Question
The tendency of , and to behave as Lewis acid decreases in the sequence:
The Lewis acid character of boron trihalides depends on the extent of back bonding. In , the fully filled 2p orbital of fluorine effectively overlaps with the vacant 2p orbital of boron, significantly reducing boron's electron deficiency. As the size of the halogen atom increases down the group ( for F, for Cl, for Br), the orbital overlap between its filled p-orbital and the empty 2p orbital of boron becomes progressively weaker and less effective due to the size mismatch. Consequently, the electron deficiency of the central boron atom is least compensated in and most compensated in . Therefore, is the strongest Lewis acid, and the correct decreasing order of Lewis acidic strength is .
This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.
More Equilibrium Questions
Consider the following reaction: $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$. At equilibrium, the concentrations of $[\text{A}_2] = 3.0 \times 10^{–3} \text{ M}$; $[\text{B}_2] = 4.2 \times 10^{–3} \text{ M}$ and $[\text{AB}] = 2.8 \times 10^{–3} \text{ M}$. The value of $K_c$ for the above-given reaction in a sealed container at $527^\circ\text{C}$ is:
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What is the molarity of the saturated solution if the solubility product for a salt of type AB is $4 \times 10^{-8}$?
In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$. Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$. What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ($K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$, $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$)
The value of equilibrium constant of the reaction $HI(g) \rightleftharpoons \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g)$ is $8.0$. The equilibrium constant of the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:
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