the sequence:}$

Options:

1. $\text{BCl}_3 > \text{BF}_3 > \text{BBr}_3$
2. $\text{BBr}_3 > \text{BCl}_3 > \text{BF}_3$
3. $\text{BBr}_3 > \text{BF}_3 > \text{BCl}_3$
4. $\text{BF}_3 > \text{BCl}_3 > \text{BBr}_3$

Solution:

$\textbf{Hint: } \text{Lewis acid strength} \propto \text{Tendency to accept electron pair}$ $\textbf{Explanation:}$ $\text{The acidic strength of the boron halides increase with the increase in the size of the halogen atom. In } \text{BF}_3, \text{ there is a p - p back bonding present between the fully-filled unutilized 2p orbitals of F and vacant 2p orbitals of boron. This renders the molecule less electron deficient. Therefore, it is the weakest Lewis acid.}$ $\text{This type of back donation is not possible in } \text{BCl}_3 \text{ or } \text{BBr}_3 \text{ as there is a large energy difference present between the orbitals of boron and chlorine or boron and bromine. This is because on going down the group, the energy difference increases. Therefore, they are strong Lewis acid.}$ $\text{Thus, the tendency to behave as Lewis acid is-}$ $\text{BBr}_3 > \text{BCl}_3 > \text{BF}_3$

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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
}