Question

In the reaction:$BC{l}_3+P{H}_3\rightleftharpoons C{l}_{3}B·P{H}_3$ Lewis base is:

• (A) $BC{l}_3$
• (B) $P{H}_3$
• (C) $C{I}_{3}B-P{H}_3$
• (D) None of these

Answer 1

The Correct Option is B

Explanation:
Given reaction:$BC{l}_3+P{H}_3\rightleftharpoons C{l}_{3}B·P{H}_3$Atomic number of $P = 15$Electronic configuration of $P=1s^{2}2s^{2}2p^{6}3s^{2}3p^3$Atomic number of $B = 5$Electronic configuration of $B=1s^{2}2s^{2}2p^1$Therefore, B is electron deficient.Now, 'P' forms three bonds with hydrogen using its three electrons present in 3p-orbitals and 2 electrons remain with P as a lone pair, which it donates to the boron atom.Therefore, $C{l}_{3}B-P{H}_3$ complex is formed by the donation of a pair of electrons from the outer shell of phosphorus of PH to the boron atom of $BC{l}_3$. Thus, $P{H}_3$is acting as a Lewis base and $BC{l}_3$ is acting as a Lewis acid.Hence, the correct option is (B).